4exam

plant cell

2011-10-14T04:20:00.001-07:00

Heart

2011-10-02T04:43:00.000-07:00

The heart is a myogenic muscular organ found in all animals with a circulatory system (including all vertebrates), that is responsible for pumping blood throughout the blood vessels by repeated, rhythmic contractions. The term cardiac (as in cardiology) means "related to the heart" and comes from the Greek καρδιά, kardia, for "heart".
The vertebrate heart is composed of cardiac muscle, which is an involuntary striated muscle tissue found only in this organ, and connective tissue. The average human heart, beating at 72 beats per minute, will beat approximately 2.5 billion times during an average 66 year lifespan. It weighs approximately 250 to 300 grams (9 to 11 oz) in females and 300 to 350 grams (11 to 12 oz) in males.^[1]
In invertebrates that possess a circulatory system, the heart is typically a tube or small sac and pumps fluid that contains water and nutrients such as proteins, fats, and sugars. In insects, the "heart" is often called the dorsal tube and insect "blood" is almost always not oxygenated since they usually respirate (breathe) directly from their body surfaces (internal and external) to air. However, the hearts of some other arthropods (including spiders and crustaceans such as crabs and shrimp) and some other animals pump hemolymph, which contains the copper-based protein hemocyanin as an oxygen transporter similar to the iron-based hemoglobin in red blood cells found in vertebrates.

In fish

Schematic of simplified fish heart

Primitive fish have a four-chambered heart, but the chambers are arranged sequentially so that this primitive heart is quite unlike the four-chambered hearts of mammals and birds. The first chamber is the sinus venosus, which collects de-oxygenated blood, from the body, through the hepatic and cardinal veins. From here, blood flows into the atrium and then to the powerful muscular ventricle where the main pumping action will take place. The fourth and final chamber is the conus arteriosus which contains several valves and sends blood to the ventral aorta. The ventral aorta delivers blood to the gills where it is oxygenated and flows, through the dorsal aorta, into the rest of the body. (In tetrapods, the ventral aorta has divided in two; one half forms the ascending aorta, while the other forms the pulmonary artery).^[8]
In the adult fish, the four chambers are not arranged in a straight row but, instead form an S-shape with the latter two chambers lying above the former two. This relatively simpler pattern is found in cartilaginous fish and in the ray-finned fish. In teleosts, the conus arteriosus is very small and can more accurately be described as part of the aorta rather than of the heart proper. The conus arteriosus is not present in any amniotes, presumably having been absorbed into the ventricles over the course of evolution. Similarly, while the sinus venosus is present as a vestigial structure in some reptiles and birds, it is otherwise absorbed into the right atrium and is no longer distinguishable.^[8]

Sound

2011-06-16T04:50:00.001-07:00

From Wikipedia, the free encyclopedia

This article is about audible acoustic waves. For other uses, see Sound (disambiguation).

In a drum sound is produced when its membrane vibrates

Sound is a mechanical wave that is an oscillation of pressure transmitted through a solid, liquid, or gas, composed of frequencies within the range of hearing and of a level sufficiently strong to be heard, or the sensation stimulated in organs of hearing by such vibrations.^[1]

[hide]

Propagation of sound

Sound is a sequence of waves of pressure that propagates through compressible media such as air or water. (Sound can propagate through solids as well, but there are additional modes of propagation). During propagation, waves can be reflected, refracted, or attenuated by the medium.^[2]

The behavior of sound propagation is generally affected by three things:

A relationship between density and pressure. This relationship, affected by temperature, determines the speed of sound within the medium.
The propagation is also affected by the motion of the medium itself. For example, sound moving through wind. Independent of the motion of sound through the medium, if the medium is moving, the sound is further transported.
The viscosity of the medium also affects the motion of sound waves. It determines the rate at which sound is attenuated. For many media, such as air or water, attenuation due to viscosity is negligible.

When sound is moving through a medium that does not have constant physical properties, it may be refracted (either dispersed or focused).^[2]

Perception of sound

Human ear

The perception of sound in any organism is limited to a certain range of frequencies. For humans, hearing is normally limited to frequencies between about 20 Hz and 20,000 Hz (20 kHz)^[3], although these limits are not definite. The upper limit generally decreases with age. Other species have a different range of hearing. For example, dogs can perceive vibrations higher than 20 kHz, but are deaf to anything below 40 Hz. As a signal perceived by one of the major senses, sound is used by many species for detecting danger, navigation,predation, and communication. Earth's atmosphere, water, and virtually any physical phenomenon, such asfire, rain, wind, surf, or earthquake, produces (and is characterized by) its unique sounds. Many species, such as frogs, birds, marine and terrestrial mammals, have also developed special organs to produce sound. In some species, these produce song and speech. Furthermore, humans have developed culture and technology (such as music, telephone and radio) that allows them to generate, record, transmit, and broadcast sound. The scientific study of human sound perception is known as psychoacoustics.

Physics of sound

The mechanical vibrations that can be interpreted as sound are able to travel through all forms of matter:gases, liquids, solids, and plasmas. The matter that supports the sound is called the medium. Sound cannot travel through a vacuum.

Longitudinal and transverse waves

Sinusoidal waves of various frequencies; the bottom waves have higher frequencies than those above. The horizontal axis represents time.

Sound is transmitted through gases, plasma, and liquids as longitudinal waves, also calledcompression waves. Through solids, however, it can be transmitted as both longitudinal waves and transverse waves. Longitudinal sound waves are waves of alternating pressuredeviations from the equilibrium pressure, causing local regions of compression andrarefaction, while transverse waves (in solids) are waves of alternating shear stress at right angle to the direction of propagation.

Matter in the medium is periodically displaced by a sound wave, and thus oscillates. The energy carried by the sound wave converts back and forth between the potential energy of the extra compression (in case of longitudinal waves) or lateral displacement strain (in case of transverse waves) of the matter and the kinetic energy of the oscillations of the medium.

Sound wave properties and characteristics

Sound waves are often simplified to a description in terms of sinusoidal plane waves, which are characterized by these generic properties:

Sometimes speed and direction are combined as a velocity vector; wavenumber and direction are combined as a wave vector.

Transverse waves, also known as shear waves, have the additional property, polarization, and are not a characteristic of sound waves.

Speed of sound

U.S. Navy F/A-18 breaking the sound barrier. The white halo is formed by condensed water droplets thought to result from a drop in air pressure around the aircraft (see Prandtl-Glauert Singularity).^[4]^[5]

Main article: Speed of sound

The speed of sound depends on the medium the waves pass through, and is a fundamental property of the material. In general, the speed of sound is proportional to the square root of theratio of the elastic modulus (stiffness) of the medium to its density. Those physical properties and the speed of sound change with ambient conditions. For example, the speed of sound in gases depends on temperature. In 20 °C (68 °F) air at the sea level, the speed of sound is approximately 343 m/s (1,230 km/h; 767 mph) using the formula "v = (331 + 0.6 T) m/s". In fresh water, also at 20 °C, the speed of sound is approximately 1,482 m/s (5,335 km/h; 3,315 mph). In steel, the speed of sound is about 5,960 m/s (21,460 km/h; 13,330 mph).^[6] The speed of sound is also slightly sensitive (a second-order anharmonic effect) to the sound amplitude, which means that there are nonlinear propagation effects, such as the production of harmonics and mixed tones not present in the original sound (see parametric array).

Acoustics

Main article: Acoustics

Acoustics is the interdisciplinary science that deals with the study of all mechanical waves in gases, liquids, and solids including vibration, sound, ultrasound and infrasound. A scientist who works in the field of acoustics is an acoustician while someone working in the field of acoustics technology may be called an acoustical or audio engineer. The application of acoustics can be seen in almost all aspects of modern society with the most obvious being the audio and noise control industries.

Noise

Main article: Noise

Noise is a term often used to refer to an unwanted sound. In science and engineering, noise is an undesirable component that obscures a

Project Tiger

2011-06-14T12:13:00.001-07:00

Project Tiger is a wildlife conservation movement initiated in India in 1973 to protect tigers. The project aims at tiger conservation in specially constituted tiger reserves representative of various regions throughout India and strives to maintain viable populations of Bengal tigers in their natural environment.

In 2008 there were more than 40 Project Tiger reserves covering an area over 37,761 km² (14,580 sq mi). Project Tiger helped to increase the population of these tigers from 1,200 in the 1970s to 3,500 in 1990s. However, a 2008 census held by the Government of India revealed that the tiger population had dropped to 1,411. Since then the government has pledged US$153 million to further fund the project, set-up a Tiger Protection Force to combat poachers, and fund the relocation of up to 200,000 villagers to minimize human-tiger conflicts.

The number of tigers in India's wild has gone up by 20%, according to the latest(2011) tiger census, which has surveyed the whole of India for the first time. The census puts the population of the big cat at 1,706. There were 1,706 tigers including tigers in the Sunderbans at the last count.

FROM WIKIPEDIA.COM

pH scale

2011-06-12T12:04:00.000-07:00

In 1909, Soreson devised a scale known as p^H scale on which the strength of acid solutions as well as basic solutions could be represented by making use of hydrogen ion concentration. Sorenson linked the hydrogen ion concentrations of acid and base solutions to the simple numbers 0 to 14 on the pH scale.

pH of a solution is defined as the negative logarithm of the hydrogen ion concentration in moles per litre.

Mathematically: pH = log₁₀

In pure water [H⁺] = 1.0 x 10^-7

\pH of pure (neutral) water = - log (10^-7) = 7

Thus, the pH value of pure water is equal to 7.

Acid SolutionsWe know that all acidic solutions have H⁺ ion concentration greater than 1.0 x 10^-7. The H⁺ ion concentration in an acidic solution may be 10^-5 , 10^-6, 10^-4, etc.

Consider an acidic solution whose H⁺ ion concentration = 10^-6

\Its pH = - log (H⁺) = - log(10^-6) = 6

Clearly, pH values of all acidic solutions will be less than 7.

Basic SolutionsAll basic solutions have H⁺ ion concentration less than 10^-7. If the solution is basic, its OH^- ion concentration will be more than 10^-7. By knowing the concentration of OH^- ions, the concentration of H⁺ ion can be calculated.

[H⁺] =

[H⁺] = 10^-7 or pH = 7, the solution is neutral.

[H⁺] > 10^-7 or pH < 7, the solution is acidic.

The complete pH scale is given below:

It may be noted that (i) solutions having pH between 0 to 2 are strongly acidic: (ii) solutions having pH between 2 to 4 are moderately acidic: and (iii) solutions having pH between 4 to 7 are weakly acidic.

Similarly, solutions have pH value

(a) between 7 and 10 are weakly basic;

(b) between 10 and 12 are moderately basic, and

Importance of pH Value
1. pH has a great importance in agriculture. Soil is often tested to determine whether acidic or basic fertilisers are required for a particular crop.

2. It has a great importance in biochemical reactions such as digestion of food, etc. Human blood has pH 7.4. If it changes by 0.2 pH units, death results.

3. Large number of qualitative and quantitative analysis is carried out at definite pH values.

For example, pH must be adjusted between 0.4 to 0.6 for the complete precipitation of cations of group II.

4. Food preservation also needs a definite pH value.

digestive system diagram

2011-06-12T12:02:00.000-07:00

Acids, Bases and salts extra qus

2011-06-12T12:00:00.000-07:00

1. What is an acid?

Solution:An acid is a hydrogen-containing chemical compound which, when dissolved in water, gives hydrogen ion (H⁺) or hydrated hydrogen ion (H₂O. H⁺) or hydronium ion (H₃O⁺).

2. What are bases and alkalies?

Solution:Oxides and hydroxides of metals and metal like radicals (e.g., NH4⁺ ions) are called bases. Bases ionise to give OH^- ions in aqueous solution. Bases may be soluble or insoluble in water. The soluble bases are called alkalies. Thus all alkalies are bases but all bases are not alkalies.
Examples
NaOH and Cu (OH)₂ both are bases, but, since NaOH is soluble in water, it is an alkali. On the other hand, since Cu (OH)₂ is insoluble in water, it is not an alkali. Other examples of alkalies are KOH, Ca (OH)₂ and NH₄OH.

3. Define pH.

Solution:pH of a given solution is the negative logarithm to the base 10 of the hydrogen ion concentration, [H⁺] expressed in g ions/lit or moles/lit. Thus
pH=- log1- [H⁺].

4. What are the practical applications of neutralisation reactions?

Solution:Being alkaline in nature, cold milk is used to neutralise the acidity produced by HCl present in the gastric juice in the stomach.
Astronauts in space ships use this reaction to neutralise the dangerous levels of CO₂.

Farmers add slaked lime (calcium hydroxide) to reduce acidity of soil.

Sting of ants and bees contains formic acid. This can be neutralised by rubbing soap, which contains free sodium hydroxide.

Persons suffering from acidity are given antacid tablets, containing magnesium hydroxide which neutralises excess HCl produced, in stomach. Alternately, they are advised to sip cold milk, which neutralises HCl.

5. Why the salts solutions of strong acid and strong alkali are neutral?

Solution:Let us take the example of potassium sulphate, which is a salt of strong acid [sulphuric acid] and strong base [potassium hydroxide solution].

Neutral
Ionised Feebly
Ionised Highly
Ionised

From the above equation, it is clear that water is always feebly ionised and hence solution of potassium sulphate is neutral in nature.

6. What is an universal indicator?

Solution:Universal indicator is a solution, which undergoes several colour, changes over a wide range of pH. The colour is used to ‘indicate’ pH directly. Universal indicators are usually mixtures of several indicators.

7. Why common indicators cannot determine pH value of a solution?

Solution:Common indicators, such as litmus; methyl orange and phenolphthalein can easily tell, us whether a solution is acidic or alkaline, but they cannot easily tell how much a given acidic solution is stronger than another acidic solution. It means they cannot really tell us the pH value of different acidic or alkaline solutions.
In case of litmus, we cannot tell about pH values of 5,6 and 7. Similarly, in case of methyl orange pH values between 3 and 8 will not be very accurate.

8. What are the general characteristics of acids?

Solution:They have a sour taste

They turn blue litmus to red and methyl orange to red.
They react with active metals to give hydrogen

They react with bases to form salt and water

They decompose carbonates to produced carbon dioxide and water..

9. What is called deliquescent? Give examples.

Solution:Compounds that take up enough water from the air to dissolve in the water they have taken up are called deliquescent. Calcium chloride (CaCl₂) and Sodium hydroxide (NaOH) are the examples of deliquescent.

10. Write the uses of chlorines.

Solution:Chlorine is used in the production of bleaching powder.
Chlorine is used for making solvents for dry cleaning.

Chlorine is used to sterilise drinking water supply, and the water in swimming pools.

It is used in the production of hydrochloric acid.

Types and Distribution of Forest and Wildlife Resources

2011-06-11T02:12:00.001-07:00

Types and Distribution of Forest and Wildlife Resources

In India, forest and wildlife resources are owned and looked after by the government of India through its Forest Department.

Forests are classified under three categories by the Forest Department

Reserved forests

The most valuable forests are classified as Reserved forests. These forests are protected under the conservation acts. More than half of the total forest land in India comes under Reserved Forests.

Protected Forests

Forests that are protected from further destruction are called Protected Forests. One-third of the total forest areas in India are classified as Protected forests.

Unclassed forests

Forests and wastelands belonging to both government and private individuals and communities are classified as Unclassed Forests.

Forest density map

Madhya Pradesh has the largest area under permanent forests.
Andhra Pradesh, Uttarakhand, Kerala, Tamil Nadu, West Bengal, and Maharashtra have large percentages of Reserved forests.
Bihar, Haryana, Punjab, Himachal Pradesh, Orissa and Rajasthan have large percentage of Protected forests.
All Northeastern states and parts of Gujarat have a very high percentage of Unclassed forests managed by local communities.

forest and wildlife

2011-06-11T02:02:00.000-07:00

Conservation of Forest and Wildlife in India

Conservation of forest and wildlife means measures taken to preserve and protect forests and wildlife.
As we know forests and wildlife are very essential for the existence of mankind and the depletion of
forests and extinction of wildlife is a cause for concern.

Conservation preserves the ecological diversity and natural resources like water, air and soil.

It preserves the genetic diversity of plants

Conservation of forest and wildlife help in the growth of animal species and in their breeding.

In the year, 1972, the Indian Wildlife (Protection) Act was implemented. It made protecting specific habitats as law.
A list of wildlife species that had to be protected was published and hunting these animals was against the law.

National Parks and Wildlife Sanctuaries were set up in many states to protect endangered species.

Wildlife Sanctuaries

The Kashmir stag, three types of crocodiles, the Asiatic lion, the Indian elephant, black buck,
the great Indian bustard and the snow leopard are protected under several projects initiated
by the Central Government of India. These animals are protected against hunting and trading.

Under the Wildlife Act of 1980 and 1986, several insects have also been included in the list of
protected species. Butterflies, moths, beetles, dragonflies and even certain plants are included
in the protected list.

Project Tiger

Project tiger stamp

The tiger is India’s national animal. It is the key wildlife species in the faunal web. The population of the Indian tigers came down from an estimated number of 55,000 to a mere 1, 827 due to various reasons like poaching for trade, shrinking forests, depletion of prey species and growing human population. Tigers were hunted for their skin and bones. The bones were used to make traditional medicines.

Realizing the gravity of the situation the "Project Tiger" campaign was initiated in the year, 1973 by the government of India. It is one of the most well publicized wildlife campaigns in the world.

There are 27 tiger reserves in India covering an area of 37,761 sq km
Due to ‘Project Tiger’ the tiger population went up to 4,002 in 1985 and 4,334 in 1989
In 1993, the tiger population again dropped to 3,600

Corbett National Park in Uttaranchal, Sunderbans National Park in West Bengal, Bandhavgarh National Park in Madhya Pradesh, Sariska Wildlife Sanctuary in Rajasthan, Manas Tiger Reserve in Assam and Periyar Tiger Reserve in Kerala are some of the tiger reserves of India.

Project Tiger Reserves

The Himalayan Yew in trouble

Himalayan Yew

The Himalayan Yew is a medicinal plant that is found in Himachal Pradesh and Arunachal Pradesh.
From the bark, needles, twigs and roots of this tree, a chemical compound called ‘taxol’ is extracted.
This chemical is used to make drugs that are used to cure certain types of cancers, but the species is
becoming extinct due to over-exploitation.

Are colonial forest policies to be blamed?

"Enrichment plantation" was carried out during the colonial period in India. When a particular species of trees which are commercial profitable are planted after the removal of other species in the area is called "enrichment plantation".

Examples of "enrichment plantation" are

Teak trees planted in South India after cutting down natural forests
Chirr Pine plantations which have replaced the Himalayan oak in the Himalayas


Teak Plantation	Chirr Pine plantations

Introduction to Trigonometry

2011-06-10T02:56:00.000-07:00

1. In D ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) SinA , CosA
(ii) SinC , CosC.

Solution:

2. In the figure, find tan P – cot R.

Solution:In a right angled D PQR
PR² = PQ² + QR²13² = 12² + QR²QR² = 169 – 144 = 25
QR = 5
\tan P - cot R = = 0

3. sin A =,calculate cos A and tan A.

Solution:Let the right angled D ABC be right angled at B.

In a right angled D ABC

AC² = AB² + BC²16 = AB² + BC
AB² = 16 – 9 = 7
AB =
sin A =
cos A = =

tan A = .

4. Given 15 cot A = 8, find sin A and sec A.

Solution:15 cot A = 8
Þ cot A =

Let ABC be the D right angled at B
Cot A = 8/15
Þ AB = 8, BC = 15
AC² = AB² + BC²
      = 8² + 15²
      = 64 + 225
      = 289
AC = = 17
sin A = =
sec A =  =

5. Given secq = calculate all other trigonometric Ratios.

Solution:

In a right angled D ABC,
AC² = AB² + BC²13² - 12² = AB²AB² = 169 - 144
AB = 5

6. If Ð A and Ð B are acute angles such that cos A = cos B,then show that Ð A = Ð B.

Solution:If cosA = cosB
Þcos A – cos B = 0
only when = 45⁰.

7. If cot q = ,evaluate : (i) (ii) cot²q.

Solution:If cot q =
(i) = =

(ii) cot²q = .

8. If 3 cotA = 4, Check whether =cos²A – sin²A or not.

Solution:If 3 cot A = 4
=  =  =  ………………… (1)
cos²A – sin²A = sin²A(cot²A - 1) Dividing and multiplying by sin²A
                   = sin²A
                   = sin²A= sin²A
                   =  =  …………………………. (2)
Since (1) = (2)
LHS = RHS.

9. In triangle ABC, right-angled at B, if tan A =find the value of (i) sin A cos C +cos A sin C (ii)cos A cos C – sin A sin C.

Solution:If tan A =
In a right angled D ABC, right angled at B
AC² = AB² + BC²
\AC = = 2
sin A = , cos A = sin C = cos C = 1/2

(i) sin A cos C + cos A sin C
= = = 1

(ii) cos A cos C – sin A sin C
= 0.

10. In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution:
PR + QR = 25
PQ = 5 cm
In a right angled D PQR, PQ² + QR² = PR² …………………… (1)
Let PR = x
QR = (25 – x)
From 1 x² = 5² + (25 – x)²
x² = 25 + (25)² – 2 ´ 25x + x²
    = 625 + 25 – 50x
50x = 650
x =  = 13
PR = 13 cm
QR = 12 cm
PQ = 5 cm
sin P =  =
cos P =  =
- tan P = = .
- 11. State whether the following are true or false. Justify your answer.
  (i) The value of tan A is always less than 1.
  (ii) sec A = for some value of angle A.
  (iii) cos A is the abbreviation used for the cosecant of angle A.
  (iv) cot A is the product of cot and A.
  (v) sin q =.
  Solution:(i) False. The value of tan 90^◦is grater than 1
  
  (ii) True. sec A = Þ cos A = as 12 the hypotenuse is the largest side of triangle.
  
  (iii) False. cos A is the abbreviation for cosine of angle A
  
  (iv) No,False. cos A is not the product of cot and A
  
  (v) False, Because the hypotenuse is the longest side whereas in sin q = , 3 which is the denominator is cannot be the hypothenuse.
  12. Evaluate the following:
  (i) sin 60° cos 30° + sin 30° cos 60°
  
  (ii) 2 tan2 45° + cos2 30° – sin2 60°
  (iii)
  (iv)
  (v) .
  Solution:
  
  =
  13. Choose the correct option and justify your choice:(i) =(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
  
  (ii) =(A) tan 90° (B) 1 (C) sin 45° (D) 0
  
  (iii) sin 2A = 2 sin A is true when A =
  (A) 0° (B) 30° (C) 45° (D) 60°
  
  (iv)=(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°.
  Solution:(i) =  =
                        =
                        =  =
                        = sin 60⁰
  Answer: A
  
  (ii)  =  = 0
  Answer: D
  (iii) sin 2A = 2 sin A is true when A = 45⁰
  Answer: A
  
  (iv)  =  =
                       =
                       =
                       =
                       =´
                       =  =  = tan 60⁰.
  
  Answer: C
  14. tan (A + B) = and tan (A – B) =; 0° < A + B 90°; A > B, find A and B.
  Solution:If tan(A + B) =
  tan(A - B) = 1/
  If tan A+B =
  Þ A + B = 60⁰ ………………………(1)
  tan (A – B) =
  A – B = 30⁰ ……………………..(2)
  Adding (1) ´ (2) 2A = 90⁰A = 45⁰
  \B = 15^0.
  15.  State whether the following are true or false. Justify your answer.
  (i) sin (A + B) = sin A + sin B.
  
  (ii) The value of sin  increases as  increases.
  
  (iii) The value of cos increases as  increases.
  
  (iv) sin  = cos  for all values of .
  
  (v) cot A is not defined for A = 0°.
  Solution:(i) sin(A+B) = sin A + sin B
  False
  
  (ii) Value of sin q as q increase
  True
  
  (iii) Value of cos q as q ncreases
  False
  
  (iv) sinq = cosq for all values of q
  False
  
  (v) cot A is not defined for A = 0⁰
  True.
  16. Evaluate:
  (i)
  (ii)
  
  (iii) cos48⁰ – sin 42⁰(iv)cosec 31⁰ – sec 59⁰.
  Solution:(i)
      =
      =  = 1
  
  (ii)  =  =  = 1
  
  (iii) cos48⁰ – sin 42⁰cos48⁰ = cos(90 - 42⁰)
            = sin 42⁰
  \cos 48⁰ – sin 42⁰ = sin 42⁰ – sin 42⁰                            = 0
  
  (iv) cosec 31⁰ – sec 59⁰
  cosec 31⁰ = cosec(90⁰ - 59⁰)
                = sec 59⁰
  LHS = sec 59⁰ – sec 59⁰ = 0.
  
  21.  A, B and C are interior angles of a triangle ABC, then show that sin = cos .
  
  Solution:
  
  22.  Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
  
  Solution:
  
  23.  Express the trigonometric ratios sin A, cos A and tan A in terms of sec A.
  
  Solution:
  
  24.  Write all the other trigonometric ratios of Ð A in terms of sec A.
  
  Solution:sin A =
             =
  cos A =
  tan A =  =  ¸
          =  ´ sec A.
              =
  
  25.  Evaluate:(i) (ii) sin 25⁰. cos 65⁰ + cos 25⁰ sin 65^0.
  
  Solution:
  
  26.  Choose the correct option. Justify your choice.
  (i) 9 sec² A – 9 tan² A =
  (A) 1 (B) 9 (C) 8 (D) 0
  
  (ii) (1 + tan q + secq ) (1 + cot q – cosec q ) =
  (A) 0 (B) 1 (C) 2 (D) –1
  
  (iii) (sec A + tan A) (1 – sin A) =
  (A)sec A (B) sin A (C) cosec A (D) cos A
  
  (iv)
  (A) sec² A (B) –1 (C) cot² A (D) tan² A.
  
  Solution:(i) 9 sec²A - 9 tan²A
     = 9 (1 + tan²A) – 9 tan²A
     = 9 + 9 tan²A - 9 tan²A
     = 9
  Answer (B)
  
  (ii)
  
  (iii) (sec A + tan A) (1 – sin A)
  sec A – sin A sec A + tan A – tan A sin A
  +  - .sinA
  
  (iv)  =  =  = tan²A.
  
  27.  Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
  (i) (cosecq - cot q )² =
  
  (ii)= 2 secA
  
  (iii)= 1 +secq cosecq
  
  (iv) =
  (v)= Cosec A +cot A
  
  Solution:(i) (cosecq - cot q )² =
  RHS  =
  
                    =
                    = cosec^2q- 2cosecq . cotq + cot^2q                  = (cosec q - cotq )²
  Hence Proved.
  (ii)
                    =  =
                    =  =
                    =  = 2 sec A
  
  (iii)
  
  (iv)  =
  LHS
  RHS  =
              = LHS
  
  (v)
  Using the identity cosec²A = 1 + cot²A.
  Taking the LHS of the given equation
  
  Dividing the numerator and denominator by sin A
  
  = (Changing = cot A ´
  =
  
  (Replacing cosec²A - cot²A = (cosec A + cot A) (cosec A - cot A) using the identity cosec²A - cot²A = 1
  
  =
  =
  = cosec A + cotA.
  
  28.  Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
  (i)  = secA + tanA
  (ii) (sin A + cosec A)² + (cos A + sec A)²= 7 +tan²A + Cot²A
  (iii) (cosecA –sinA)(secA- cosA) =
  
  (iv)= .
  
  Solution:(i)
        =  Taking the conjugate of (1-sinA) + multiplying the numerator and denominator by it.
        =
        =  (Since 1 - sin²A = cos²A)
        =  =
        =  = sec A + tan A (As  = sec A,  = tan A)
  
  (ii) (sin A + cosec A)² + (cos A + sec A)²= sin²A + cosec²A + 2sinA.cosec A + cos²A + sec²A + 2cos A. sec A
        = sin²A + cos²A + cosec²A + sec²A + 2sinA .  + 2cos A.
        = 1 + 2 + 2 + cosec²A + sec²A (using cosec A =  secA = )
        = 5 + 1 + tan²A + 1 + cot²A
        = 7 + tan²A + cot²A (1 + tan²A = sec²A) 1 + cot²A = cosec²A
  
  (iii) (cosec A - sin A) (secA - cosA)
  LHS
  (cosec A - sin A) (sec A - cos A)
  
  (since 1 - sin²A = cos²A and 1 - cos²A = sin²A)
  =  = sin A cosA
  RHS  =
                         =  (Taking LCM as cosA. sinA)
                         =  (since sin²A + cos²A = 1)
  LHS = RHS
  Hence the result.
  
  (iv)
  LHS
  
  RHS
  
  Thus, LHS = RHS.
  17. Show that tan 48⁰ tan 23⁰ tan 42⁰ tan 67⁰ = 1.
  Solution:(i) tan 48⁰ tan23⁰ tan 42⁰ tan67⁰ = 1
  LHS = tan 48⁰. tan 42⁰ tan 23⁰ tan67⁰= tan 48⁰. tan(90⁰ - 48⁰). tan(90⁰ - 67⁰) tan 67⁰
  = tan 48⁰. cot 48⁰. cot67⁰tan 67⁰ .= 1 ´ 1 = 1.
  18. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
  Solution:tan 2A = cot (A – 18) ………………. (1)
  where 2A is an acute angle
  tan 2A = cot(90 – 2A) = cot(A – 18)
  cot 90 – 2A and A - 18 are acute angles.
  90 – 2A = A – 18
  3A = 108
  A = 36⁰.
  19. If tan A = cot B, prove that A + B = 90°.
  Solution:If tan A = cot B …………… (1)
  tan A = cot (90 – A)
  From (1)
  cot (90 - A) = cot B
  Þ 90 – A = B
  or A + B = 90⁰.
  20. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
  Solution:If sec 4A = cosec (A – 20⁰)
  sec 4A = cosec (90 – 4A)
  \cosec(90 – 4A) = cosec (A – 20⁰)
  Since (90 – 4A) and A – 20⁰ are acute angles
  90 – 4A = A – 20⁰
  90 + 20 = 5A
  A =
  = 22⁰.

Statistics WORKSHEET

2011-06-10T02:32:00.000-07:00

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants	0 - 2	2 - 4	4 - 6	6 - 8	8 - 10	10 - 12	12 - 14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

Solution:Let us solve the problem using direct method

Number of plants	Number of houses (f_i)	Class mark (x_i)	f_i x_i
0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14	1 2 1 5 6 2 3	1 3 5 7 9 11 13	1 6 5 35 54 22 39
	å f_i = 20		å f_ix_i = 162

Mean ( = = = 8.1

Thus the mean number of plants per house = 8.1

The method used here is direct method because the numerical values of x_i and f_i are small.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs)	100 – 120	120 – 140	140 – 160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:(Direct method)
The following problem is solved using direct method

Daily wages (in Rs.)	Number of workers (f_i)	Class mark (x_i)	f_i x_i
100 - 120 120 - 140 140 - 160 160 - 180 180 - 200	12 14 8 6 10	110 130 150 170 190	1320 1820 1200 1020 1900
Total	å f_i = 50		å f_ix_i = 7260

The mean of the given data = = = 145.2
The mean daily wages of the workers = Rs. 145.20.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily pocket allowance (in Rs)	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Number of workers	7	6	9	13	f	5	4

Solution:The mean pocket allowance = Rs. 18
Let us solve the problem using assumed mean method

Daily pocket allowance (in Rs.)	Number of children	x_i	d_i = x_i- A = x_i - 18	f_id_i
11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25	7 6 9 13 f 5 4	12 14 16 18 20 22 24	-6 -4 -2 0 2 4 6	-42 -24 -18 0 2f 20 24
	å f_i = 44 + f			å f_id_i = - 40 + 2f

Here å f_id_i = -40 + 2f
          å f_i = 44 + f
Mean  = A +  = 18
              = 18 +  = 18
Þ  = 18 - 18
     = 0
   -40 + 2f = 0
        Þ 2f = 40
     f =  = 20.

4. Thirty women are examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, clothing a suitable method.

Numbers of heart beats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-36
Number of women	2	4	3	8	7	4	2

Solution:The given problem is solved using assumed mean method

Number of heart beats per minute	Number of women (f_i)	Class mark (x_i)	d_i = x_i - A = x_i - 75.5	f_id_i
65-68 68-71 71-74 74-77 77-80 80-83 83-86	2 4 3 8 7 4 2	66.5 69.5 72.5 75.5 78.5 81.5 84.5	-9 -6 -3 0 3 6 9	-18 -24 -9 0 21 24 18
	å f_i = 30			å f_id_i = 12

Mean(x) = 75.5 +
            = 75.5 + 0.4
            = 75.9
(x) = A +  where A = 75.5

The mean heart beats per minute = 75.9.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50 – 52	53 – 55	56 – 58	59 – 61	62 - 64
Number of boxes	15	110	135	115	25

Solution:

Number of mangoes	Number of mangoes	Number of boxes (f_i)	Class marks (x_i)	di = x_i - A = x_i - 57	f_id_i
50 - 52 53 - 55 56 - 58 59 - 61 62 - 64	49.5 - 52.5 52.5 - 55.5 55.5 - 58.5 58.5 - 61.5 61.5 - 64.5	15 110 135 115 25	51 54 57 60 63	-6 -3 0 3 6	-90 -330 0 345 150
		å f_i = 400			å f_id_i = 75

Mean () = A +  = 57 +
              = 57 + 0.1875
              = 57.1875
The mean number of mangoes in a packing box = 57.19

The method adopted for solving this problem is assumed mean method.

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily Expenditure (in Rs.)	100 – 150	150 - 200	200 – 250	250 – 300	300 – 350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution:

Daily expenditure (in Rs)	Number of households (f_i)	Class Mark (x_i)	d_i =	f_iu_i
100 - 150	4	125	-2	-8
150 - 200	5	175	-1	-5
200 - 250	12	225	0	0
250 - 300	2	275	1	2
300 - 350	2	325	2	4
	å f_i = 25			å f_iu_i = -7

Mean  = A +
              = 225 +
              = 225 - 14
              = 211

\ The mean of daily expenditure on food = Rs. 211.

7. To find out the concentration of SO₂ in the air(in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO₂ (in ppm)

Frequency

0.00 - 0.04

0.04 - 0.08

0.08 - 0.12

0.12 - 0.16

0.16 - 0.20

0.20 - 0.24
4

9

9

2

4

2

Find the mean concentration of SO₂ in the air.

Solution:Let us solve the problem by Step-deviation method

Concentration of SO₂(in ppm)	Frequency (f_i)	x_i	d_i = x_i - 0.1	u_i =	f_iu_i
0.0 - 0.04 0.04 - 0.08 0.08 - 0.12 0.12 - 0.16 0.16 - 0.20 0.20 - 0.24	4 9 9 2 4 2	0.02 0.06 0.1 0.14 0.18 0.22	-0.08 -0.04 0 0.04 0.08 0.12	-2 -1 0 1 2 3	-8 -9 0 2 8 6
Total	å f = 30				å f_iu_i = -1

Mean x = A +  =0.1 +
            = 0.1 - (0.00133)
            = 0.09867

\ The mean concentration of SO₂ in the air is 0.09867 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of Days	0 – 6	6 – 10	10 – 14	14 – 20	20 – 28	28 – 38	38 – 40
Number of students	11	10	7	4	4	3	1

Solution:

Number of day	Number of students (f_i)	x_i	d_i = x_i – A = x_i – 17	U_i =	f_iu_i
0 – 6	11	3	-14	-7	-77
6 – 10	10	8	-9	-4.5	-45
10 – 14	7	12	-5	-2.5	-17.5
14 – 20	4	17	0	0	0
20 – 28	4	24	7	3.5	14
28 – 38	3	33	16	8	24
38 – 40	1	39	22	11	11
	å f_i =40				å f_iu_i = -90.5

Mean (x) = A + ´ h = 17+
= 17 – 4.525
= 12.475

The mean number of days a student was absent = 12.48 days.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45 – 55	55 – 65	65 – 75	75 – 85	85 – 95
Number of cities	3	10	11	8	3

Solution:

Literacy rate (in %)	Number of cities f_i	Class mark (x_i)	ui = =	f_iu_i
45 – 55	3	50	-2	-6
55 – 65	10	60	-1	-10
65 – 75	11	70	0	0
75 – 85	8	80	1	8
85 – 95	3	90	2	6
	å f_i = 35			å f_iu_i = -2

Mean (x) = A + ´ h
             = 70 +
             = 70 – 0.571 = 69.428
             = 69.43%

The mean literacy rate (in percentage) = 69.43%.

10. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5 – 55	15 – 25	25 – 35	35 – 45	45 – 55	55 – 65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

Age (in years)	Number of patients (f_i)	x_i	u_i = u_i =	f_iu_i
5 – 15	6	10	-3	-18
15 – 25	11	20	-2	-22
25 – 35	21	30	-1	-21
35 – 45	23	40	0	0
45 – 55	14	50	1	44
55 – 65	5	60	2	10
	å f_i = 80			å f_iu_i = -37

Mean  = A + ´ h
              = 40 +
              = 40 – 4.625
              = 35.375
Mode = l + ´ h

Here the maximum class frequency is 23, and the class corresponding frequency is 35 – 45. So the modal class is 35 – 75.

Now, lower limit(l) of modal class = 35
Class size(h) = 10
Frequency (f₁) of the modal class = 23
Frequency (f₀) of class preceding the model class = 21
Frequency (f₂) of class succeeding the model class = 14
Mode = l + ´ h
         = 35 +
         = 35 +
         = 35 +
         = 35 + 1.82
         = 36.82.

11. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components

Lifetime (in hours)	0 – 20	20 – 40	40 – 60	60 – 80	80 - 100	100 – 120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Solution:

Life time
(in hours)

Frequency (f_i)

0 – 20

10

20 – 40

35

40 – 60

52

60 – 80

61

80 – 100

38

100 – 120

29

Here maximum frequency is 61, the class corresponding to this 60 – 80. So, the model class is 60 – 80.

Now lower limit of the model class (l) = 60
frequency(f₁) of the modal class = 61
frequency(f₀) of the preceding the model class = 52
frequency(f₂) of class succeeding the model class = 38
class size (h) = 20
Mode = + ´ h
        = 60 + ´ 20
        = 60 + ´ 20
        = 60 +
        = 60 + 5.625
        = 65.625

\ Modal lifetime of the components = 65.63.

12. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in Rs)	Number of families
1000 – 1500	24
1500 – 2000	40
2000 – 2500	33
2500 – 3000	28
3000 – 3500	30
3500 – 4000	22
4000 – 4500	16
4500 – 5000	7

Solution:

Expenditure (in Rs.)	Number of families	Mid value (x_i)	u_i = =	f_iu_i
1000 - 1500	24	1250	-4	-96
1500 - 2000	40	1750	-3	-120
2000 - 2500	33	2250	-2	-66
2500 - 3000	28	2750	-1	-28
3000 - 3500	30	3250	0	0
3500 - 4000	22	3750	1	22
4000 - 4500	16	4250	2	32
4500 - 5000	7	4750	3	21
	å f_i = 200		å u_i = -4	f_iu_i = 235

Mode l + ´ h

The highest frequency is 40. The modal class is 1500 - 2000.

Lower limit (l) of the modal class = 1500

Class size (h) = 500

Frequency of the modal class (f₁) = 40
Frequency of the preceding modal class (f₀) = 24
Frequency of the succeeding modal class (f₂) = 33
Mode = l + ´ h
Mode = 1500 + ´ 500
        = 1500 +
        = 1500 +
        = 1500 + 347.83
        = 1847.83
\ The model monthly expenditure of the family = Rs. 1847.83
Mean = A +
        = 3250 -
        = 3250 - 587.50
        = 266.65

\ Mean monthly expenditure = Rs. 2662.5.

13. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of student per teacher	Number of states/ U.T.
15 – 20	3
20 – 25	8
25 – 30	9
30 – 35	10
35 – 40	3
40 – 45	0
45 – 50	0
50 – 55	2

Solution:Let us first find the mode.

In this problem, the maximum frequency is 10, and the class corresponding to this frequency is 30-35. So, the modal class is 30 - 35.

Now
Lower limit (l) of modal class = 30
Class size (h)= 5
frequency (f₁) of model class = 10
frequency (f₀) proceeding the modal class = 9
frequency (f₂) succeeding the modal class = 3
Mode = l +

Substituting the values above in the given formula,

        = 30 +
        = 30 +
        = 30 +  =  = 30.625

\ The mode of the given data is 30.6

Now calculate mean by applying direct method.

Number of students / teacher	Number of states/U.T (f_i)	Classmark (x_i)	di = x_i - A = x_i - 325	f_id_i
15-20	3	17.5	-15	-45
20-25	8	22.5	-10	-80
25-30	9	27.5	-5	-45
30-35	10	32.5	0	0
35-40	3	37.5	5	15
40-45	0	42.5	10	0
45-50	0	47.5	15	0
50 - 55	2	52.5	20	40
	å f_i = 35			å f_id_i = -11.5

= A +
= 32.5 +
= 32.5 - 3.29 = 29.21

\ The mean of the given data = 29.21

Thus we interpret that the most states/ U.T. student teacher ratio of 30.6 and on an average this ratio is 29.2.

14. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsman
3000 – 4000	4
4000 – 5000	18
5000 – 6000	9
6000 – 7000	7
7000 – 8000	6
8000 – 9000	3
9000 – 10000	1
10000 – 11000	1

Find the mode of the data.

Solution:The maximum frequency is 18, and the class corresponding to the maximum frequency is 4000 – 5000.
Now the model class is 4000 – 5000
Class size(h) = 1000
lower limit (f) of the model class = 4000
frequency(fo) of the class proceeding the modal class = 4
frequency(f₁) of the modal class = 18
frequency(f₂) of the class succeeding the modal class = 9
mode = l +
         = 4000 +
         = 4000 +
         = 4000 +
         = 4000 + 608.7
         = 4608.7

Mode = 4608.7 runs.

15. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Number of cars	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 – 80
Frequency	7	14	13	12	20	11	15	8

Solution:In this problem the maximum frequency is 20, the class corresponding to the maximum frequency is 40 – 50.
Now,
The model class is 40 – 50
Class size (h) = 10
lower limit (l) of the model class = 40
frequency(f₁) of the modal class = 20
frequency(f₀)of the class proceeding the modal class = 12
frequency(f₂)of the class succeeding the modal class = 11
Mode = l +
        = 40 +
        = 40 +
        = 40 +
        = 40 + 4.71
        = 44.71 cars

\ Mode = 44.71 cars.

16. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)

Number of consumers

65 - 85

4

85 - 105

5

105 - 125

13

125 - 145

20

145 - 165

14

165 - 185

8

185 - 205

4

Solution:

Monthly consumption (in units)	Number of consumers	Cumulative frequency
65 - 85	4	4
85 - 105	5	4 + 5 = 9
105 - 125	13	9 + 13 = 22
125 - 145	20	22 + 20 = 42
145 - 165	14	42 + 14 = 56
165 - 185	8	56 + 8 = 64
185 - 205	4	64 + 4 = 68

n = 68, = 34. Here 125 - 145 is the class whose cumulative frequency 42 is greater than 34.

Here
L - lower limit of median class = 125
= 34 where n = number of observations
c.f - cumulative frequency of class proceeding the median class = 22
f - frequency of median class = 20
h - class size = 20
Median = l +
           = 125 +
           = 125 + 12
           = 137

\ The median of the given data is 137.

The mean can be calculated using assumed mean method.

Monthly consumption (in units)	Number of consumer (f_i)	Class mark (x_i)	d_i = x_i- f_i = x_i - 135	f_id_i
65-85	4	75	-60	-240
85-105	5	95	-40	-200
105-125	13	115	-20	-260
125 - 145	20	135	0	0
145 - 165	14	155	20	280
165 - 185	8	175	40	320
185 - 205	4	195	60	240
	å f_i = 68			å f_id_i = 140

Mean( = A +
            = 135 +
            = 135 + 2.06 = 137.06

\ The mean of the given data = 137.06

Let us find the mode as follows.

The highest frequency is 20 which corresponds to the class interval 125 - 145. So the modal class is 125 - 145.

Here
Lower limit (l) of the modal class = 125
Class size (h) = 20
frequency (f₁) of the modal class = 20
frequency (f₀) of the class succeeding the modal class = 13
frequency (f₂) of the class succeeding the modal class = 14
mode = l +
        = 125 +
        = 125 +
        = 125 + 10.77
        = 135.77

\ The mode of the data = 135.77

The three measures are approximately the same in this case.

17. If the median of the distribution given below is 28.5, find the values of x and y.

Class Interval	Frequency
0 – 10	5
10 – 20	x
20 – 30	20
30 - 40	15
40 – 50	y
50 – 60	5
Total	60

Solution:

Class Interval	Frequency (f_i)	Cumulative frequency
0 - 10	5	5
10 - 20	x	5 + x
20 - 30	20	25 + x
30 - 40	15	40 + x
40 - 50	y	40 + x + y
50 - 60	5	45 + x + y
Total	å f_i= 60

Here in this problem = n = 45 + x + y

It is given that total frequency is 60, thus n = 60
\ 45 + x + y = 60
Þ x + y = 60 - 45
= 15(1)

The median is 28.5 which lies in the class 20-30
So,
l (lower limit) of the median class = 20
f (frequency) of the median class = 20
cf(cumulative) frequency of the class = 5 + x

proceeding 20 - 30
h (class size) = 10
median = l +
28.5 = 20 +
28.5 = 20 +
28.5 - 20 +
8.5 =
Þ 17.0 = 25 - x
Þ x = 25 - 17
= 8

substitute x = 8 in (1)
x + y = 15
8 + y = 15
y = 15 - 8
= 7

\ The value of x and y are 8 and 7.

18. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year

Age (in years)	Number of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Solution:

Age (in years)	Number of policy holders	Cumulative frequency
Below 20	2	2
Below 25	6	6
Below 30	24	24
Below 35	45	45
Below 40	78	78
Below 45	89	89
Below 50	92	92
Below 55	98	98
Below 60	100	100

Here n = 100,  =  = 50

The observation lies in the class 35 – 40

Lower limit (l) = 35

cf – cumulative frequency of the class proceeding the interval 35 – 40 = 45.

frequency (f) of the median class 35 – 40 = 33

class size (h) = 5
median = l +
           = 35 +
           = 35 +
           = 35 + 0.758
           = 35.76

\ The median age is 35.76 years.

19. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm)	Number of leaves
118 – 126	3
127 – 135	5
136 – 144	9
145 – 153	12
154 – 162	5
163 – 171	4
172 – 180	2

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Solution:

Length (in mm)	Length (in mm) continuous class	Number of leaves (f_i)	Cumulative frequency
118 - 126	117.5 - 126.5	3	3
127 - 135	126.5 - 135.5	5	3 + 5 = 8
136 - 144	135.5 - 144.5	9	8 + 9 = 17
145 - 153	144.5 - 153.5	12	17 + 12 = 29
154 - 162	153.5 - 162.5	5	29 + 5 = 34
163 - 171	162.5 - 171.5	4	34 + 4 = 38
172 - 180	171.5 - 179.5	2	38 + 2 = 40
		å f = 40

Now n = 40, = = 20

This observation lies in the class 144.5 - 153.5

l (lower limit of the classes) = 144.5

cf (cumulative frequency of the class proceeding 144.5 - 153.5 = 17

f (frequency of the median class) = 12

h (class size) = 9
median = l +
           = 144.5 +
           = 144.5 +
           = 144.5 +
           = 144.5 + 2.25
           = 146.75 mm

\ The median length of the leaves = 146.75 mm.

20. The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500 – 2000	14
2000 – 2500	56
2500 – 3000	60
3000 – 3500	86
3500 – 4000	74
4000 – 4500	62
4500 – 5000	48

Find the median life time of a lamp.

Solution:

Life time (in hours)	Number of lamps (f_i)	Cumulative frequency
1500 - 2000	14	14
2000 - 2500	56	14 + 56 = 70
2500 - 3000	60	70 + 60 = 130
3000 - 3500	86	130 + 86 = 216
3500 - 4000	74	216 + 74 = 290
4000 - 4500	62	290 + 62 = 352
4500 - 5000	48	352 + 48 = 400
	å f_i = 400

n = å f_i = 400

= = 200

Thus the observation lies in the class 3000 - 3500
median = l +

where l - lower limit of the median class = 3000

f = frequency of the median class = 86

cf = cumulative frequency proceeding the median class = 216

h - class size = 500
                    = 3000 +
                    = 3000 +
                    = 3000 + 406.98
                    = 3406.98

Thus the median life time of a lamp is 3406.98.

21. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	1 – 4	4 – 7	7 – 10	10 – 13	13 – 16	16 – 19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution:

Number of letters	Number of surnames	Cumulative frequency
1 - 4	6	6
4 - 7	30	6 + 30 = 36
7 - 10	40	36 + 40 = 76
10 - 13	16	76 + 16 = 92
13 - 16	4	9 + 4 = 96
16 - 19	4	96 + 4 = 100
	100

Here n = 100,

Here the observation lies in the class 7 - 10

l - lower limit of the class = 7

f - frequency of the median class (7 - 10) = 40

cf - (cumulative frequency of the class proceeding (7 - 10) = 36

h - class size = 3
median = l +
           = 7 +
           = 7 +
           = 7 + 1.05
           = 8.05

\ median = 8.05

Median number of letters in the surnames = 8.05

Let us calculate the mean using step deviation method

Number of letters	Number of surnames f_i	Class size (x_i)	u_i = =	f_iu_i
1 - 4	6	2.5	-3	-18
4 - 7	30	5.5	-2	-60
7 - 10	40	8.5	-1	-40
10 - 13	16	11.5	0	0
13 - 16	4	14.5	1	4
16 - 19	4	17.5	2	8
				å f_iu_i = -106

Mean = A +
            = 11.5 +
            = 11.5 - 3.18
            = 8.32

This mean number of letters in the surname = 8.32

Now we have to find the modal size of the surnames

Number of letters	Numbered surnames (f)
1 - 4	6
4 - 7	30
7 - 10	40
10 - 13	16
13 - 16	4
16 - 19	4

Here the maximum frequency is 40. The class corresponding to this frequency is 7 - 10.

l = lower limit of the model class = 7

frequency(f₁) of the modal class = 40

frequency(f₀) of the class proceeding the modal class = 30

frequency(f₂) of the class succeeding the modal class = 16

size of the class (h) = 3
mode = l +
         = 7 +
         = 7 +
         = 7 +
         = 7 +  = 7 + 0.88
         = 7.88

\ The model size of the surname = 7.88.

22. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)	40 – 45	45 – 50	50 – 55	55 – 60	60 – 65	65 – 70	70 – 75
Number of students	2	3	8	6	6	3	2

Solution:

Weight (in kg)	Number of students (å f_i)	Cumulative frequency
40 – 45	2	2
45 – 50	3	2+3 = 5
50 – 55	8	5+8 = 13
55 – 60	6	13+6 = 19
60 – 65	6	19 + 6 = 25
65 – 70	3	25+3 = 28
70 – 75	2	28+2 = 30

n = 30,

Here the observation lies in the class 55 – 60

The lower limit (l) = 55

f (frequency of the median class) = 6

cf (cumulative frequency of the class proceeding the median class = 13
Median = l +
           = 55 +
           = 55 +
           = 55 + 3.46
           = 58.46 kg.

23. The following distribution gives the daily income of 50 workers of a factory.

Daily Income (in Rs.)	100-120	120-140	140-160	160-180	180-200
Number of students	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:The less than type cumulative frequency distribution is given as follows:

Daily income (in Rs.)	Number of students	Cumulative frequency
less than 120	12	12
less than 140	14	12 + 14 = 26
less than 160	8	26 + 8 = 34
less than 180	6	34 + 6 = 40
less than 200	10	40 + 10 = 50

Draw the ogive curve by plotting the points(120, 12), (140, 26), (160, 34), (180, 40), (200, 50)

First draw the coordinate axes with upper limits of daily income along the horizontal axis and the cumulative frequency along the vertical axis.

24. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:Draw a less than type ogive curve, by plotting the points(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
Here n = 35,

The x-coordinate of the point 17.5 will be the median.

25. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg/ha)	50-55	55-60	60-65	65-70	70-75	75-80
Number of fans	2	8	12	24	38	16

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Production yield kg / ha	Number of forms	Cumulative frequency
More than or equal to 50	2	100
More than or equal to 55	8	98
More than or equal to 60	12	90
More than or equal to 65	24	78
More than or equal to 70	38	54
More than or equal to 75	16	16

Now draw the ogive by plotting the points(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16).

WORKSHEET Management of Natural Resources

2011-06-08T12:10:00.000-07:00

1. What changes would you suggest in your home in order to be environment-friendly?

Solution:The following can be adhered to, to be environment-friendly.
(i) Taking a bus, rather than using our personal vehicle or walking/cycling.

(ii) Using bulbs or fluorescent tubes in our homes.

(iii) Instead of using the lift, we can take the stairs.

(iv) Wearing an extra sweater rather than using a heating device on cold days.

2. Can you suggest some changes in your school, which would make it environment friendly?

Solution:By following the given points, we can make our school environment friendly.
(i) We can plant a number of trees.

(ii) Toxic substances like plastic should be avoided, in canteens.

(iii) Students must make it a point to throw the plastic covers in one place, rather than throwing everywhere.

3. We saw in this chapter that there are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management of forest produce? Why do you think so?

Solution:Management of forest resources has to take into account the interests of various stakeholders. Among the various stakeholders "The Forest Department of the Government" have the authority to decide the management of forest and its products. They have the authority because the land belongs to the government and only they can decide on the forest product of the particular area.

4. How can you as an individual contribute or make a difference to the management of

(a) forests and wildlife,

(b) water resources and

(c) coal and petroleum?

Solution:(a) Forests and Wildlife
·  Educating the people about the importance of forest.

·  Using less forest products and avoid wasting forest resources.

·  Changing our life style and adapting eco-friendly methods.

(b) Water Resources

·  Usage of recycled water for gardens.

·  Avoid using water directly from tap; instead collect water in containers and use for various purposes.

·  Create awareness about rainwater harvesting.

(c) Coal and Petroleum

·  Taking a bus, rather than using our personal vehicle.

·  If the distance for traveling is short walking will be better option rather than taking a vehicle.

·  Be conservative while using LPG at homes.

5. What can you as an individual do to reduce your consumption of the various natural resources?

Solution:As an individual we reduce the consumption of petrol either by walking to places close by. If the place is in a far of distance then we ca use the public transport instead of using our own vehicle. Changing our life style and adapting eco-friendly methods.

6. On the basis of the issues raised in this chapter, what changes would you incorporate in your life-style in a move towards a sustainable use of our resources?

Solution:As an individual the changes which we can incorporate in our life-style is a move towards a sustainable use of our resources are:
(i) Taking a bus, rather than using our personal vehicle or walking/cycling.

(ii) Using bulbs or fluorescent tubes in our homes.

(iii) Instead of using the lift, we can take the stairs.

(iv) Wearing an extra sweater rather than using a heating device on cold days.

(v) Be conservative while using LPG at homes.

(vi)Changing our life style and adapting eco-friendly methods.

7. What would be the advantages of exploiting resources with short-term aims?

Solution:Exploitation of resources for short-term aims will be more advantageous than long term needs because only fewer amounts will be exploited while the remaining resource will be retained for future generation.

8. How would these advantages differ from the advantages of using a long-term perspective in managing our resources?

Solution:Exploitation of resources for short term is more advantageous because the amount exploited will be comparatively less than long term. Using fewer amounts gives ample time for the depleted resource to replenish

9. Why do you think there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?

Solution:The resources should be equally distributed because concentration of resources in one area would deprive others of the same resources. If the resources has to be transported it may not be economic. The forces working against the equitable distribution are excess usage of one particular resource in one area will lead to the depletion of that resource in that area. It is very essential to use the resource in an economic way.

10. Why should we conserve forests and wildlife?

Solution:Forests are ‘biodiversity hot spots’. It is also the home for the wild life. One measure of the biodiversity of an area is the number of species found there. However, the range of different life forms (bacteria, fungi, ferns, flowering plants, nematodes, insects, birds, reptiles and so on) is also important. One of the main aims of conservation is to try and preserve the biodiversity we have inherited. Experiments and field studies suggest that a loss of diversity may lead to a loss of ecological stability.

11. Suggest some approaches towards the conservation of forests.

Solution:Some of the approaches towards the conservation of forest are as follows

1.The Chipko movement in which the women of the village reached the forest quickly and clasped the tree trunks thus preventing the workers from felling the trees.
2. Active and willing participation of the local community has made the Sal forests of Arabari to undergo remarkable recovery.

WORKSHEET 1 Control and coordination

2011-06-08T12:08:00.001-07:00

1. What is the function of receptors in our body?

Solution:Receptors are usually located in our sense organs, such as the inner ear, the nose, the tongue, and so on. So gustatory receptors will detect taste while olfactory receptors will detect smell.

2. Draw the structure of neuron and explain its function.

Solution:The specialised tips of some nerve cells detect all information from our environment. These receptors are usually located in our sense organs, such as the inner ear, the nose, the tongue, and so on. So gustatory receptors will detect taste while olfactory receptors will detect smell. This information, acquired at the end of the dendritic tip of a nerve cell, sets off a chemical reaction that creates an electrical impulse. This impulse travels from the dendrite to the cell body, and then along the axon to its end. At the end of the axon, the electrical impulse sets off the release of some chemicals. These chemicals cross the gap, or synapse, and start a similar electrical impulse in a dendrite of the next neuron. This is a general scheme of how nervous impulses travel in the body. A similar synapse finally allows delivery of such impulses from neurons to other cells, such as muscles cells or gland. It is thus no surprise that nervous tissue is made up of an organized network of nerve cells or neurons, and is specialised for conducting information via electrical impulses from one part of the body to another.

(a) Structure of neuron,

(b) Neuromuscular junction

3. How does phototropism occur in plants?

Solution:Phototropism is a growth movement induced by a light stimulus. Growth towards a source of light is called positive phototropism, that away from the source is termed negative phototropism. The tips of shoots are usually positively, that of roots negatively phototropic.

Charles Darwin and his son Francis discovered (in 1880) that the phototropic stimulus is detected at the tip of the plant.
The Darwins used grass seedlings for some of their experiments. When grass seeds germinate, the primary leaf pierces the seed coverings and the soil while protected by the coleoptile, a hollow, cylindrical sheath that surrounds it. Once the seedling has grown above the surface, the coleoptile stops growing and the primary leaf pierces it.

The Darwins found that the tip of the coleoptile was necessary for phototropism but that the bending takes place in the region below the tip.

If they placed an opaque cover over the tip, phototropism failed to occur even though the rest of the coleoptile was illuminated from one side.

However, when they buried the plant in fine black sand so that only its tip was exposed, there was no interference with the tropism - the buried coleoptile bent in the direction of the light.

From these experiments, it seemed clear that

The stimulus (light) was detected at one location (the tip)

The response (bending) was carried out at another (the region of elongation).

This implied that the tip was, in some way, communicating with the cells of the region of elongation.

4. How does chemical coordination occur in plants?

Solution:It has been found that the growth of plants is regulated by certain chemical substances which are synthesized by the plants in very small amounts. These are known as plant hormones or phytohormones. They are the organic substances which either promote or inhibit growth. A phytohormones can be defined as a chemical substances which are produced naturally in plants and are capable of translocation and regulating one or more physiological processes when present in low concentration. Main categories of plant hormones are:

(i) Auxins

(ii) Gibberellins
(iii) Cytokinins

(iv) Ethylene

(v) Abscisic acid

Auxins and Gibbrerellins stimulate cell elongations, cytokinins stimulate cell division ethylene promotes transverse or isodiametric growth and abscisic acid is a growth inhibitor.

5. What is the need for a system of control and coordination in an organism?

Solution:Co-ordination in this sense refers to the regulation or control of body activity.

Plants need very little in the way of a control system. Since growth and reproduction are about the only things that are regulated, a rapid control system is not required and hormonal control is all they possess.

Animals are continually moving through new environments that may pose all types of changes and threatening situations to the organism. This requires the rapid and precise control of a nervous system. Hormones regulate slower activities, such as growth, development and reproduction.

6. How are involuntary actions and reflex actions different from each other?

Solution:All reflex actions are involuntary in nature but all involutary actions are not reflexes as the beating of heart is an involuntary action but is not reflex action.

Reflex actions are very quick but all involutary actions may not be very fast as in heart beating.

A reflex action may involve any muscle or a gland as we withdraw our hand on touching a hot object but all involuntary actions involve only smooth i.e., involuntary or cardiac muscles.

Reflex actions are at the level of spinal cord whereas the involuntary actions generally involve brain too.

Nerves and autonomious nervous system can increase or decrease the rate of involuntary actions but reflex actions can be controlled by great will only and are not usually controllable.
Reflex actions are done to meet emergencies where as an inv.action may or may not be for just meeting an emergency but may be a critical lie process as circulation of blood, swallowing of food, movement of food in food pipe, etc.

Digestion in Humans

2011-06-08T12:00:00.001-07:00

Digestion in Humans

Digestion occurs through a series of connected organs whose purpose is to break down or digest the food we eat. Food is made up of large complex substances which the digestive system breaks down into smaller simpler substances that can be absorbed into the bloodstream. The simple molecules travel through the bloodstream to all the body cells which use them for growth, repair, and energy.

All animals have a digestive system, a feature that distinguishes them from plants. Animals including man must take in food in the form of organic matter, either from plants or other animals.

Digestion generally involves two phases:

A mechanical phase and
A chemical phase.

In the mechanical phase, teeth or other structures physically break down large pieces of food into smaller pieces. In the chemical phase, digestive chemicals called enzymes break apart individual molecules of food to yield molecules that can be absorbed and distributed throughout the body. Glands in the body secrete these enzymes.

The digestive system of most animals consists mainly of a long continuous tube called the alimentary canal, or digestive tract. This canal has a mouth at one end, through which food is taken in, and an anus at the other end, through which digestive wastes are excreted. Muscles in the walls of the alimentary canal move the food along. Most digestive organs are part of the alimentary canal. However, two accessory digestive organs, the liver and pancreas, are located outside the alimentary canal. These organs contribute to chemical digestion by releasing digestive juices into the canal through tubes called ducts.

WORKSHEET -1-Life Processes

2011-06-08T11:59:00.001-07:00

1. How are fats digested in our bodies? Where does this process take place?

Solution:The small intestine is the site of the complete digestion of carbohydrates, proteins and fats. It receives the secretions of the liver and pancreas for this purpose. The food coming from the stomach is acidic and has to be made alkaline for the pancreatic enzymes to act. Bile juice from the liver accomplishes this in addition to acting on fats. Fats are present in the intestine in the form of large globules, which make it difficult for enzymes to act on them. Bile salts break them down into smaller globules increasing the efficiency of enzyme action. The pancreas secretes pancreatic juice, which contains enzymes like trypsin for digesting proteins and lipase for breaking down emulsified fats. The walls of the small intestine contain glands, which secrete intestinal juice. The enzymes present in it finally convert the proteins to amino acids, complex carbohydrates into glucose and fats into fatty acids and glycerol.

2. What is the role of saliva in the digestion of food?

Solution:When we eat something we like, our mouth ‘waters’. This is actually not only water, but also a fluid called saliva secreted by the salivary glands. Another aspect of the food we ingest is its complex nature. If it is to be absorbed from the alimentary canal, it has to be broken into smaller molecules. This is done with the help of biological catalysts called enzymes. The saliva contains an enzyme called salivary amylase that breaks down starch, which is a complex molecule to give sugar. The food is mixed thoroughly with saliva and moved around the mouth while chewing by the muscular tongue.

3. What are the necessary conditions for autotrophic nutrition and what are its byproducts?

Solution:Carbon and energy requirements of the autotrophic organism are fulfilled by photosynthesis. It is the process by which autotrophs take in substances from the outside and convert them into stored forms of energy. This material is taken in the form of carbon dioxide and water, which is converted into carbohydrates in the presence of sunlight and chlorophyll. Carbohydrates are utilised for providing energy to the plant. The carbohydrates, which are not used immediately, are stored in the form of starch, which serves as the internal energy reserve to be used as and when required by the plant.

4. What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.

Solution:The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. Diverse organisms do this in different ways – some use oxygen to breakdown glucose completely into carbon dioxide and water; some use other pathways that do not involve oxygen. In all cases, the first step is the breakdown of glucose, a six-carbon molecule, into a three-carbon molecule called pyruvate. This process takes place in the cytoplasm. Further, the pyruvate may be converted into ethanol and carbon dioxide. This process takes place in yeast during fermentation. Since this process takes place in the absence of air (oxygen), it is called anaerobic respiration. Breakdown of pyruvate using oxygen takes place in the mitochondria. This process breaks up the three-carbon pyruvate molecule to give three molecules of carbon dioxide. The other product is water. Since this process takes place in the presence of air (oxygen), it is called aerobic respiration. The release of energy in this aerobic process is a lot greater than in the anaerobic process.

5. How are the alveoli designed to maximise the exchange of gases?

Solution:Within the lungs, the passage divides into smaller and smaller tubes, which finally terminate in balloon-like structures, which are called alveoli. The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels. As we have seen in earlier years, when we breathe in, we lift our ribs and flatten our diaphragm, and the chest cavity becomes larger as a result. Because of this, air is sucked into the lungs and fills the expanded alveoli. The blood brings carbon dioxide from the rest of the body for release into the alveoli, and the oxygen in the alveolar air is taken up by blood in the alveolar blood vessels to be transported to all the cells in the body. During the breathing cycle, when air is taken in and let out, the lungs always contain a residual volume of air so that there is sufficient time for oxygen to be absorbed and for the carbon dioxide to be released.

6. Describe double circulation in human beings. Why is it necessary?

Solution:
The double circulatory system of blood flow refers to the separate systems of pulmonary circulation and the systemic circulation.
The adult human heart consists of two separated pumps, the right side with the right atrium and ventricle which pumps deoxygenated blood into the pulmonary circulation.
The oxygenated blood re-enters the left side of the heart through the pulmonary vein into the left atrium and passes to the left ventricle where it is pumped to the rest of the body. This part of the circulation is called as systemic circulation. This type of circulation is called double circulation. The advantage of a double circulatory system is that blood can be pumped to the rest of the body at a higher pressure.

2 - Forest and Wildlife Resources

2011-06-08T03:52:00.001-07:00

1. What is biodiversity? Why is biodiversity important for human lives?

Solution:Biodiversity is the degree of variation of life forms within a given ecosystem, or on an entire planet. There are millions of living organisms on planet earth. All these living organisms, including man, are interdependent on each other.

2. How have human activities affected the depletion of flora and fauna? Explain.

Solution:Cutting down of forests for agricultural expansion, large scale developmental projects, grazing and fuel wood collection and for urbanization has led to the depletion of flora and fauna.

3. Describe how communities have conserved and protected forests and wildlife in India?

Solution:In India many traditional communities still live in the forests and depend on their livelihood for forest produce. These communities are working hand in hand with the government to conserve forests.
In Sariska Tiger Reserve, Rajasthan, villagers fought against mining activities. In Alwar district of Rajasthan, local communities belonging to five villages have set their own rules and regulations in 1,200 hectares of forest land. They have named it as the Bhairodev Dakav ‘Sonchuri’. Hunting is not allowed in these lands and outside encroachments are prohibited.

The famous Chipko movement was started in the Himalayan region to stop deforestation. People belonging to the local community took to afforestation in a big way. Indigenous species were cultivated and protected.

Involving local communities in protecting the environment, and stopping degradation of forests has reaped many benefits.

4. Write a note on good practices towards conserving forest and wildlife.

Solution:In 1972, the Indian Wildlife (Protection) Act was implemented. It made protecting specific habitats a law. A list of wildlife species that had to be protected was published and hunting these animals was against the law.
National Parks and Wildlife Sanctuaries were set up in many states to protect endangered species.

Under the Wildlife Act of 1980 and 1986, several insects have also been included in the list of protected species. Butterflies, moths, beetles, dragonflies and even certain plants are included in the protected list.

"Project Tiger" was initiated in 1973 by the government of India to protect tigers. It is one of the most well publicized wildlife campaigns in the world.

1 - Resources and Development

2011-06-08T03:51:00.001-07:00

Types of Resources

On the basis of origin

Biotic resources - All the resources in the biosphere having life are called biotic resources. Eg. Human beings, animals. plants etc.

Abiotic resources - all the resources in the nature which dont have life are called abiotic resources. Eg. Water, air, rocks etc.

On the basis of exhaustiblity

Renewable resources - All the resources which get replenished by natural processes at a rate comparable or faster than its rate of consumption by humans. Eg. Water, wind, forests etc.

Water Forests Solar energy

Non-Renewable resources
All the resources which cannot be produced, re-grown, regenerated, or reused on a scale which can its consumption rate. These resources often exist in a fixed amount, or are consumed much faster than nature can recreate them. Eg. Fossil fuels.

On the Basis of Ownership

Individual ResourcesAll the resources privately owned by an individual are called individual resources. Eg. House, car, farmland etc.

Individual resources

Community Owned resources - the resources owned be a group of people or the members of a community are called community resources. Eg. Burial grounds, park, ponds.

Community owned resources

National resources- these are the resources which belong to a nation or the government. Eg. Forests, mountains, wildlife, rivers, etc.

National resources

International resources - these are the resources which are under the control of international organisations. Eg. Open ocean , resources in the Antarctic.

On the basis of status of development.

Potential resource - Potential resources are those which exist in a region and may be used in the future. For example, Petroleum may exist in many parts of India having sedimentary rocks but till the time it is actually drilled out and put into use, it remains a potential resource.

Developed resources - Those which have been surveyed, their quantity and quality determined and are being used in present times.

For example, the petroleum and the natural gas which is obtained from the Bombay High Fields.

The development of an actual resource depends upon the technology available and the involved.

Stock - the resources which have been identified but cannot be used due to lack of technology are called stock.

Reserve - The part of the stock which can be developed profitably with the available technology is called a reserve.

worksheet 1 Triangles

2011-06-08T03:48:00.000-07:00

1. Give two different examples of pair of (i) similar figures. (ii) non-similar figures.

Solution:Two square of sides 4 cm and 8 cm each.
A rhombus and a trapezium .

2. State whether the following quadrilaterals are similar or not

Solution:Similar

3. In the figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:(i)

In DABC
DE is parallel to BC
By Basic Proportionality Theorem
------------------- (1)
Given: AD = 1.5 cm, DB = 3 cm, AE = 1 cm
Let EC = ‘x’ cm
Applying in (1)

1.5x = 3
x =
x = 2 cm
EC = 2cm

(ii)

Since DE || BC, using BPT
…………………………. (1)
Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm
Let AD be = x
sub. in (1)

x =
=
\ AD = 2.4 cm

4. E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution:

(i) PE = 3.9 cm, EQ = 3 cm
PF = 3.6 cm, FR = 2.4 cm
=
=
\
EF is not parallel to QR by convene of BDT

(ii) PE= 4cm QE = 4.5cm PF = 8 RF = 9cm
=

\
EF || QR by convene of BPT.

(iii) PQ = 1.28 cm ,  PR = 2.56 cm,  PE = 0.18 cm,  PF = 0.36 cm
EQ = PQ – PE
     = 1.28 – 0.18
     = 1.10
FR = PR – PF
     = 2.56 – 0.36
     = 2.20

Þ
Þ EF is  parallel to QR

5. In the figure, if LM || CB and LN || CD, prove that

Solution:

Given: LM || CB and LN || CD
To prove:
ProofIn D ABC
LM || BC using basic proportionality Theorem
\  ………………………….. (1)
Also in D ADC
LN ||  DC
\  using basic proportionality Theorem …………….. (2)

from (1) and D

Here proved .

6. In the figure, DE || AC and DF || AE. Prove that

Solution:

Given: ABC is a triangle and DE || AC and DF is parallel to AE
To prove:
ProofIn D ABC,
DE || AC (given)
(By BPT) …………………….. (1)

In D AEB,
\ DF || AE
(By BPT) ………………………. (2)
comparing equation (1) and equation (2)

Here proved.

7. In the figure, DE || OQ and DF || OR. Show that
EF || QR.

Solution:

Given: DE || OQ and DF || OR
To prove : EF || QR
Proof
In D POQ
DE || OQ (given)
By using BPT
………………………… (1)
In D POR
………………………….(2)
By comparing equations (1) and (2)

By using inverse of BPT
EF || QR
Here proved.

8. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR.

Solution:

Given: A, B and C are points on OP, OQ and OR respectively such that AB || PQ, AC || PR
Proof
In D OPQ
AB || PQ (Given)
\ (By using BPT) ……………………………. (1)
In D OPR
Since AC || PQ
(By using BPT) ………………….. (2)

By comparing (1) and (2)

By using converse of BPT BC || QR
Here proved.

9. Using Theorem , prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution:

Given: ABCD is a trapezium and the diagonals AC and BD intersect at 0.

To prove: The ratio
Construction : Draw OM || AB meeting BC at ‘M’
Proof: In D ACB OM || AB
\ By using BPT ………………………………………. (1)
|||ly In D BDC
OM || CD [ \(OM || AB AND AB || CD Þ OM || CD)]
\ using BPT
Taking the reciprocal
…………………………….(2)
from (1) and (2)

(or)
Here proved.

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that. Show that ABCD is a trapezium.

Solution:

Given: ABCD is a quadrilateral and the diagonals AC and BD intersect of ‘O’ such that

To prove: ABCD is a trapezium
Construction: Draw OM || AB
ProofIn D ACB \ OM || AB (By construction)
………………………. (1)
Given:
(or)
sub = in equation (1)
\ =
or =
In D BDC by converse of BPT
OM || DC. But OM || AB
Þ AB || CD
Þ ABCD is a trapezium.
11. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Solution:(i) Similar
AAA D ABC ~ D PQR
(ii) Similar
SSS, D ABC ~ D QRP
(iii) Not similar
(iv) Not smilar
(v) Similar
(vi) Not similar
12. DODC ~ D OBA, Ð BOC = 125°and Ð CDO = 70°. Find Ð DOC, Ð DCO and Ð OAB.
Solution:

Given: D ODC ~ D OBA
Ð BOC= 125⁰
Ð CDO = 70⁰
To find Ð DOC, Ð DCO and Ð OAB
Ð DOC= 180 – 125 = 55
Ð DCO = 125 – 70 = 55
Ð OAB =55 ( Q D ODC ~ D OBA).
13. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles show that
Solution:

Given. ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at ‘O’.

To prove: =

Proof
In the figure consider the triangle OAB and OCD

Ð DOC = Ð AOB (Vertically opposite angles are equal)

Ð DCO = Ð OAB (Alternate angles are equal)

\ By AA corollary of similar triangles.
\ D OAB ~ D OCB When the two triangle are similar, the side are proportionally.
Þ =
Here proved.
14. In the figure, = and Ð 1 = Ð 2. Show that Δ PQS ~ Δ TQR.
Solution:Given: =
and Ð 1 = Ð 2
To prove: D PQS ~ D TQR
Proof
In D PQR Ð 1 = Ð 2 Ð PQR = Ð PRQ
Þ PQ = PR
=
=
= ( \ QP = PR proved above )Ð 1 is a common angle
By SAS similarity D PQS ~ D TQR Hence proved.
15. S and T are points on sides PR and QR of D PQR such that Ð P = Ð RTS. Show that D RPQ ~ D RTS.
Solution:
Given: S and T are the points on sides. PR and QR of a. D PQR such that Ð P = Ð RTS
To prove: D RPQ ~ D RTS
Proof:
D S PQR and D T TRs
Ð QRP = Ð TRS (common angle) Ð STR = Ð QPR (given)
\ By similarly
D S RPQ ~ D RTS
Here proved.
16. In the figure, if ΔABE @ Δ ACD, show that Δ ADE ~ Δ ABC.
Solution:Given: D ABE @ D ACD
To prove: Δ ADE ~ Δ ABC
Prove: D ABE @ D ACD
\ AB = AC …………………… (1) (since corresponding parts of D ^s are equal)
and AD = AE …………………… (2)

and ÐA is a common angle \ÐDAE = ÐBAC
By SAS similarity condition
D ADE ~ D ABC .
17. In the Figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:
(i) Δ AEP ~ Δ CDP
(ii) Δ ABD ~ Δ CBE
(iii) Δ AEP ~ Δ ADB
(iv) Δ PDC ~ Δ BEC.
Solution:

Given: D ABC in which the altitudes AD and CE intersect at P.
To prove:
(i) D AEP ~ D COP (ii) D ABD ~ D CBE
(iii) D AEP ~ D AOB (iv) D PDC ~ D BEC
Proof :(i) D AEP and D CDP
Ð APE = Ð CPD                      (Vertically opposite angle and equal)
Ð AEP = Ð CDP                      (Both are equal to 90⁰ since CE and AD are altitudes)
\ BY AA similarity
D AEP ~ D CDP
Here proved.

(ii) To prove D ABD ~ D CBE
In D ABD and D CBE
Ð CBE = Ð ABD                              (Common angle)
  Ð ADB = Ð CEB                             (Both are equal to 90^o)
\ By AA similarity
D ABD ~ CDE

(iii) To prove. D AEP ~ ADB
In D AEP and D ADB
Ð PAE = Ð DAB                             (common angle)
Ð PAE = Ð DAB                      (Both are equal to 90^o)
Þ By AA similarity
D AEP ~ D ADB

(iv) To prove: D PDC ~ D BEC
Proof: In D PDC and D BEC
Ð PCD = Ð ECB                             (common angle)Ð PDC = Ð BLC                             (Both are equal to 90⁰)
D PDC ~ BFC (AA similarity).
18. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) D ABC ~ D AMP
(ii)
Solution:

(i) To prove D ABC ~ D AMP
In D ABC and D AMP
(Both are equal to 90⁰)
(common angle)
\ BY AA corollary
&D ABC ~ AMP

(ii) To prove

Proof
As proved above
D ABC ~ D AMP
Since the corresponding sides of the similar triangles are proportional

Þ
Here proved.
19. CD and GH are respectively the bisectors of Ð ACB and Ð EGF such that D and H lie on sides AB and FE of D ABC and D EFG respectively. If D ABC ~ D FEG, show that:
(i)
(ii) D DCB ~ D HGE
(iii) D DCA ~ D HGF
Solution:

Given: D ABC ~ D FEG and CD and GH are bisectors of the Ð ACB and Ð EGF of the D ACB and D EGF respectively.
To prove:
(i)
(ii) D DCB ~ D HGE
(iii) D DCA ~ D HGF
Proof: (i)
To prove:

consider D ADC and D FGH
Ð BAC = Ð EFG (Q D ABC ~ D FEG) is given)
(or)Ð DAC = Ð HFG
Also
Ð BCA = Ð EGF (Q D ABC ~ D FEG is given)
´ by
Ð BCA = Ð EGF
=> Ð DCA = Ð HGF

\ By AA corollary
D DAC ~ D HFG
\

(ii) To prove: D DCB ~ D HGE
In D DBC and D HEG
Ð ABC = Ð FEG [QD ABC ~ D FEG]
(or) Ð DBC = Ð HEG [ Q D and H lie on the line AB and EF respectively]
Ð BCA = Ð EGF …………………..(1) [Q D ABC ~ D FEG]
multiplying by
Ð BCA = Ð EGF
(or) Ð DCB = Ð HGE …………………………..(2) [Q CD and GH are the bisectors of the Ð ACB and Ð EGF of the D ACB and D EGF respectively]

from (1) and (2)
D DCB ~ D HGE [By AA corollary]

(iii) Already proved in (i)
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC prove that D ABD ~ D ECF.

Given: D ABC is isosceles with AB = AC. E is a point as CB such that EF perpendicular to AC. AD is perpendicular to BC.
To prove: D ABD ~ D ECF
Proof :
D ABC is isosceles with AB = AC (Given)
\ ÐABC = ÐACB ………………. (1) Property of an isosceles D
and ÐADB = ÐEFC ……………………(2) Both are equal to 90⁰Now, in D ABD and D ECFÐABD = ÐECF From equation (1)
and ÐADB = ÐEFC From equation (2)
\ D ABD ~ D ECF .
20. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC prove that D ABD ~ D ECF.
Solution:

Given:
Replacing BC = 2BD and QR = 2QM
Since AD and PM are the medians from A and P respectively.

Þ
Since the sides of D ABD ~ D PQM ÐABD = ÐPQM
ÐABC = ÐPQR -------------------------- (1)
Now from given data
ÐABC = ÐPQR
Þ D ABC ~ D PQR (By SAS similarity condition).

21.  A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

Since the shadow is cast of the since from the angle made by the sun’s ray is same
2 D ABC and PQR
ÐACB = ÐPRQ
ÐABC = ÐPQR          (Q both are equal to 90⁰)
Þ D ABC ~ D PQR (By AA corollary)
\ =

or 6 ´  = 42
Height of the tower = 42 m .

22.  If AD and PM are medians of triangles ABC and PQR, respectively where D ABC ~ D PQR, prove that

Solution:
Given: D ABC ~ D PQR
To prove :
Proof:
In D ABC ~ D PQR
\ = ………………………………(1)
and Ð B = Ð Q ……………………………..(2)
replacing BC = 2BD and QR = 2QM in equation (1)
……………………………..(3)
from (2) and (3)
D ABD ~ D PQM By SAS similarly
\
Here proved .

23.  Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of D PQR. Show that DABC ~ D PQR.

Solution:
Given: D ABC and AD is the median from A to BC and D PQR, PM is the median from P to, QR. Also

=

To prove: D ABC ~ D PQR

Construction : Draw DE || AC and DF || AB in D ABC .

Draw MN || PR and MO || PQ in D PQR .

Proof:

In D ABC, since, AD in the median D is the midpoint of BC and DF is parallel to AC by construction E is the midpoint of AB(using midpoint theorem)

Similarly N is the midpoint of PQ

In D ABC and PQR, given that
=
=  ………………………………… (1)
In D AEO AND PNM, the sides are proportional (from (1)
D AED ~ D PNM
Þ ÐDAE = ÐMPN ……………………….. (2)
|||ly we can prove that D AFD ~ D POM are similar. Hence
ÐDAF = ÐMPO …………………….(3)

Adding equation (2) + Equation (3)

ÐDAE + ÐDAF = ÐMPN + ÐMPO

ÐEAF = ÐNPO

ÐBAC = ÐQPR   …………………….(4)

As  and from equation (4)

\ By SAS similarity criterion

D ABC ~ D QPR

Here the result.

24.  Let D ABC ~ D DEF and their areas be, respectively, 64 cm² and 121 cm2. If EF = 15.4cm, find BC.

Solution:Since D ABC ~ D DEF
=

(or)  =
=
BC =  ´ 15.4
BC = 11.2 cm.

25.  Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:
In D AOB and COD
ÐCOD = ÐAOB  (vertically opposite angles)ÐCDO = ÐOBA (Alternate interior angles)
By AA similarity
D COD ~ D AOB
=  =
\  =

26.  In the figure ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Solution:

Construction: From A and D draw perpendiculars to BC meeting BC at E and F respectively.
Proof:In D AFO and D DEO

ÐAFO = ÐDEO = 90⁰ ÐAOE = ÐDOE   (vertically opposite angle)
Þ D AOE ~ D DOF (By AA similarity)
\  =  =  ………………………. (1)
\ =
Here proved.

27.  If the areas of two similar triangles are equal, prove that they are congruent.

Solution:Let ABC and DEF be two similar triangles where areas are equal.
(ie) ar(D ABC) = ar(D DEF)
=  =  =
since ar. D ABC = ar.D DEF
=  = 1
\  = 1 or AB² = DE²Þ AB – DF
|||ly BC = EF
and AC = DF
By SSS congruency
D ABC @ D DEF
Here the areas of two similar D ^s are equal, those two triangles are congruent.

28.  D, E and F are respectively the mid-points of sides AB, BC and CA of D ABC. Find the ratio of the areas of D DEF and D ABC.

Solution:
Since D and F are the midpoints of the AB and AC, DF || AC or DF || BE
Similarly EF || AB or EF || DB
AEFD is a parallelogram as both pairs of opposite sides are parallel.
\ By the property of parallelogram

ÐDBE = ÐDFE = (In an parallelogram opposite angle are equal)
or ÐDFE = ÐABC   -------------------(1)
Similarly ÐFEB = ÐACB   …………………………..(2)
In D DEF and D ABC, from
Equation (1) and (2)
D DEF ~ D CAB
\  =  =
\ ar D DEF: ar D CAB = 1 : 4 .

29.  Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:
Given: AM and DN are medians of the D ABC and D DEF and D ABC ~ D DEF
To prove:
=
Proof : since D ABC ~ D DEF

or
and  ( QD ABC ~ D DEF)
\ By SAS similarity
D ABM ~ D DEN
or  ………………………... (1)
=
=  (From (1)
Hence proved.

30.  Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Given: ABCD is a square and two equilateral triangle are drawn with BC and DB as are units side respectively.
To Prove: ar D PBC =
Proof:
Consider D ^s PBC and D BQD
Since both these triangles are equilateral triangles, by the property of equilateral D s,
both the D ^s are equiangular with each angle equal to 60^o.
Þ D PBC ~ D QBD
\  =  ………………………… (1)
Consider D BCP
Since it is a right angled triangle
BC² + DC² = BD² ( By using Pythagoras theorem)
But BC = DC
QBC² = BD²
Substituting in (1)
=
\ ar D PBC =  or D BDQ
Here the results.

35.  ABC is an isosceles triangle with AC = BC. If AB² = 2 AC², prove that ABC is a right triangle.

Solution:
Given AB² = 2AC² and AC = BC
To prove: ABC is a right triangle
Proof:
Given: AB² = 2AC²
(or) AB² = AC² +AC² ………………(1)
since AC = BC given, equation (1) becomes
AB² = AC² + BC²
\ By the converse of Pythagoras theorem D ABC is a right triangle with  .

36.  ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution:
Given: ABC is an equilateral triangle with each side equal to 2a
Construction: Draw AD ^ r to BC
To find: Length of AD
Construction D ADB and ADC
ÐADB = ÐADC = 90⁰
AC = AB (QABC is isosceles)
AD = AD (common side)
\ D ADB @ D ADB (By RHS property)
Þ BD = DC (CPCT)
BC = BD + DC
= BD + BD
2a = 2BD
BD = a
2D ABD
AB² = AD² + BD²
(2a)² = AD² + a²
4a² = AD² + a²
a² – a² = AB²
3a² = AD²
AD = a
Similarly the other altitudes will also be a only.

37.  Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution:
Given: ABCD is a rhombus and the diagonals AC and BD intersect at 0
To prove: AB² + BC² + DC² + AD² = AC² + BD²
Proof: The diagonals of C rhombus intersect at right angle.
\ = 90⁰
In D AOB
AO² + OB² = AB²
+ = AB²
Hence AC² + BD² = 4AB²
Since 4AB² can be taken as sum of the square of the 4 side3 of the rhombus as each side of the rhombus equal, the required result in proved.

38.  In the figure O is a point in the interior of a triangleABC, OD perpendicular to BC, OE perpendicular to AC and OF perpendicular to AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²,(ii) AF² + BD² + CE² = AE² + CD² + BF².

Solution:
(i) Given : In D ABC OD ^ BC, OE ^ AC and OF ^ AB
Construction: Join OC, OA and OB.
In D OBD, OB² = OD² + BD²
In D OFA, OA² = OF² + AF²In D OCE, OC² = OE² + CE²\ OB² + OA² + OC² = OD² + OF² + OE² + BD² + AF² + CE²
(or) OB² + OA² + OC² – OD² – OF² – OE²
= BD² + AF² + CF²(ii) From the result above
AF² + BD² + C² = OB² + OA² + OC – OD² – OF² – OE²
                      = (OA² – OE²) + (OC² – OD²) +(OB² – OF²)
                      = AE² + CD² + BF²

39.  A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution:
In the figure AC = ladder = 10 m
AB = Height of the window = 8m
BC = Distance from the foot of the wall to the foot of the ladder = x m
ABC is a right angle D
AB² + BC² = AC²
(or) 8² + x² = 10²
x² + 10² - 8²x² = (10 + 8) (10 – 8)
    = (18)(2)
    = 36 x =  = 6 m
\ The distance of the foot of the ladder from the wall = 6 m

40.  A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution:
In D ACB, AC = length of the wire = 24m and the largest of the tower = 18 m
18² + x² = 24² (By using Pythagoras theorem)
x² + 24² = 18²
x²= 576 – 324
   = 252
x =
   = 6m
Therefore, the station should be driven to a distance of 6m so that the wire will be taunt.

41.  An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 hours?

Solution:
Distance traveled by the aeroplane in the North directions is
AB = 1000 ´  = 1500km. (Distance = speed ´ time)
And the Distance traveled by the Distance traveled by the aeroplane in the was direction is AB = 1200 ´  = 1800 km
BC is the distance between the two 11/2 hours.
AB² + AC² = BC²
(1500)² + (1800)² = BC²
2250000 + 3240000 = BC²
5490000 = BC²
BC = 100km.

42.  Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution:Let AD be a pole of height 11m and BC be a pole of height 6m
Construction: Join CD and draw ^ r from C to AD meeting AD at E.
Proof: ABCD is rectangle
\ AB = CE = 12m
and BC = AE = 6m
\ DE = AD – AE
11 – 6 = 5m
In D PEC
DE² + EC² = DC²
or 5² + 12² = DC²
25 + 144 = DC²
DC =  = 13 m
the distance between the two poles = 13m.

43.  D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.Prove that AE² + BD² = AB² + DE².

Solution:
To prove: AC² + BD² = AB² + DE²
Proof:
In D AEC
AE² = EC² + AC² ………………………….. (1)
In D BDC
BD² = BC² + CD² ………………………….. (2)
(1) + (2)
A ²+ BD² = EC² + BC² + AC² + CO²
               = (EC² + CO²) + (AC² + BC²)
               = DE² + AB²
Here proved.

44.  The perpendicular from A on side BC of a D ABC intersects BC at D such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC².

Solution:
From DABD

From DADC

From (1) and (2)

BD = 3DC (given)\ BC = BD + DC = 4DC

Now

45.  In an equilateral triangle ABC, D is a point on side BC such that BD =9 AD² = 7 AB².

Solution:
Given: D ABC is an equilateral D . ‘D’ is a point on BC such that BD =  BC.
To prove: 9AD² = 7AB²
construction: Draw AE ^ r to BC meeting C and E
Proof:
In an equilateral triangle the altitude and median are same line regret. Therefore A is the median of the D ABC.E is the midpoint of BC
BC + EC = BC
2EC = BC
BE = EC =  BC ………………………….. (1)
BD = BC (Given)……………………(2)
DC = BC – BD
     = BC - BC .  BC ………………… (3)
DE = BE – B
     = BC -
     = BC
     =  = BC
In D ADE
D ADE
AD² = AE² + ED² …………………(1)
But in D AEC
AC² = AE² + ED²(or) AE² = AC² - EC²substituting in (1)
AD² = AC² – EC² + ED²
      = AC² – DE² - CE²
      = AC² + (DC+CE)(DE – CE)
      = AC² + (DC)(DE – CE)
      = AC² + DC.DE – DC.CE
      = AC² +
      = AC² +
      = AC² +
      = AC² +
      = AC² -
      =
AD² = AB²

46.  In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution:
Let the side of the equilateral D ABC be ‘a’
Let AE be the altitude from A to BC meeting
BC at E.
E is the midpoint of is C
Therefore BE =  BC = a
In D ABF AB² = AB² + BE²
AE² = AB² – BE²
      = a² -
      = a² -
      =  =
AE =  ………………………….. (1)
Now,
Three times, square one of its side = 3a²Four times the square of y
Its one of the altitude = 4 = 3a²
Therefore it is proved that in an equilateral triangle three times the square of one side is equal to four times the square of its altitude.

47.  Tick the correct answer and justify : In Ä ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle B is :
(A) 120° (B) 60°
(C) 90° (D) 45°

Solution:In D ABC AB = 6cm AC = 12 cm
BC = 6 cm

AB² = (6)² = 36 ´ 3 = 108
BC² = 6² = 36
AC² = 12² = 144
AB² + BC² = 108 + 36 = 144 = AC²
AB² + BC² = AC²
Þ = 90Answer = (c) = 90^0.

WORKSHEET 1 Arithmetic Progressions

2011-06-08T03:36:00.000-07:00

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

(iii)The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.

(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.

Solution:(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
The taxi fare will be 15, 15 + 8, 15 + 16, 15 + 24, 15 + 32, 15 + 40 etc.

(i.e.,) 15, 23, 31, 39, 47, 55 etc.

This is an AP as every succeeding term is obtained by adding 8 in its preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

Let amount of air present in the cylinder be "a".

The vacuum removes of air remaining in the cylinder.

\ The air present in the cylinder will be a, a - = , -== ,

a, , , …

This is not an AP because the difference between second term and first term is not the same as the difference between the third term and the second term.

(iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.

The cost of digging the well will be Rs. 150, Rs. (150 + 50), Rs. (150 + 100), Rs. (150 + 150) etc.

(i.e.,) Rs. 150, Rs. 200, Rs. 250, Rs. 300, Rs. 350 etc

This is an AP as every succeeding term is obtained by adding Rs. 50 in its preceding term.

(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.

Initial amount of money = Rs. 10,000

Compound Interest = 8% p.a

Amount every year increase by 8% Compound Interest p.a.

In the second year amount increases to 10,000

In the 3^rd year amount increases to 10,000

In the 4^th year amount increases to 10,000

This is not an AP as the difference is not the same in this problem.

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3 (iv) a = -1, d =

(v) a = -1.25, d = -0.25

Solution:(i) a = 10, d = 10
Let a₁, a₂, a₃, a₄ be the first four terms of the AP

Here a = a₁= 10

a₂= a₁ + d = 10 + 10 = 20

a₃ = a₂ + d = 20 + 10 = 30

a₄= a₃ + d= 30 + 10 = 40

\ The first four terms of the AP are 10, 20, 30, 40.

(ii) a = -2, d = 0

Let a₁, a₂, a₃, a₄ be the first four terms of the AP

Here a = a₁= -2

a₂= a₁ + d = -2 + 0 = -2

a₃ = a₂ + d = -2 + 0 = -2

a₄= a₃ + d= -2 + 0 = -2

Thus first four terms of the AP are -2, -2, -2, -2

(iii) a = 4, d = -3

Let a₁, a₂, a₃, a₄ be the first four terms of the AP

Here a = a₁= 4

a₂= a₁ + d = 4 + (-3) = 1

a₃ = a₂ + d = 1 + (-3) = -2

a₄= a₃ + d= -2 + (-3) = -5

The first 4 terms of the AP are 4, 1, -2, -5

(iv) a = -1 , d =

Let a₁, a₂, a₃, a₄ be the first four terms of the AP

Here a = a₁= -1

a₂= a₁ + d = -1 +  = -

a₃ = a₂ + d = - +  = 0

a₄= a₃ + d= 0 +  =

The first 4 terms of the AP are –1, -, 0, .

(v) a = -1.25, d = -0.25

Let a₁, a₂, a₃, a₄ be the first four terms of the AP

Here a = a₁= -1.25

a₂= a₁ + d = -1.25 + (-0.25) = -1.50

a₃ = a₂ + d = -1.50 + (-0.25) = -1.75

a₄= a₃ + d= -1.75 + (-0.25) = -2.0

The first 4 terms of the AP are -1.25, -1.50, -1.75, -2.0.

3. For the following APs, write the first term and the common difference:

(i) 3, 1, -1, -3, … (ii) –5, -1, 3, 7, … (iii) … (iv) 0.6, 1.7, 2.8, 3.9, …

Solution:(i) The first term is 3 and common difference = a₂ – a₁= 1 - 3 = -2.
(ii) The first term is -5 and common difference = a₂ – a₁= (-1) – (-5) = 4.

(iii) The first term is and common difference = a₂ – a₁= - = .

(iv) The first term is 0.6 and common difference = a₂ – a₁= (1.7) – (0.6) = 1.1.

4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, … (ii) 2, , … (iii) –1.2, -3.2, -5.2, -7.2, …

(iv) –10, -6, -2, 2, … (v) 3, 3 +, 3 + 2, 3 + 3, …

Solution:(i) 2, 4, 8, 16, …
Here a₂ – a₁ = 4 – 2 = 2

a₃ – a₂ = 8 – 4 = 4

a₄ – a₃ = 16 – 8 = 8

Since a_k+1 - a_k is not the same throughout, it is not an AP.

(ii) 2, , …

Here

a₂ – a₁ =  – 2

=

a₃ – a₂ = 3 –  =

a₄ – a₃ =  – 3 =

As a_k+1 - a_k is the same throughout, it is an AP with first term = 2 and common difference =

The next three terms are 4, 4 +  = ,  +  = = 5.

(iii) –1.2, -3.2, -5.2, -7.2, …

Here

a₂ – a₁ = -3.2 – (-1.2)

= -3.2 + 1.2 = -2

a₃ – a₂ = -5.2 – (-3.2)

= -2

a₄ – a₃ = -7.2 – (-5.2)

= -2

Since a_k+1 – a_k is the same throughout it is an AP with first term (a) = -1. 2 and common difference, (d) = -2

Thus the next three terms are

a₅ = a₄ + d = –7.2 – 2 = -9.2

a₆ = a₅ + d = -9.2 – 2 = -11.2

a₇ = a₆ + d = - 11.2 – 2 = - 13.2

(iv) –10, -6, -2, 2, …

Here

a₂ – a₁ = -6 – (-10)

= 4

a₃ – a₂ = -2 – (-6)

= 4

a₄ – a₃ = 2 – (-2)

= 4

As a_k+1 – a_k is the same throughout it is an AP with 1^st term as -10 and common difference as 4.

Thus the next three terms are

a₅ = a₄+ d = 2 + 4 = 6

a₆ = a₅ + d = 6 + 4 = 10

a₇ = a₆ + d = 10 + 4 = 14

(v) 3, 3 +, 3 + 2, 3 + 3, …

Here a₂ – a₁ = 3 + -3 =

a₃ – a₂ = 3 + 2-(3+) =

a₄ – a₃ = 3 + 3-(3+2) =

In this problem, a_k+1 –a_k is the same throughout thus the given list is an AP with 1^st term as 3 and common difference (d) as .

Thus the next three terms are

a₅ = a₄+ d = (3 + 3) +  = 3 + 4

a₆ = a₅ + d = (3 + 4) +  = 3 +5

a₇ = a₆ + d = (3 +5) +  = 3 + 6

5. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 0.2, 0.22, 0.222, 0.222, … (ii) 0, -4, -8, -12, … (iii), … (iv) 1, 3, 9, 27, … (v) a, 2a, 3a, 4a, …

Solution:(i) 0.2, 0.22, 0.222, 0.2222, …
Here

a₂ – a₁ = 0.22 – 0.2 = 0.02

a₃ – a₂ = 0.222 – 0.22 = 0.002

a₄ – a₃ = 0.2222 – 0.222 = 0.0002

Hence a_k+1 – a_k is not the same throughout, hence it is not an AP.

(ii) 0, -4, -8, -12

Here

a₂ – a₁ = -4 – 0 = -4

a₃ – a₂ = -8 – (-4) = -4

a₄ – a₃ = -12 – (-8) = -4

Here a_k+1 – a_k is the same throughout hence it is an AP with 1^st term = 0

and common difference = -4

Thus the next three terms are

a₅ = a₄+ d = -12 + (-4) = -16

a₆ = a₅ + d = -16 + (-4) = -20

a₇ = a₆ + d = -20 + (-4) = -24

(iii) , …

Here

a₂ – a₁ = - = 0

a₃ – a₂ = –  = 0

a₄ – a₃ = –  = 0

Here a_k+1 – a_k = 0 is the same throughout hence the list is an AP with 1^st term =  and common difference d = 0.

Thus the next three terms are

a₅ = a₄+ d =  + 0 =

a₆ = a₅ + d =  + 0 =

a₇ = a₆ + d =  + 0 = .

(iv) 1, 3, 9, 27, …

Here the 1^st term (a) = 1

a₂ – a₁ = 3 – 1 = 2

a₃ – a₂ = 9 - 3 = 6

a₄ – a₃ = 27 – 9 = 18

Since a_k+1 – a_k is not the same throughout this is not an AP.

(v) a, 2a, 3a, 4a

Here

a₂ – a₁ = 2a - a = a

a₃ – a₁ = 3a – 2a = a

a₄ – a₃ = 4a – 3a = a

Since a_k+1 – a₁ is the same throughout this is a AP with first term = a and common difference = a

Thus the next three terms are

a₅ = a₄+ d = 4a + a = 5a

a₆ = a₅ + d = 5a + a = 6a

a₇ = a₆ + d = 6a + a = 7a

6. Which of the following are APs? If they form an AP, find the common difference d and write three more terms. (i) a, a², a³, a⁴, … (ii) ,,, , …

(iii) , … (iv) 1², 3², 5², 7², … (v) 1², 5², 7², 73, …

Solution:(i) a, a², a³, a⁴, …
Here

a₂ – a₁ = a² - a = a(a – 1)
a₃ – a₂ = a³ – a² = a²(a – 1)

a₄ – a₃ = a⁴ – a³ = a³(a – 1)

Since a_k+1 – a_k is not the same throughout this is not an AP.

(ii) ,,,, …

Here

a₂ – a, =  -  = 2 -  =

a₃ – a₂ =  -  = 3 - 2 =

a₄ – a₃ =  - = 4 - 3 =

Since a_k+1 – a_k is the same throughout it is an AP with the first term as  and common difference d = .

Thus the next three terms are

a₅ = a₄+ d = +  = 4 +  = 5

a₆ = a₅ + d = 5 +  = 6

a₇ = a₆ + d = 6 +  = 7.

(iii)

Here

a₂ – a₁ =  = ( - 1)

a₃ – a₂ =  = 3 -= (- )

a₄ – a₃ = = 2 - 3 = (2 - ).

Here a_k+1 – a_k is not the same throughout hence the list is not an AP.

(iv) 1², 3², 5², 7²

Here

a₂ – a₁ = 3² – 1² = 9 – 1 = 8

a₃ – a₂ = 5² – 3² = 25 – 9 = 16

a₄ – a₃ = 7² – 5² = 49 – 25 = 24

Since a_k+1 – a_k is not the same throughout hence the list not an AP.

(v) 1², 5², 7², 73, …

Here

a₂ – a₁ = 5² – 1² = 25 – 1 = 24

a₃ – a₂ = 7² – 5² = 49 – 25 = 24

a₄ – a₃ = 73 – 7² = 73 – 49 = 24

Since a_k+1 – a_k = 24 is the same throughout this list is an AP with first term = 1² and common difference = 24.

Thus the next three terms are

a₅ = a₄+ d = 73 + 24 = 97

a₆ = a₅ + d = 97 + 24 = 121

a₇ = a₆ + d = 121 + 24 = 145.

7. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_nthe nth term of the AP:

	a	d	n	a_n
(i) (ii) (iii) (iv) (v)	7 -18 …. -18.9 3.5	3 …. -3 2.5 0	8 10 18 … 105	... 0 -5 3.6 ...

Solution:(i) a = 7, d = 3 , n = 8, a_n = ?
a_n = a + (n – 1) ´ d

= 7 + (8 - 1) ´ 3

= 7 + 7 ´ 3 = 7 + 21 = 28.

(ii) a = -18

n = 10

a_n = 0

d = ?

a_n = a + (n – 1) ´ d

0 = -18 + (10 – 1) ´ d

0 = -18 + 9 ´ d

0 = -18 + 9d

9d = 18

d = 2.

(iii) a = ?, d = -3, n = 18, a_n = -5

a_n = a + (n – 1) ´ d

-5 = a + (18 – 1) ´ (-3)

-5 = a + 17 ´ (-3)

-5 = a - 51

a = 51 – 5 = 46.

(iv) a = -18.9

d = 2.5

a_n= 3.6

n = ?
a_n = a + (n – 1) ´ d

3.6 = -18.9 +(n – 1) ´ 2.5

(n – 1) ´ 2.5 = 3.6 + 18.9

(n – 1) = = ( To remove the decimal multiply and divide by 10)

=> n –1 = 9

\ n = 10.

(v) a = 3.5

d = 0

n = 105

a_n = ?

a_n = a + (n – 1) ´ d

= 3.5 +(105 – 1) ´ 0

a_n = 3.5.

8. Choose the correct choice in the following and justify:
(i) 30^th term of the AP: 10, 7, 4, …, is

(A) 97 (B) 77 (C) –77 (D) -87

(ii) 11^th term of the AP: -3, , 2, …, is

(A) 28 (B)22 (C) –38 (D) -48

Solution:(i) 30^th term of the AP: 10, 7, 4, …, is
n^th term = a + (n – 1) ´ d, where n = 30

30^th term = 10 + (30 – 1) ´ (-3)

              = 10 + (29 ´ - 3)

              = 10 – 87 = - 77

(C) is the correct answer.

(ii) 11^th term of the AP: -3,  , 2, …, is
Here n = 11

a = -3

d =  – (-3)

= = = 2

a_n = a + (n – 1) ´ d

a_n = -3+ (n – 1) ´ (2)

    = -3 +(11 – 1) ´  = -3 + 10 ´

    =– 3 + 25 = 22

a₁₃ = 22.

(B) is the correct answer.

9.  In the following APs, find the missing terms in the boxes:
(i) 2,  ,26 (ii)  , 13,  , 3 (iii) 5, , , 9
(iv) -4,, , , , 6    (v) , 38, , , , -22

Solution:(i) Let x be the missing term
x - 2 = 26 - x

2x = 28

x = 14

\ Missing term = 14

(ii)  , 13,  , 3

Let missing term in 1st box be x , missing term in the 2^nd box be y,

13 - x = y - 13 = 3 - y

Now considering the last two ,

y - 13 = 3 - y

2y = 16

y = 8.

The value of y = 8.

Here 13 - x = y - 13

=> 13 - x = 8 - 13

x = 26 - 8

= 18.

(iii) 5, , , 9

Here a = 5

a₄ = 9

n = 4

d = ?

a_n = a + (n- 1) ´ d

9 = 5 + (4 - 1) ´ d

= 5 + 3d

3d =  - 5 =  =

d =  =

\ If a₁ = 5

a₂ = a₁ + d = 5 + == = 6

a₃ = 6  +  = 8

5 ,

(iv) -4,  ,  ,  ,  , 6

a_n = 6

a = -4

d = ?

a_n = a + (n -1) ´ d

6 = -4 + (6 -1) ´ d

6 = -4 + 5d
5d = 6 + 4 = 10

d =  = 2

\ -4,  ,  ,  ,  , 6 is the AP series.

(v)  , 38,  ,  ,  , -22

a₂ = 38

n = 6

a₆ = -22

a_n = a + (n - 1) ´ d if n = 6,

-22 = a + 5d ……. (1)

if n = 2,

38 = a + d ………. (2)

- (2) => 4d = -60

\ d = = -15.

Thus a₂ = a + d = a + (-15)

38 = a - 15

a = 38 + 15 = 53.

a₃ = 38 + (-15) = 23

a₄= 23 + (-15) = 8

a₅ = 8 + (-15) = -7

\ 53 , 38 , 23 , , , -22.

10. Which term of the AP : 3,8, 13, 18, …,is 78

Solution:Let 78 be the n^th term
Here a = 3

d = 5, a_n = 78

a_n = a + (n –1) ´ d

78 = 3 + (n – 1) ´ d

78 – 3 = (n –1) ´ 5

75 = (n – 1) 5

= (n – 1)

n – 1 = 15

n = 16

78 is the 16^th term.

11. Find the number of terms in each of the following APs:
(i) 7, 13, 19, …,205 (ii) 18, 15 , 13, …, 47
- Solution:(i) Let the number of terms be n
  Here a =7
  
  d = 6
  
  a_n = 205
  
  a_n = a + (n – 1) d
  
  205 = 7 + (n – 1) ´ 6
  
  (n – 1)6 = 205 – 7
  n – 1 =
  
  n = 33 +1 = 34
  
  \ Number of terms = 34.
  
  (ii) 18, 15  , 13, …, 47
  
  Here a = 18
  
  d = 18 – 15 = -2
  
  a_n = -47
  
  Let the number of terms be n
  
  a_n = a + (n – 1) d
  
  -47 = 18 + (n –1) (-2  )
  
  -47 – 18 = (n – 1) ()
  
  -65 ´  = (n – 1)
  
  26 + 1 = n
  
  \ Number of terms = 27
12. Check whether – 150 is a term of the AP: 11, 8 , 5, 2 ………
- Solution:Here a = 11
  d = -3
  
  If a_n = -150 be the n^th term of the AP
  
  a_n = a + (n – 1) ´ d
  
  -150 = 11 + (n – 1) ´ (-3)
  
  -150 - 11 = (n – 1) ´ (-3)
  
  = (n – 1)
  
  n = = 53 + 1
  
  which is not a whole number
  
  \ -150 is not a term of the AP.
13. Find the 31stterm of an AP whose 11th term is 38 and the 16th term is 73.
- Solution:11th term is 38 and
  16th term is 73
  
  a₁₁ = a +(n –1)d
  
  38 = a + 10d …………….(1)
  
  a₁₆ = a +(n –1)d
  
       = a +15d ………………..(2)
  
  a + 10d = 38
  
  a + 15d = 73
  
  -5d = -35
  
  d = 7
  
  sub d =7 in (1)
  
  a + 10 d = 38
  
  a + 10(7) = 38
  
  a = -70 + 38
  
  = - 32
  
  Thus 31st term = a + (31 – 1) ´ d
  
                        = -32 + 30 ´ 7
  
                        = – 32 + 210
  
                        = 178.
14. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
- Solution:3rd term is a₃ = a + (3 – 1)d = 12 ………………….. (1)
  50th term is a₅₀ = a + (50 – 1)d = 106 ………………….. (2)
  
  (1) Þ a + 2d = 12 ……………….. (3)
  
  (2) Þ a + 49d = 106 ……………….. (4)
  
  -----------------
  
  (3) – (4) => 47d = 94
  
  d = 2
  
  sub d = 2 in (3)
  
  a + 2(2) = 12
  
  a = 12 – 4 = 8
  
  \ 29th term of an AP is a + (29 – 1)d ….(5)
  
  substituting for a = 8 and d = 2 in (5),
  
  a + 28d = 8 + 28 ´ 2
  
  = 8 + 56
  
  = 64.
15. If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?
- Solution:3rd term of the AP is a + (3 –1)d = (i.e.,) a + 2d = 4 …………. (1)
  
  9th term of the AP is a + (9 – 1)d = -8
  
  a + 8d = -8 …………………….(2)
  
  a + 2d = 4…….(1)
  
  a + 8d = -8…..(2)
  
  (1) – (2) => 6d = -12
  
  d = -2
  
  substituting d = -2 in (1) => a – 4 = 4
  
  a = 8
  
  Let the nth term be 0 then a + (n – 1) d = 0
  
  (i.e.,) 8 + (n – 1) (-2) = 0
  
  -2n + 2 = -8
  
  -2n = -8 –2
  
  = -10
  
  n = 5
  
  Thus n = 5. Hence the 5th term is zero.
16. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
- Solution:17th term of the AP
  = a +(17 – 1)d
  
  = a + 16 d ………………………….. (1)
  
  10th term of the AP
  
  = a + (10 – 1)d = a + 9d ------------- (2)
  
  Given (1) exceeds (2) by 7
  
  [a + 16d] – [a+9d] = 7
  
  16d – 9d = 7
  
  7d = 7
  
  d = 1
17. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
- Solution:54th term of the AP is a + (54 - 1)d
  Here a = 3
  
  d = a₂ – a₁ = 15 – 3 = 12
  
  \ 54th term = 3 + (54 – 1) ´ 12
  
                    = 3 + 53 ´ 12
  
                    = 3 + 636 = 639
  
  132 more than 54th term = 639 + 132 = 771
  
  To find out which term of the AP is 771, let us assume it is the nth term
  
  a + (n – 1) ´ d = 771
  
  substituting for a = 3 and d = 12
  
  3 + (n – 1) ´ 12 = 771
  
  (n – 1) ´ 12 = 771 - 3
  
  (n – 1) =  = 64
  n = 65
  
  Hence the 65th term is 132 more than its 54th term.
18. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
- Solution:Let a₁ and a₂ be the 1st term of the 2 APs and "d" be the common difference.
  100th term of 1st AP = a₁ + (100 – 1)d ………………………. (1)
  
  100th term of 2nd AP = a₂ + (100 – 1)d ………………………. (2)
  
  Difference between (1) and (2) = 100
  
  a₁ + 99d – (a₂ + 99d) = 100
  
  (i.e.,) a₁ – a₂ = 100 …………….(3)
  
  Difference between then 1000^th terms = (a₁ + 999d) – (a₂ + 999d)
  
  = a₁ – a₂ = 100 (from (3))
  
  Thus the difference between their 1000th terms = 100.
19. How many three-digit numbers are divisible by 7?
- Solution:The list of 3 digit numbers divisible by 7 are 105, 112, 119, …, 994.
  This is an A.P. with a = 105, d = 7
  
  a_n = 994
  
  \ 994 = a+(n – 1) ´ d
  
           = 105+(n – 1) ´ 7
  
  7(n – 1) = 994 – 105
  
              = 889
  
  n – 1 =  = 127
  
  n = 128
  
  Thus the number of three-digit numbers divisible by 7 are 128.
20. How many multiples of 4 lie between 10 and 250?
- Solution:Multiples of 4 between 10 and 250 are 12, 16, 20, …, 248.
  Here
  
  a = 12
  
  d = 4
  
  a_n = 248
  
  \ 248 = 12 + (n – 1) ´ 4
  
  (n – 1) ´ 4 = 248 - 12
  
  (n – 1) =
  
  n = 59 + 1
  
  = 60
  
  \ Number of multiples of 4 between 10 and 250 = 60.
  21.  For what value of n, are the nth terms of two APs: 63, 65, 67, …, and 3, 10, 17, …, equal?
  
  Solution:nth term of the 1st AP
  = a + (n –1) ´ d
  
  Here a = 63, d = 2
  
  = 63 + (n - 1) ´ 2 …………….(1)
  
  nth term of the 2nd AP = a + (n –1) ´ d
  
  Here a = 3, d = 7
  
  = 3 + (n – 1) ´ 7 ……………….(2)
  
  63 + ( n – 1) ´ 2 = 3 + (n – 1) ´ 7 (Given)
  
  63 – 3 = 7n – 7 – 2n + 2
  
  60 + 5 = 5n
  
  5n = 65
  
  n =  = 13.
  
  13th term of the 2 APs are equal.
  
  22.  Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
  
  Solution:3rd term = a + (3 - 1) ´ d
  = a + 2d = 16 ……………. (1)
  
  7th term = a + 6d
  
  5th term = a + 4d
  
  Given (a + 6d) – (a + 4d) = 12
  
  2d = 12 Þ d = 6
  
  substitute d = 6 in (1)
  
  a + 2(6) = 16
  
  a + 12 = 16
  
  a = 16 – 12
  
  = 4.
  
  a = 4, d = 6
  
  The AP is 4, 10, 16, …
  
  23.  Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
  
  Solution:Here
  a = 3
  
  d = 5
  
  Last term of the AP is 253
  
  Let 253 be the nth term of AP
  
  = a + (n – 1) ´ d
  
  = 3 + 5 ´ (n – 1)
  
  5n – 5 = 250
  
  5n = 255
  
  n = 51
  
  \ 253 is the 51st term.
  
  20th term from the 51st term is 32nd term
  
  32nd term = a + (32 – 1) ´ d
  
  = 3 + 31 ´ 5
  
  = 3 +155
  
  = 158.
  
  24.  The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
  
  Solution:4th term = a + 3d
  8th term = a + 7d
  
  Sum of the 4th term and 8th term = a + 3d + a + 7d = 24
  
  Þ 2d + 10d = 24
  
  a + 5d = 12 …………………………..(1)
  
  Sum of 6th term and 10th term = 44
  
  Þ a + 5d + a + 9d = 44
  
  2a + 14d = 44
  
  a + 7d = 22 ……………………..………(2)
  
  a + 5d = 12 ……. (1)
  
  a + 7d = 22 …….. (2)
  
  --------------
  
  subtracting (1) – (2) => 2d = 10
  
  d = 5
  
  substituting d = 5 in (1)
  
  a + 5d = 12
  
  a + 5(5) = 12
  
  a = 12 – 25
  
  = -13.
  
  The 1st 3 terms are –13, -13 + 5, -13 + 10
  
  (i.e.,) -13, -8, -3
  
  25.  Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
  
  Solution:Annual salary = Rs. 5000
  Increment = Rs. 200
  
  Total salary after n years =Rs. 7000
  
  If the A.P. was of the form 5000, 5200, 5400, …
  
  When 1^st term a = 5000
  
  d = 200
  
  nth term = a + (n – 1) d
  
  = 5000 +(n – 1) ´ 200 = 7000
  
  200(n – 1) = 7000 – 5000
  
  = 2000
  
  (n – 1) =
  
  n = 10 + 1
  
  His income reached Rs. 7000 in the 11th year
  
  26.  Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs.1.75. If in the nth week, her weekly savings become Rs.20.75, find n.
  
  Solution:Here
  a = 5
  
  d = 1.75
  
  In the nth week saving’s = 5 + 1.75(n – 1) = 20.75
  
  1.75(n – 1) = 20.75 – 5
  
  n - 1 =
  
  n -1 = 9
  
  n = 10.
  
  27.  In an AP: (i) given a = 5, d = 3, a_n= 50, find n and S_n.
  
  (ii) given a = 7, a₁₃= 35, find d and S₁₃.
  
  (iii) given a₁₂ = 37, d = 3, find a and S₁₂.
  
  (iv) given a₃ = 15, S₁₀= 125, find d and a₁₀.
  
  (v) given d = 5 , S₉ = 75, find a and a₉.
  
  Solution:(i) given a = 5, d = 3, a_n= 50, find n and S_n.
  a = 5
  
  d = 3
  
  a_n = 50
  
  n =  =  =  = 16
  
  S_n =
  
  S₁₆=
  
  = 8 ´ 55 = 440.
  
  Thus n = 16 and S₁₆ = 440.
  
  (ii) given a = 7, a₁₃= 35, find d and S₁₃.
  
  a = 7, a₁₃ = 35 find d and S₁₃
  
  a = 7
  
  l = 35
  
  n = 13
  
  We know that n =
  
  13 =
  
  13 – 1 =
  
  d =  =
  
  S_n =
  
  S₁₃ =
  
  =  = 273.
  
  Thus d =  and S₁₃ = 273.
  
  (iii) given a₁₂ = 37, d = 3, find a and S₁₂.
  
  a₁₂ = 37, d = 3
  
  l = 37
  
  d = 3
  
  n = 12 a = ?
  
  n =
  
  12 =
  
  37 – a = 11 ´ 3
  
  -a = 33 - 37
  
  = -4
  
  \ a = 4
  
  S_n =
  
  S₁₂ =
  
  =  = 246.
  
  Thus a = 4 and S₁₂ = 246.
  
  (iv) given a₃ = 15, S₁₀= 125, find d and a₁₀.
  
  a₃ = 15, S₁₀ = 125
  
  a +2d = 15 …………………….. (1)
  
  S₁₀= 125
  
  => (2a + 9d) = 125
  
  => 5 (2a + 9d ) 125
  
  2a + 9d =  = 25 ……………… (2)
  
  (1) ´ (2) => 2a + 4d = 30 …..(3)
  
  2a + 9d = 25…..(4)
  
  (3) – (4)=> -5d = 5
  d = -1
  
  Substituting d= -1 in (1), we get,
  
  a + 2d = 15
  
  a + 2(-1) = 15
  
  a = 17
  
  a₁₀ = a + 9d = 17 + 9(-1)
  
  = 17 – 9
  
  = 8.
  
  Thus d = -1 and a₁₀ = 8.
  
  (v) given d = 5 , S₉= 75, find a and a₉.
  
  d = 5
  
  S₉= 75 , n = 9
  
  a_n = a +(n – 1)5
  
  a₉ = a + 8 ´ 5 Þ a₉ = 40 + a …………….. (1)
  
  S_n =
  
  S₉ =  = 75
  
  2a + 40 = 75 ´  …………………………………………………(2)
  
  2a =  – 40
  
  2a =  =
  
  a =  =
  
  Substituting a =  in (1), we get,
  
  a₉ = 40 +  = =
  
  Thus a =  and a₉ =  .
  
  28.  In an AP:
  (i) given a = 2 , d = 8, S_n= 90, find n and a_n.
  (ii) given a = 8, a_n = 62, S_n = 210, find n and d.
  
  (iii) given a_n = 4, d = 2, S_n = -14, find n and a.
  
  (iv) given a = 3, n = 8, S = 192, find d.
  
  (v) given l = 28, S = 144, and there are total 9 items. Find a.
  
  Solution:(i) given a = 2 , d = 8, S_n= 90, find n and a_n.
  a = 2
  
  d = 8
  
  S_n = 90, find n and a_n
  
  S_n =
  
  =
  
  =
  
  But S_n = 90,
  
  => n[8n – 4] = 90 ´ 2
  
  8n² – 4n – 180 = 0
  
  Dividing throughout by 4, we get
  
  2n² – n – 45 = 0
  
  2n² –10n + 9n – 45 = 0
  
  2n(n - 5) + 9(n - 5) = 0
  
  \ (2n + 9)(n – 5) = 0
  
  n = 5 (or) n =
  
  \ Number of terms could be positive only, thus
  
  n = 5 , and n = -9 not possible
  
  \ a_n = [a + (n – 1) ´ d]
  
          = [2 + (5 – 1) ´ 8] = 2 + 32 = 34.
  
  Thus n = 5 and a_n= 34.
  
  (ii) given a = 8, a_n = 62, S_n = 210, find n and d.
  
  Here,
  
  a = 8, a_n = 62, S_n = 210
  
  Now we have to find n and d,
  
  a_n = a + (n – 1) ´ d
  
      = 8 + (n – 1) ´ d
  
  (n – 1) ´ d = 62 – 8
  
  = 54 ……………………..(1)
  
  S_n =
  
  210 =
  
  420 =  ……………………. (2)
  
  Substituting for (n – 1) ´ d = 54 from (1) in (2)
  
  = n[16 +54]
  
  n =  = 6
  
  Substitute n = 6 in (1), we get,
  
      (n -1) ´ d = 54
  
  Þ (6 - 1) ´ d = 54
  
  Þ 5d = 54
  
  Þ d =
  Thus n = 6 and d = .
  
  (iii) given a_n = 4, d = 2, S_n = -14, find n and a.
  
  Here,
  
  a_n = 4, d = 2, S_n = -14.
  
  a_n = a +(n – 1)d
  
  4 = a +2(n – 1)
  
  Þ a = 4 – [2(n -1)] …………………… (1)
  
  S_n =
  
  -14 =
  
  =
  
  Substitute a = 4 – [2(n -1)] from (1)
  
  = n[4 – 2(n – 1) + (n – 1)]
  
  -14 = n[4 – (n – 1)]
  
  = 4n – n² + n
  
  n² – 5n – 14 = 0
  
  n² – 7n + 2n – 14 = 0
  
  n(n – 7) + 2(n - 7) = 0
  
  (n - 7)(n+2) = 0
  
  n = 7, n = -2
  
  Number of terms cannot be negative thus n = 7 9; 9;
  
  Substitute n = 7 in (1), we get,
  
  a = 4 – [2 (7 – 1)]
  
  = 4 – (2 ´ 6)
  
  = 4 – 12 = -8
  
  n = 7, a = -8
  
  Thus n = 7 and a = -8.
  
  (iv) given a = 3, n = 8, S = 192, find d.
  
  a = 3, n = 8, S = 192 find d
  
  S_n =
  
  192 =
  
  192 = 4[(2 ´ 3) + (8 – 1)d]
  
  192 = 4(6 + (8 – 1)d)
  
  6 + 7d =
  
  6 + 7d = 48
  
  7d = 42
  
  d = = 6
  
  Thus d = 6.
  
  (v) given l = 28, S = 144, and there are total 9 items. Find a.
  
  l = 28
  
  S = 144
  
  n = 9
  
  a = ?
  
  S =  (a+l)
  
  144 =
  
  144 ´  = a + 28
  
  32 = a + 28
  
  Þ a = 4
  
  Hence a = 4.
  
  29.  How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
  
  Solution:The given AP is 9, 17, 25, …
  S_n = 636
  
  In this AP, a = 9
  
  d = a₂ – a₁ = 17 – 9 = 8
  
  Let n be the number of terms
  
  S_n =
  
  636 =
  
  1272 =
  
  =
  
  = 8n² + 10n
  
  8n² + 10n – 1272 = 0
  
  Dividing throughout by 2, we get
  
  4n² + 5n – 636 = 0
  
  4n² + 53n – 48n - 636 = 0
  
  n(4n + 53) – 12(4n + 53) = 0
  
  (n - 12)(4n + 53) = 0
  
  Þ 4n = -53 Þ n = ; n = 12
  
  Only one value n = 12 is admissible.
  
  Thus the number of terms is 12.
  
  30.  The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
  
  Solution:a = 5
  a_n = 45
  
  S_n = 400
  
  S_n =  where l - the last term, a - first term and n - number of terms
  
  400 =
  
  =
  
  400 = 25n
  
  n =  = 16
  
  \ Number of terms = 16
  
  a_n = a + (n-1)d
  
  45 = 5 +(16 – 1) d
  
  45 – 5 = 15d
  
  d =
  Hence the number of terms and the common difference is 16 and  .
  
  31.  The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
  
  Solution:In this problem,
  a = 17
  
  l = 350
  
  d = 9
  
  Let n be the no. of terms and S_n their sum
  
  Let a_n = 350
  
  a_n = a + (n – 1)d
  
  350 = 17 + (n – 1) 9
  
  350 – 17 = (n – 1) 9
  
  = (n – 1)
  
  (n – 1) = 37
  
  n = 38
  
  S_n =
  
  =
  
  =  = 6973.
  
  Thus the number of terms and its sum is 38 and 6973.
  
  32.  Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
  
  Solution:d = 7
  a₂₂ = 149
  
  n = 22
  
  As a_n = a + (n – 1)d
  
  a₂₂ = a + (n – 1) ´ 7
  
  149 = a + 21 ´ 7
  
  = a + 147
  
  a = 149 - 147
  
  a = 2
  
  \ S_n =
  
  =
  
  = 11 ´ 151
  
  \ The sum of first 22 terms = 1661
  
  33.  Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
  
  Solution:a₂ = 14
  a₃= 18
  
  a₂ = 14 = a + (2-1)d
  
  Þ a + d = 14 …………………………(1)
  
  a₃ = 18
  
  Þ a + 2d = 18 ………………………….(2)
  
  a + d = 14
  
  a + 2d = 18
  
  (1) – (2) Þ -d = -4 Þ d = 4
  
  Substituting d = 4 in (1), we get,
  
  a + d = 14
  
  a + 4 = 14
  
  a = 14 – 4
  
  = 10
  
  S_n =
  
  S₅₁ =
  
  =
  
  =
  
  =  = 5610.
  
  \ The sum of first 51 terms = 5610.
  
  34.  If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
  
  Solution:S₇ = 49
  S₁₇ = 289
  
  S_n =
  
  S₇ =
  
  49 =
  
  49 ´  = (2a + 6d)
  
  7= a + 3d ------------------- (1) (Dividing by 2)
  
  S₁₇ = [2a + (17 – 1) d]
  
  289 = [2a + (17 – 1) d]
  
  289 ´  = (2a + 16d)
  
  = a + 8d ……………………….. (2) (Dividing by 2)
  
  from (1) and (2) Þ a + 3d = 7
  
  a + 8d = 17
  
  (1) – (2) Þ -5d = -10
  
  d = 2
  
  Substituting d = 2 in (1), we get,
  
  a + 3d = 7
  
  a + 3(2) = 7
  
  a = 7 – 6
  
  a = 1.
  
  S_n =
  
       =  =
  
       = n²
  
  Hence the sum of first n terms is n².
  
  35.  Show that a₁, a₂, …, a_n, … form an AP where a_n is defined as below:
  (i) a_n = 3 + 4n                                            (ii) a = 9 - 5n
  
  Solution:When n = 1, a₁ = 3 + 4(1)
  = 3 + 4
  
  = 7
  
  When n = 2,
  
  a₂ = 3 + 4(2)
  
  = 3 + 8 = 11
  
  When n = 3, a₃ = 3 + 4(3)
  
  = 3 +12
  
  = 15
  
  Since a₂ – a₁ = 11 – 7 = 4 and a₃ – a₂ = 15 – 11 = 4.
  
  i.e., a_k+1- a_k is the same every time, the given list forms an AP with the common difference (d) = 4
  
  S_n =
  
  S₁₅=
  
  =
  
  =
  
  =
  
  =  = 525
  
  This is an AP with a = 7, d = 4, S₁₅ = 525.
  
  (ii) a = 9 - 5n
  
  When n = 1,
  
         a₁ = 4
  
  When n = 2,
  
          a₂ = 9 - 10
  
              = -1
  
  When n = 3 ,
  
          a₃ = 9 – 15
  
              = -6
  
  Since a₂ – a₁ = -1 – 4 = -5 and a₃ – a₂ = -6 + 1 = -5.
  
  i.e., a_k+1- a_k is the same every time, the given list forms an AP with the common difference (d) = -5
  
  S_n =
  
  S₁₅=
  
  =
  
  =
  
  =
  
  = - 465.
  
  This is an AP with a = 4, d = -5, S₁₅ = -465
  
  36.  If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
  
  Solution:S_n = (4n – n²)
  a₁ = S₁ = 4(1) – (1)²
  
       = 4 – 1 = 3
  
  S₂ = 4(2) – 2² = 4,
  
  S₃ = 4(3) – 3²
  
       = 12 – 9
  
       = 3
  
  a₂ = S₂– S₁ = 4 – 3 = 1
  
  a₃ = S₃– S₂ = 3 – 4 = -1
  
  a₁, a₂, a₃ is given by 3, 1, -1
  
  Here a = 3 a₁₀ = a + (n – 1)d
  
            = 3 + 9 ´ (-2)
  
            = 3 – 18
  
            = -15
  
  a_n = 3 + (n – 1) (– 2)
  
  a_n = -2n + 5 = 5 – 2n
  
  Thus a₁, a₂, a₃ is given by 3, 1, -1 , the 10th term is –15 , and the nth term is 5 – 2n.
  
  37.  Find the sum of the first 40 positive integers divisible by 6.
  
  Solution:First term a = 6
  
  d = 6
  n = 40
  
  Sum of the 40 positive integers divisible by 6
  
  S_n =
  
  S₄₀ =
  
  = 20[12 + (39 ´ 6)] = 20 [12 + 234]
  
  = 20 ´ 246 = 4920
  
  38.  Find the sum of the first 15 multiples of 8.
  
  Solution:First term a = 8
  d = 8
  
  n = 15
  
  Sum of 1st 15 multiples of 8:
  
  S_n =
  
  S₁₅ =
  
       =
  
       =
  
       = 960
  
  39.  Find the sum of the odd numbers between 0 and 50.
  
  Solution:The list of odd numbers between 0 and 50: 1, 3, 5, …, 49.
  First term (a) = 1
  
  Common difference (d) = 2
  
  Last term (l) = 49
  
  S_n = ………………….. (1)
  
  Last term is given by
  
  a_n = 1 + (n – 1) ´ d
  
  = 1 + (n – 1) ´ 2
  
  Þ 2n – 2 = 49 – 1
  
  2n = 48 + 2
  
  2n = 50
  
  n = 25
  
  Substituting n = 25 in (1) , we get,
  
  Sum =  = 625 .
  
  \ The sum of odd numbers between 0 and 50 = 625
  
  40.  A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
  
  Solution:a₁ = 200
  a₂ = 250
  
  a₃ = 300
  
  (i.e.,) a = 200, d = 50
  
  If n = 30
  
  S_n =
  
      =
  
      =
  
      =
  
      = 1850 ´ 15 = Rs. 27,750
  
  If the contractor delays the work by 30 days, he has to pay Rs. 27,750 as penalty.

- WORKSHEET 1 - Quadratic Equations

2011-06-08T03:34:00.000-07:00

1. Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x – 3) (ii) x² – 2x = -2(3 – x)

(iii) (x - 2)(x + 1) = (x - 1)(x + 3)

Solution:(i) (x + 1)² = 2(x – 3)
x² + 2x + 1 = 2x – 6

x² + 7 = 0

This is of the form ax² + bx + c = 0

\ It is a quadratic equation.
(ii) x² – 2x = -2(3 – x)

x² – 2x = -6 + 2x

x² – 2x -2x + 6 = 0

x² – 4x + 6. This is of the form ax² + bx + c = 0

It is a quadratic equation.
(iii) (x - 2)(x + 1) = (x - 1)(x + 3)

x² – 2x + x - 2 = x² + 3x – x - 3

– x – 2x +1 = 0

- 3x + 1 = 0

It is not of the form ax² + bx + c

It is not a quadratic equation.

2. Check whether the following are quadratic equations:
(i) (x – 3) (2x + 1) = x (x + 5)
(ii) (2x – 1) (x – 3) = (x + 5) (x – 1) (iii) x² + 3x + 1 = (x – 2)²

(iv) (x+2)³ = 2x(x² – 1) (v) x³ – 4x² – x + 1 = (x – 2)³

Solution:(i)(x – 3) (2x + 1) = x (x + 5)
2x² + x – 6x – 3 = x² + 5x

x² - 3 = 0

This is of the form ax² + bx + c = 0

Hence it is a quadratic equation.
(ii)(2x – 1) (x – 3) = (x + 5) (x – 1)

2x² – 6x – x + 3 = x² – x + 5x – 5

2x² – 7x + 3 - x² –4x + 5 = 0

x² – 11x + 8 = 0

This is of the form ax² + bx + x = 0

Hence it is a quadratic equation.
(iii) x² + 3x + 1 = (x – 2)²

x² + 3x + 1 = x² – 4x +4

7x – 3 = 0

This is not of the form ax² + bx + c = 0

\ It is not a quadratic equation.
(iv) (x+2)³ = 2x(x² – 1)

x² + 2³ + 3x² ´ 2 + 3x ´ 4 = 2x³ – 2x

x³ + 8 + 6x²+ 12x = 2x³ – 2x

x³ – 6x² – 14x – 8 = 0

Its not of the form ax² + bx + c = 0

It is not a quadratic equation.
(v) x³ – 4x² – x + 1 = (x – 2)³

x³ – 4x² – x + 1 = x³ – 8 - 3x² ´ 2 + 3x ´ 2²

x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x

2x² –13x + 1 = 0

This of the form ax² + bx + c = 0

Hence it is a quadratic equation.

3. Represent the following situations in the form of quadratic equations:
(i)The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii)The product of two consecutive positive integers is 306. We need to find the integers.

Solution:(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let l be length (in metres) and b be the breadth (in metres) of the rectangle.

Given

l = 1 + 2b …………………(1)

lb = 528 m² ----- (2)
(Area of rectangle = length ´ breadth)

Substituting (1) in (2)

(1+2b)b = 528

b +2b²l = 528

2b² + b – 528 = 0

This is the quadratic equation where the breadth is in metres.

(ii) The product of two consecutive positive integers is 306. We need need to find the integers

Let the 2 consecutive positive integer be x, x + 1

x(x + 1) = 306

x²+ x - 306 = 0 is the quadratic equation where x is the smallest

positive integer

4. Represent the following situations in the form of quadratic equations: (i) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(ii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:(i) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Let the Rohans present age be x

Rohans mothers present age = x + 26

After 3 years,

Rohans age = x + 3

Rohans mothers age = x + 26 + 3

= x + 29

(x + 3)(x + 29) = 360 (given)

x² + 3x + 29x + 87 = 360

x² + 32x – 273 = 0 where x is Rohans present age (in years)

(ii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Let the speed of the train be x km / hr

\ Time taken =

If speed = (x – 8) km/ h

Time taken for the same distance = hr

Given 9; - = 3

480x – 480 (x – 8) = 3(x – 8)x

480x – 480x + 480 x 8 = 3x² – 24x

3x² – 24x – 3840 = 0

x² – 8x - 1280 = 0 is the required quadratic equation where x is the speed of the train in km / hr.

5. Find the roots of the following quadratic equations by factorization:
(i) x²– 3x – 10 = 0 (ii) 2x² + x – 6 = 0
(iii) + 7x + 5 = 0 (iv) 2x² – x + = 0

(v) 100x² – 20x + 1 = 0

Solution:(i)x²– 3x – 10 = 0
x²– 5x + 2x - 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5)(x + 2) = 0

\ x = -2, x = 5

Thus –2, 5 are the roots of the equation.
(ii) 2x² + x – 6 = 0

2x² + 4x – 3x – 6 = 0

2x(x + 2) – 3(x + 2) = 0

(x + 2) (2x - 3) = 0

Þ x = -2 (or) x = are the roots of the equation
(iii) + 7x + 5 = 0

x²+ 5x + 2x + 5 = 0

x() + (x + 5) = 0

(x + )(x + 5) = 0

\ x = -, x = -

\ The roots are -, -.
(iv) 2x² – x + = 0

16x² – 8x + 1 = 0

16x² – 4x – 4x + 1 = 0

4x(4x – 1) – (4x – 1) = 0

(4x –1)² = 0

x = ,

\ The root of the equation are , .

(v) 100x² – 20x + 1 = 0

100x² – 10x – 10x + 1 = 0

10x(10x – 1) –1 (10x – 1) = 0

(10x – 1)² = 0

x = ,

\ The roots of the equation are, .

6. Solve the following situation mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy(in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

Solution:(i) Let the number of marbles John had be x
Then the number of marbles Jivanti had = 45 – x.

The number of marbles left with John, when he lost 5 marbles = x – 5

The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5

                                                                                           = 40 – x

Therefore, their product = (x – 5)(40 – x)

                                 =40x – x² – 200 + 5x

                                 = -x² + 45x – 200

So, -x² + 45x – 200 = 124

i.e., -x² + 45x – 324 = 0

i.e., x² - 45x + 324 = 0

Therefore, the number of marbles John had, satisfies the quadratic equation

x² – 45x + 324 = 0

Which is the required representation of the problem mathematically.

Now by solving we get, x² – 9x – 36 x + 324 = 0

x(x – 9) - 36(x – 9) = 0

(x - 9)(x - 36) = 0

\ x = 9 (or) 36.

\ The number of marbles John had is 9 (or) 36.

(ii) Let the number of toys produced on that day be x.

Therefore, the cost of production (in rupees) of each toy that day = 55 – x

So, the total cost of production (in rupees) that day = x(55 – x)

Therefore, x(55 – x) = 750

i.e., 55x – x² = 750

i.e., -x² + 55x – 750 = 0

i.e., x² - 55x + 750 = 0

Therefore, the number of toys produced that day satisfies the quadratic equation

x² – 55x + 750 = 0

which is the required representation of the problem mathematically.

Now by solving we get, x² – 25x – 30x + 750 = 0

x(x – 25)- 30(x – 25) = 0

(x - 25)(x - 30) = 0

x = 25, x = 30.

The number of toys produced in a day = 25 (or) 30.

7. Find two numbers whose sum is 27 and product is 182.

Solution:Let the two numbers be x and y
x + y = 27 …………………….. (1)

xy = 182 …………………….. (2)

from (1) x = 27 – y

Substituting in (2) we get,

(27 – y)y = 182

27y – y² = 182

y² – 27y + 182 = 0

y² – 13y - 14y + 182 = 0

y(y – 13) – 14(y – 13) = 0

(y – 14)(y - 13) = 0

y = 14 (or) y = 13

If y = 14, x = 27 – 14

= 13

If y = 13, x = 27 – 13

= 14

Hence the two numbers are 13 and 14.

8. Find two consecutive positive integers, sum of whose squares is 365.

Solution:Let the 2 consecutive positive nos be x, x+1
x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x – 364 = 0

x ² + x – 182 = 0 (Dividing by 2)

Þ x² + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x – 13)(x + 14) = 0

Þ x = 13 (or) x = -14.

Since the positive integers cannot be -14

Hence the positive integers are 13 and 14.

9. The altitude of a right triangle is 7 cm less that its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:In the right angled D
Let (BC) the base be x cm and altitude (AB) be (x – 7) cm

In the right angled D ,

(AB)² + (BC)² = (AC)² (Pythagoras theorem)

(x - 7)² + x² = 13²

x² – 14x + 49 + x² = 169

2x² – 14x – 120 = 0

x² – 7x – 60 = 0 (Dividing by 2 )

x² – 12x +5x - 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

=> x = -5 (or) x = 12

Since the side (base) cannot be negative, x ¹ -5

We assume x = 12

=> Base (BC) = 12 cm

Altitude = x – 7 = 5 cm

Thus the two sides of the triangle are 5 cm and 12 cm.

10. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article(in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

Solution:Let the number of articles be x and the cost of each article be y
Given:

y = 2x + 3 ………………..(1)

Total cost of production be xy = 90 …………….. (2)

Substuting y = 2x + 3 in (2), we get

x(2x+3) = 90

2x² + 3x – 90 = 0

2x² + 15x – 12x – 90 = 0

2x² + 15x – 12x – 90 = 0

x(2x + 15) - 6(2x + 15) = 0

(2x + 15) (x – 6) = 0

2x + 15 = 0 ; x – 6 = 0

=>x = ; 6

Since the number of articles cannot be negative as well as a fraction x ¹

Hence the number of articles is 6 (i.e.,) x = 6

y = 2x + 3 from (1)

= 2(6) + 3 = Rs. 15

(i.e.,) Number of articles = 6

cost of each article = Rs. 15

11. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² – 7x + 3 = 0 (ii) 2x² + x – 4 = 0

(iii) 4x² +4x + 3 = 0 (iv) 2x² + x + 4 = 0
- Solution:(i)2x² – 7x + 3 = 0
  Dividing by the coefficient of x², we get
  
  x² -
  
  Adding and subtracting the square of (coefficient of x) we get,
  
  x² – 2 ´ ´ x+ = 0
  
  = 0
  
  = 0
  
  = 0
  
  Taking the square root we get,
  
  When
  
  x =  =
  
  When
  
  (or) x =
  (ii) 2x² + x – 4 = 0
  
  Dividing by the coefficient of x², we get
  
  x² +  = 0
  
  Adding and subtracting the square of (coefficient of x) we get,
  
  x² + 2(x) ´  +
  
  = 0
  
  Taking the square root we get,
  
  x =
  
  \ x = ;
  (iii) 4x² +4x + 3 = 0
  
  (2x)² + 2(2x) ´  + ()² = 0
  
  (2x +)² = 0
  
  2x =-
  
  x = -
  
  \ Roots of equation are x = -
  (iv) 2x² + x + 4 = 0
  
  x² + x/2 + 2 = 0
  Adding and subtracting the square of (coefficient of x) we get,
  
  x² + 2(x) ´
  
  = 0
  
  The square of a number cannot be negative .Hence roots do not exist.
12. Find the roots of the quadratic equations given in Q.1. above by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0 (ii) 2x² + x – 4 = 0

(iii) 4x² + 4 = 0 (iv) 2x² + x + 4 = 0
- Solution:(i) 2x² – 7x + 3 = 0
  According to the quadratic formula: ax² + bx + c = 0
  
  a = 2
  
  b = -7
  
  c = 3
  \ The roots are =
  
                        =
  
                        = =
  
                        =
  
  =  = 3 ;
  
  The roots are 3, .
  (ii) 2x² + x – 4 = 0
  
  According to the quadratic formula: ax² + bx + c = 0
  
  a = 2, b = 1, c = -4
  
  \ The roots are
  
                    =
  
                    =
  
  \ The roots are ,
  
  (iii) 4x² + 4 = 0
  
  According to the quadratic formula: ax² + bx + c = 0
  
  a = 4, b = 4, c = 3
  
  \ The roots are =
  
                        =
  
                        =
  
                        =
  
                        = ,
  
  The roots are  =
  (iv)2x² + x + 4 = 0
  
  According to the quadratic formula: ax² + bx + c = 0
  
  a = 2, b = 1, c = 4
  
  \ The roots are =
  
                        =
  
                        =
  
                        =
  
  This is not possible, hence the roots do not exists.
13.  Find the roots of the following equations:
(i)  x ¹ 0

(ii)  x ¹ -4, 7
- Solution:(i) x ¹ 0
  x² – 1 = 3x
  
  x² – 3x – 1 = 0
  
  a = 1, b = -3, c = -1
  
  The roots are x =
  
  x =
  
  =
  
  (i.e.,) x = and
  
  (ii)
  
  (x+4)(x – 7) = -30
  
  x² – 3x – 28 = -30
  
  x² – 2x - x + 2 = 0
  
  x(x – 2) –1(x – 2) = 0
  
  (x – 2) (x – 1) = 0
  
  x = 1 (or) 2 are the roots of the equation.
14. The sum of the reciprocals of Rehman’s ages(in years) 3 years ago and 5 years from now is Find his present age.
- Solution:Let Rehman’s present age be x years.
  His age 3 years ago = x – 3
  
  5 years from now = x + 5
  
  =
  
  Taking LCM,
  
  3(2x + 2) = (x – 3) (x + 5)
  
  6x + 6 = x² + 2x -15
  
  x² – 4x – 21 = 0
  
  x² – 7x + 3x - 21 = 0
  
  x(x – 7) + 3(x – 7) = 0
  
  Þ x = 7 or x = -3
  
  Age cannot be negative thus x = 7.
  
  \ His present age = 7 years
15. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
- Solution:Let her marks in Maths = x and in English = y
  Given x + y = 30 …………………. (1)
  
  Given, (x + 2)(y – 3) = 210 …… (2)
  
  From (1) : x = 30 - y ……(3)
  
  Substututing in (2) we get
  
  (30 – y + 2) (y – 3) = 210
  
  (32 – y) (y – 3) = 210
  
  32y – 96 – y² + 3y = 210
  
  y² – 35y + 306 = 0
  
  y² – 17y – 18y + 306 = 0
  
  y(y – 17) – 18(y – 17) = 0
  
  (y – 18) (y – 17) = 0
  
  y = 18 (or) 17
  
  Þ x = 13 (when y= 17 by substituting in (3))
  
  x = 12 (when y= 18 by substituting in (3))
  
  Her marks in Maths is 13 and Science is 17 (or) Maths is 12 and Science is 18.
16. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
- Solution:Let x be the length of the shorter side
  Longer side = (x + 30)
  
  In the right angled D BCD,
  
  (BD)² = (BC)² + (CD)² (Pythogoras theorem)
  
  (60 + x)² = (x + 30)² + x²
  
  (x + 60)² = x² + x² + 60x + 900
  
  2x² + 60x + 900 = x² + 120 x + 3600
  
  x² - 60x – 2700 = 0
  
  x² - 90 x + 30x – 2700 = 0
  
  x(x - 90) + 30 (x - 90) = 0
  
  (x + 30)(x – 90) = 0
  
  x = -30; x = 90
  
  Since the length of a side cannot be negative , x = 90
  
  \ Shorter side = 90 m
  
  \ Longer side = 90 + 30 = 120 m
17. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers
- Solution:Let the two numbers be x, y
  x² – y² = 180 ………………. (1)
  
  y² = 8x ………………. (2)
  
  Substitute (2) in (1)
  
  x² – 8x = 180
  
  x² – 8x – 180 = 0
  
  x² + 10x – 18x – 180 = 0
  
  x(x + 10) – 18(x + 10) = 0
  
  (x + 10)(x – 18) = 0
  
  \ x = -10 ; x = 18
  
  If x = -10 from (2)
  
  y² = 8x
  
  = 8(-10)
  
  = -80
  
  This is not possible as square of any cannot be negative.
  
  If x = 18 from (2)
  
  y² = 8x
  
  = 8(18)
  
  = 144
  
  \ y = 12
  
  Thus the numbers are 18, 12 (or) 18, -12.
18. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train
- Solution:Let the speed of the train = x km/hr
  Distance traveled = 360 km
  
  Time taken = =
  
  If the speed increases by 5km/hr the time required to reach the place reduces to one hour
  
  Time taken =
  
  = 1
  
  360
  
  360
  
  1800 = x² + 5x
  
  x² + 5x – 1800 = 0
  
  x² + 45x – 40x - 1800 = 0
  
  x(x + 45) – 40(x + 45) = 0
  
  (x + 45)(x – 40) = 0
  
  \ x = 40 (or) x = -45
  
  But speed of the train can only be positive, thus x = 40 km/hr
19. Two water taps together can fill a tank in 9 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
- Solution:Let the time taken by smaller tap = x hrs
  In 1 hr  of the tank can filled by smaller tap.
  Let the time taken by larger tap = (x – 10) hrs
  
  In 1 hr  of the tank can be filled by larger tap.
  
  &Together they fill the tank in
  
  =
  
  =
  
  75(2x - 10) = 8(x² – 10x)
  
  8x² – 80x – 150x + 750 = 0
  
  8x² – 230x + 750 = 0
  
  4x² – 115x + 375 = 0 (Dividing by 2)
  
  4x² – 100x – 15x + 375 = 0
  
  4x(x – 25) – 15(x – 25) = 0
  
  4x – 15 = 0 (or) x – 25 = 0
  
  x =  (or) x = 25
  
  x =  cannot satisfy the condition
  
  Small tap will fill the tank in 25 hrs and larger tap in 15 hrs.
20. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore(without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
- Solution:
  Let the average speed of the passenger train be x km/hr and the express train be (x + 11) km/hr
  Time taken by passenger train =
  
  Time taken by express train =
  
  According to the question,
  
  = 1
  
  21.  Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m, find the sides of the two squares.
  
  Solution:Let the side of the two squares be a, b.
  Area of first square = a²
  
  Area of second square = b²
  
  \ a² + b² = 468 …………….. (1)
  
  Perimeter of first square = 4a
  
  Perimeter of second square = 4b
  
  4a - 4b = 24
  
  Þ a – b = 6
  
  a = b + 6 ……………….. (2)
  
  substitute a in (1)
  
  (b+6)² + b² = 468
  
  b² + 12b + 36 + b² = 468
  
  2b² + 12b – 432 = 0
  
  Dividing by 2,
  
  b² + 6b – 216 = 0
  
  b² + 18b – 12b – 216 = 0
  
  b(b + 18) –12 (b +18) = 0
  
  (b + 18)(b – 12) = 0
  
  \ b = -18, b = 12
  
  Side cannot be negative thus b = 12 m
  
  Substitute b = 12 in (2)
  
  a = b + 6
  
  = 12 + 6
  
  = 18 m
  
  Side of first square = 18 m
  
  Side of second square = 12 m
  
  22.  Find nature of roots of the following quadratic equation. If real roots exists find them.
  (i) 2x² – 3x + 5 (ii) 3x² - 4x + 4 (iii) 2x² – 6x + 3 = 0
  
  Solution:(i) 2x² – 3x + 5
  Here a = 2, b = -3, c = 5
  
  Discriminant D = b² – 4ac = 9 - (4 ´ 2 ´ 5)
  
                       = 9 – 40
  
                       = -31
  
  Equation 2x² – 3x + 5 has imaginary roots.
  (ii) 3x² - 4x + 4
  
  Here
  
  a = 3, b = -4, c = 4
  
  D = b² – 4ac
  
     = (16 ´ 3) – (4 ´ 3 ´ 4)
  
     = (16 ´ 3) – (16 ´ 3)
  
     = 0
  
  The roots are real and equal.
  
  The roots are x =
  
                        =
  
                        =
  
                        =
  
                        =,
  
                        =,
  
                        = ,
  
  (iii) 2x² – 6x + 3 = 0
  
  a = 2 , b = -6, c = 3
  
  Discriminant D = b² – 4ac
  
                       = 36 – (4 ´ 2 ´ 3)
  
                       = 36 – 24
  
                       = 12
  
  D > 0
  
  \ Equation has real and distinct roots
  
  x =
  
  =
  
  =
  
  x =
  
  x =  (or) x =
  
  23.  Find the value of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x² + kx + 3 = 0
  
  (ii) kx(x – 2) + 6 = 0
  
  Solution:(i) 2x² + kx + 3 = 0
  As the equation has equal roots the value of D = 0
  
  a = 2 , b = k, c = 3
  
  D = b² – 4ac
  
  Z = k² – (4 ´ 2 ´ 3)
  
  0 = k² – 24
  
  k² = 24
  
  k =
  
  k = 2
  (ii) kx² – 2kx + 6 = 0
  
  As the equation has equal roots the value of D = 0
  
  a = k , b = -2k, c = 6
  
  D = b² – 4ac
  
  0 = 4k² – (4 ´ k ´ 6)
  
  0 = 4k² – 24k
  
  0 = k² – 6k
  
  k(k – 6) = 0
  
  k = 0 (or) k = 6
  
  Thus k = 6.
  
  24.  Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
  
  Solution:Let Length be x m and breadth be y m
  x = 2y (1)
  
  xy = 800 m²
  
  substituting value of x from (1)
  
  2y ´ y = 800
  
  2y² = 800
  
  => y² – 400 = 0
  
  y² = 400
  
  y = 20 m Þ x = 40 m
  
  Yes, it is possible to design a rectangular mango grove with length = 40m and breadth = 20 m.
  
  25.  Is the following situtation possible, If so determine the present ages.The sum of ages of 2 friends is 20 years. Four years ago, the product of their ages in year was 48.
  
  Solution:Let the present ages of friends be x, y.
  x + y = 20 (1)
  
  x = 20 –y
  
  The product of their ages 4 yrs ago,
  
  (x – 4)(y – 4) = 48
  
  xy – 4y – 4x +16 = 48
  
  xy – 4y – 4x = 32
  
  (20 – y)y – 4y – 4(20 – y) = 32
  
  20y – y² – 4y – 80 + 4y = 32
  
  -y² + 20y – 112 = 0
  
  D = b² – 4ac
  
     = 400 – 4 ´ 1 ´ -112
  
     = 400 – 448
  
     = -48
  
  As D < 0 is not possible.
  
  26.  Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so find its length and breadth.
  
  Solution:Area = Length ´ Breadth = l ´ b = 400 m²
  Perimeter = 2(l+b) = 80 m
  
  2(l + b) = 80
  
  l+b = 40 (1)
  
  b =
  substituting value of b in equation (1) as .
  
  l +  = 40
  
  l² + 400 = 40l
  
  l² – 40l + 400 = 0
  
  D = b² – 4ac
  
     = (-40)² – (4 ´ 400 ´ 1)
  
     = 1600 – 1600
  
     = 0
  
  Equation has equal roots.
  
  As the equation has equal roots we can only design a square plot.
  
  132
  
  1452 = x² + 11x
  
  x² +11 x – 1452 = 0
  
  x² + 44x – 33x – 1452 = 0
  
  x(x + 44) – 33(x + 44) = 0
  
  (x – 33)(x + 44) = 0
  
  x = 33, x = -44
  Since the speed of train cannot be negative, x = 33 km / hr
  
  Speed of the passenger train = 33 km/hr
  
  Speed of the express train = 22km/hr

WORKSHEET 1 Pair of Linear Equations in Two variables

2011-06-08T03:32:00.000-07:00

1. Aftab tells his daughter, "Seven years ago, I was se ven times as old as you were then. Also, three years from now, I shall be three times as old as you will be,"(Isn’t this interesting?) Represent this situation algebraically and graphically.

Solution:Let the present age of Aftab be x years and her daughter age be y years
Seven years ago

Aftab’s age = (x – 7)
Her daughter’s age =(y – 7)

(x – 7) =7(y – 7) ……………………….. (1)

Three years from now

Aftab’s age = x + 3

Her daughter age = y + 3

x + 3 = 3(y + 3) ……………………….. (2)

(1) becomes,

x – 7 = 7(y – 7)

x – 7 = 7y – 49

x – 7y = -42 ……………………….. (3)

(2) becomes

x + 3 = 3(y + 3)

= 3y + 9

x - 3y = 6 ………………………(4)

Algebraically the two situations can be represented as follows:

x – 7y + 42 = 0 , x - 3y - 6 = 0 where x and y are respectively the ages of Aftab and his daughter.

Graphic Representation :

x – 7y = -42 (1)

x – 3y = 6 (2)

=> x – 7y = -42

-7y = -42 – x

y =

when x = 0,

y =  = = 6

when x = 7,

y =  =  = 7

when x = 14,

y =  =  = 8

when x = -7,

y = =  = 5

when x = -14,

y =  =  = 4

x

-14

-7

0

7

14

y

4

5

6

7

6

=> x – 3y = 6

-3y = 6 – x

y =

when x = 0,

y =  = = -2

when x = 3,

y =  = =  = -1

when x = 6,

y =  = = 0

when x = -3,

y =  = =  = -3

when x = -6,

y =  = = -4

x

-6

-3

0

3

6

y

-4

-3

-2

-1

0

2. The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another but and 2 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.

Solution:Let the cost of each bat be Rs.x
Let the cost of each ball be Rs.y
Algebraically

3x + 6y = 3900 …………………(1)

(1) Þ x + 2y = 1300

x + 2y = 1300 …………..(2)

X

1000

700

y =

150

300

3. The cost of 2kg of apples and 1kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.

Solution:Cost per kg of apple = Rs. x
Cost per kg of grapes = Rs. y

Algebraically 2x + y = 160 ……………….(1)

4x + 2y = 300 ……………….(2)

from (1) y = 160 – 2x

Graphically

x

50

60

Y =

60

40

4. Form the pair of linear equations in the following problem, and find their solutions graphically.10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution:Let the number of boy be x and the number of girls be y
x + y = 10 ………………………. (1) (given)

y = x + 4 ……………………. (2) (given)

x + y = 10 …..(1)

=> y = 10 – x

When x = -1

y = 10 – (-1)

= 11

When x = 0

y = 10 – 0

= 10

When x = 1

y = 10 – 1

= 9

when x = 2

y = 10 – 2

= 8

when x = 3

y = 10 – 3

= 7

x

-1

0

1

2

3

y = 10 - x

11

10

9

8

7

y = x + 4 …………………….(2)

Let x = -1

y = x + 4

= -1 + 4

= 3

Let x = 0

y = x + 4

= 0 +4 = 4

Let x = 1

y = x + 4

= 1 + 4

= 5

Let x = 2

y = x + 4

= 2 + 4

= 6

Let x = 3

y = x + 4

= 3 + 4

= 7

x

-1

0

1

2

3

y = x + 4

3

4

5

6

7

The solution thus obtained graphically is (3, 7). Number of girls = 7 and number of boys = 3.

5. Form the pair of linear equations in the following problem, and find their solutions graphically. 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs.46. Find the cost of one pencil and that of one pen.

Solution:Let the cost of one pencil be Rs. x
Let the cost of one pen be Rs. y

5x + 7y = 50

y =  …………… (1)

When x = 3

y =

=  =

= 5

When x = 10

y =

=  =

= 0

When x = -4

y =

=  =  =10

x

-4

3

10

Y

10

5

0

7x + 5y = 46

y =

When x = 0

y =

=  =

= 9.2

When x = 8

y =

=  =

=  = -2

x

0

3

8

Y

9.2

5

-2

By solving graphically , cost of one pencil = Rs. 3, cost of one pen = Rs. 5.

6. On comparing the ratios and , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0
7x +6y – 9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y +24 = 0

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0

Solution:(i) 5x – 4y + 8 = 0
7x +6y – 9 = 0

\

\Lines are intersecting.

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

Since , the lines are coincident.

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

But

Since , the lines are parallel.

7. On comparing the ratios and , find out whether the lines representing the following pairs of linear equations are consistent , or incoincident.
(i) 3x + 2y = 5

2x – 3y = 17

(ii) 2x – 3y = 8

4x – 6y = 9

(iii)

9x – 10y = 14

(iv) 5x – 3y = 11

-10x + 6y = -22

(v)

2x + 3y = 12

Solution:(i) 3x + 2y = 5
2x – 3y = 17

Since , equations are consistent.

(ii) 2x – 3y = 8

4x – 6y = 9

Here

\ The equations are inconsistent.

(iii)

9x – 10y = 14

=

=

=

\ The equations are consistent.

(iv) 5x – 3y = 11

-10x + 6y = -22

=

=

=

Since , the equations are consistent.

(v)

2x + 3y = 12

a₁ = , a₂ = 2, c₁ =- 8

a₁ = 2, b₂ = 3, c₂ = -12

= =

=

Since

The equations are consistent.

8. Which of the following pairs of linear equations are consistent/inconsistent ?If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv)2x – 2y – 2 = 0, 4x – 4y – 5 =0

Solution:x + y = 5, 2x + 2y = 10

Equations are consistent .

We have to draw the graphs of both the given equations

x + y = 5

x

0

1

y = 5 - x

5

4

2x + 2y = 10

x

0

1

y =

5

4

Hence the lines represented by both the equations are coincident. Thus both the equations have infinitely many solutions.

(ii) x – y = 8, 3x – 3y = 16

=

=

\The equation are inconsistent.

2x + y – 6 = 0

4x – 2y – 4 = 0

=

Thus as , the equations are consistent.

2x + y – 6 = 0

when x = 1 , y = 6 – 2x

= 4

x = 0 , y = 6

x = -1 , y = 8

x

2

0

-1

y = 6 – 2x

2

6

8

4x – 2y – 4 = 0

4x – 2y = 4

y =

= 2x – 2

4x – 2y – 4 = 0

x

2

0

-1

y = 2x – 2

2

-2

-4

When x = 1, y = 2 – 2 = 0

x = 0, y = -2

x = -1, y = -2 -2 = -4

4x – 4y – 5 = 0

4x – 4y = 5

(iv) 2x – 2y – 2 = 0

4x – 4y - 5= 0

Equations are inconsistent

9. Half perimeter of a rectangular garden , whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:Let l be the length and b be the breadth
l = b+ 4

Þ l - b = 4 …………..(1)

Perimeter = 2(l + b)

´ 2(l + b) = 36(given) ……………….(2)

l – b = 4 ………… (1)

l + b = 36 ………..(2)

– (2) => 2l = 40

l = = 20 m

Substitute l = 20 in (1)

l - b = 4

20 – b = 4

b = 20 – 4

= 16.

10. Given the linear equation 2x +3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines (ii) parallel lines (iii) coincident lines

Solution:2x + 3y – 8 = 0
3x + 2y – 7 = 0 (intersecting lines)

2x + 3y + 12 = 0 (parallel lines)

4x + 6y – 16 = 0 (coincident lines)

11. Draw the graphs of the equations x – y +1 =0 and 3x +2y –12 =0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis , and the triangular region.
- Solution:x – y + 1 = 0
  y = x +1
  
  x
  
  -1
  
  0
  
  1
  
  2
  
  y = x + 1
  
  0
  
  1
  
  2
  
  3
  
  3x +2y – 12 =0
  
  2y =12-3x
  
  y =
  
  when x = 0, y =
  
  =
  
  =
  
  when x = 2, y =
  
  ==
  
  = = 3
  
  when x = 4, y =
  
  ==
  
  = = 0
  
  when x = -2, y =
  
  ==
  
  = = 9
  
  x
  
  -2
  
  0
  
  2
  
  4
  
  y = x + 1
  
  9
  
  6
  
  ]
  
  0
12. Solve the following pair of linear equation by the substitution method.
x + y = 14

x - y = 4
- Solution:x + y = 14 ……………. (1)
  x - y = 4 ……………. (2)
  
  from 2
  
  x = y + 4
  
  sub in (1) y + 4 + y = 14
  
  2y = 10 Þ y = 5
  
  x = 9
  
  x = 9, y = 5
13. Solve the following pair of linear equation by the substitution method.
s – t = 3
- Solution:s – t = 3 ………………….. (1)
  
  = 6
  
  Þ 2s + 3t = 36 ……………(2)
  
  from (1) s = t + 3 …………..(3)
  
  sub in (1) 2(t + 3) + 3t = 36
  
  2t + 6 + 3t = 36
  
  5t = 30 Þ t = 6
  
  \ from (3), s = 6 + 3 = 9
14. Solve the following pair of linear equations by the substitution method 3x – y = 3

9x – 3y = 9
- Solution:3x – y = 3 …………………. (1);
  from(1) y = 3x - 3
  
  Substitute y= 3x – 3 in (2)
  
  9x – 3y = 9 ………………….(2)
  
  9x – 3(3x – 3) = 9
  
  9x – 9x + 9 = 9
  
  This statement is true for all values of x. However, we do not have a specific value of x as a solution. Therefore, we cannot obtain a specific value of y. This situation has arisen because both the given equations are the same . Therefore , Equations (1) and (2) have infinitely many solutions.
15. Solve the following pair of linear equations by the substitution method
0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
- Solution:0.2x + 0.3y = 1.3 ……………….. (1)
  0.4x + 0.5y = 2.3 ……………….. (2)
  
  (1) ´ 5
  
  1x + 1.5y = 1.3 ´ 5
  
  x = 6.5 – 1.5 y
  
  sub in (2)
  
  0.4 (6.5 – 1.5y) + 0.5y = 2.3
  
  2.6 – 0.6y + 0.5y = 2.3
  
  -0.1y = -0.3
  
  y = 3
  
  Substitute y = 3 in x = 6.5 – 1.5 y
  
  x = 6.5 – 1.5(3 )
  
  = 6.5 – 4.5 = 2
  
  x = 2, y = 3
16. Solve the following pair of linear equations by the substitution method
= 0

= 0
- Solution: = 0 ……………..(1)
  = 0 ……………..(2)
  
  (1) ´ : 2x + = 0
  
  x =
  
  sub in (2)
  
  = 0
  
  y( = 0
  
  y( = 0
  
  y( = 0
  
  Þ y = 0
  
  Hence x = 0
  
  \ x = 0, y = 0.
17. Solve the following pair of linear equations by the substitution method
- Solution: ……………………. (1)
  …………………….(2)
  
  Multiply (1) and (2) by 6
  
  3x ´ 3 – 5y ´ 2 = -12
  
  (i.e.,) 9x – 10y = -12 ……………………….. (3)
  
  2x + 3y = 13 ……………..(4)
  
  (3) => 9x = 10y – 12
  
  x =
  
  sub in (4)
  
  = 13
  
  20y – 24 + 27y = 13 ´ 9
  
  47y = 117 + 24
  
  = 141
  
  y = 3
  
  x =
  
  == 2
  
  x = 2, y = 3
18. Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ’m’ for which
y = mx + 3
- Solution:2x + 3y = 11 …………………(1)
  2x - 4y = -24 …………………(2)
  
  From (1)
  
  x =
  
  sub in (2)
  
  = - 24
  
  11 – 7y = -24
  
  7y = 24 + 11
  
  = 35
  
  y = 5
  
  Substitute y =5 in x =
  
  x =
  
  = -2
  
  x = -2, y = 5.
  
  For the equation y = mx + 3
  
  Substitute x = -2, y = 5 then
  
  5 = -2m + 3
  
  -2m = 2
  
  m = -1
19. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i)The difference between two numbers is 26 and one number is three times the other. Find them.

(ii)The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes , if 2 is added to both the numerator and the denominator. If , 3 is added to both the numerator and the denominator it becomes . Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son . Five years ago, Jacob’s age was seven times that of his son. What are their present ages
- Solution:(i) Let the 2 numbers be x, y
  x – y = 26 …………….(1)
  
  x = 3y ……………… (2)
  
  By substitution method, substituting (2) in (1)
  
  3y – y = 26
  
  y =  = 13
  
  Substituting y =13 in (2)
  
  x = 39, y = 13
  
  (ii) Let the 2 angles be x and y
  
  x + y = 180⁰ ……………………..(1)
  
  x - y = 18⁰ ……………………..(2)
  
  from (1) x = 180 – y
  
  Substituting in (2) 180 – y – y = 18
  
  - 2y = 18 – 180
  
  y =  = 81⁰
  
  Substituting y = 81 in x = 180 – y
  
  x = 180 – 81
  
  = 99⁰
  
  (iii) Let the price of a bat be x and that of a ball be y
  
  7x + 6y = 3800 ………………….(1)
  
  3x + 5y = 1750 ………………….(2)
  
  Let us express x in terms of y
  
  From (1) 7x = 3800 – 6y
  
  x =  ……………….. (3)
  
  Substitute in (2)
  
  11,400 – 18y + 35y = 1750 ´ 7
  
  17y = 12250 - 11400
  
  = 850
  
  y =
  
  from(3) x =  =  = 500
  
  \ Cost of bat is Rs. 500
  
  \ Cost of ball is Rs. 50
  
  (iv) Let the fixed charge be Rs. x and charge per km be Rs. y
  
  Total charges =
  
  x + 10y = 105 ………………… (1)
  
  x + 15y = 115 …………………..(2)
  
  from (1) x = 105 – 10y
  
  Substituting in (2)
  
  x + 15y = 115
  
  105 – 10y + 15y = 115
  
  5y = 115 – 105
  
  y =  = 2
  
  x = 105 – ( 2 ´ 10)
  
  = 85
  
  Fixed charges = Rs. 85/-
  
  Charge per km = Rs. 5/-
  
  For travelling a distance of 25km the person has to pay x + 25y = 85 + 25(2)
  
  = Rs. 135
  
  (v) Let the fraction be , x the numerator and y the denominator
  
  ………………………. (1)
  
  ………………………. (2)
  
  from (1) (x+2)11 = 9(y+2)
  
  11x + 22 = 9y + 18
  
  11x - 9y = -4
  
  x =   ………………… (3)
  
  from (2) (x+3)6 = (y+3)5
  
  6x + 18 = 5y + 15
  
  6x – 5y = -3 …………………(4)
  
  Substituting (3) in (4)
  
  6() – 5y = -3
  
  54y – 24 –55y = -33
  
  -y = -9
  
  y = 9.
  
  Substituting y =9 in (3) we get,
  
  x =
  
  =
  
  = =
  
  = 7
  
  \ The fraction is .
  
  (vi) Let Jacob’s age be x and his son’s age be y
  
  5 years hence,
  
  Jacob’s age = x + 5
  
  Jacob’s son’s age = y + 5
  
  (x+5) = 3(y+5) ………………………. (1)
  
  5 years ago,
  
  Jacob’s age = x - 5
  
  Jacob’s son’s age = y - 5
  
  (x – 5) = 7(y - 5) ………………(2)
  
  (1) becomes x+5 = 3y + 15
  
  x – 3y = 10 Þ x = 3y + 10 ………………(3)
  
  (2) becomes x – 5 = 7y – 35
  
  x – 7y = -30 ……………………..(4)
  
  sub for x from (3) in (4)
  
  (3y + 10) –7y = -30
  
  -4y = -10-30
  
  = -40
  
  y = 10
  
  x = 3y + 10 = 40
  
  = 30 +10 = 40
  
  \ Jacob age = 40 and his son’s age = 10.
20. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x - 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) and
- Solution:(i) x + y = 5 and 2x - 3y = 4
  Elimination method:
  
  x + y = 5……… (1)
  
  2x - 3y = 4………. (2)
  
  ´ 2 : 2x + 2y = 10…… (3)
  
  (2): 2x - 3y = 4
  
  – (2) => 5y = 6
  
  y =
  
  Substitute y =  in (1)
  
  x + y = 5
  
  x = 5 -
  
  =
  
  =
  
  \ x = , y =
  
  Substitution method:
  
  From (1) : x = 5 – y
  
  Substituting in (2) : 2(5 – y) – 3y = 4
  
  10 – 2y – 3y = 4
  
  -5y = 4 – 10
  
  y =
  
  x = 5 – y
  
  = 5 - () =
  
  \ x = , y =
  (ii) 3x + 4y = 10 and 2x – 2y = 2
  
  Solution
  
  Elimination method:
  
  3x + 4y = 10 ……………….. (1)
  
  2x – 2y = 2 ……………………(2)
  
  Multiply (2) with 2 and adding with (1) we get,
  
  3x + 4y = 10 ……………….. (1)
  
  4x – 4y = 4 …………………… (3)
  
  7x = 14
  
  x =  = 2
  
  Substitute x = 2 in (2)
  
  2x – 2y = 2
  
  2(2) –2y = 2
  
  2y = 4 – 2
  
  = 2
  
  \ y =1
  
  Thus x = 2, y = 1.
  
  Substitution method:
  
  3x + 4y = 10 ……………….. (1)
  
  2x – 2y = 2 ……………………(2)
  
  from (2) x – y = 1
  
  => x = 1 + y
  
  Substitute in (1): 3(1+y) + 4y = 10
  
  3 + 3y + 4y = 10
  
  7y = 7 Þ y =1
  
  Substitute in (1) 3x + 4 = 10
  
  3x = 6
  
  x = 2
  
  Thus x = 2, y = 1.
  (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
  
  Solution
  
  Elimination method:
  
  3x - 5y = 4 ……………..(1)
  
  9x - 2y = 7 ……………….(2)
  
  Multiply (1) with 3 and adding with (2) we get,
  
  9x – 15y = 12……………….(1)
  
  9x - 2y = 7 ……………….(2)
  
  -13 y = 5
  
  y =
  Substitute y =  in (1)
  
  3x - 5y = 4
  
  3x – 5() = 4
  
  3x = 4 -
  
  =
  
  x =  =
  
  \ x = , y = .
  
  Substitution method:
  
  3x = 5y + 4 ……………..(1)
  
  9x = 2y + 7 ……………….(2)
  
  from (1) x =
  
  sub in (2)
  
  15y + 12 – 2y – 7 = 0
  
  13y = -5
  
  y =
  
  x =  =  =  = .
  
  \ x = , y = .
  
  (iv)  and
  Solution
  
  Elimination method:
  
  …………………. (1)
  
  ……………………(2)
  
  Multiply by 6 on both sides in equation (1)
  
  Multiply by 3 in both sides in equation (2)
  
  3x + 4y = -6 ……………… (3)
  
  3x - y = 9 ……………… (4)
  
  – (4) => 5y = -15
  
  y = -3
  
  Substituting y = -3 in (3) we get,
  
  3x + 4y = -6
  
  3x + 4(-3) = -6
  
  3x = -6 +12
  
  = 6
  
  x = 2
  
  Thus x = 2, y = -3.
  
  Substitution method:
  
  Multiply by 6 on both sides in equation (1)
  
  Multiply by 3 in both sides in equation (2)
  
  3x + 4y = -6 ……………… (3)
  
  3x - y = 9 ……………… (4)
  
  =>y = 3x – 9 from (4)
  
  Substitute in (3)
  
  3(x) + 4(3x – 9) = -6
  
  3x + 12x – 36 = -6
  
  15x = 30
  
  x = 2
  
  y = 3x – 9
  
  y = 3(2) – 9
  
  y = 6 – 9
  
  = -3.
  
  Thus x = 2, y = -3.
  
  21.  Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
  
  (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes  if we only add 1 to the denominator. What is the fraction?
  (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
  (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
  Solution:(i) Let the fraction be x is the numerator and y is the denominator
  …………….. (1)
  
  …………….. (2)
  
  becomes x + 1 = y - 1
  
  (i.e.,) x - y = -2
  
  Þ x = y - 2 ……………….. (3)
  
  (2) becomes,
  
  2x = y + 1 ……………………. (4)
  
  sub x in (4)
  
  2(y - 2) = y + 1
  
  2y - 4 = y + 1
  
  2y - y = 4 + 1
  
  y = 5
  
  \ x = y - 2 from (3)
  
  = 3
  
  \ The fraction is .
  
  (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
  
  Nuri's age = x
  
  Sonu's age = y
  
  5 years ago, x - 5 = 3(y - 5) ………………. (1)
  
  10 years later, x + 10 = 2(y + 10) ……….. (2)
  
  x - 5 = 3y - 15
  
  x = 3y - 10 ……….. (3)
  
  sub x in (2)
  
  3y - 10 + 10 = 2y + 20
  
  3y - 2y = 20
  
  y = 20
  
  Nuri's age x = 3y - 10
  
  = 3(20) - 10
  
  = 60 - 10 = 50
  
  Nuri's age x = 50 years
  Sonu's age y = 20 years
  
  (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
  
  Let x, y be the 2 digits of the number
  
  x + y = 9 ………………(1)
  
  9(10x + y) = 2 (10y + x) ………………(2)
  
  (2) => 90x + 9y = 20y + 2x
  
  88x - 11y = 0
  
  Þ 8x - y = 0 …………………….. (3)
  
  Solving (1) and (3) we get,
  
  x +y = 9 ………………(4)
  
  8x - y = 0 …………………….. (3)
  
  => 9x = 9
  
  x = 1
  
  Substituting x = 1 in (1)
  
  x + y = 9
  
  1+ y = 9
  
  y = 8.
  
  The two-digit number is 18.
  22.  Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: (i) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
  
  (ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
  
  Solution:(i) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
  Let the number of Rs. 50 notes be x and the number of Rs. 100 notes be y
  
  x + y = 25 …………….. (1)
  
  50x + 100y = 2000 ………(2)
  
  simplifying (2)
  
  x + 2y = 40 …………(3)
  
  x + y = 25
  
  x + 2y = 40
  
  y = 15
  
  sub for y = 15 in (1)
  
  x + 15 = 25
  
  x = 10
  
  Number of Rs. 50 notes = 10
  
  Number of Rs. 100 notes = 15
  
  (ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
  
  Let x be the fixed charge for first three days and y be the additional charge.
  
  Saritha had the book for 7 days, so she pays additional charges for 4 days.
  
  Susy had the book for 5 days so she pays additional charges for 2 days. Saritha paid
  
  x + 4y = 27 …………………… (1)
  
  Susy paid, x + 2y = 21 …………… (2)
  
  Subtracting (1) and (2)
  
  2y = 6
  
  y = 3
  
  sub in (1) x + 4(3) = 27
  
  x + 12 = 27
  
  x = 15
  
  Fixed charges = Rs. 15
  
  Additional charges = Rs. 3 per day
  
  23.  Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
  (i) x – 3y – 3 = 0
  
  3x – 9y – 2 = 0
  
  (ii) 2x + y = 5
  
  3x +2y = 8
  
  (iii) 3x – 5y = 20
  
  6x – 10y = 40
  
  (iv) x - 3y – 7 = 0
  
  3x - 3y – 15 = 0
  
  Solution:(i)  =
  =  =
  
  =
  
  Since  =  ¹ , there is no solution.
  
  (ii) 2x + y = 5
  
  3x +2y = 8
  
  =
  
  =
  
  =
  
  Since  ¹ , we have a unique solution.
  
  By cross multiplication method
  
  2x + y = 5
  
  3x +2y = 8
  
  x y 1
  
  1 -5 2 1
  
  2 -8 3 2
  
  =  =
  
  =>  =  =
  
  => x = 2, y =1.
  
  (iii) 3x – 5y = 20
  
  6x – 10y = 40
  
  = =
  
  ==
  
  =  =
  
  Since  = = , there are infinitely many solutions
  
  (iv) x - 3y – 7 = 0
  
  3x - 3y – 15 = 0
  
  = 1
  
  \ It has a unique solution
  
  By cross multiplication method
  
  x - 3y – 7 = 0
  
  3x - 3y – 15 = 0
  
  x y 1
  
  -3 -7 1 -3
  
  -3 -15 3 -3
  
  =  =
  
  =>  =  =
  
  => x = 4 , y = -1.
  
  24.  (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
  2x + 3y = 7
  
  (a – b)x + (a + b)y = 3a + b – 2
  
  (ii) For which value of k will the following pair of the linear equations have no solution?
  
  3x + y = 1
  
  (2k – 1)x + (k – 1)y = 2k + 1
  
  Solution:(i) If the above equation should have an infinite number of solution.
  
  =
  (i.e.,)
  (i.e.,) 2(a+b) = 3(a-b)
  
  2a – 3a = -3b – 2b
         -a = -5b
          a = 5b
  ………..(1)
  
  Substitute a = 5b in (1)
  
  16b – 2 = 14 b
         2b = 2
           b = 1
  Substitute b =1 in a= 5b
  a = 5(1)
     = 5
  a = 5, b = 1.
  
  (ii) For which value of k will the following pair of the linear equations have no solution?
  3x + y = 1
  (2k – 1)x + (k – 1)y = 2k + 1
  If the two equation should have no solution then =
  (i.e.,)  =
  
  (i.e.,) 3(k – 1) = (2k – 1)
  
  3k –3 = 2k – 1
  3k – 2k = 3 - 1
           k = 2
  Hence k = 2 satifies the above condition
  
  25.  Solve the following pair of linear equations by the substitutions and cross multiplication methods:
  
  8x + 5y = 9
  3x + 2y = 4
  
  Solution:By substitution method
  8x + 5y = 9 ………………… (1)
  
  3x + 2y = 4 ………………… (2)
  
  from(2) y =  ……………..(3)
  
  Substitute in (1)
  
  8x + 5 = 9
  
  16x + 20 – 15x = 18
  
  x = 18 – 20
  
  = -2
  
  Substitute x = -2 in (3)
  
  y =
  
  =
  
  = 5
  
  \ x = -2
  
  y = 5
  
  By cross multiplication method:
  
  8x + 5y = 9
  
  3x + 2y = 4
  
  =  =
  
  =  = 1
  
  Þ x = -2
  
  y = 5
  
  26.  Form the pair of linear equations in the following problem and find it's solution (if they exist) by any algebraic method:
  
  A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.
  
  Solution:Let the fixed charges be x and cost of food per day be y
  In the case of student A
  
  x + 20y = 1000 ……………………….. (1)
  
  In the case of student B
  
  x + 26y = 1180 ……………………….. (2)
  
  Subtracting
  
  (1) from (2)
  
  20y – 26y = 1000 – 1180
  
  -6y = -180
  
  y = 30
  
  substituting y = 30 in (1) we get,
  
  x + 20(30) = 1000
  
  x = 1000 – 600 = 400
  
  \ Fixed charges = Rs. 400 and cost of food per day = Rs. 30
  
  27.  Form the pair of linear equations in the following problem and find it's solution (if they exist) by any algebraic method:
  A fraction becomes  when 1 is subtracted from the numerator and it becomes  when 8 is added to its denominator. Find the fraction.
  
  Solution:Let  be the fraction
  =  ……………….(1)
  
  =  ……………….(2)
  
  from (1) 3(x – 1) = y
  
  3x – y = 3 ……………….(3)
  
  from (2) 4x = y + 8
  
  4x – y = 8 …………………… (4)
  
  subtracting (3) from (4)
  
  (3x – y) – (4x – y) = 3 – 8
  
  -x = -5
  
  x = 5
  
  sub x = 5 in (3)
  
  3x – y = 3
  
  3 ´ 5 – y = 3
  
  y = -3 + 15 = 12
  
  The fraction is
  
  28.  Form the pair of linear equations in the following problem and find it's solution (if they exist) by any algebraic method:
  Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
  
  Solution:Let x be the no of right answers and y be the no of wrong answers written by Yash respectively.
  3x – y = 40 …………… (1)
  
  4x – 2y = 50 ……….…(2)
  
  (1) ´ (2) 6x – 2y = 80
  
  subtract from (2) 4x – 2y = 50
  
  2x = 30
  
  x = 15
  
  From (1)
  
  3(15) - y = 40
  
  y = 45 – 40
  
  = 5
  
  \ Number of right answers be 15 and number of wrong answers be 5.
  
  29.  Form the pair of linear equations in the following problem and find it's solution (if they exist) by any algebraic method: Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
  
  Solution:Let x be the speed of 1^st car and y be the speed of the 2^nd car
  If car (1) meets (2) at C
  
  Distance traveled by car(1) is 5 x
  
  Distance traveled by car(2) is 5 y
  
  We know that
  
  5x – 5y = 100
  
  when they travel in the opposite direction (towards each other) they meet at C in 1 hour.
  
  Since distance between A and B = 100
  
  AB = AC - BC
  
  Distance traveled by car (1) = x
  
  Distance traveled by car (2) = y
  
  We know that x + y = 100 (AC + BC = 100)
  
  5x – 5y = 100 …………….(1)
  
  x + y = 100 ………(2)
  
  (1) + 5 ´ (2)
  
  5x – 5y = 100
  
  5x + 5y = 500
  
  10x = 600
  
  x = 60
  
  from (1) 5y = 5x – 100
  
                  = 5(60) – 100 = 300 – 100
  
                  = 200
  
  \ y = 40
  
  \ Speed of car (1) = 60 km/ hr
  
  Speed of car (2) = 40 km/ hr
  
  30.  Form the pair of linear equations in the following problem and find it's solution (if they exist) by any algebraic method:
  The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
  
  Solution:Area of rectangle = lb
  Let l be the length and b be the breadth of the rectangle.
  
  (l – 5) (b + 3) = lb – 9 ……………(1)
  
  (l + 3) (b + 2) = lb + 67 ……………(2)
  
  becomes
  
  (l - 5)(b + 3) = lb – 9
  
  lb + 3l – 5b – 15 = lb – 9
  
  3l – 5b = 6 ………………….(3)
  
  becomes
  
  (l + 3)(b + 2) = lb + 67
  
  lb + 2l + 3b + 6 = lb + 67
  
  2l + 3b = 61 …………………..(4)
  
  Multiply (3) by 2 and (4) by 3,
  
  6l – 10b = 12
  
  6l + 9b = 183
  
  - 19 b = -171
  
  b = 9
  
  from (3) #9; 3l – 9(5) = 6
  
  3l = 45 + 6
  
  l =  = 17
  
  \ l = 17, b = 9.
  
  Thus the length and breadth of the rectangle are 17units and 9 units respectively.
31.  Solve the following pairs of equations by reducing them to a pair of linear equation:
(i)

(ii)  = 2

= -1
Solution:(i)  ………………………….(1)
…………………..(2)

Let  = a

= b

+ = 2 ……………………. (3)
+ =  ……………………. (4)

Multiplying (3) and (4) by 6

Þ 3a + 2b = 12 …………….. (5)

Þ 2a + 3b = 13 …………….. (6)

(5) ´ 2 and (6) ´ 3

6a + 4b = 24

6a + 9b = 39

Subtracting -5b = -15

b = 3

Þ y =  =

put b = 3 in (5)

3a + 2(3) = 12

3a = 6

a = 2

x =  =

\ x = , y =

(ii)  = 2

= -1

= 2 ……………. (1)

= -1 ……………(2)

Let  be a

be b

2a + 3b = 2 ……………………… (3)

4a – 9b = -1 ……………………… (4)

(3) ´ 2: 4a + 6b = 4

4a – 9b = 1

15b = 5

b = =

Þ b =  =

\  = 3

y = 9

(3) Þ 2a + 3() = 2

2a +1 = 2

2a = 2 – 1

a =

a = =

Þ  = 2 , x = 4

\ x = 4, y = 9
32.  Solve the following pairs of equations by reducing them to a pair of linear equation:(i)  = 14
= 23
(ii)  = 2

= 1
Solution:(i)  = 14
= 23

Let

4a + 3y = 14 …………….. (3)

3a - 4y = 23 ………………(4)

4 ´ (3) - (4) ´ 3

16a + 12y = 56

9a – 12y = 69

Adding 25a = 125

a = =

a =  =5

\ x =

Substituting in (3), a =5 we get

4(5) + 3y = 14

3y = 14 – 20

    = -6

y = -2

\ x = , y = -2

(ii)  = 2

= 1

= 2 ………………. (1)

= 1 ………………. (2)

Let

Substituting a, b in (1) and (2) we get,

5a + b = 2 ………….(3)

6a – 3b = 1 ………….(4)

(3) ´ 3 : 15a + 3b = 6

6a – 3b = 1

21a = 7

a =

=

Substituting in eq. (3)

b = 2 –

=

a = , b =

Substitute in  = a

=

x - 1 = 3 Þ x = 4

= b

=

Þ y - 2 = 3

Þ y = 5

Hence x = 4, y = 5.
33. Solve the following pairs of equations by reducing them to a pair of linear equation:

(i)

(ii) 6x + 3y = 6xy

2x + 4y = 5xy
Solution:(i)

……………………. (1)

……………………. (2)

Separating the fraction

= 5

= 5

= 15

= 15

- = 5

= 15

Let

-2a + 7b = 5 …….(1)

7a + 8b = 15 …… (2)

Multiply (1) with 7, (2) with 2, we get.

-14a + 49b = 35

14a + 16b = 30

         65b = 65

            b = 1

b = = 1

=>y = 1

Substitute b =1 in (1)

-2a + 7(1) = 5

-2a = 5 – 7

a =

= 1

a = = 1

=>x = 1

\ x =1, y = 1.

(ii) 6x + 3y = 6xy

2x + 4y = 5xy

6x + 3y = 6xy

2x + 4y = 5xy

¸ by xy

=

= 6 ………………..(1)

2x + 4y = 5xy

¸ by xy

……. (2)

substituting  and  in eq. (1) and (2)

a + 2b = 2 …… (3)

4a + 2b = 5 ……(4)

-3a       = -3

a = 1

Substituting a = 1 in (3) we get,

1 + 2b = 2

2b = 1

b =

Thus if a =  = 1

=> x = 1.

Thus if b =  =
y = 2

Hence x =1, y=2.
34.  Solve the following pairs of equations by reducing them to a pair of linear equation:
(i)  = 4
= -2

(ii)  =

= -
Solution:(i)  = 4
= -2

= 4

= -2

Let ,

10a + 2b = 4 ……. (1)

15a - 5b = -2 ………(2)

Multiply (1) with 3 and (2) with 2, we get

30a + 6b = 12

30a - 10b = -4

16b = 16

b = 1

Substituting b = 1 in (1) we get,

10a + 2(1) = 4

10a = 4 - 2

10a = 2

a =

=

as  =

x + y = 5 ………….. (3)

and

= 1

x – y = 1 ………………….. (4)

Solving (3) and (4)

x + y = 5

x – y = 1

2x = 6

x = 3

when x = 3

x + y = 5

3 + y = 5

y = 5 – 3

= 2

Thus x = 3, y = 2.

(ii)  =
= -

Let  and

a + b =

4a + 4b = 3 ………………… (1)

a – b = =

4a + 4b = 3

4a - 4b = -1

8a = 2

a =
a =

substituting in (1)

4() + 4b = 3

4b = 2

b =  =

3x + y = 4 ……………….. (3)

3x – y = 2 ………………….. (4)

(3) + (4)

3x + y = 4

3x - y = 2

6x = 6

x = 1

substituting x = 1 in (3) we get

3(1) + y = 4

y = 1

Thus x = 1, y = 1.
35. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 woman and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Solution:(i) Let the speed of rowing in still water be ‘x’ and the speed of the current be y km/hr.
\ Speed upstream = (x – y) km/hr

\ Speed downstream = (x + y) km/hr

case 1

Where Ritu rows 20 km downstream in 2 hours, the equation is

2(x + y) = 20

x + y = 10 …………..(1)

case 2

When Ritu rows 4 km downstream in 2 hours, the equation is

2(x – y) = 4

x – y = = 2

x – y = 2 …………………. (2)

solving (1) and (2)

2x = 12

x = 6

Substitute x = 6 in (1)

x + y = 10

6 + y = 10

y = 10 – 6

= 4

Therefore speed of the rowing in still water = 6 km/hr

And speed of the current = 4 km/hr.

(ii) 2 woman and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Let a woman take x days to complete the work.

One women’s one day work =

Similarly let one man take ‘y’ days to complete the work.

Therefore one man’s one day work =

\

Let ,

2a + 5b =

8a + 20b = 1 ………………… (1)

=

Let ,

3a + 6b =

9a + 18b = 1 ………… (2)

72a + 180 b = 9 ((1) ´ 9)

72a + 144 b = 8 ((2) ´ 8)

36b = 1

b =

Substitute b = in (1)

3a + 6

3a = -

3a =

3a =

a =

when a =

or x = 18

when b =

y = 36

\ Time taken by 1 woman to do the work = 18 days.

\ Time taken by 1 man to do the work = 36 days.
36. Formulate the following problem as a pair of equations, and hence find it's solution
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:Let the spread the train be ‘x’ km/hr
Let the spread the bus be ‘y’ km/hr

Case (i)

Distance traveled by train = 60 km

Distance traveled by bus = 300 - 60 = 240 km

Time taken by Roohi in train =  (Time taken = )

Time taken by Roohi in bus =

Total time = 4 hrs

=  …………………………. (1)

case (ii)

Distance traveled by train = 100 km

Distance traveled by bus = 300 – 100

= 200 km

Time taken by Roohi in Train =  (Time taken = )

Time taken by Roohi in Bus =

Total time is 10 more minutes

Longer than time taken in case (i) :\ 4 +  = 4 +

=

\  ………………….. (2)

Solving (1) and (2)

-------------- (1)

-------------- (2)

Let  = a and  = b

Rewrite (1) and (2) as follows

60 a + 240 b = 4

100 a + 200 b =

Multiply (1) with 5 , and (2) with 6

300a + 1200b = 20 …….. (3)

600a + 1200b = 25 …….(4)

(3) –(4) =>-300a = -5

a =  = in (1)

Substituting a =  in (1)

60 ´  + 240b = 4

240 b = 4 – 1

240 b = 3

b =

when a =

(or)  ……………….. x = 60 km / hr

when b =

(or)  ……………….. y = 80 km / hr

\ speed of train = 60 km/hr

and speed of bus = 80 km/hr

WORKSHEET 1 - Polynomials

2011-06-08T03:29:00.000-07:00

1. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).

Solution:no zeroes.

2. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).

Solution:One

3. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).

Solution:Three

4. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).

Solution:Two

5. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).

Solution:Four

6. The graphs of y = p (x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p (x).

Solution:Three.

7. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8 (ii) 4s² – 4s + 1 (iii) 6x² – 3 – 7x
(iv) 4u² + 8u (v) t² – 15 (vi) 3x² – x – 4.

Solution:(i) x² – 2x – 8
= x² – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x – 4)(x + 2)

Therefore the zeroes of the polynomial x² – 2x – 8 are {4, -2}.

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = -

= -

Also sum of the zeroes of the polynomial = 4 – 2 = 2.

Product of the zeroes = =

Also product of the zeroes = 4 x –2 = -8

Hence verified.

(ii) 4s² – 4s + 1

= 4s² – 2s – 2s + 1

= 2s(2s – 1) – 1(2s – 1)

= (2s – 1)(2s – 1)

= 2(s - 2(s -

Therefore the zeroes of the polynomial are

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = -  = -  = 1

Also sum of the zeroes = = 1

Product of the zeroes = =

Also product of the zeroes =

Hence verified.

(iii) 6x² – 3 – 7x

= 6x² – 7x – 3

= 6x² – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x –3)(3x + 1)

= 2(x - 3(x +

= 6(x - (x +

The zeroes of the polynomials are {

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = -  = -  =

Also sum of the zeroes =

Product of the zeroes = =

Also product of the zeroes =

Hence verified.

(iv) 4u² + 8u

= 4u(u + 2)

= 4[u – 0][u –(- 2)]

The zeroes of the polynomials are {0, -2}

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = -  = -

Also sum of the zeroes =

Product of the zeroes = =

Also product of the zeroes =

Hence verified.

(v) t² – 15

= (t +

The zeroes of the polynomials are {

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = -  = -

Also sum of the zeroes =

Product of the zeroes = =

Also product of the zeroes =

Hence verified.

(vi) 3x² – x – 4.

= 3x² – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4)(x + 1)

The zeroes of the polynomials are {

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = -  = -

Also sum of the zeroes =

Product of the zeroes = =

Also product of the zeroes =

Hence verified.

8. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) (ii) (iii) 0, (iv) 1, 1 (v) -, (vi) 4,1

Solution:(i)
Let the quadratic polynomial be ax² + bx + c, and its zeroes be a and b

Given a + b =  =

a b = -1 =

If a = 4 , b = -1 and c = -4

The quadratic polynomial is 4x² - x - 4.

(ii)

Let the quadratic polynomial be ax² + bx + c, and its zeroes be a and b

Given a + b =  =

a b =  =

If a = 1, b = - and c =

The quadratic polynomial x² - x + (or) 3x² - 3x + 1 .

(iii) 0,

Let the quadratic polynomial be ax² + bx + c and the zeroes be a + b

Given a + b = 0 =

a b =  =

If a = 1, b = 0 and c =

The quadratic polynomial is x² +

(iv) 1, 1

Let the quadratic polynomial be ax² + bx + c , and its zeroes be a + b

Given a + b =  = 1

a b =  = 1

\ If a = 1 , b = -1 and c = 1

\ The quadratic polynomial is x² - x + 1.

(v)  ,

Let the quadratic polynomial be ax² + bx + c , and its zeroes be a + b
Given a + b = =

a b == .

If a = 4, b = 1 and c = 1

The quadratic polynomial is 4x² + x + 1.

(vi) 4, 1

Let the quadratic polynomial be ax² + bx + c

Given a + b = = 4

a b = = 1

If a = 1, b = -4 and c = 1

\ The quadratic polynomial is x² – 4x +1.

9. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x³ – 3x² + 5x + 3, g(x) = x² – 2

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Solution:(i) p(x) = x³ – 3x² + 5x + 3, g(x) = x² – 2

Quotient is (x - 3)
Remainder is 7x – 9

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

Rearrange g(x) as x² - x + 1

The Quotient is x² + x – 3

Remainder is 8

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Rearrange g(x) as –x² + 2

Quotient is –x² – 2

Remainder is –5x + 10

10. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
(ii) x² + 3x +1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12

\ 1^st is a factor of the second polynomial.

(ii) x² + 3x+1, 3x⁴ + 5x³ – 7x² + 2x + 2

Hence 1^st polynomial is a factor of the second polynomial

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

No it is not a solution

11. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are and -.
- Solution:Two zeros are  and -.
  Since the two zeros are  and-
  
  (x -) (x +) = x² –  is a factor of the polynomial (i.e.,) (3x² – 5)
  
  \ Divide the given polynomial by 3x² – 5.
  
  (3x⁴ + 6x³ – 2x² – 10x – 5) = (3x² – 5) (x² + 2x + 1)
  
  = (3x² – 5) (x + 1)²
  
  Its zeroes are given by x = -1, x = -1
  
  The zeroes of the given polynomial are , -, -1, -1.
12. On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).
- Solution:By the division algorithm,
  Dividend = Divisor ´ Quotient + Remainder.
  
  In this problem f(x)= x³ – 3x² + x + 2, since (-2x + 4 ) is the remainder, subtract (-2x + 4) from f(x) then divide the result by (x - 2).
  
  Thus g(x) =x² – x + 1.
13. Give examples of polynomials p(x), g(x), q(x), r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) =0.
- Solution:If p(x) denotes the dividend, g(x) denotes the divisor, q(x) denotes the quotient, r(x) denotes the remainder.
  p(x) = 3x² – 6x + 12 , g(x) = 3 , q(x) = x²- 2x + 4, r(x) = 0.
  
  p(x) = x³ + x² + x +1, g(x) = x² – 1, q(x) = x + 1, r(x) = 2x + 2.
  
  p(x) = 2x³ + 2x² – 2x +2, g(x) = x² – x –1, q(x) = 2x + 4, r(x) = 6.

WORKSHEET -1 - Electricity

2011-06-07T12:24:00.000-07:00

1. What does an electric circuit means?

Solution:The continuous and closed path of an electric current is called an electric circuit. If the circuit is broken anywhere the current stops flowing.

2. Define the unit of current.

Solution:The unit of current is ampere(A). 1 ampere is constituted by the flow of 1 coulomb of charge per second, .

3. Calculate the number of electrons constituting one coulomb of charge.

Solution:The S.I unit of electric charge is coulomb (C), which is equivalent to the charge contained in nearly 6x10¹⁸ electrons.

4. Name a device that helps to maintain a potential difference across a conductor.

Solution:Voltmeter is a device that helps to maintain a potential difference across a conductor.

5. What is meant by saying that the potential difference between the two points is 1 V?

Solution:1 V is the potential difference between the two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to other.
Therefore,
1 V=1 JC^-1

6. How much energy is given to each coulomb of charge passing through a 6 V battery?

Solution:Energy, W = VQ = 6 V × 1.6 × 10^-19C = 9.6 × 10^-19 C

Therefore, 9.6 × 10^-19 C of energy is given to each coulomb of charge passing through a 6 V battery.

7. On what factors does the resistance of a conductor depend?

Solution:The resistance of a conductor depends on its Length, Area of cross-section and on the nature of its material.

8. Will current flow more easily through a thick wire or a thin wire of the same material, When connected to the same source? Why?

Solution:The current flow more easily through a thick wire because are Area µ thickness
Resistance µ
Current µ
If thickness more area is more, area is more, resistance is less and hence current flow is more.

9. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Solution:

2I₂=I₁
therefore,
So that the current decreases to half of its former value.

10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Solution:The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise readily at high temperatures. For this reason alloy is commonly used in electric toasters and electric irons.11. Use the data in the table to answer the following:

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?
- Solution:(a) Iron is the better conductor than mercury.
  (b) Silver is the best conductor.
12. Draw a schematic diagram of a circuit consisting of a battery of four 2 V cells, a 5 ohm resistor, an 8 ohm resistor and a 12 ohm resistor and a plug key all connected in series.
- Solution:
13. Redraw the circuit of the previous problem, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the 12 ohm resistor. What would be the readings in the ammeter and the voltmeter.
- Solution:
  
  Total voltage applied V = 4 ´ 2 = 8 volts
  Total resistance R = 5 + 8 + 12 = 25 W
  Therefore reading of the ammeter I = = 0.32 A
  Reading of voltmeter V = IR = 0.32 ´ 12 = 3.84 V.
14. Judge the equivalent resistance when the following are connected in parallel
(a) 1 W and 10⁶ W

(b) 1 W and 10³Wand 10⁶W
- Solution:(a) R₁=1 W , R₂=10⁶ W
  
  R_T =1 W.
  (b) R1=1 W , R2=10³W, R₃=10⁶W
  
  =1
  Therefore, R_T=1 W.
15. An electric lamp of 100 W , a toaster of resistance 50 W , and a water filter of resistance 500 W are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all the three appliances, and What is the current through it?
- Solution:R₁= 100 W , R₂= 50 W , and R₃= 500 W
  
  
        = 0.032 W
  
  We know that, V=220 V
  Total current,
  = 220 V x 0.032 W
  = 7.04 V
  resistance of electric iron =
                                      =
                                      = 31.25 W.
16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
- Solution:A parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has different resistance and requires different to operate properly.
17. How can three resistors of resistances 2W , 3W and 6W be connected to give a total resistance of (a) 4 W (b) 1 W
- Solution:(a) Resistance 3 W and 6 W are connected in parallel and 2 W is connected in series with this combination.
  
  (b) All the three resistance are connected in parallel.
18. What is (a) the highest, (b) the lowest total resistance that can be secured by combination of four coils of resistance 4 W , 8 W , 12 W , 24 W ?
- Solution:(a) Highest resistance =4 + 8 + 12 + 24 = 48 W
  All resistance should be connected in series.
  (b) If all the resistance are connected in parallel the lowest resistance
  = 0.5
  R_T = = 2 W.
19. Why does the cord of an electric heater not glow while the heating element does?
- Solution:The cord of an electric heater will not glow because it is made of insulating material. An insulator never conducts electricity.
20. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
- Solution:Q = 96000 coulomb, V = 50 V
  
  H = VxIxt = V x Q x t
                         t
  
      = V x Q
      = 50 V x 96000 C
      = 480000 J.
- 21.  An electric iron of resistance 20 W takes a current of 5 A. Calculate the heat developed in 30 s.
  
  Solution:R = 20 W , I = 5 A, t = 30 s.
  H = I²Rt
     = 25 × 20 × 30
     = 15000 J
  The heat developed in 30 s = 15000 J.
  
  22.  What determines the rate at which energy is delivered by a current?
  
  Solution:The rate at which electric energy is consumed in an electric circuit is termed as electric power. The power P is given by
  P=VI
  Or, P=I²R =
  
  23.  An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hr.
  
  Solution:V = 220 V, I = 5 A
  Power, P = VI = 220 V x 5 A = 1100 W
  Energy=Pt = 1100 W x 2 hr = 2200 W hr.
  
  24.  A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is(a) 1/25 (b) 1/5 (c) 5 (d) 25
  
  Solution:(d) 25
  
  25.  Which of these following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d)
  
  Solution:(b) IR²
  
  26.  An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be (a) 100 W (b) 75 W (c) 50 W (d) 25 W
  
  Solution:(c) 50 W
  
  27.  Tow conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
  
  Solution:(c) 1:4
  
  28.  How is a Voltmeter connected in the circuit to measure the potential difference between two points?
  
  Solution:A Voltmeter should be connected between the two points in the circuit to measure the potential difference.
  
  29.  A copper wire has diameter 0.5 mm and resistivity of 1.6x10^-8Wm. What will be the length of this wire to make its resistance 10 W ? How much does the resistance change if the diameter is doubled?
  
  Solution:R=10 W , r=
  
  l=1.23
  The length of this wire is 1.23
  If d₁=2d, r₁=2r
  
  Therefore the resistance reduces by quarter of the original value.
  
  30.  The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below
  
  I (amperes) 0.5 1.0 2.0 3.0 4.0
  
  V (Volts) 1.6 3.4 6.7 10.2 13.2
  
  Solution:
  
  =2 W.
  
  31.  When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
  
  Solution:V=12 V, I=2.5x10^-3mA
  
  R=4.8x10^-3 W
  
  32.  A battery of 9 V is connected in series with resistors of 0.2 W , 0.3 W , 0.4 W , 0.5 W and 12 W , respectively. How much current will flow through the 12 W resistor?
  
  Solution:Current through 12 W resistor= =  = 0.67 A.
  
  33.  How many 176 W resistors are required to carry 5 A on a 220 V line?
  
  Solution:V = 220 V, I = 5 A
  =  = 44 W
  
  = 4
  Therefore no of resistors = 4.
  
  34.  Several electric bulbs designed to be used on a 220 V electric supply line , are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
  
  Solution:V=220 V
  P=10 W
  
  =4840 W
  R_T=220        5
  =44
  
  No of lamps=
  
  =110.
  
  35.  A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 W resistance, which may be used separately, In series, or in parallel. What are the currents in the three cases?
  
  Solution:V=220 V, R=24 W
  When the resistance coils are used separately, current
  
  =9.17 W
  When the resistance are connected in series,  =4.58 W
  When the resistance are connected in parallel, I=
  =18.33 W
  
  36.  Compare the power used in the 2 W resistor in each of the following circuits
  (i) A 6 V battery in series with 1 W and 2 W resistors, and
  
  (ii) A 4 V battery in parallel with 12 W and 2 W resistors.
  
  Solution:(i)The voltage across 2 W resistor=6x
  =4 V
  Power used in the 2 W resistor==
  =8 W
  (ii) The voltage across 2 W resistor=4 V
  Power used in the 2 W resistor==
  =8 W.
  
  37.  Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to a 220 V supply. What current is drawn from the supply line?
  
  Solution:Let the current drawn from the supply be I and the individual currents drawn by each lamp be I₁ and I₂ respectively. Since they are connected in parallel the voltage drop ‘V’ across each lamp is the same, which is also equal to the supply voltage 220 V.
  Then current in lamp 1
  I₁ = = 0.4545 A
  Similarly current in lamp 2
  I₂ = = 0.2727 A
  Therefore the net current drawn I = I₁ + I₂ = 0.7272 A
  
  38.  Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
  
  Solution:Energy=Pt
  Energy in TV set =250x1
  =250 Whr
  Energy in toaster =1200x
  =200 Whr
  Therefore TV set consumes more energy than toaster.
  
  39.  An electric heater of resistance 8 W draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
  
  Solution:
  
  I=15A, R=8 W
  
  =28.125
  The rate at which heat is developed in the heater is 28.125
  
  40.  Explain the following:
  (i) Why is tungsten used almost exclusively as filament for incandescent lamps?
  (ii) Why are the conductors of electric heating devices, such as toasters and electric irons, made of an alloy rather than a pure metal?
  (iii) Why is the series arrangement not used for domestic circuits?
  (iv) How does the resistance of a wire vary with its cross-sectional area?
  
  Solution:(i) A thin wire of tungsten has high resistance and does not get oxidized easily at high temperatures. Hence a tungsten filament is used for making filaments of electric bulbs.
  
  (ii) The resistance of all pure metals increases on raising the temperature and decreases on lowering the temperature. But the resistance of alloys like manganin, constantan and nichrome is almost unaffected by temperature.
  
  (iii) If we connect all the electrical appliances like bulbs, fans and sockets, etc., in a series, then if one appliance is switched off or gets fused then all the appliances will also stop working because their electricity supply will be cut-off.
  
  (iv) When the area of a cross-section of a wire is doubled, its resistance gets halved; and if the area of a cross-section of wire is halved, then its resistance will get doubled. Hence a thick wire which has a greater cross-section has less resistance and a thin wire which has a smaller area of cross-section has high resistance.

WORKSHEET -1 - metals and non metals

2011-06-07T12:20:00.000-07:00

1. What are amphoteric oxides? Give two examples of amphoteric oxides.

Solution:Amphoteric oxides are the oxides, which react with both acids and bases to form salt and water. E.g. ZnO and Al₂O₃.

2. Name two metals, which will displace hydrogen from dilute acids, and two metals which will not.

Solution:Very reactive metals like Zn and Mg displace hydrogen from dilute acids. On the other hand less reactive metals like Cu, Ag, etc. do not displace hydrogen from dilute acids.

3. In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?

Solution:Anode is impure, thick block of metal M.
Cathode is a thin strip/wire of pure metal M.

Electrolyte is a suitable salt solution of metal M.

4. State two ways to prevent the rusting of iron.

Solution:By coating the surface of iron by rust proof paints.
By applying oil or grease to the surface of iron objects so that supply of air consisting of moisture is cut off form the surface.

5. What types of oxides are formed when non-metals combine with oxygen?

Solution:When non-metals combine with oxygen it forms either neutral or acidic oxides. CO is a neutral oxide; N₂O₅ or N₂O₃ is an acidic oxide.

6. Give reason

i. Metals replace hydrogen from dilute acids, where as non-metals do not.

ii. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Solution:i. Metals are electropositive in nature. They readily lose electrons. These electrons reduce the protons liberated from the acid to liberate hydrogen gas, where as non-metals possess a tendency to gain electrons and hence they do not furnish electrons to protons liberated from acids. Hence H₂gas is not liberated.
ii. As it is easier to reduce metal oxides to metal, prior to reduction, metal sulphides and carbonates must be converted to oxides.

7. Differentiate between metals and non-metals on the basis of their chemical properties.

Solution:

Metals	Non-metals
Metals on heating with oxygen form ionic oxides, which are basic in nature and which dissolve in water to form bases, which turn red litmus blue.	Non-metals on heating with oxygen form covalent oxides which are acidic in nature and dissolve in water to form acids, which turn blue litmus red.
Lustrous	Non-lustrous except graphite
Conductor of electricity and heat	Non-conductor of heat and electricity, except graphite
All are solid except mercury	Solid-liquid-gaseous
Electro positive. Loses electrons readily and becomes a positive ion	Electro negative: gain electrons and become negative ions.
Metals are reducing agents	Non-metals are good oxidizing agents

8. Explain why the surface of some metals acquires a dull appearance when exposed to air for a long time.

Solution:This is due to the surface oxidation of metals when exposed to moist air. For e.g. copper turns green on its surface due to the formation of basic copper carbonate Cu(OH) ₂. CuCO₃. Similarly silver becomes black due to the formation of black Ag₂S and Aluminium forms a white coating of Al2O3 on its surface.

9. State which of the following metals would give hydrogen when added to dilute hydrochloric acid. i. Iron, ii. Copper iii. Magnesium

Solution:Iron and magnesium react with dilute hydrochloric acid to give H₂. Magnesium reacts more readily than iron with dilute hydrochloric acid to liberate hydrogen.
Fe + 2HCl  FeCl₂ + H₂

Mg + 2HCl  MgCl₂+ H₂

Cu + HCl  No reaction

Copper does not react with dilute hydrochloric acid at all. This shows that copper is even less reactive than iron.

10. Name a non-metallic element, which conducts electricity.

Solution:Carbon in the form of graphite conducts electricity, as there is a free electron in each carbon atom, which moves freely in between the hexagonal layers.

11. Which metals do not corrode easily?

Solution:Gold and platinum and other noble metals do not corrode in air.

12. What are alloys?

Solution:Alloys are homogeneous mixtures of two or more metals, or a metal and a non-metal.E.g. steel, brass, bronze, etc.

13. Define the following terms.

(i) Minerals

(ii) Ores

(iii) Gangue

Solution:(i) MineralsAll compounds or elements, which occur naturally in the earth's crust, are called minerals. Example: Alums, K₂SO₄.Al₂(SO₄)₃ . 24 H₂O, Bauxite Al₂O₃.2H₂O
(ii) OresThose minerals from which a metal can be profitably extracted are called ores. Bauxite (Al₂O₃.2H₂O) is the ore of Al, copper pyrite CuFeS₂. All minerals are not ores but all ores are minerals.

(iii) GangueWhen an ore is mined from the earth, it is always found to be contaminated with sand rocky materials. The impurity of sand and rock materials present in the ore is known as gangue.

14. Name two metals that are found in nature in the free state.

Solution:Gold and platinum are found in the free state in nature.

15. What is chemical process used for obtaining a metal from its oxide?

Solution:The moderately reactive metals such as iron, zinc and copper, etc. in the form of their oxides are reduced to the respective metals by heating with carbon (coke) which is know as carbon reduction method.
ZnO + C Zn + CO

PbO + C Pb + CO

16. Name two metals, which can form hydrides with metals.

Solution:Sodium and calcium form stable hydrides on reacting with hydrogen.

17. Does every mineral have a definite and a fixed composition? Explain.

Solution:Yes, every mineral has a definite and a fixed composition. Minerals are widely distributed in the earth’s crust in the form of oxides, carbonates, sulphides, sulphates, nitrates, etc. These minerals are formed as a result of chemical changes taking place during the formation of earth.

18. Explain the meaning of malleable and ductile.

Solution:Malleable is being able to be beaten/hammered into thin sheets.

Ductile is being able to be drawn into thin wires.

19. i. Write the electron dot structures for sodium, oxygen and magnesium.

ii. Show the formation of MgO and Na₂O by the transfer of electrons.

iii. What are the ions present in these compounds?

Solution:i. Sodium: Na·
Oxygen:

Magnesium:

ii. Formation of Magnesium oxide
When magnesium reacts with oxygen, the magnesium atom transfers its two outermost electrons to an oxygen atom. By losing 2 electrons, the magnesium atoms form a magnesium ion (Mg²⁺) and by gaining 2 electrons, the oxygen atom forms an oxide ion (O^2-).

Formation of Sodium oxide
Two sodium atoms transfer their 2 outermost electrons to an oxygen atom. By losing two electrons, the two sodium atoms form two sodium ions (2Na⁺). And by gaining two electrons, the oxygen atom forms an oxide ion (O^2-.)

iii. The ions present in sodium oxide compound (Na₂O) are sodium ions (2Na⁺) and oxide ions (O^2-).

The ions present in Magnesium oxide compound (MgO) are magnesium ions Mg²⁺ and oxide ions (O^2-).

20. You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.

Solution:The sour substances such as lemon (or tamarind juice) contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and make them shining red-brown again.
21. Give an example of a metal which

i. is a liquid at room temperature.

ii. can be easily cut with a knife.

iii. is the best conductor of heat.

iv. is a poor conductor of heat.
- Solution:i. Mercury is in liquid state at room temperature.
  ii. Sodium and potassium are soft metals which can be easily cut with a knife.
  
  iii. Silver is the best conductor of electricity.
  
  iv. Mercury is a poor conductor of heat.
22. Why is sodium kept immersed in kerosene?
- Solution:Sodium metal is kept immersed in kerosene to prevent their reaction with oxygen, moisture and carbon dioxide of air.
23. Why do ionic compounds have high melting points?
- Solution:These compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositively charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compounds. This is why ionic compounds have high melting points.
24. A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
- Solution:Aqua regia (By volume, this contains 3 parts of concentrated hydrochloric acid and 1 part of concentrated nitric acid) is the solution, which is used to sparkle the bangles like new, but their weight will be reduced drastically.
25. Write equations for the reactions of
(i) iron with water
(ii) calcium and potassium with water
- Solution:(i) Iron reacts with steam to form magnetic oxide of Fe with the liberation of H_2.
  3Fe(s) + 4 H₂O(g)  Fe₃O₄(s) + 4H₂(g)
  
  (ii) Calcium reacts with water to form calcium hydroxide and hydrogen.
  
  Ca(s) + 2H₂O(l)  Ca(OH)₂(aq) + H₂(g)
  
  Potassium reacts with cold water violently immediately with evolution of H2 which catches fire.
  
  2K(s) + 2H₂O(l)  2KOH(aq) + 2H₂(g)
26. What would you observe when zinc is added to a sodium of iron(II) sulphate? Write the chemical reaction that takes place?
- Solution:Zinc is more reactive (more electro positive) than iron. Therefore it displaces iron from its salt solution. The colour of ferrous sulphate is pale green which becomes colourless.
  FeSO₄ + Zn ZnSO₄ + Fe_(s)Light green Zinc sulphate
27. Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test-tube over the burning sulphur.
What will be the action of this gas on:
Dry litmus paper?
Moist litmus paper?
Write a balanced chemical equation for the reaction taking place.
- Solution:a) When sulphur is brunt in air then sulphur dioxide gas is formed.
  (i) Sulphur dioxide gas has no action on dry litmus paper.
  
  (ii) Sulphur dioxide gas turns moist blue litmus paper to red.
  
  b) S_(s) + O_2(g) SO_2(g)

WORKSHEET -1 - Acids, Bases and salts

2011-06-07T12:12:00.000-07:00

1. You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?

Solution:The contents of each test tube would be identified by change in colour of red litmus paper. For example, when we wet the red litmus paper with the basic solution, it changes into blue colour. Put the changed blue litmus paper in the solution which turns the blue to red will be the acidic solution. The solution, which has no effect on any litmus paper, will be neutral and hence it will be distilled water.

2. Why should curd and sour substances not be kept in brass and copper vessels?

Solution:Curd and other sour foodstuffs contain acids, which can react with the metal of the vessel to form poisonous metal compounds which can cause food poisoning and damage our health.

3. Which gas is usually liberated when an acid reacts with a metal?

Solution:When an acid reacts with metal, a salt and hydrogen gas is formed. i.e

Metal + Acid Salt + Hydrogen gas.

4. Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.

Solution:The gas that extinguishes a burning candle is carbon dioxide, which is formed by the action of dilute hydrochloric acid on a metal carbonate and produces effervescence. Now, since one of the compounds formed is calcium chloride, it shows that the metal compound is calcium carbonate. Thus, the metal compound A is calcium carbonate (CaCO₃). Calcium carbonate reacts with dilute hydrochloric acid to form calcium chloride, carbon dioxide and water. This can be written as:
CaCO_3(s) + 2HCl_(aq) CaCl_2(aq)+ CO_2(g) + H₂O_(l)

5. Why do HCl, HNO₃, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?

Solution:An acid is a substance, which dissociates on dissolving in water to produce hydrogen ions [H⁺(aq) ions]. The acids like HCl, H₂SO₄, HNO₃ and CH₃COOH, etc., show acidic character because they dissociate in aqueous solutions to produce hydrogen ions, H+(aq) ions.
The compounds such as glucose and alcohol also contain hydrogen but they do not show acidic character. The aqueous solutions of glucose and alcohol do not show acidic character because the hydrogen in them does not separate out as hydrogen ions [H⁺ (aq) ions] on dissolving in water.

6. Why does an aqueous solution of acid conduct electricity?

Solution:The aqueous solution of an acid conducts electricity due to the presence of charged particles called ions in it.

7. Why does dry HCl gas not change the colour of the dry litmus paper?

Solution:Dry HCl gas does not contain any hydrogen ions in it, so it does not show acidic behaviour. In fact, dry HCl gas does not change the colour of dry litmus paper as it has no hydrogen ions [H⁺(aq) ions] in it.

8. While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?

Solution:Diluting an acid should be done by adding concentrated acid to water gradually with stirring and not by adding water to concentrated acid.

The heat is evolved gradually when a concentrated acid is added to water for diluting an acid and the large amount of water is easily absorbed.

If, however, water is added to concentrated acid to dilute it, a large amount of heat is evolved at once. The heat generated may cause the mixture to splash the acid on our face or clothes and cause acid burns.

9. How is the concentration of hydronium ions (H₃O+) affected when a solution of an acid is diluted?

Solution:When the concentrated solution of an acid is diluted by mixing water, the concentration of hydronium ions H₃O+ per unit volume decreases.

10. How is the concentration of hydroxide ions (OH-) affected when excess base is dissolved in water?

Solution:When the solution of a base is diluted by mixing more water in it, the concentration of hydroxide ions (OH^- ions) per unit volume decreases.

11. You have two solutions A and B. The pH of solution A is 6 and pH of solution B is 8.

i. Which solution has more hydrogen ion concentration?

ii. Which of this is acidic and which one is basic?

Solution:The pH of a solution is inversely proportional to its hydrogen ion concentration. This means that the solution having lower pH will have more hydrogen ion concentration. In this case, solution A(having a lower pH of 6) will have more hydrogen ion concentration.Solution A is acidic and solution B is basic.

12. What effect does the concentration of H⁺(aq) ions have on the nature of the solution?

Solution:Acids produce hydrogen ions in water. So, when an acid is added to water, the concentration of hydrogen ions in water increases. The solution of acid thus formed will have more of hydrogen ions and it will be acidic in nature.

13. Do basic solutions also have H⁺(aq) ions? If yes, why are these basic?

Solution:No, the basic solution doesn’t have H⁺ ions as the solution has excess of hydroxide ions.

14. Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?

Solution:Most often the soil in the fields is too acidic. If the soil is too acidic (having low pH), it is treated with materials like quicklime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate). Thus, a farmer should add lime or slaked lime in his fields when the soil is too acidic.

15. What is the common name of the compound CaOCl₂?

Solution:The common name of the compound CaOCl₂ is bleaching powder.

16. Name the substance that on treatment with chlorine yields bleaching powder.

Solution:Calcium hydroxide is the substance that on treatment with chlorine yields bleaching powder.

17. Name the sodium compound, which is used, for softening hard water.

Solution:Sodium carbonate (washing soda) is used for softening hard water.

18. What will happen if a solution of sodium hydro carbonate is heated? Give the equation of the reaction involved.

Solution:
Sodium Sodium Hydrogen Carbonate
Carbonate

Sodium carbonate and carbon dioxide are evolved when sodium hydro carbonate is heated.

19. Write an equation to show the reaction between plaster of Paris and water.

Solution:Plaster of Paris has a very remarkable property of setting into a hard mass on wetting with water. So, when water is added to plaster of Paris, it sets into a hard mass in about half an hour. The setting of plaster of Paris is due to the hydration crystals of gypsum, which set to form a hard, solid mass.
CaSO₄. ½ H₂O + 1 ½ H₂O CaSO₄ .2H₂O
Plaster of Paris Gypsum

20. Why does distilled water not conduct electricity, whereas rainwater does?

Solution:Distilled water does not conduct electricity because it does not contain any ionic compound (like acids, bases or salts) dissolved in it. On the other hand, rain water conducts electricity. This can be explained as follows: Rain water, while falling to the earth through the atmosphere, dissolves an acidic gas carbon dioxide from the air and forms carbonic acid (H₂CO₃). Carbonic acid provides hydrogen ions, H⁺(aq), and carbonate ions, CO^2-₃(aq), to rain water. So, due to the presence of carbonic acid (which provides ions to rain water), the rain water conducts electricity.

21. Why do acids not show acidic behaviour in the absence of water?

Solution:The acidic behaviour of acid is due to the presence of hydrogen ions. The acids will not show its acidic behaviour in the absence of water, this is because the acids produce hydrogen ions only in the presence of water.

22. Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9 respectively. Which solution is?

(i) Neutral?

(ii) Strongly alkaline?

(iii) Strongly acidic?

(iv) Weakly acidic?

(v) Weakly alkaline? Arrange the pH in increasing order of hydrogen-ion concentration.

Solution:

Solutions

pH

Neutral (D)

7

Strongly alkaline (C)

11

Strongly acidic (B)

1

Weakly acidic (A)

4

Weakly alkaline (E)

9

Arrangement of pH in the increasing order of hydrogen ion concentration is as follows:

pH (11) pH (9) pH (7) pH (4) pH (1)

C E D A B

pH in the increasing order of hydrogen ion concentration

23. Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test-tube A while acetic acid (CH₃COOH) is added to test-tube B. In which test-tube will fizzing occur more vigorously and why?

Solution:Acetic acid (CH₃COOH) is a weak acid whereas hydrochloric acid (HCl) is a strong acid. Fizzing occurs in the test tube due to the evolution of hydrogen gas by the action of acid on magnesium ribbon. Since hydrochloric acid is a strong acid a large amount of hydrogen gas is liberated in the test tube A. So fizzing occurs more vigorously in test tube A .

24. Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain.

Solution:The pH will change to below 6, as lactic acid is formed when milk turns into curd.

25. Plaster of Paris should be stored in a moisture-proof container. Why?

Solution:The presence of moisture can affect the slow setting of plaster of Paris by bringing about its hydration. This will make the plaster of Paris useless after some time.

26. What is a neutralization reaction? Give two examples.

Solution:The reaction of an acid and a base, giving rise to the corresponding salt and water is called neutralization reaction.
Examples:

NaOH + HCl NaCl + H₂O

Mg(OH)₂ + H₂CO₃ MgCO₃ + 2H₂O

27. Give two important uses of washing soda and baking soda.

Solution:Washing soda
1. It is often used as an electrolyte.

2. Domestically it is used as a water softener during laundry.

Baking soda

1. It is used to test garden soil for acidity. If it develops bubbles, the soil is too acidic.

2. Washing a car with it can remove dead bug bodies without damaging the paint.

Concentration of SO₂ (in ppm)	Frequency
0.00 - 0.04 0.04 - 0.08 0.08 - 0.12 0.12 - 0.16 0.16 - 0.20 0.20 - 0.24	4 9 9 2 4 2

Life time (in hours)	Frequency (f_i)
0 – 20	10
20 – 40	35
40 – 60	52
60 – 80	61
80 – 100	38
100 – 120	29

Monthly consumption (in units)	Number of consumers
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

Solutions	pH
Neutral (D)	7
Strongly alkaline (C)	11
Strongly acidic (B)	1
Weakly acidic (A)	4
Weakly alkaline (E)	9

pH (11)	pH (9)	pH (7)	pH (4)	pH (1)
C	E	D	A	B
pH in the increasing order of hydrogen ion concentration

4exam

plant cell

Heart

In fish

Sound

Contents

Propagation of sound

Perception of sound

Physics of sound

Longitudinal and transverse waves

Sound wave properties and characteristics

Speed of sound

Acoustics

Noise

Project Tiger

pH scale

digestive system diagram

Acids, Bases and salts extra qus

Types and Distribution of Forest and Wildlife Resources

forest and wildlife

Introduction to Trigonometry

Statistics WORKSHEET

WORKSHEET Management of Natural Resources

WORKSHEET 1 Control and coordination

Digestion in Humans

WORKSHEET -1-Life Processes

2 - Forest and Wildlife Resources

1 - Resources and Development

worksheet 1 Triangles

WORKSHEET 1 Arithmetic Progressions

- WORKSHEET 1 - Quadratic Equations

WORKSHEET 1 Pair of Linear Equations in Two variables

WORKSHEET 1 - Polynomials

WORKSHEET -1 - Electricity

WORKSHEET -1 - metals and non metals

WORKSHEET -1 - Acids, Bases and salts