KAAC

SOLVED TIME AND DISTANCE

2018-01-31T12:20:00.000+05:30

FROM THE CLUES GIVEN FINDING THE UNKNOWN

2018-01-31T11:11:00.001+05:30

In solving the mathematics one has to analyse the given clues and finding the unknown. Also all the data should be deduced in common unit. Like kilometers, seconds, minutes, hours etc.

WHY WE FORGET AN ANIMAL IS THE SAME AS HUMANS ? EARTH IS EQUAL TO ALL.

2018-01-06T18:21:00.003+05:30

Elementary math again with Speed and Time relationship. Math is fun and finding unknown from the given clues , Enjoy Math

2018-01-06T18:18:00.006+05:30

Series on elementary mathematics on time and speed.

2018-01-06T14:14:00.002+05:30

WHEN INSTEAD OF SPEED TIME TAKEN BY THE TRAIN IS GIVEN LENGTH OF THE STATION CAN BE DETERMINED.NEXT MATH.

A series of solved elementary mathematics on time and speed

2018-01-06T13:18:00.002+05:30

CONVERT MINUTES INTO SECONDS

SURFACE AREA INCREASED WHEN SPHERES ARE REDUCED IN SIZE

2016-12-10T12:50:00.001+05:30

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One math in finding surface area . Do know water, surface, container , kaacconsultant@gmail.com

2016-12-09T11:35:00.005+05:30

DEPRECIATION

2016-12-09T10:19:00.004+05:30

UNDERSTANDING DEPRECIATION ON ACCOUNTING PRINCIPLE

EVERY EQUIPMENT OR ITEM HAS ITS OWN USEFUL LIFE AFTER THAT IT CEASES TO BE USEFUL ONE.
IN THIS LOGIC , WHEN WE USE ANY EQUIPMENT , ITS ORIGINAL PURCHASE PRICE ALSO REDUCES THROUGHOUT ITS USEFUL LIFE TILL ITS PRICE REDUCES TO ZERO.

THAT IS THE EQUIPMENT DEPRECIATES ITS PRICE. IT DEPENDS ON ITS USEFUL LIFE OF THAT PARTICULAR ITEM. THE PRICE OF THE EQUIPMENT IS DIVIDED BY THE USEFUL LIFE AND THAT AMOUNT IS REDUCED FROM ITS PREVIOUS PRICE.
THERE ARE SEVERAL METHODS OF CALCULATING DEPRECIATION.

A EXAMPLE IS GIVEN LIKE

ISHAAN BOUGHT A CAR FOR RUPEES 300000. ITS PRICE DEPRECIATES AT THE RATE OF 10 % PER YEAR.

AFTER 1 YEAR 10 % OF RUPEES 300000 WILL BE REDUCED FROM RS 300000

SO AMOUNT WILL BE 300000* 10/100= RUPEES 30000

SO, DEPRECIATED PRICE OF THE CAR AFTER 1 YEAR WILL BE 300000 - 30000 = RS.270000

NOW AFTER 2 YEAR AGAIN 10% WILL BE DEPRECIATED FROM RS. 270000

SO, 10 % OF RUPEES 270000 IS 270000*10/100= RS. 27000

RUPEES 27000 WILL BE DEDUCTED FROM RS.270000

ONE METHOD CALCULATION OF DEPRECIATION IS LIKE THIS.

PROBLEM SOLVER ON QUADRILATERAL

2016-12-08T10:47:00.002+05:30

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An interesting problem in trapezium

2016-12-08T09:42:00.002+05:30

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Again returned to Operations on numbers

2016-11-27T18:34:00.002+05:30

PROBLEM GIVEN

What least no must be subtracted from 1672 to obtain a number which is completely divisible by 17 ?

LET US DIVIDE

17| 1672 | 98
153
_________
142
136
__________
6

SO 6 needs to be subtracted from 1672 to get exactly divisible by 17.

ANOTHER INTERESTING MATH>
FIND THE TOTAL NUMBER OF PRIME FACTORS IN THE

((4)¹¹ X 7 ⁵ X (11)² ) SOLUTION:

((2 X 2)¹¹ X 7 ⁵X ( 11 )²

((2²)¹¹ X 7 ⁵X ( 11 )²

(2)²² X 7 ⁵ X (11) ²

REQUIRED NO OF FACTORS = 22 +5 +2 = 29

This problem is interesting

2016-11-27T09:39:00.002+05:30

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Maths are interesting , when you grow interested

2016-11-22T21:35:00.001+05:30

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mathematics is fun, area of the rhombus

2016-11-22T21:07:00.002+05:30

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mathematics drawn on pictorial

2016-11-20T21:14:00.004+05:30

Mensuration on cylinders and cubes

2016-11-20T10:15:00.006+05:30

Which has a greater volume , a cylinder of height 21 Cm, and a base of radius 10 Cm.
or, a cube with an edge by measuring 19 cm.

solution
volume of the cylinder = π * Radius ² * Height = π1010 *21=				6600	cm ³

Volume of the cube = 1919 19=		6859	cm ³

The cube is greater than by = 259	cm ³

An interesting math applied on practical life

2016-11-20T09:21:00.001+05:30

Find the area covered by a road roller in 50 revolutions if it's diameter is 140 Cm. and it,s width is 120 Cm. Convert the area in Square Meter

Before going to the mathematics solution, this formulas needs to be memorized

1. Volume of a cube ( length L * Breadth B * Height H ) , if the cube is of equal dimensions then it is L *L*L cubic unit ,

2. Surface area of a equal unit of cube		6 L ²
3 .For pure water	1 ml = 1 cm³
4. surface area of a cuboid = 2 ( LB + BH + L*H) where L =length, B= Breadth, H= Height
5. Volume of a cuboid = (LBH) where L =length, B= Breadth, H= Height

solution

When the wheel of the road roller moves it covers an area equal to its lateral surface area per revolution

Given data	width of the road roller wheel L = 120 Cm
	Radius of the road roller wheel R = diameter/2 = 140/2= 70 Cm

	on per revolution , it covers an area = 2 π R* L		52800	Sq Cm

Now the road roller wheel has taken 50 revolutions , so total lateral area is 50*52800 = 264000 square Cm

The area covered in square Meter 264000/(100*100)		26.4	square meter	M²

save energy , fuel of life

2016-11-20T08:58:00.000+05:30

SAVE ENERGY SAVE FUEL

TIPS ON ENERGY SAVING

IN HOME APPLIANCES AND ELECTRICITY SAFETY

The Domestic Sector accounts for 30% of total energy consumption in the country. There is a

tremendous scope to conserve energy by adopting simple measures.

This information is a guide, which offers easy, practical solutions for saving energy in Home

Appliances. Please, take a few moments to read the valuable tips that will save energy & money

and ultimately help conserve our natural resources.

It would be useful to know which gadget consumes how much electricity. Economic use of home

appliances can help in reducing electricity bills.

By following these simple tips one can save energy to a large extent.

Lighting

Ø Turn off the lights when not in use

Ø Take advantage of daylight by using light-colored, loose-weave curtains on your windows

to allow daylight to penetrate the room. Also, decorate with lighter colors that reflect

daylight

Ø De-dust lighting fixtures to maintain illumination

Ø Use task lighting; instead of brightly lighting an entire room, focus the light where you need

it

Ø Compact fluorescent bulbs are four times more energy efficient than incandescent bulbs

and provide the same lighting

Ø Use electronic chokes in place of conventional copper chokes

Fans

Ø Replace conventional regulators with electronic regulators for ceiling fans

Ø Install exhaust fans at a higher elevation than ceiling fans

Electric iron

Ø Select iron boxes with automatic temperature cutoff

Ø Use appropriate regulator position for ironing

Ø Do not put more water on clothes while ironing

Ø Do not iron wet clothes

Kitchen Appliances

Ø Mixers

§ Avoid dry grinding in your food processors ( mixers and grinders) as it takes longer

time than liquid grinding

Ø Microwaves ovens

§ Consumes 50 % less energy than conventional electric / gas stoves

§ Do not bake large food items

§ Unless you're baking breads or pastries, you may not even need to preheat

§ Don't open the oven door too often to check food condition as each opening leads to a

temperature drop of 25°C

Ø Electric stove

§ Turn off electric stoves several minutes before the specified cooking time

§ Use flat-bottomed pans that make full contact with the cooking coil

Tips on Energy Savings Page 3 of 5

Ø Gas stove

§ When cooking on a gas burner, use moderate flame settings to conserve LPG

§ Remember that a blue flame means your gas stove is operating efficiently

§ Yellowish flame is an indicator that the burner needs cleaning

§ Use pressure cookers as much as possible

§ Use lids to cover the pans while cooking

§ Bring items taken out of refrigerators (like vegetables, milk etc) to room temperature

before placing on the gas stove for heating

Ø Use Solar Water Heater – a good replacement for a electric water heater

Electronic Devices

Ø Do not switch on the power when TV and Audio Systems are not in use i.e. idle operation

leads to an energy loss of 10 watts/device

Computers

Ø Turn off your home office equipment when not in use. A computer that runs 24 hours a

day, for instance, uses - more power than an energy-efficient refrigerator.

Ø If your computer must be left on, turn off the monitor; this device alone uses more than half

the system's energy.

Ø Setting computers, monitors, and copiers to use sleep-mode when not in use helps cut

energy costs by approximately 40%.

Ø Battery chargers, such as those for laptops, cell phones and digital cameras, draw power

whenever they are plugged in and are very inefficient. Pull the plug and save.

Ø Screen savers save computer screens, not energy. Start-ups and shutdowns do not use

any extra energy, nor are they hard on your computer components. In fact, shutting

computers down when you are finished using them actually reduces system wear - and

saves energy

Refrigerator

Ø Regularly defrost manual-defrost refrigerators and freezers; frost buildup increases the

amount of energy needed to keep the motor running.

Ø Leave enough space between your refrigerator and the walls so that air can easily

circulate around the refrigerator

Ø Don't keep your refrigerator or freezer too cold.

Ø Make sure your refrigerator door seals are airtight

Ø Cover liquids and wrap foods stored in the refrigerator. Uncovered foods release moisture

and make the compressor work harder.

Ø Do not open the doors of the refrigerators frequently

Ø Don't leave the fridge door open for longer than necessary, as cold air will escape.

Ø Use smaller cabinets for storing frequently used items

Ø Avoid putting hot or warm food straight into the fridge

Tips on Energy Savings Page 4 of 5

Washing machines

Ø Always wash only with full loads

Ø Use optimal quantity of water

Ø Use timer facility to save energy

Ø Use the correct amount of detergent

Ø Use hot water only for very dirty clothes

Ø Always use cold water in the rinse cycle

Ø Prefer natural drying over electric dryers

Air Conditioners

Ø Prefer air conditioners having automatic temperature cut off

Ø Keep regulators at “low cool” position

Ø Operate the ceiling fan in conjunction with your window air conditioner to spread the

cooled air more effectively throughout the room and operate the air conditioner at higher

temperature

Ø Seal the doors and windows properly

Ø Leave enough space between your air conditioner and the walls to allow better air

circulation

Ø A roof garden can reduce the load on Air Conditioner

Ø Use windows with sun films/curtains

Ø Set your thermostat as high as comfortably possible in the summer. The less difference

between the indoor and outdoor temperatures, the lower will be energy consumption.

Ø Don't set your thermostat at a colder setting than normal when you turn on your air

conditioner. It will not cool your home any faster and could result in excessive cooling.

Ø Don't place lamps or TV sets near your air-conditioning thermostat. The thermostat senses

heat from these appliances, which can cause the air conditioner to run longer than

necessary.

Ø Plant trees or shrubs to shade air-conditioning units but not to block the airflow. A unit

operating in the shade uses as much as 10% less electricity than the same one operating

in the sun.

Tips on Energy Savings Page 5 of 5

Electrical Safety Tips for Homes

Electrical Hazards

Ø Shocks

§ Electric Shock causes an involuntary grip which prolongs the period of contact.

§ More the period of contact, more the damage

§ Passage of current through the heart , stops the blood supply to the brain , resulting in

loss of consciousness and termination of breathing

§ When a person standing at a height receives an electrical shock , it is most likely that

he will fall

§ Personal sensitivity to electrical shock varies from person to person

Ø Burns

§ Whenever an electrical flash appears, and if a part of a body comes within flashing

distance, burns can be caused

§ Burns may be caused by short circuits as well, because a short circuit could create an

electrical fire
Fuels are NOT EVERLASTING. ENERGY SAVING IS RETAINING NATIONAL AND INTERNATIONAL PROPERTY SAVING
Preventive Measures

Ø Allow only a qualified person to attend to your electrical repairs

Ø Service your electrical equipment at frequent intervals through a competent electrician

Ø In case of a short circuit or a fire, switch off the mains instantly Make sure that you have

easy access to switch off the supply source quickly, in case of an emergency

Ø Make sure your extension cords are free from cuts, improper insulation, or joints

Ø Ensure pins of your plugs are tight and not loose

Ø Use switches of the correct current rating and preferably with indicators to indicate whether

the switch is ON/OFF

Ø Use appliances with 3 pin plugs and connect them to 3 pin sockets

Ø Do not overload electrical outlets or use extension cords in place of additional outlets

Ø Switch off electrical appliances when not in use

Ø Provide proper earthing for the building/house

Mathematics on unitary methods

2016-11-20T08:26:00.000+05:30

A can complete a work in 45 days while B can complete the work in 50 days and C can do in 55 days.

After 8 days of working by A , B joins with A and on 20 day C joined for till completion of rest of the work.

How much time it took for the total work ?

SOLUTION

Let the total work is 1

A can complete the work in 45 days when does alone.

So in 1 day A can do 1/45 parts of the work.

A worked till 8 days. So in 8 days 8/45 parts of the work is completed=

0.17778

parts

Now B can complete the work in 50 days

So in 1 day B can do 1/50 parts of the work.

A and B did work upto 20 th. day.

In one day A and B in together does (1/45+ 1/50 ) parts of the work = 19/450 parts

For 12 days total work done is =12* 19/450 =

0.50667

Total work completed = 0.17778+0.50667=

0.68445

Work left = 1 - 0.68445 =

0.31555

parts

Now, this 0.31555 parts of the work A, B and C worked together

C can do 1/55 parts of the work in one day.

In one day 1/45 +1/55+ 1/60 =

0.057071

parts of the work can be executed

Work left is 0.31555 , so time will be

5.529078

days

Total time taken A alone 8 days

And A and B 12 days

A,B and C together 5.53 days

25.53

days.

ANS

A water tank can be filled up with a filling line by 10 Hrs. and other by 14 Hrs.

How much time it will take when both are filling simultaneosly

solution :

Let we consider the filled up tank quantity is 1

In 10 hours the tank is filled up by one line

so , in 1 hour the tank is filled up by 1/10 parts quantity

By other line in 14 hrs. the tank can be filled

so in 1 hour the tank is filled up by 1/14 parts

Now , when both the filling lines are being used ,

simultaneously 1/10 + 1/14 parts are being filled up in 1 hour

0.1714

parts are filled up in 1 hour when both the filling lines are opened.

Now let us calculate

0.1714 parts are filled up in 1 hour

1 that is full tank is filled up 1/0.1714 hr=

5.8343

hrs.

the tank will be filled up in 5.8384 hrs.

A water tank can be filled up with a filling line by 10 Hrs. and other by 14 Hrs.

How much time it will take when first filling line is opened for 5 hrs . And second line started

to fill the tank. calculate the time taken for filling the tank

solution

First filling line was opened for 5 hours.

Let us suppose that the filling volume of the tank is 1

first filling line takes 10 hrs. to completely fill up the tank

so in 1 hr 1/10 parts of the tank is filled up

In five hours 5*1/10 parts

0.5

parts of the tank is filled up

remaining part of the tank will be 1-0.5 = 0.5 parts

now let us take the second line which takes 14 hrs to fill up the full tank

but in this case 0.5 parts that is half of the tank needs to be filled up

so , in mathematically to deduce

1 volume of tank is filled up in 14 hours

0.5 volume of tank will require 14 * 0.5 = 7 Hrs.

A water tank can be filled up with a filling line by 10 Hrs. and other by 14 Hrs.

How much time it will take when second filling line is opened for 5 hrs . And first line started

to fill the tank. calculate the time taken for filling the tank

solution

let us consider the volume, that is filled up tank is 1 unit

the second filling line takes 14 hrs to fill up the tank

so in one hour second filling line fills up 1/14 volume of the tank

Second filling line opened for five hours

total filled up volume by the second line till five hours is 5/14 unit

0.3571

unit

we have taken

total volume as 1

so empty tank is 1- 0.3571

0.6429

unit

now the first line is started which fills up the tank in 10 hrs. so in one hour

tank filled up by 1/10 unit

now we can say that 1/10 unit is filled up in 1 hour

1 unit will be filled up in 10 hours

0.6429 unit will be filled 10*0.6429

6.429

hrs.

on compound interest comparison to simple interest

2016-11-14T20:33:00.000+05:30

In how much time will Rs 10000 amounts to Rs. 10,600:00 with simple interest .rate of interest is 6%. With the same time period find the compound interest rate.

amount in Rs.	year	interest rate= 6/100=0.06	interest	total
10000	after 1st year	0.06	600	10600
10000	after 2 year	0.06	600	10600
10000	after 3 year	0.06	600	10600
			1800

After three years Rs 10000.00 will become Rs.11800 .00. Total interest received is Rs. 1800.00

While calculating with compound interest
amount in Rs.	year	interest rate= 6/100=0.06	interest	total
10000	after 1st year	0.06	600	10600
10600	after 2 year	0.06	636	11236
11236	after 2 year	0.06	674.16	11910.16

With compound interest rate , interest received Rs.			1910.16

On mathematics, it is real fun. Mathematics on compound interest

2016-11-14T20:12:00.001+05:30

Mathematics is real fun.Generally teachers failed to extract the fun from the mathematics and percolate this to their students. Objectives and usefulness of the particular math. is not known to the students and they failed to grasp where they will apply those stuffs in real life. So they felt the burden of mathematics and the fear of burden is felt life long.
For some mathematics is a hobby and they can take this as their profession.

A practical aspect on mathematics. If students know the power of compounding they will be much matured at early stage of life on financial freedom.

In how much time will Rs. 50000 amount to Rs. 560000 at 4 % compound interest

amount	interest rate	interest	year	total amount
50000	4/100= 0.04	2000	after 1 st year	52000
52000	0.04	2080	2 year end	54080
54080	0.04	2163.2	3 year end	56243.2
56243.2	0.04	2249.728	4 year end	58492.928
58492.928	0.04	2339.71712	5 year end	60832.64512
60832.6451	0.04	2433.3058	6 year end	63265.95092
63265.9509	0.04	2530.63804	7 year end	65796.58896
65796.589	0.04	2631.86356	8 year end	68428.45252
68428.4525	0.04	2737.1381	9 year end	71165.59062
71165.5906	0.04	2846.62362	10 year end	74012.21425
74012.2142	0.04	2960.48857	11 year end	76972.70282
76972.7028	0.04	3078.90811	12 year end	80051.61093
80051.6109	0.04	3202.06444	13 year end	83253.67537
83253.6754	0.04	3330.14701	14 year end	86583.82238
86583.8224	0.04	3463.3529	15 year end	90047.17528
90047.1753	0.04	3601.88701	16 year end	93649.06229
93649.0623	0.04	3745.96249	17 year end	97395.02478
97395.0248	0.04	3895.80099	18 year end	101290.8258
101290.826	0.04	4051.63303	19 year end	105342.4588
105342.459	0.04	4213.69835	20 year end	109556.1572
109556.157	0.04	4382.24629	21 year end	113938.4034
113938.403	0.04	4557.53614	22 year end	118495.9396
118495.94	0.04	4739.83758	23 year end	123235.7772
123235.777	0.04	4929.43109	24 year end	128165.2082
128165.208	0.04	5126.60833	25 year end	133291.8166
133291.817	0.04	5331.67266	26 year end	138623.4892
138623.489	0.04	5544.93957	27 year end	144168.4288
144168.429	0.04	5766.73715	28 year end	149935.166
149935.166	0.04	5997.40664	29 year end	155932.5726
155932.573	0.04	6237.3029	30 year end	162169.8755
162169.876	0.04	6486.79502	31 year end	168656.6705
168656.671	0.04	6746.26682	32 year end	175402.9373
175402.937	0.04	7016.11749	33 year end	182419.0548
182419.055	0.04	7296.76219	34 year end	189715.817
189715.817	0.04	7588.63268	35 year end	197304.4497
197304.45	0.04	7892.17799	36 year end	205196.6277
205196.628	0.04	8207.86511	37 year end	213404.4928
213404.493	0.04	8536.17971	38 year end	221940.6725
221940.673	0.04	8877.6269	39 year end	230818.2994
230818.299	0.04	9232.73198	40 year end	240051.0314
240051.031	0.04	9602.04126	41 year end	249653.0727
249653.073	0.04	9986.12291	42 year end	259639.1956
259639.196	0.04	10385.5678	43 year end	270024.7634
270024.763	0.04	10800.9905	44 year end	280825.7539
280825.754	0.04	11233.0302	45 year end	292058.7841
292058.784	0.04	11682.3514	46 year end	303741.1354
303741.135	0.04	12149.6454	47 year end	315890.7809
315890.781	0.04	12635.6312	48 year end	328526.4121
328526.412	0.04	13141.0565	49 year end	341667.4686
341667.469	0.04	13666.6987	50 year end	355334.1673
355334.167	0.04	14213.3667	51 year end	369547.534
369547.534	0.04	14781.9014	52 year end	384329.4354
384329.435	0.04	15373.1774	53 year end	399702.6128
399702.613	0.04	15988.1045	54 year end	415690.7173
415690.717	0.04	16627.6287	55 year end	432318.346
432318.346	0.04	17292.7338	56 year end	449611.0798
449611.08	0.04	17984.4432	57 year end	467595.523
467595.523	0.04	18703.8209	58 year end	486299.3439
486299.344	0.04	19451.9738	59 year end	505751.3177
505751.318	0.04	20230.0527	60 year end	525981.3704
525981.37	0.04	21039.2548	61 year end	547020.6252
547020.625	0.04	21880.825	62 year end	568901.4502
568901.45	0.04	22756.058	63 year end	591657.5082

IT is called UNIFICATION method on Mathematics

2016-11-13T22:42:00.005+05:30

Problem with Chintu and Mintu

Chintu can do a task in 24 days while Mintu can do it in 30 days. In how many days can they complete the task when work to gether?

Now let us take the complete work as 1

Chintu can complete the work in 24 days.
So in one day Chintu can complete 1/24 part of the work

Similarly Mintu is capable of doing the work in 30 days.
So in one day Mintu completes 1/30 parts of work.

Now when they both works in one day total work completion is 1/24 + 1/30 parts of work
1/24+ 1/30
= 54/720= 3/40 parts of the total work

Now we can say that

3/40 parts of work is being completed in 1 day

so 1 work(complete) will be completed in 40/3 days= 13.33 days

Mathematics on Time and Distance

2016-11-13T08:34:00.003+05:30

Problem given

One car has a speed of 204.2 Km/hr and other car has the speed 200 km/hr. Both the car started simultaneously , but the second car reached the same distance after 7 seconds of the first car. What is the distance and what will be time required for both the cars .

Solution:

Let , the distance be X km and let the time taken for the first car be T hr. to cover the entire distance.
Time taken for the first car T = X/ 204.2 Hr.
Naturally other car has taken T Hr + 7 secs to cover that distance

Now , for the first car , at the speed of 204.2 km/hr at T hr it will traverse 204.2 T Km.

and for the second car at the speed of 200 Km/hr. at (T+ 7/3600) Hrs. it will cover the same distance of 200 *( T + 7/3600 ) Km , which we have already assumed as X km.

204*T Km = 200 *(T+ 7/3600 ) KM

204 T = 200 T + 200 x 7/3600
4 T = 7/ 18
T= 28/18 Hrs. = 1.5 Hrs.

Distance covered by the first car in 1.5 Hrs. at the speed of 204.2 Km/Hr 204.2 x 1.5= 306.30 Km

Another problem
Two train starts , one from station A towards station B at morning 8:00 AM and reached at station B at 10:30 AM.
Another train starts from station B at 9:00 AM and reached station B at 12:00 noon. When both the trains will meet?

Solution

Let the distance between the stations A and B be X km. and let after a time T both the trains meet at a same point.
Speed of train A is ( X /2.5 ) Km/hr and speed of train B is ( X/3) Km/hr
So in time T train A travels (X/2.5)*T Km.
and in time T train B travels (X/3)* T Km.

Now , we know that to total distance is X Km.

(X/2.5)*T +(X/3)*T = X
or, XT/2.5 + XT/3= X
or,X(T/2.5+ T/3)= X
or, T/2.5+ T/3= 1
or,( 3T +2.5T)/7.5= 1
or,5.5 T= 7.5
or,T = 7.5/5.5 Hr= 15/11 Hr=

One simple mathematics

Factorise
4 X2 - 12 X + 9 = 0

OR, (2X)2 - 2. 2X. 3 + (3)2 = 0
OR, (2X -3)(2X-3)
OR, X = 3/2

PROBLEM

3 ^(X-Y) = 27 = 3 ^{3 ______________________}1

3^(X+Y)= 243= 3 ^{5________________________}2

OR, X- Y = 3

X+ Y= 5 OR , 2 X = 8 OR X = 4 When X = 4 , Y = 1

some mathematics on percentage

2016-11-07T22:28:00.004+05:30

problem

In an examination total students appeared are 40% girls and rest boys. Now 80% of the boys and 60% of the girls has succeeded in examination. How many % of the total have failed in the examination.

Solution:
When 40 % are the girls , so 60 % are boys. { In totality it is 100%}

When the answer is asking in % number, let us assume that total no os girls and boys are 100.

Now what is happening that out of 60 boys, 80% of the boys have passed.

No of boys have passed is = 60* 80/100 = 48 No of boys or 48 % of boys have passed.
No of girls passed is = 40* 60/100 = 24 % of the girls have passed.

so no of boys and girls failed = 100 - (48 + 24) = 28%

Other problem

A person makes a profit of 12 % on selling an article at 10 % discount on the printed price. Find out the ratio of the cost price and printed price

Cost price is the price in which the seller has purchased the article on which he has to make the profit.
Printed price is the price which is printed on the packaging of the article. like MRP ( maximum retail price)

Let us assume the printed price is Rs. x

the person sold the article at 10 % discount

Selling price x - (10/100)*x

= Rs. ( x- 0.1x)= Rs.0.9x

Now the other statement is showing that on selling at Rs. 0.9 he makes a 12 % profit.
So his purchasing price is 12 % less of Rs. 0.9 x

so cost price or the sellers purchasing price is =Rs. 0.9 x- (12/100)0.9x = Rs. 0.792 x

so in terms of ratio of cost price/ printed price = 0.792 x/ x= 0.792:1

other problem

By selling an umbrella for Rs. 500 a shopkeeper gains 20 % . During a clearance sale the shopkeeper allows a discount of 10 % on the marked price. What is his gain percentage.

Here marked price is Rs. 500

Cost price of the umbrella is Rs. 500 - (20/100)*500 = Rs. 400

Now on clearance sale it is discounted to 10 % , so the new selling price is Rs. 500- (10/100)*500= Rs 450

Now the item is sold at Rs. 450 against cost price of Rs. 400

percentage(%) gain (450-400)/400*100 = 50/400*100 =12.5 %