RAJ EDUCATION HUB

noreply@blogger.com (Unknown) — Fri, 19 Sep 2025 10:19:00 +0530

1. Which village in India received the "Best Tourism Village 2023" award?

(a) Mawlynnong, Meghalaya

(b) Kongthong, Meghalaya

(d) Pochampally, Telangana

2. Which state granted a 32% quota for tribals in schools?

(a) Madhya Pradesh

(b) Gujarat

(C)Chhattisgarh

(d) Jharkhand

3 What is the name of India's first toll-free National Anti-Narcotics Helpline?

(a) Nasha Mukt Bharat

(b) MANAS

(d) Nasha Vimochana

4. Recently, Prem Singh Tamang became the Chief Minister of which Himalayan state?

(a) Uttarakhand

(b) Sikkim

(d) Meghalaya

5 .What is the name of India's longest cable stayed bridge

(a) Dhola-Sadlya Bridge

(b) Vidyasagar Setu

(C)Sudarshan Setu

(d) Bogibeel Bridge

6. Which state recently launched the Mukhya Mantri Nijut Moina (MMNM) scheme to promote girl education in the state?

(a) Manipur

(b) Nagaland

(d). Asam

7. Which of the following states won the Best Horticulture Award in 2024

(a)Nagaland

(b) Uttar Pradesh

(d) Kerala

8. Hakkı Pikki tribe, recently seen in the news, is predominantly found in which state?

(a) Karnataka

(b) Bihar

(d) Telangana

9. Which country hosts the famous Frankfurt Book Fair

(a) France

(b) Germany

(d) UK

10. in the 2024 Global Gender Gap index, what is India's ranking out of 146 countries?

(a) 127th

(b) 112th

(d) 129th

11. Which country was the third-largest fish producer in the world in 20237

(a) United States

(b) Japan

(C)India

(d) Norway

12. Recently, India announced a medical e-visa facility for citizens of which country?

(a)Bangladesh

(b) Nepal

(d) Myanmar

13. Which country will host the multinational air exercise Tarang Shakti 2024' in August?

(a) Germany

(b) France

(d)-India

14. Which country hosted the 50th G7 leaders summit?

(a) Italy

(b) France

(d) Canada

15. Barsingsar Solar Power Plant, recently seen in the news, is located in which state?

(a) Gujarat

(b) Rajasthan

(d) Odisha

16. Which company has unveiled its new open source Al model Llama 3.17

(a) Meta

(b) Google

(d) Facebook

17. Lyme disease, recently seen in the news, is caused by which one of the following agents?

(a) Bacteria

(b) Virus

(d) Helminths

18. Who is the first Indian to secure a gold medal at the World Athletics Championships?

(a) Abhinav Bindra

(b) Anju Bobby George

(d) Muhammed Anas

19 Which ministry launched the Swachh Sujal Shakti Samman 20237

(a) Ministry of Labour

(b) Ministry of Communications

(d) Ministry of Health

20 Swadesh Darshan 2.0 Programme' was implemented by the

(a) Ministry of Railways

(b) Ministry of Home Affairs

(d) Ministry of Culture

21 A historic city in Japan was designated as a UNESCO Creative City for its cultural heritage and traditions in 2023. Which city is this?

(a) Kyoto

(b) Tokyo

(d) Hiroshima

22 .Which of the following countries hosted the Badminton Asia Championship 2024?

(a) India

(b) China

(d) Japan

23. Which Indian table tennis player won the men's singles event at the WTT Feeder Beirut 2024 tournament?

(a) Sharath Kamal

(b) Sathiyan Gnanasekaran

(d) Manav Thakkar

24. Who won two bronze medals for India at the Archery World Cup 20247

(a) Pravin Jadhav

(b) Abhishek Verma

(d) Limba Ram

25. What is the theme of 'International Olympics Day 20247

(a) Together, For a Peaceful World

(b) Let's Move and Celebrate

(d) Moving Forward: United by Emotion

Tables from 2 to 40

noreply@blogger.com (Unknown) — Tue, 26 Aug 2025 18:34:00 +0530

Tables from 2 to 40

Tables from 1 to 10

Tables from 11 to 20

Tables from 21 to 30

Tables from 31 to 40

CLASS 8 TEST PAPER EXERCISE 1 A, 1B, 1C RS AGARWAL

noreply@blogger.com (Unknown) — Tue, 10 Jun 2025 11:05:00 +0530

CLASS 8 TEST PAPER EXERCISE 1 A, 1B, 1C RS AGARWAL

QESTION 1

Fill in the blanks with the correct symbol out of

$>, = and <$ :
$\begin{matrix} (i) \frac{- 3}{7} ... ... . \frac{6}{- 13} & (i i) \frac{5}{- 13} ... . . \frac{- 35}{91} & (i i i) - 2 ... ... . \frac{- 13}{5} (i v) \frac{- 2}{3} ... ... . . \frac{5}{- 8} \end{matrix}$

QUESTION 2

Arrange the following rational numbers in descending order:

$(i) - 2, \frac{- 13}{6}, \frac{8}{- 3}, \frac{1}{3}$
$(i i) \frac{- 3}{10}, \frac{7}{- 15}, \frac{- 11}{20}, \frac{17}{- 30}$
$(i i i) \frac{- 5}{6}, \frac{- 7}{2}, \frac{- 13}{18}, \frac{23}{- 24}$
$(i v) \frac{- 10}{11}, \frac{- 19}{11}, \frac{- 23}{33}, \frac{- 39}{44}$

QUESTION 3

Which of the following statements are true and which are false?
(i) $\frac{- 3}{5}$ lies to the left of 0 on the number line.
(ii) $\frac{- 12}{7}$ lies to the right of 0 on the number line.
(iii) The rational numbers $\frac{1}{3} and \frac{- 5}{2}$ are on opposite sides of 0 on the number line.
(iv) The rational number $\frac{- 18}{- 13}$ lies to the left of 0 on the number line.

QUESTION 4

Verify the following:

(i) $(\frac{3}{4} + \frac{- 2}{5}) + \frac{- 7}{10} = \frac{3}{4} + (\frac{- 2}{5} + \frac{- 7}{10})$
(ii) $(\frac{- 7}{11} + \frac{2}{- 5}) + \frac{- 13}{22} = \frac{- 7}{11} + (\frac{2}{- 5} + \frac{- 13}{22})$

QUESTION 5

Subtract:

(i) $\frac{3}{4}$ from $\frac{1}{3}$
(ii) $\frac{- 5}{6}$ from $\frac{1}{3}$
(iii) $\frac{- 8}{9}$ from $\frac{- 3}{5}$
(iv) $\frac{- 9}{7}$ from -1

QUESTION 6

The sum of two rational numbers is $- 2$ . If one of the numbers is $\frac{- 14}{5}$ , find the other.

QUESTION 7

What number should be subtracted from

\frac{- 2}{3}

to get

\frac{- 1}{6}

QUESTION 8

What number should be added to

$\frac{- 5}{8}$ so as to get $\frac{- 3}{2}$ ?

QUESTION 9

Express

$\frac{- 42}{98}$ as a rational number with denominator $7$

QUESTION10

Write tables from 15 to 30

Write square from 11 to 30

Formulas Surface area and volume

noreply@blogger.com (Unknown) — Mon, 2 Jun 2025 19:37:00 +0530

1.Cube

Surface Area: $6 a^{2}$
Volume: $a^{3}$
(where $a$ is the edge length)

2. Cuboid

Surface Area: $2 (l b + b h + h l)$
Volume: $l \times b \times h$
(length $l$ , breadth $b$ , height $h$ )

3. Cylinder

Curved Surface Area (CSA): $2 π r h$
Total Surface Area (TSA): $2 π r (h + r)$
Volume: $π r^{2} h$
(radius $r$ , height $h$ )

4. Cone

CSA: $π r l$
TSA: $π r (l + r)$
Volume: $\frac{1}{3} π r^{2} h$
(slant height $l$ , vertical height $h$ )

5. Sphere

Surface Area: $4 π r^{2}$
Volume: $\frac{4}{3} π r^{3}$

6. Hemisphere

CSA: $2 π r^{2}$
TSA: $3 π r^{2}$
Volume: $\frac{2}{3} π r^{3}$

7. Frustum of a Cone (optional in some syllabi)

CSA: $π (r_{1} + r_{2}) l$
TSA: $π (r_{1} + r_{2}) l + π r_{1}^{2} + π r_{2}^{2}$
Volume: $\frac{1}{3} π h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$
(where $r_{1}$ , $r_{2}$ = radii of circular ends; $l$ = slant height, $h$ = height )

New Tuition classes

noreply@blogger.com (Unknown) — Sun, 23 Mar 2025 11:00:00 +0530

RJ18 TUITION CLASSES

NEW ONLINE AND OFFLINE BATCH STARTING ON RJ 18 TUITION CLASSES

Importance of water

noreply@blogger.com (Unknown) — Sun, 7 May 2023 17:57:00 +0530

Water is one of the most essential elements for life on earth. Without water, life as we know it would not be possible. It is a natural resource that is indispensable for the survival of all living beings. However, despite its importance, many people take water for granted and fail to recognize its value. In this blog, we will discuss the importance of water and its various uses.

Water is the basis of life on earth. All living beings, including humans, plants, and animals, require water to survive. Water is needed for various bodily functions, such as digestion, absorption, and excretion. It is also essential for the maintenance of body temperature and the transportation of nutrients and oxygen throughout the body.

Water is also used for various purposes in daily life. It is used for drinking, cooking, bathing, washing, and cleaning. Water is also used in industries for manufacturing, processing, and cooling. In agriculture, water is used for irrigation, which is necessary for the growth of crops.

Apart from its uses for human and animal life, water is also an important element in the ecosystem. It plays a vital role in maintaining the balance of the environment. Water bodies like rivers, lakes, and oceans are home to various aquatic life forms. They provide habitat and nourishment for fish, turtles, crabs, and other marine animals. These water bodies are also important breeding grounds for many aquatic species.

Moreover, water is a natural resource that is limited and needs to be conserved. Many parts of the world are facing water scarcity due to overuse, pollution, and climate change. The scarcity of water not only affects human and animal life but also has an adverse impact on agriculture, industries, and the economy.

Therefore, it is crucial to conserve water and use it wisely. We can adopt various water conservation techniques like rainwater harvesting, drip irrigation, and recycling of water. We should also avoid wastage of water by repairing leaking taps and using water-efficient appliances.

In conclusion, water is a precious and irreplaceable natural resource that is essential for life on earth. Its various uses in daily life, ecosystem, and industries make it a valuable commodity. We should recognize the importance of water and take steps to conserve and use it wisely. By doing so, we can ensure the availability of water for future generations and preserve the balance of the environment.

Current Affairs Today 6 May 2023

noreply@blogger.com (Unknown) — Sun, 7 May 2023 07:33:00 +0530

06 May 2023 Current Affairs

➼ Assamese Braille dictionary adjudged largest by Guinness World Records
गिनीज वर्ल्ड रिकॉर्ड्स द्वारा असमिया ब्रेल डिक्शनरी को सबसे बड़ा शब्दकोश घोषित किया गया।

➼ Karnataka ranked the most ‘innovative’ State by the National Manufacturing Innovation Survey
नेशनल मैन्युफैक्चरिंग इनोवेशन सर्वे द्वारा कर्नाटक को सबसे 'इनोवेटिव' राज्य का दर्जा दिया गया।

➼ UGC launches 'CU-Chayan', a unified faculty recruitment portal for central universities
यूजीसी ने केंद्रीय विश्वविद्यालयों के लिए एकीकृत संकाय भर्ती पोर्टल 'सीयू-चयन' लॉन्च किया।

➼ US Food and Drug Administration (FDA) Approves First RSV Vaccine for Older Adults
यूएस फूड एंड ड्रग एडमिनिस्ट्रेशन (एफडीए) ने वृद्ध वयस्कों के लिए पहले आरएसवी वैक्सीन को मंजूरी दी।

➼ Olympic silver medallist Nijel Amos gets 3-year ban for doping
ओलंपिक रजत पदक विजेता निजेल अमोस पर डोपिंग के लिए 3 साल का प्रतिबंध लगा।

➼ South Africa fast bowler Shabnim Ismail quits international cricket
दक्षिण अफ्रीका के तेज गेंदबाज शबनीम इस्माइल ने अंतरराष्ट्रीय क्रिकेट से संन्यास ले लिया।

➼ Geoffrey Emmanuel first Indian to compete in FIM JuniorGP
जेफ्री इमैनुएल एफआईएम जूनियरजीपी में प्रतिस्पर्धा करने वाले पहले भारतीय बने।

➼ Air Marshal Ashutosh Dixit appointed as new Deputy Chief of Air Staff
एयर मार्शल आशुतोष दीक्षित को वायु सेना का नया उप प्रमुख नियुक्त किया गया।

➼ Polavarapu Mallikharjuna Prasad selected as the next Chairman and Managing Director of Coal India (CIL)
पोलावरापु मल्लिकार्जुन प्रसाद को कोल इंडिया (CIL) के अगले अध्यक्ष और प्रबंध निदेशक के रूप में चुना गया।

➼ 2016 Olympic Gold Medalist Tori Bowie Dies At 32
2016 ओलंपिक स्वर्ण पदक विजेता टोरी बॉवी का 32 वर्ष की आयु में निधन।

➼ International Firefighter’s Day 2023: 04 May
अंतर्राष्ट्रीय अग्निशामक दिवस 2023: 04 मई

➼ World Portuguese Language Day 2023: 05 May
विश्व पुर्तगाली भाषा दिवस 2023: 05 मई

➼ Himachal Cabinet approves monthly incentive of Rs 1,500 for women of Spiti
हिमाचल मंत्रिमंडल ने स्पीति की महिलाओं के लिए 1,500 रुपये मासिक प्रोत्साहन राशि को मंजूरी दी।

Current Affairs Today 7 May 2023

noreply@blogger.com (Unknown) — Sun, 7 May 2023 07:31:00 +0530

07 मई 2023 Current Affairs in Hindi Today

यहां आज के प्रमुख करंट अफेयर्स की जानकारी दी जा रही है, जिन्हें आप नीचे दिए गए बिंदुओं में देख सकते हैं:-

➼ हाल ही में फ्रांस की ‘बैस्टल डे परेड’ में सम्मानित अतिथि के रूप में ‘प्रधानमंत्री नरेंद्र मोदी’ को आमंत्रित किया गया है।

➼ हाल ही में देवेंद्र फडणवीस ने ‘मॉरीशस’ देश में छत्रपति शिवाजी महाराज की प्रतिमा का अनावरण किया है।

➼ हाल ही में ‘इंटरनेट इन इंडिया रिपोर्ट 2022’ के अनुसार भारत में ’52 प्रतिशत’ आबादी के पास इंटरनेट सुविधा है।

➼ हाल ही में विश्व का सबसे बड़ा ‘फ्रेंचाइजी शतरंज लीग’ दुबई में शुरू होगा।

➼ हाल ही में पूर्व इंडियन नेवी चीफ ‘करमबीर सिंह’ को जापान देश ने ‘ऑर्डर ऑफ द राइजिंग सन, गोल्ड एंड सिल्वर स्टार’ से सम्मानित किया है।

➼ हाल ही में ‘माइकल डगलस’ को 76वें कान्स फिल्म फेस्टिवल में मानद ‘पाल्में डी पुरस्कार’ से सम्मानित किया जाएगा।

➼ हाल ही में ‘उत्तर प्रदेश’ राज्य सरकार ने डिज्नीलैंड के आधार पर ‘रामालैंड’ विकसित करने की योजना बनाई है।

➼ हाल ही में ADB के निदेशक मंडल की 56वीं वार्षिक आम बैठक ‘साउथ कोरिया’ में आयोजित की गई है।

➼ हाल ही में अमेरिका की एथलीट ‘टोरी बॉवी’ का निघन हो गया है।

➼ हाल ही में वायुसेना उपप्रमुख ‘आशुतोष दीक्षित’ बने है।

➼ हाल ही में केंद्रीय मंत्री डॉ मनसुख मंडाविया ने ‘फ़ूड स्ट्रीट प्रोजेक्ट’ की समीक्षा की है।

➼ हाल ही में ‘बिहार हाईकोर्ट’ ने जाति आधारित सर्वे को तत्काल बंद करने के आदेश दिए है।

➼ हाल ही में FIM जूनियर GP में प्रतिस्पर्धा करने वाले पहले भारतीय ‘जेफ्री इमैनुएल’ बनेंगे।

➼ हाल ही में एकीकृत संकाय भर्ती पोर्टल ‘सीयू चयन’ यूनिवर्सिटी ग्रांट कमीशन (UGC) ने लॉन्च किया है।

➼ हाल ही में ‘नेपाल’ देश ने अपना पहला ‘इतालवी सीरी A’ ख़िताब जीता है।

➼ हाल ही में ‘आयुष्मान खुराना’ को ‘वेकफीट’ के ब्रांड एंबेसडर के रूप में नियुक्त किया गया है।

➼ हाल ही में रूस देश ने अपना खुद का ‘ChatGPT गीगाचैट’ लॉन्च किया है।

➼ हाल ही में ‘भोगापुरम एयरपोर्ट’ का शिलान्यास आंध्र प्रदेश राज्य में किया गया है।

➼ हाल ही में नई दिल्ली में ‘रिवर सिटीज एलायंस ग्लोबल सेमिनार’ का आयोजन किया गया है।

➼ हाल ही में ‘पोलावरापु एम प्रसाद’ को कोल इंडिया के अगले अध्यक्ष के रूप में नियुक्त किया गया है।

Computer Keyboard Shortcut Keys List (A to Z)

noreply@blogger.com (Unknown) — Mon, 24 Apr 2023 07:27:00 +0530

Computer Keyboard Shortcut Keys

To be a pro in navigating and executing commands in computer software, one must have knowledge of Computer Shortcut keys. Learning the computer shortcut keys would help you to make your work easier and faster. No doubt, these days more than 90% population is a user on the computer system and to be a frequent user do learn the below complete Computer Keyboard Shortcut keys List. Computer Shortcut keys help the user to perform functions in a fast way while using any computer application. In this article, we are providing the all basic computer keyboard shortcut keys and shortcut keys used in Microsoft Windows, MS Office programs like MS Office, MS Word, MS Excel and PowerPoint that everyone should know. If you are regularly working on computer software or want to learn the basic keyboard shortcut keys, then the information is a must to read.

Shortcut Keys of Computer A to Z (Basic)

When a user works on the computer system, selecting text, copying, pasting, and deleting are the basic and frequently done tasks. Instead of using your cursor, you must be practicing Computer Keyboard Shortcut Keys. All the basic computer shortcut keys to be used on daily basis are mentioned in the table below. Check out the table given below to know the all basic computer keyboard shortcut keys from A to Z.

Basic Computer Keyboard Shortcut Keys A to Z
Shortcuts	Uses of Shortcut keys
Alt + F	File menu options in the current program
Alt + E	Edits options in the current program
F1	Universal help (for any sort of program)
Ctrl + A	Selects all text
Ctrl + X	Cuts the selected item
Ctrl + Del	Cut selected item
Ctrl + C	Copy the selected item
Ctrl + Ins	Copy the selected item
Ctrl + V	Paste the selected item
Shift + Ins	Paste the selected item
Home	Takes the user to the start of the current line
Ctrl + Home	Go to the beginning of the document
End	Go to the end of the current line
Ctrl + End	Go to the end of a document
Shift + Home	Highlight from the prevailing position to the start of the line
Shift + End	Highlight from the prevailing position to end of the line
Ctrl + (Left arrow)	Move one term to the left at a time
Ctrl + (Right arrow)	Move one term to the right at a time

Computer Shortcut Keys for Microsoft Windows

Computer shortcuts are keys or combinations of keys that assist you in an alternative way to do something that you typically do with a mouse or cursor. All the computer shortcut keys for Microsoft Windows are given below. Let's have a look at the computer keyboard shortcut keys for Microsoft Windows from the below table.

Shortcut Keys	Uses of Shortcut keys
Ctrl + Plus Key	Adjust the widths of all columns automatically, in Windows Explorer
Alt + Enter	Open the properties window for the selected icon or program
Alt + Print Screen	Take a screenshot of the current page.
Ctrl + Alt + Del	Reboot/Windows task manager
Ctrl + Esc	These keys allow you to activate the start menu
F4	Its purpose in Windows 95 to XP is to open the locate window
F5	Refresh the contents of your windows system
F3	Find anything from your system’s desktop
Alt + Esc	Switch between desktop applications on the taskbar
F2	Rename the selected icon
Alt + Shift + Tab	It allows you to switch back between ongoing applications
Alt + Tab	Switch between open applications/ programs.
Shift + Delete	When you press the Shift and Delete keys together, your program or files will be deleted permanently.
Alt + F4	It is used to close the ongoing program
Ctrl + F4	It’s used to swiftly close a document or a file that’s currently open.

Computer Shortcut Keys for MS Excel

Preparing a lengthy sheet in MS Excel and then revising the data or editing any single piece of information in the complete sheet would be time-consuming if you are not aware of the shortcut keys for MS Excel. All Computer Keyboard Shortcut Keys for MS Office programs have been discussed below. Check out the table given below to know the shortcut keys used in MS Excel.

Shortcut keys	Uses of shortcut keys
Alt + Shift + F1	When you wanna insert new worksheet use these keys together
Shift + F3	With the help of these keys, you can open the MS-Excel formula window
Shift + F5	When you press these keys together it will open the search box
F11	The F11 key is used to create a chart in MS-Excel
Ctrl + Shift +;	With the help of these keys, you can enter the current time
Ctrl +;	Use these keys together to enter the current date
Ctrl + K	When you want to Insert a link, you can use these keys together
Ctrl + F	These keys are used to open find and replace options in MS-Excel
Ctrl + G	Use these keys together to open go-to options
Ctrl + B	When you press these keys together it will bold highlighted selection.
F2	When you want to edit the selected cell using this key
F5	With the help of this key, you can go to a specific cell
F7	With the help of this key, you can check the spell of selected text
Ctrl + I	These commands are used to Italicize highlighted selection.
Ctrl + Space	Use these keys together to select the entire column
Shift + Space	Use these keys together to select the entire row
Ctrl + W	Use these keys together to close the document
Ctrl + H	Use these keys to open find and replace options
Ctrl + U	With help of these keys, you can underline highlighted text.
Ctrl + Y	With help of these keys, you can underline highlighted text.
Ctrl + Z	With the help of these keys, you can undo the last deleted action
Ctrl + F9	Use these keys to minimize a current window in MS-Excel
Ctrl + F10	Use these keys to maximize the currently selected window in MS-Excel
Ctrl + Tab	With the help of these keys, you can move between two or more open MS-Excel files
Alt + =	With the help of these keys, you can initiate the formula to add all of the above cells
Ctrl +	With the use of these keys together you can insert the value in the current cell from the above cell.
Ctrl + (Right arrow)	With the help of these keys, you can jump on to the next section of text
Ctrl + O	Use these keys to open options in MS-Excel
Ctrl + N	Use these keys together to open the document in MS-Excel
Ctrl + P	Use these keys together to open the print dialogue box in MS-Excel

Computer Shortcut Keys for MS Word

Though MS Word is the basic computer software and seems to be the easiest one to work on it. However, there is a complete list of Computer Shortcut Keys that makes typing on MS Word easier and enjoyable. All Computer Shortcut Keys related to MS Word are given below. Check out the table given below to know the shortcut keys used in MS Word.

Shortcuts	Uses of Shortcut keys
Ctrl + B	Bold highlighted selection
Ctrl + C	Copy selected text
Ctrl + X	Cut selected text
Ctrl + N	Open new/blank document
Ctrl + O	Open options
Ctrl + P	Open the print window
Ctrl + F	Open find box
Ctrl + I	Italicize highlighted selection
Ctrl + K	Insert link
Ctrl + U	Underline highlighted selection
Ctrl + V	Paste
Ctrl + G	Find and replace options
Ctrl + H	Find and replace options
Ctrl + J	Justify paragraph alignment
Ctrl + L	Align selected text or line to the left
Ctrl + Q	Align selected paragraph to the left
Ctrl + E	Align selected text or line to the centre
Ctrl + R	Align selected text or line to the right
Ctrl + M	Indent the paragraph
Ctrl + T	Hanging indent
Ctrl + D	Font options
Ctrl + Shift + F	Change the font
Ctrl + Shift + >	Increase selected font +1
Ctrl + ]	Increase selected font +1
Ctrl + [	Decrease selected font -1
Ctrl + Shift + *	View or hide non printing characters
Ctrl + (Left arrow)	Move one word to the left
Ctrl + (Right arrow)	Move one word to the right
Ctrl + (Up arrow)	Move to the beginning of the line or paragraph
Ctrl + (Down arrow)	Move to the end of the paragraph
Ctrl + Del	Delete the word to the right of the cursor
Ctrl + Backspace	Delete the word to the left of the cursor
Ctrl + End	Move the cursor to the end of the document
Ctrl + Home	Move the cursor to the beginning of the document
Ctrl + Space	Reset highlighted text to the default font
Ctrl + 1	Single-space lines
Ctrl + 2	Double-space lines
Ctrl + 5	1.5-line spacing
Ctrl + Alt + 1	Change text to heading 1
Ctrl + Alt + 2	Change text to heading 2
Ctrl + Alt + 3	Change text to heading 3
Shift + F3	Change case of selected text
Shift + Insert	Paste
F4	Repeat the last action performed (Word 2000+)
F7	Spell check selected text and/or document
Shift + F7	Activate the thesaurus
F12	Save as
Ctrl + S	Save
Shift + F12	Save
Alt + Shift + D	Insert the current date
Alt + Shift + T	Insert the current time
Ctrl + W	Close document
Ctrl+=	Set chosen text as a subscript.
Ctrl+Shift+=	Set chosen text as superscript.

Computer Shortcut Keys for PowerPoint

Beginning to work on Microsoft Powerpoint and wants to learn the shortcut keys to navigate and perform commands quickly, then have a look at the computer keyboard shortcut keys used in the PowerPoint program. All Computer Keyboard Shortcut Keys for PowerPoint programs are given below.

Shortcuts Keys	Uses of Shortcut Keys
Ctrl+Shift+<	Reduce the font size of the selected text by one size.
CTRL + G	Group things together
Ctrl+M	Create a new slide just after your chosen slide.
CTRL + D	Create a duplicate of your current slide.
Esc	Exit the ongoing slide show and it will redirect you to the earlier live view.
Ctrl+K	When you want to enter a hyperlink use these keys together.
Ctrl+Shift+>	These commands are used to maximize the font size of the selected text by one size.
F5	With the help of F5, you can start the presentation from the initial slide.
Ctrl+N	These commands are used in a different Powerpoint software window, create a new, blank slide.

महत्वपूर्ण 50 प्रश्नोतर

noreply@blogger.com (Unknown) — Thu, 15 Dec 2022 21:28:00 +0530

🔻सभी परीक्षा महत्वपूर्ण 50 प्रश्नोतर 🔻

1) सेब में कौनसा अम्ल पाया जाता है ?
Ans. मैलिक अम्ल

2) इमली में कौन सा अम्ल पाया जाता है ?
Ans. टार्टरिक अम्ल

3) दूध,दही में कौन सा अम्ल पाया जाता है ?
Ans. लैक्टिक अम्ल

4) सिरके में कौनसा अम्ल पाया जाता है v
Ans. एसिटिक अम्ल

5) लाल चींटी के डंक में कौन सा अम्ल होता है ?
Ans. फॉर्मिक अम्ल

6) नीम्बू व खट्टे पदार्थों में कौनसा अम्ल पाया जाता है ?
Ans. साइट्रिक अम्ल

7) टमाटर के बीज में कौनसा अम्ल पाया जाता है ?
Ans. आक्सैलिक अम्ल

8) किडनी की पथरी को क्या कहते है ?
Ans. कैल्शियम ऑक्सलेट

9) प्रोटीन पाचन में कौनसा अम्ल सहयोगी है ?
Ans. हाइड्रोक्लोरिक अम्ल

10) साइलेंट वैली कहाँ स्थित हैं ?
Ans. केरल

11) अंतर्राष्ट्रीय सौर गठबंधन का मुख्यालय कहाँ स्थित है ?
Ans. गुरुग्राम (हरियाणा)

12) विक्रम साराभाई अंतरिक्ष केन्द्र कहाँ स्थित है ?
Ans. तिरुवनंतपुरम

13) सतीश धवन अंतरिक्ष केंद्र कहाँ स्थित है ?
Ans. श्री हरिकोटा

14) भारतीय कृषि अनुसंधान केंद्र कहाँ स्थित है ?
Ans. नई दिल्ली

15) केंद्रीय चावल अनुसंधान संस्थान कहाँ स्थित है ?
Ans. कटक (ओड़िसा)

16) हॉकी वर्ल्ड कप 2023 का आयोजन किस देश में होगा ?
Ans. भारत

17) हॉकी का जादूगर किसे कहा जाता है ?
Ans. मेजर ध्यानचंद

18) क्योटो प्रोटोकॉल का संबंध किससे है ?
Ans. ग्रीन हाउस गैस

19) मांट्रियल प्रोटोकॉल का सम्बंध किससे है ?
Ans. ओजोन परत संरक्षण

20) रामसर कन्वेन्शन का सम्बंध किससे है ?
Ans. आद्रभूमि के संरक्षण

21) स्कॉटहोम सम्मेलन कब हुआ था ?
Ans. 1912 में हुआ

22) विश्व बैंक का मुख्यालय कहाँ स्थित है ?
Ans. वाशिंगटन डीसी

23) एशियाई विकास बैंक (ADB) का मुख्यालय कहाँ स्थित है ?
Ans. मनीला

24) संयुक्त राष्ट्र पर्यावरण कार्यक्रम (UNEP) का मुख्यालय कहाँ स्थित है ?
Ans. नैरोबी (केन्या)

25) विश्व व्यपार संगठन का मुख्यालय कहाँ स्थित है ?
Ans. जिनेवा

26) यूनेस्को का मुख्यालय कहाँ स्थित है ?
Ans. पेरिस

27) एमनेस्टी इंटरनेशनल का मुख्यालय कहाँ स्थित है ?
Ans. लंदन

28) पेट्रोलियम उत्पादक देशों के संगठन (OPEC) का मुख्यालय कहाँ स्थित है ?
Ans. वियना

29) आर्थिक सहयोग और विकास संगठन (OECD) का मुख्यालय कहाँ स्थित है ?
Ans. पेरिस

30) अंतरराष्ट्रीय श्रम संगठन (ILO) का मुख्यालय कहाँ स्थित है ?
Ans. जिनेवा

31) falcon 9 रॉकेट किस प्स्पेस एजेंसी ने लॉन्च किया ?
Ans. Space-X

32) होप (HOPE) मिशन किस देश के द्वारा लॉन्च किया गया है ?
Ans. संयुक्त अरब अमीरात (UAE)

33) 2017 में भारत ने 104 सैटेलाइट किस व्हीकल द्वारा लॉन्च किए थे ?
Ans. PSLV C37

34) शिपकिला दर्रा कहाँ पर स्थित है ?
Ans. हिमाचल प्रदेश

35) किस दर्रे से सतलुज नदी भारत में प्रवेश करती है ?
Ans. शिपकिला दर्र

36) नाथूला दर्रा किस राज्य में स्थित है ?
Ans. सिक्किम

37) बोमडिला दर्रा किस राज्य में स्थित है ?
Ans. अरुणाचल प्रदेश

38) तुजू दर्रा किस राज्य में स्थित है ?
Ans. मणिपुर

39) टाइगर स्टेट किसे कहते है ?
Ans. मध्यप्रदेश

40) सिमलीपाल टाइगर रिजर्व कहाँ स्थित है ?
Ans. ओडिसा

41) नागरहोल टाइगर रिजर्व किस राज्य में स्थित है ?
Ans. कर्नाटक

42) पलामू टाइगर रिजर्व किस राज्य में स्थित है ?
Ans. झारखंड

43) टडोबा अंधेरी टाइगर रिजर्व किस राज्य में स्थित है ?
Ans. महाराष्ट्र

44) खजुराहो के मंदिर का निर्माण किसने करवाया था ?
Ans. चंदेल शासक (छतर, मध्यप्रदेश)

45) खजुराहो के मंदिर किस शेली में निर्मित है ?
Ans. पँचायतन शैली

46) हुमायूँ का मकबरा किस शेली में निर्मित है ?
Ans. चारबाग शैली

47) पूर्व का ताजमहल किसे कहते है ?
Ans. हुमायूँ का मकबरा

48) वृहदेश्वर मंदिर किस शेली में निर्मित है ?
Ans. द्रविड़ शैली

49) वृहदेश्वर मंदिर का निर्माण किस शाशकों ने करवाया था ?
Ans. चोल शासक

50) वृहदेश्वर मंदिर कहाँ स्थित है ?
Ans. तंजौर

हिन्दी प्रश्नोत्तर

noreply@blogger.com (Unknown) — Thu, 15 Dec 2022 21:25:00 +0530

✅हिन्दी प्रश्नोत्तर भाग 01

[ आगामी परीक्षाओं के लिए महत्वपूर्ण ]

प्रश्‍न – (1) वर्णो के समूह को कहते है।
उत्‍तर – शब्‍द

प्रश्‍न – (2) विस्‍तृत तथ्‍यों को संक्षिप्‍त में लिखना कहलाता है।
उत्‍तर – सारांश

प्रश्‍न - (3) ‘सुवाच्‍य‘ का सही अर्थ है।
उत्‍तर – सुपठनीय लेख

प्रश्‍न - (4) शिक्षण की श्रेष्‍ठ प्रविधि है।
उत्‍तर – प्रश्‍न प्रविधि

प्रश्‍न - (5) शब्‍द भण्‍डार में वृद्धि करने की प्रमुख विधि है।
उत्‍तर – वार्तालाप

प्रश्‍न - (6) बाल्‍यावस्‍था में भाषा सीखने का मुख्‍य आधार होता है।
उत्‍तर – अनुसरण

प्रश्‍न - (7) दु:ख की अनुभूति पर वाचक का स्‍वर भारी होना है।
उत्‍तर – ध्‍वनिवश्‍यता

प्रश्‍न - (8) ‘ कामायनी ’ किसकी प्रसिद्ध रचना है।
उत्‍तर – जयशंकर प्रसाद

प्रश्‍न – (9) ‘ गीतांजलि ’ के रचनाकार है।
उत्‍तर – रवीन्‍द्रनाथ टैगोर

प्रश्‍न – (10) ‘ गोदान ’ के लेखक है।
उत्‍तर – मुंशी प्रेमचन्‍द

प्रश्‍न – (11) जो धातु या शब्‍द के अंत में जोडा जाता है। उसे क्‍या कहते है।
उत्‍तर – प्रत्‍यय

प्रश्‍न – (12) प्रख्‍यात में कौन सा उपसर्ग है।
उत्‍तर – प्र

प्रश्‍न – (13) हिन्‍दी में ‘कृत’ प्रत्‍ययों की संख्‍या कितनी है।
उत्‍तर – 42

प्रश्‍न – (14) ‘पुरोहित’ शब्‍द में प्रयुक्‍त उपसर्ग है।
उत्‍तर – पुर:

प्रश्‍न – (15) ‘संस्‍कार’ शब्‍द में प्रयुक्‍त उपसर्ग है।
उत्‍तर – सम्

प्रश्‍न – (16) ‘विज्ञान’ शब्‍द में प्रयुक्‍त उपसर्ग है।
उत्‍तर – वि

प्रश्‍न – (17) ‘चिरायु’ शब्‍द में प्रयुक्‍त उपसर्ग है।
उत्‍तर – चिर

प्रश्‍न – (18) ‘निर्वाह’ शब्‍द में प्रयुक्‍त उपसर्ग है।
उत्‍तर – निर

प्रश्‍न – (19) ‘धुंधला’ शब्‍द में प्रयुक्‍त प्रत्‍यय है।
उत्‍तर –ला
*भाग-2*
प्रश्‍न – (1) ‘जिसके पास कुछ भी न हो’ उसे कहते है।
उत्‍तर – अकिंचन

प्रश्‍न – (2) ‘ जो व्‍यतीत हो गया हो ’ उसे कहते है।
उत्‍त्‍र - अतीत

प्रश्‍न – (3) ‘ जिसकी थाह न मिले ’ उसे कहते है।
उत्‍तर – अथाह

प्रश्‍न – (4) ‘ रात में पहरा देने वाला व्‍यक्ति ’ उसे कहते है।
उत्‍तर – यामिक

प्रश्‍न – (5) ‘ जिस वस्‍तु का मूल्‍य न आंका जा सके ’ उसे कहते है।
उत्‍तर – अमूल्‍य

प्रश्‍न – (6) ‘ श्रीगणेश करना ’ मुहावरे का अर्थ है।
उत्‍तर – शुभारम्‍भ करना

प्रश्‍न – (7) ‘ हाथ पैर मारना ’ मुहावरे का अर्थ है।
उत्‍तर – काफी प्रयत्‍न करना

प्रश्‍न – (8) ‘ हाथ मलना ’ मुहावरे का अर्थ है।
उत्‍तर – पछताना

प्रश्‍न – (9) ‘ चेहरे पर हवाइयॉ उडना ’ मुहावरे का अर्थ है।
उत्‍तर – घबराना

प्रश्‍न – (10) ‘ टका – सा मॅुह लेकर रहना ’ मुहावरे का अर्थ है।
उत्‍तर – शर्मिन्‍दा होना

प्रश्‍न – (11) हिन्‍दी खडी बोली किस अपभ्रंस से विकसित हुई है।
उत्‍तर – शौरसेनी

प्रश्‍न – (12) भाषा के आधार पर भारतीय राज्‍यों की पुन: संरचना कब की गई थी।
उत्‍तर – 1956 ई में

प्रश्‍न – (13) भाषाई आधार पर सर्वप्रथम किस राज्‍य का गठन हुआ।
उत्‍तर – आंन्‍ध्र प्रदेश

प्रश्‍न – (14) ‘बघेली’ बोली का संबंध किस उपभाषा से है।
उत्‍तर – पूर्वी हिन्‍दी

प्रश्‍न – (15) भारतीय संविधान की 8वीं अनुसूची में शामिल भाषाओं की संख्‍या है।
उत्‍तर – 22

प्रश्‍न – (16) हिन्‍दी की आदि जननी है।
उत्‍तर – संस्‍कृत

प्रश्‍न – (17) भाषा की सबसे छोटी इकाई है।
उत्‍तर – वर्ण

प्रश्‍न – (18) हिन्‍दी वर्णमाला में ‘अयोगवाह’ वर्ण कौन से है।
उत्‍तर – अं, अ:

प्रश्‍न – (19) ‘ए’ ‘ऐ’ वर्ण क्‍या कहलाते है।
उत्‍तर – कंठ-तालव्‍य

प्रश्‍न – (20) ‘क’, ‘ग’, ‘ज’, ‘फ’ ध्‍वनियॉ किसकी है।
उत्‍तर – अरबी-फारसी

कारक

noreply@blogger.com (Unknown) — Thu, 15 Dec 2022 21:15:00 +0530

🌿🌿【हिंदी व्याकरण -कारक】🌿🌿

1) 'जिसे देखता हूँ, वही स्वार्थी निकलता है',कौनसा कारक है ?
अ) कर्ता           ब) करण
स)अपादान       द)संबंध

द)संबंध✓
2) गरीबों के निमित धन इकट्ठा करो में कौनसा कारक है ?
अ)करण           ब)अपादान
स)सम्प्रदान        द)कर्ता

स)सम्प्रदान✓
3)'मोहन ,कहाँ जा रहे हो |'वाक्य में कारक है -
अ)अपादान             ब)सम्बन्ध
स)अधिकरण           द)सम्बोधन

द)सम्बोधन✓
4)मुझे इस कार्यालय ..................सभी जानकारियाँ चाहिए |उपयुक्त विभक्ति चिह्न से रिक्त स्थान की पूर्ति कीजिए |
अ) पर    (ब)की    (स)में (द)द्वारा

ब)की✓
5) उत्तम पुरूष,बहुवचन,संबंध कारक है -
अ)तुम्हारा
ब)उसका
स)मेरा
द)हमारा

द)हमारा✓
6) "पूजा को "नौकरी नहीं करनी है |इस वाक्य में उद्धरण चिह्न में बंद अंश में कौनसा कारक है ?
अ)कर्म कारक
ब)कर्ता कारक
स)कर्म और कर्ता
द)सम्प्रदान

अ)कर्म कारक✓
7 )विकास अपने गुरू ................... श्रद्धा रखता है |रिक्त स्थान भरे |
अ)पर   (ब)में   (स)को    (द)से

अ)पर✓
8)बच्चा साँप को देखकर डर गया |वाक्य में कारक है -
अ)करण
ब)सम्प्रदान
स)अपादान
द)सम्बन्ध

स)अपादान✓
9)राम मोहन से अधिक चतुर है ,वाक्य में कारक ?
अ)करण
ब) अपादान
स)सम्बन्ध
द)सम्प्रदान

ब)अपादान✓
10)निम्न में से कौनसे कारक चिह्न तत्पुरुष समास में नहीं आते हैं ?
अ )कर्ता ,करण
ब)कर्ता, संबोधन
स)अपादान,संबंध
द)कर्ता, अधिकरण

ब)कर्ता, संबोधन✓
11) क्रिया के साथ कौन अथवा किसने शब्द लगाकर प्रश्न पूछने पर जो उत्तर आता है ,उसे क्या कहते है ?
अ)कर्म (ब)कर्ता   (स)क्रिया   (द)कारक

ब)कर्ता✓
प्रश्न 12. कारक का संबंध होता है?
अ. संज्ञा
ब. सर्वनाम
स. विशेषण
द. क्रिया

द. क्रिया✓
प्रश्न 13. किस विकल्प में करण कारक है?
अ. हस्तलिखित
ब. द्वार द्वार
स. हवन सामग्री
द. विद्यालय

अ. हस्तलिखित✓
प्रश्न 14 किस विकल्प मैं अपादान कारक नहीं है?
अ. भयमुक्त
ब. पथभ्रष्ट
स. रेखांकित
द. रोग विहीन

स. रेखांकित✓
प्रश्न 15. "वहाँ एक सभा हो रही थी" इस वाक्य में कौन सा कारक है?
अ. कर्म
ब. अधिकरण
स. संबंध
द. करण

ब. अधिकरण✓
16.राजा *सेवक को* कंबल देता है , गहरा काला शब्द के लिए कारक का सही क्रमांक है-
A. कर्म कारक
B. संबंध कारक
C. संप्रदान कारक
D. कर्ता कारक

C. संप्रदान कारक ✓
17. 'झगड़ा मेरे और उसके मध्य में था' में कौन सा कारक है
A.संबंध कारक
B. अधिकरण कारक
C. करण कारक
D. कर्म कारक

B. अधिकरण कारक ✓
18. 'देवेंद्र मैदान में खेल रहा है' वाक्य में कौन सा कारक है-
A. कर्म कारक
B. संबंध कारक
C. अपादान कारक
D. अधिकरण कारक

D. अधिकरण कारक✓
19. 'मैं साँप से डरता हूँ' वाक्य मे कारक है-
A. करण कारक
B. अपादान कारक
C. कर्ता कारक
D. संबंध कारक

B. अपादान कारक ✓
20. 'मैंने गुरुजी से हिन्दी व्याकरण सीखा' ,वाक्य मे कौनसा कारक है-
A. करण कारक
B. अपादान कारक
C. सम्प्रदान कारक
D. संबंध कारक

B. अपादान कारक✓
21-निम्नलिखित मेँ से अपादान कारक है|
(a)राधिका चाकू से फल काटती है ।
(b)वह गेंद से खेल रहा है
(c)राम खिलौनों को फेंक रहा है।
(d)हाथ से गिलास छूट गया

(d)हाथ से गिलास छूट गया✓
22-'वह निर्धंन के लिये धन देता है ।'वाक्य मेँ किस कारक का प्रयाग हुआ है ?
(a)सम्प्रदान कारक
(b)करन कारक
(c)अपादान कारक
(d)कर्म कारक

(a)सम्प्रदान कारक ✓
23-'पेड़ से फल गिरते है ।' वाक्य मेँ निहित कारक है ?
(a)अधिकरण
(b) अपादान कारक
(c)सम्बंध कारक
(d)सम्बोधन कारक

(b) अपादान कारक✓
24-मुख्यत कारक है ।
(a)छः     (b)सात    (c)आठ    (d)पाँच

(a)छः ✓
25-' चारपाई पर भाईसाहब बैठे है । ' वाक्य मेँ चारपाई शब्द किस कारक मेँ है ?
(a)अधिकरण     (b)सम्प्रदान
(c)करन            (d)सम्बंध

(a)अधिकरण✓
26-इन फूलों का रंग सुहावना है । वाक्य मेँ ' का' किस कारक का चिन्ह है ?
(a)कर्म                  (b)सम्बन्ध
(c)अधिकरण         (d)अपादान

(b)सम्बन्ध✓
27-वह घर से बाहर गया ।' इस वाक्य मेँ कौनसा कारक है ?
(a)कर्म                (b)कर्ता
(c)करन                (d)अपादान

(d)अपादान ✓
28-'वह अगले साल आयेगा।'वाक्य मेँ कौनसा कारक है ?
(a)सम्बोधन              (b)कर्म
(c)अधिकरण            (d)सम्बंध

(c)अधिकरण✓
29-'सीता ने गीता को अपनी कलम दे दी।'वाक्य मेँ 'को' किस कारक का चिन्ह है ?
(a)कर्म                   (b)करन
(c)संम्प्रदान             (d) सम्बन्ध

(c)संम्प्रदान✓
30-' निशा विकास की बहन है। ' वाक्य मेँ 'की' किस कारक का चिन्ह है ?
(a)करन            (b)सम्प्रदान
(c)कर्ता             (d)सम्बंध
(d)सम्बंध✓

प्रधानमंत्री की शक्तियां

noreply@blogger.com (Unknown) — Thu, 1 Dec 2022 07:27:00 +0530

प्रधानमंत्री की शक्तिय में प्रधानमंत्री से संबंधित समस्त शक्तियां एवं कार्य के अंतर्गत प्रधानमंत्री की कार्यकाल | प्रधानमंत्री की नियुक्ति कौन करता है | प्रधानमंत्री की नियुक्ति प्रक्रिया | प्रधानमंत्री की शक्तियां एवं कार्य | प्रधानमंत्री और मंत्रिपरिषद के बीच संबंध |प्रधानमंत्री एवं संसद के बीच संबंध इत्यादि टॉपिक पर विस्तार से पढ़ने वाले है।

प्रधानमंत्री की नियुक्ति, शक्तियां एवं कार्य

संसदीय शासन प्रणाली की सबसे महत्वपूर्ण संस्था मंत्रिपरिषद होती है जो कि प्रधानमंत्री के नेतृत्व में कार्य करते हैं। यह कार्य मंत्रीपरिषद करेगी। संविधान के अनुच्छेद 74 में मंत्रीपरिषद के प्रधान के रूप में प्रधानमंत्री का उल्लेख किया गया है। संविधान द्वारा भारत में संसदीय शासन प्रणाली की स्थापना की गई है तथा कार्यपालिका की सर्वोच्च शक्ति राष्ट्रपति में निहित की गई है, परंतु व्यवहारिक तौर पर उसकी समस्त शक्तियों का प्रयोग प्रधानमंत्री द्वारा किया जाता है। प्रधानमंत्री सत्ताधारी दल का नेता होता है तथा सरकार का प्रमुख भी होता है।

प्रधानमंत्री का कार्यकाल

प्रधानमंत्री को मुख्यत: 5 वर्षों के लिए चुना जाता है, परंतु उसका कार्यकाल निश्चित नहीं होता है, क्योंकि प्रधानमंत्री अपने पद पर तब तक बना रहता है जबतक उसे लोकसभा में पूर्ण बहुमत का समर्थन प्राप्त होता है। यदि प्रधानमंत्री लोकसभा में बहुमत खो देता है, तो उसे अपना त्यागपत्र देना पड़ता है।

प्रधानमंत्री की न्युक्ति कौन करता है?

राष्ट्रपति संविधान के अनुच्छेद 75(1) के तहत प्रधानमंत्री की नियुक्ति करता है। लेकिन राष्ट्रपति किसे प्रधानमंत्री के पद पर न्युक्ति करेगा, यह स्पष्ट नहीं किया गया है। सामान्य तौर पर राष्ट्रपति सबसे बड़े दल के नेता को अथवा सबसे बड़े गठबंधन वाले दलों के नेता को प्रधानमंत्री के पद पर न्युक्ति करता है।

सामान्य परंपरा यह है कि लोकसभा में बहुमत दल के नेता को प्रधानमंत्री के रूप में न्युक्ति किया जा सकता है, परंतु लोकसभा में किसी भी दल को बहुमत प्राप्त न होने की स्थिति में, प्रधानमंत्री की नियुक्ति में राष्ट्रपति अपना विवेक का प्रयोग कर सकता है।

प्रधानमंत्री की नियुक्ति की प्रक्रिया

प्रधानमंत्री की नियुक्ति के लिए वही योग्यताएं होनी चाहिए, जो लोकसभा का सदस्य बनने के लिए होती हैं, क्योंकि प्रधानमंत्री लोकसभा का सदस्य और बहुमत प्राप्त दल का नेता होता है।
प्रधानमंत्री पद के लिए यह अनिवार्य नहीं है कि प्रधानमंत्री लोकसभा का सदस्य हो।
प्रधानमंत्री पद पर ऐसा भी व्यक्ति को नियुक्त किया जा सकता है जो संसद के किसी भी सदन का सदस्य न हो, बशर्ते उस व्यक्ति को लोकसभा के बहुत वाले दल अपना नेता चुने। लेकिन 6 माह के अंदर उस व्यक्ति को लोकसभा का सदस्य बनना जरूरी है।

प्रधानमंत्री की शक्तियां एवं कार्य

प्रधानमंत्री द्वारा मंत्रियों की नियुक्ति एवं पद्धति के अनुसार सा राष्ट्रपति को ही की जाती है।
लोकसभा में बहुमत दल के नेता होने के कारण वह लोकसभा में शासन की प्रमुख नीतियों एवं कार्यों की घोषणा करता है तथा लोकसभा के सदस्यों द्वारा गंभीर विषयों से संबंधित पूछे गए प्रश्नों के उत्तर देता है।
देश की वित्त व्यवस्था एवं वार्षिक वित्तीय विवरण निर्धारित करने में भी प्रधानमंत्री की मुख्य भूमिका होती है।
प्रधानमंत्री की सलाह पर मंत्री परिषद के सदस्य नियुक्त और बर्खास्त किए जाते हैं। प्रधानमंत्री मंत्रियों के विभागों का आवंटन करता है तथा प्रधानमंत्री मंत्रियों के विभागों में परिवर्तन भी कर सकता है।
प्रधानमंत्री किसी भी समय लोकसभा के विघटन के अनुशंसा राष्ट्रपति से कर सकता है।
प्रधानमंत्री मंत्री परिषद की बैठकों की अध्यक्षता करता है तथा उसके निर्णय को प्रभावित करता है।
संविधान के अनुच्छेद 78 के अनुसार प्रधानमंत्री को यह शक्ति है कि वह प्रशासन तथा विधान से संबंधित सूचना राष्ट्रपति को देता है।
प्रधानमंत्री के सलाह पर ही सभी राज्यों के राज्यपाल, महान्यायवादी, महालेखा परीक्षक, लोक सेवा आयोग के अधिपति व अन्य सदस्य, विभिन्न देशों के राजदूत, वाणिज्य दूत इत्यादि सभी उच्च स्तरीय नियुक्ति का कार्य भारत के राष्ट्रपति द्वारा किया जाता है।
प्रधानमंत्री नीति आयोग का अध्यक्ष होता है तथा भारत रत्न, पदम विभूषण, पदम भूषण एवं पदम श्री आदि उपाधियों को स्वीकृति का कार्य वास्तविक तौर पर प्रधानमंत्री द्वारा ही की जाती है।

प्रधानमंत्री और मंत्रीपरिषद के बीच संबंध

91वें संविधान संशोधन 2003 में केंद्र और राज्य मंत्री परिषद की सदस्य संख्या लोकसभा और विधानसभा की कुल संख्या की 15% से अधिक नहीं होनी चाहिए तथापि छोटे राज्यों के लिए न्यूनतम संख्या 12 निर्धारित की गई है।

प्रधानमंत्री एवं मंत्रियों के लिए यह आवश्यक है कि वे संघ की विधायिका के सदस्य हो।

मंत्रिमंडल में शामिल मंत्री आवश्यक रूप से संसद के सदस्य होते हैं परंतु ऐसे व्यक्ति को भी मंत्री बनने पर कोई प्रतिबंध नहीं है, जो संसद के सदस्य न हो। अगर वह व्यक्ति 6 माह के संसद के किसी भी सदन का सदस्यता प्राप्त नहीं करता है तो वह मंत्रीमंडल के मंत्री पद पर नहीं रह सकता है।

प्रधानमंत्री व संसद के बीच संबंध

प्रधानमंत्री संसद की गतिविधियों में भी खास भूमिका निभाता है। संसद के नेता की दृष्टि से प्रधानमंत्री विशेष है जो निम्नलिखित है :-

(i) सदन का कार्यक्रम निश्चित करना :

प्रधानमंत्री संसद का अधिवेशन बुलाता है एवं स्थगित भी कर सकता है। लोकसभा की कार्रवाई किस प्रकार चलेगी! कौन सा प्रस्ताव पहले पेश किया जाएगा एवं कौन बाद में, आदि बातों की निर्णय लोकसभा का अध्यक्ष, प्रधानमंत्री एवं विपक्ष के नेता की सलाह से करता है

(ii) महत्वपूर्ण नीतियों की घोषणा :

परंपरा के अनुसार सदन में सभी महत्वपूर्ण नीतियों की घोषणा प्रधानमंत्री स्वयं करता है। युद्ध जैसी घटनाओं के समय केवल प्रधानमंत्री ही बोल सकता है। यह अधिकार उसके पास है।

(iii) सरकार की नीतियों का बचाव :

सरकार की नीतियों का बचाओ अथवा विरोधी दल की आलोचना का जवाब देना प्रधानमंत्री की जिम्मेदारी है।

(iv) लोकसभा का भंग करना :

ब्रिटेन के परंपरा के अनुसार प्रधानमंत्री के पास यह विशेषाधिकार है कि वह लोकसभा को भंग कर सकता है और नया सभा बुला सकता है।

निष्कर्ष (Conclusion) :-

दोस्तों..आज हम इस लेख में प्रधानमंत्री की समस्त कार्य एवं शक्तियां के बारे में जानकारी प्राप्त की है। इसके साथ-साथ भारत के प्रधानमंत्री से संबंधित अन्य महत्वपर्ण टॉपिक जैसे प्रधानमंत्री की नियुक्ति कौन करता है | प्रधानमंत्री की नियुक्ति की प्रक्रिया |प्रधानमंत्री का कार्यकाल | प्रधानमंत्री की शक्तियां एवं कार्य | प्रधानमंत्री और मंत्रिपरिषद के बीच संबंध इत्यादि को कवर किया।

भारत देश का राष्ट्रपति औपचारिक या संवैधानिक शासक है परंतु प्रधानमंत्री को देश का वास्तविक शासक समझा जा सकता है, क्योंकि भारतीय संविधान के अनुच्छेद 75 के अनुसार देश की वास्तविक कार्यपालिका की शक्तियां प्रधानमंत्री के पास है।

भारतीय संविधान में प्रधानमंत्री की उपादेयता बहुआयामी एवं सर्वाधिक महत्वपूर्ण है। देश की एकता और अखंडता का सूत्रधार है। प्रधानमंत्री देश की राजनीतिक व्यवस्था की धुरी है। इसे देश का हृदय स्थल 'गुरुत्वाकर्षण का केंद्र' राजनीतिक शासक और सर्वोच्च शासक की संज्ञा दी जाती है। अतः प्रधानमंत्री को राष्ट्र का नेता अथवा नायक माना जा सकता है।

GK QUIZ

noreply@blogger.com (Unknown) — Sun, 27 Nov 2022 20:29:00 +0530

Multiple Choice Quiz Questions (MCQs) on Ancient Indian History for General Studies and GK preparation of SSC, NDA, CDS, UPSC, UPPSC and State PSC Examinations.

A chapterwise course of total 2250 Ancient Indian History MCQs divided into 40 chapters and 16 Mock Tests is available in Android app of GKToday under UPSC Prelims Category.

1.Who among the following was the ruler of Kanchi during the time of Samudragupta ?

[A] Hastivarman
[B] Mantaraja
[C] Nilaraja
[D] Vishnugopa

Show Answer

2.What was the name of Gautam Buddha’s only son ?

[A] Rahul
[B] Channa
[C] Kanthala
[D] Chunda

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3.In the Maurya empire, who was the superintendent of Ports ?

[A] Pattanadhyaksha
[B] Manadhyaksha
[C] Kosadhyaksha
[D] Ganikadhyaksha

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4.In which Mandal of Rigveda ‘Gayatri Mantra is mentioned?

[A] III
[B] IV
[C] V
[D] VI

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5.In context of Sangam Age, which among the following was the ruling class in society?

[A] Arasar
[B] Kadaisiyar
[C] Enadi
[D] Pandyas

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6.Uraiyur was an ancient Chola city famous for__?

[A] Pearls
[B] Ship Building
[C] Houses
[D] Grammar

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7.Who composed Buddhacharita?

[A] Kanishka I
[B] Kanishka II
[C] Asvaghosha
[D] Charak

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8.Kushans belongs to which tribe?

[A] Yuchi
[B] Terai
[C] Ikswaku
[D] Jnatrika

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9.Apart from Madurai, which among the following was also a capital of Pandyas?

[A] Korkai
[B] Kottaikkadu
[C] Thirukkoloor
[D] Pragasapuram

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10.Which dynasty Asoka the Great was emperor of ?

[A] Maurya Dynasty
[B] Kakatiya dynasty
[C] Nanda dynasty
[D] Pradyota dynasty

Show Answer

राष्ट्रपति की शक्तियां

noreply@blogger.com (Unknown) — Sun, 27 Nov 2022 19:48:00 +0530

राष्ट्रपति की शक्तियां

द्रोपदी मुर्मु देश की 15वीं राष्ट्रपति बन गई हैं। द्रौपदी मुर्मू आजादी के बाद पैदा होने वाली पहली और शीर्ष पद पर काबिज होने वाली सबसे कम उम्र की शख्सियत बन गई हैं। प्रतिभा पाटिल के बाद मुर्मु देश की दूसरी महिला और प्रथम आदिवासी महिला राष्ट्रपति हैं। राष्ट्रपति बनने के बाद द्रौपदी मुर्मु के साथ कई शक्तियां और अधिकार भी स्वत: आ गए हैं, आइए जानते हैं क्या है राष्ट्रपति की शक्तियां- अधिकार और कितने दिन का होता है कार्यकाल।

राष्ट्रपति की सैलरी व भत्ते

राष्ट्रपति की सैलरी 5 लाख प्रतिमाह है, राष्ट्रपति के वेतन पर कोई टैक्स नहीं लगता है। इसके अलावा आजीवन मुफ्त चिकित्सा सुविधाएं, आवास व संसद द्वारा स्वीकृत अन्य भत्ते प्राप्त होते हैं। भारतीय संविधान के अनुच्छेद 59 के अनुसार, राष्ट्रपति के कार्यकाल के दौरान सैलरी तथा भत्ते में किसी प्रकार की कमी नहीं की जा सकती है।

राष्ट्रपति की शक्तियां एवं अधिकार

भारत के राष्ट्रपति की कई प्रकार की शक्तियां एवं अधिकार होते हैं जैसे-

कार्यपालिका शक्ति

इसके तहत राष्ट्रपति कई पदों पर नियुक्ति करने का अधिकार होता है।

-प्रधानमंत्री के सलाहकार मंत्रीपरिषद के अन्य सदस्य

-सर्वोच्च एवं उच्च न्यायालय के मुख्य न्यायाधीश

-भारत के नियंत्रक एवं महालेखा परीक्षक

-भारत के महान्यायवादी

-राज्यों के राज्यपाल

-मुख्य चुनाव आयुक्त एवं अन्य चुनाव आयुक्त

-संघ लोक सेवा आयोग के अध्यक्ष और अन्य सदस्यों

-वित्त आयोग, भाषा आयोग, पिछड़ा वर्ग आयोग, राष्ट्रीय महिला आयोग, मानवाधिकार आयोग, अल्पसंख्यक आयोग एवं अनुसूचित क्षेत्रों के प्रशासन के संबंध में रिपोर्ट देने वाले आयोग के सदस्यों को नियुक्त करने का अधिकार होता है।

विधायी शक्तियां

राष्ट्रपति लोकसभा के प्रथम सत्र को संबोधित करते हैं व संयुक्त अधिवेशन बुलाकर भाषण देने की शक्तियां भी प्राप्त है। राष्ट्रपति को संसद सत्र आहूत, सत्रावसान करना एवं लोकसभा को भंग करने की शक्ति भी रखता है। नए राज्यों के निर्माण राज्य की सीमा में परिवर्तन संबंधित विधेयक, धन विधेयक या संचित निधि से व्यय करने वाला विधेयक एवं राज्य हित से जुड़े विधेयक बिना राष्ट्रपति के पूर्व अनुमति के संसद में प्रस्तुत नहीं होते हैं। राष्ट्रपति लोकसभा के लिए आंग्ल भारतीय समुदाय से 2 सदस्य तथा राज्यसभा के लिए कला, साहित्य, विज्ञान, समाज सेवा क्षेत्र के 12 सदस्यों को मनोनीत करने का शक्तियां प्राप्त है।

न्यायिक शक्तियां

राष्ट्रपति को किसी अपराधी को सजा को क्षमा करने, उसका प्रविलंवन करने, संशोधन और सजा कम करने का अधिकार प्राप्त है। राष्ट्रपति को मृत्युदंड माफ करने का भी अधिकार प्राप्त है। राष्ट्रपति, सजा या कोर्ट मार्शल की सजा को माफ कर सकता है।

राष्ट्रपति की सैन्य शक्तियां

भारत के राष्ट्रपति के पास सैन्य बलों की सर्वोच्च कमांडर होता है। राष्ट्रपति को युद्ध और शांति की घोषणा करने तथा सैन्य बलों को विस्तार करने हेतु आदेश देने की शक्ति प्राप्त है।

आपातकालीन शक्तियां

भारतीय संविधान में राष्ट्रपति को तीन स्थितियों में आपातकालीन शक्तियां प्राप्त है-

–अनुच्छेद 352 – देश में युद्ध बाहरी आक्रमण या सशस्त्र विद्रोह की स्थिति में राष्ट्रपति को यह शक्ति प्राप्त है कि पूरे भारत या किसी एक भाग की सुरक्षा खतरे में है, तो वह पूरे भारत या किसी भाग में आपातकाल घोषणा कर सकता है। हालांकि अगर यह 1 माह के बाद संसद से अनुमोदित ना हो, तो स्वत: समाप्त हो जाती है। इस तरह की घोषणा को संसद के दो तिहाई बहुमत से पास होना आवश्यक होता है।

–अनुच्छेद 356- इसके अंतर्गत यदि कोई राज्य सरकार संवैधानिक नियमों के अनुरूप कार्य नहीं कर रही है, तो वहां राष्ट्रपति शासन लगा सकता है। इसे संसद द्वारा 2 माह के भीतर अनुमोदन करना आवश्यक होता है।

–अनुच्छेद 360- अनुच्छेद 360 के अंतर्गत देश में आर्थिक संकट की स्थिति में राष्ट्रपति अपनी विशिष्ट शक्तियों का प्रयोग कर वित्तीय आपात की घोषणा कर सकता है।

राष्ट्रपति की विटों शक्तियां

पूर्व वीटो

इस वीटो शक्ति के तहत राष्ट्रपति किसी विधेयक पर अपनी अनुमति नहीं देता है, अर्थात वह अपनी अनुमति को सुरक्षित रख सकता है।

निलंबित वीटो

इस वीटो शक्ति के अंतर्गत राष्ट्रपति किसी विधेयक को संसद के पास पुनर्विचार हेतु भेज सकता है।

पॉकेट वीटो

इस वीटो शक्ति के तहत राष्ट्रपति किसी विधेयक पर न अनुमति देता है, न ही अनुमति देने से इनकार करता है और न ही पुनर्विचार हेतु संसद के पास भेजता है। यानि किसी विधेयक को अनिश्चितकाल के लिए अपने पास सुरक्षित रख सकता है।

राष्ट्रपति का कार्यकाल

देश के राष्ट्रपति का कार्यकाल 5 वर्षों का होता है, हालांकि राष्ट्रपति अपने पद पर तबतक बना रहेगा, जबतक की उसका उत्तराधिकारी पद ग्रहण नहीं कर लेता। अगर राष्ट्रपति का पद मृत्यु, त्याग पत्र या महाभियोग की वजह से खाली हो जाता है, तो इस स्थिति में नए राष्ट्रपति का चुनाव 5 वर्षों के लिए ही होता है, ना कि शेष अवधि के लिए। इसके अलावा खाली पद पर चुनाव 6 महीने के भीतर कराया जाता है। वैसे अगर राष्ट्रपति का पद मृत्यु त्याग पत्र अथवा पद से हटाए जाने के कारण रिक्त होता है, तो उपराष्ट्रपति राष्ट्रपति के रूप में कार्य करता है।

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.7

noreply@blogger.com (Unknown) — Tue, 14 Jun 2022 10:27:00 +0530

Exercise 3.7 Page: 68

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju sister Ann B/2 yrs

By using the information that is given,

Case (i)

When Ani is older than that of Biju by 3 yrs then A – B = 3 – – – – – – – – (1)

2A−B/2 = 30

4A – B = 60 – – – – – – – – – – – (2)

By subtracting the equations (1) and (2) we get,

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

Case (ii)

When Biju is older than Ani,

B – A = 3 – – – – – – – – – (1)

2A − B/2 = 30

4A – B = 60 – – – – – – – – – (2)

Adding the equation (1) and (2) we get,

3A = 63

A = 21

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

Solution:

Let Sangam have Rs A with him and Reuben have Rs B with him.

Using the information that is given we get,

A + 100 = 2(B – 100) ⇒ A + 100 = 2B – 200

Or A – 2B = -300 – – – – – – – (1)

And

6(A – 10) = ( B + 10 )

Or 6A – 60 = B + 10

Or 6A – B = 70 – – – – – – (2)

When equation (2) is multiplied by 2 we get,

12A – 2B = 140 – – – – – – – (3)

When equation (1) is subtracted from equation (3) we get,

11A = 140 + 300

11A = 440

⇒ A = 440/11 = 40

Using A =40 in equation (1) we get,

40 – 2B = -300

40 + 300 = 2B

2B = 340

B = 170

Therefore, Sangam had Rs 40 and Reuben had Rs 170 with them.

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let the speed of the train be A km/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.

Speed of the train = Distance travelled by train / Time taken to travel that distance

A = N (distance)/ X (time)

Or, N = AX – – – – – – – – – – – (1)

Using the information that is given, we get:

(A+10) = X/(N-2)

(A + 10) (N – 2) = X

AN + 10N – 2A – 20 = X

By using the equation (1) we get,

– 2A + 10N = 20 – – – – – – – – – – (2)

(A-10) = X/(N+3)

(A – 10) (N + 3) = X

AN – 10N + 3A – 30 = X

By using the equation (1) we get,

3A – 10N = 30 – – – – – – – – – (3)

Adding equation (2) and equation (3) we get,

A = 50

Using the equation (2) we get,

(-2) x (50) + 10N = 20

-100 +10N = 20

=> 10N = 120

N = 12hours

From the equation (1) we get,

Distance travelled by the train, X = AN

= 50 x 12

= 600 km

Hence, the distance covered by the train is 600km.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of rows be A and the number of students in a row be B.

Total number of students = Number of rows x Number of students in a row

=AB

Using the information, that is given,

First Condition:

Total number of students = (A – 1) ( B + 3)

Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3

Or 3A – B – 3 = 0

Or 3A – Y = 3 – – – – – – – – – – – – – (1)

Second condition:

Total Number of students = (A + 2 ) ( B – 3 )

Or AB = AB + 2B – 3A – 6

Or 3A – 2B = -6 – – – – – – – – – (2)

When equation (2) is subtracted from (1)

(3A – B) – (3A – 2B) = 3 – (-6)

-B + 2B = 3 + 6B = 9

By using the equation (1) we get,

3A – 9 =3

3A = 9+3 = 12

A = 4

Number of rows, A = 4

Number of students in a row, B = 9

Number of total students in a class = AB = 4 x 9 = 36

5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B). Find the three angles.

Solution:

Given,

∠C = 3 ∠B = 2(∠B + ∠A)

3∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

2∠A – ∠B= 0- – – – – – – – – – – – (i)

We know, the sum of all the interior angles of a triangle is 180O.

Thus, ∠ A +∠B+ ∠C = 180O

∠A + ∠B +3 ∠B = 180O

∠A + 4 ∠B = 180O– – – – – – – – – – – – – – -(ii)

Multiplying 4 to equation (i) , we get

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180O

∠A = 20O

Using this in equation (ii), we get

20O+ 4∠B = 180O

∠B = 40O

3∠B =∠C

∠C = 3 x 40 = 120O

Therefore, ∠A = 20O

∠B=40O

∠C = 120O

6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Solutions:

Given,

5x – y = 5

=> y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

y = 3x – 3

The graphical representation of these lines will be as follows:

From the above graph we can see that the triangle formed is ∆ABC by the lines and the y axis. Also the coordinates of the vertices are A(1,0) , C(0,-5) and B(0,-3).

7. Solve the following pair of linear equations:

(i) px + qy = p – q

qx – py = p + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) x/a – y/b = 0

ax + by = a2 + b2

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

(a + b)(x + y) = a2 + b2

(v) 152x – 378y = – 74

–378x + 152y = – 604

Solutions:

(i) px + qy = p – q……………(i)

qx – py = p + q……………….(ii)

Multiplying p to equation (1) and q to equation (2), we get

p2x + pqy = p2 − pq ………… (iii)

q2x − pqy = pq + q2 ………… (iv)

Adding equation (iii) and equation (iv),we get

p2x + q2 x = p2 + q2

(p2 + q2 ) x = p2 + q2

x = (p2 + q2)/ p2 + q2 = 1

From equation (i), we get

p(1) + qy = p – q

qy = p-q-p

qy = -q

y = -1

(ii) ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying a to equation (i) and b to equation (ii), we obtain

a2x + aby = ac ………………… (iii)

b2x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a2 – b2) x = ac − bc– b

x = (ac − bc– b)/ (a2 – b2)

x = c(a-b) –b / (a2+b2)

From equation (i), we obtain

ax +by = c

a{c(a−b)−b)/ (a2 – b2)} +by=c

ac(a−b)−ab/ (a2 – b2)+by=c

by=c–ac(a−b)−ab/(a2 – b2)

by=abc – b2 c+ab/a2-b2

y = c(a-b)+a/a2-b2

(iii) x/a – y/b = 0

ax + by = a2 + b2

x/a – y/b = 0

=> bx − ay = 0 ……. (i)

ax + by = a2 + b2 …….. (ii)

Multiplying a and b to equation (i) and (ii) respectively, we get

b2x − aby = 0 …………… (iii)

a2x + aby = a 3 + ab3 …… (iv)

Adding equations (iii) and (iv), we get

b2x + a2x = a 3 + ab2

x (b2 + a2) = a (a2 + b2) x = a

Using equation (i), we get

b(a) − ay = 0

ab − ay = 0

ay = ab,

y = b

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

(a + b)(x + y) = a2 + b2

(a + b) y + (a – b) x = a2− 2ab − b2 …………… (i)

(x + y)(a + b) = a 2 + b2

(a + b) y + (a + b) x = a2 + b2 ………………… (ii)

Subtracting equation (ii) from equation (i), we get

(a − b) x − (a + b) x = (a 2 − 2ab − b 2) − (a2 + b2)

x(a − b − a − b) = − 2ab − 2b2

− 2bx = − 2b (b + a)

x = b + a

Substituting this value in equation (i), we get

(a + b)(a − b) +y (a + b) = a2− 2ab – b2

a2 − b2 + y(a + b) = a2− 2ab – b2

(a + b) y = − 2ab

y = -2ab/(a+b)

(v) 152x − 378y = − 74

76x − 189y = − 37

x =(189y-37)/76……………..…(i)

− 378x + 152y = − 604

− 189x + 76y = − 302 ………….. (ii)

Using the value of x in equation (ii), we get

−189(189y−37)/76+76y=−302

− (189)2y + 189 × 37 + (76)2 y = − 302 × 76

189 × 37 + 302 × 76 = (189)2 y − (76)2y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation (i), we get

x = (189-37)/76

x = 152/76 = 2

8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Solution:

It is known that the sum of the opposite angles of a cyclic quadrilateral is 180o

Thus, we have

∠C +∠A = 180

4y + 20− 4x = 180

− 4x + 4y = 160

x − y = − 40 ……………(1)

And, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 ………..(2)

Multiplying 3 to equation (1), we get

3x − 3y = − 120 ………(3)

Adding equation (2) to equation (3), we get

− 7x + 3x = 180 – 120

− 4x = 60

x = −15

Substituting this value in equation (i), we get

x − y = − 40

-y−15 = − 40

y = 40-15

= 25

∠A = 4y + 20 = 20+4(25) = 120°

∠B = 3y − 5 = − 5+3(25) = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = 5-7x

∠D= 5− 7(−15) = 110°

Hence, all the angles are measured.

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.6

noreply@blogger.com (Unknown) — Tue, 14 Jun 2022 10:25:00 +0530

Exercise 3.6 Page: 67

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Solution:

Let us assume 1/x = m and 1/y = n , then the equation will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0…………………….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0……………………….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

Solution:

Substituting 1/√x = m and 1/√y = n in the given equations, we get

2m + 3n = 2 ………………………..(i)

4m – 9n = -1 ………………………(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………………….…..(iii)

Adding equation (ii) and (iii), we get

10m = 5

m = 1/2…………………………….…(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×1/2 + 3n = 2

3n = 1

n = 1/3

m =1/√x

½ = 1/√x

x = 4

n = 1/√y

1/3 = 1/√y

y = 9

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

Solution:

Putting in the given equation we get,

So, 4m + 3y = 14 => 4m + 3y – 14 = 0 ……………..…..(1)

3m – 4y = 23 => 3m – 4y – 23 = 0 ……………………….(2)

By cross-multiplication, we get,

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-m/125 = y/50 = -1/ 25

-m/125 = -1/25 and y/50 = -1/25

m = 5 and b = -2

m = 1/x = 5

So , x = 1/5

y = -2

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

Solution:

Substituting 1/(x-1) = m and 1/(y-2) = n in the given equations, we get,

5m + n = 2 …………………………(i)

6m – 3n = 1 ……………………….(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 …………………….(iii)

Adding (ii) and (iii) we get

21m = 7

m = 1/3

Putting this value in equation (i), we get

5×1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

⇒ x = 4

n = 1/(y-2)

⇒ 1/3 = 1/(y-2)

⇒ y = 5

Hence, x = 4 and y = 5

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

Solution:

(7x-2y)/ xy = 5

7/y – 2/x = 5…………………………..(i)

(8x + 7y)/xy = 15

8/y + 7/x = 15…………………………(ii)

Substituting 1/x =m in the given equation we get,

– 2m + 7n = 5 => -2 + 7n – 5 = 0 ……..(iii)

7m + 8n = 15 => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

m = 1

n/(-65) = 1/(-65)

n = 1

m = 1 and n = 1

m = 1/x = 1 n = 1/x = 1

Therefore, x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

Solution:

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0…………………….(i)

2x + 4y = 5xy

=> 2/y + 4/x = 5

=> 2n +4m = 5

=> 4m+2n-5 = 0……………………..(ii)

3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m/(-30 –(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

m = 1

n/-9 = 1/-18

n = 1/2

m = 1 and n = 1/2

m = 1/x = 1 and n = 1/y = 1/2

x = 1 and y = 2

Hence, x = 1 and y = 2

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

Solution:

Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,

10m + 2n = 4 => 10m + 2n – 4 = 0 ………………..…..(i)

15m – 5n = -2 => 15m – 5n + 2 = 0 ……………………..(ii)

Using cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 …………………………………………(iii)

n = 1/(x-y) = 1

x-y = 1……………………………………………(iv)

Adding equation (iii) and (iv), we get

2x = 6 => x = 3 …….(v)

Putting the value of x = 3 in equation (3), we get

y = 2

Hence, x = 3 and y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Solution:

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 …………………………….…… (1)

m/2 – n/2 = -1/8

m – n = -1/4 …………………………..…(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

2m = 1/2

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

m = 1/(3x+y) = 1/4

3x + y = 4 …………………………………(3)

n = 1/( 3x-y) = 1/2

3x – y = 2 ………………………………(4)

Adding equations (3) and (4), we get

6x = 6

x = 1 ……………………………….(5)

Putting in (3), we get

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1

2. 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 women 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.

(iii) 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.

Solutions:

(i) Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

As per the question given,

4(2/x + 5/y) = 1

(2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1

(3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1…………………(1)

3m + 6n =1/3 => 9m + 18n = 1………………….(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m/2 = 1/36

m = 1/18

m = 1/x = 1/18

or x = 18

n = 1/y = 1/36

y = 36

Therefore,

Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

1200n = 15

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/60

m = 1/x = 1/60

x = 60

And y = 1/n

y = 80

Therefore,

Speed of the train = 60 km/h

Speed of the bus = 80 km/h

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.5

noreply@blogger.com (Unknown) — Tue, 14 Jun 2022 10:23:00 +0530

Exercise 3.5 Page: 62

1. 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 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8

(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Solutions:

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

a1/a2=1/3 , b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2

(a1/a2) = (b1/b2) ≠ (c1/c2)

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8

a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x/(b1c2-c1b2) = y/(c1a2 – c2a1) = 1/(a1b2-a2b1)

x/(-8-(-10)) = y/(-15-(-16)) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a1/a2) = 3/6 = 1/2

(b1/b2) = -5/-10 = 1/2

(c1/c2) = 20/40 = 1/2

a1/a2 = b1/b2 = c1/c2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a1/a2) = 1/3

(b1/b2) = -3/-3 = 1

(c1/c2) = -7/-15

a1/a2 ≠ b1/b2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1.

2. (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 linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Solution:

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a1/a2 = 2/(a-b) , b1/b2 = 3/(a+b) , c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x + (k-1)y – 2k -1 = 0

a1/a2 = 3/(2k -1) , b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)

For no solutions

a1/a2 = b1/b2 ≠ c1/c2

3/(2k-1) = 1/(k -1) ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

k =2

Therefore, for k = 2 the given pair of linear equations will have no solution.

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y )/ 3 ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) 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.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) 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?

(iv) 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?

(v) 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.

Solutions:

(i) Let x be the fixed charge and y be the charge of food per day.

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 -26 x 30

x= Rs.400.

Therefore, fixed charges is Rs.400 and charge per day is Rs.30.

(ii) Let the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

Subtracting equation (1) from (2) , we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

y= 12

Therefore, the fraction is 5/12.

(iii) Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

(iv) Let x km/h be the speed of car from point A and y km/h be the speed of car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equation (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Therefore, the speed of car from point A = 60 km/h

Speed of car from point B = 40 km/h.

(v) Let,

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units

And breadth of the rectangle = 9 units

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.4

noreply@blogger.com (Unknown) — Tue, 14 Jun 2022 10:21:00 +0530

Exercise 3.4 Page: 56

1. 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) x/2+ 2y/3 = -1 and x-y/3 = 3

Solutions:

(i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

3x = 27/13

x =9/13

∴x = 9/13 and y = -5/13

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

x = 9/13

∴x = 9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

5y = -15

y = -3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

By the method of Substitution:

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

2. 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?

Solution:

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

3 – b = -2

-b = -5

b = 5

Hence, the fraction is 3/5.

(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?

Solution:

Let us assume, present age of Nuri is x

And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now,

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

Age of Sonu is 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.

Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

9B = 9

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) 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.

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

(v) 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:

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.3

noreply@blogger.com (Unknown) — Sun, 12 Jun 2022 19:55:00 +0530

Exercise 3.3 Page: 53

1. Solve the following pair of linear equations by the substitution method

(i) x + y = 14

x – y = 4

(ii) s – t = 3

(s/3) + (t/2) = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2 x+√3 y = 0

√3 x-√8 y = 0

(vi) (3x/2) – (5y/3) = -2

(x/3) + (y/2) = (13/6)

Solutions:

(i) Given,

x + y = 14 and x – y = 4 are the two equations.

From 1st equation, we get,

x = 14 – y

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Or y = 5

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

(ii) Given,

s – t = 3 and (s/3) + (t/2) = 6 are the two equations.

From 1st equation, we get,

s = 3 + t ________________(1)

Now, substitute the value of s in second equation to get,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

Now, substitute the value of t in equation (1)

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii) Given,

3x – y = 3 and 9x – 3y = 9 are the two equations.

From 1st equation, we get,

x = (3+y)/3

Now, substitute the value of x in the given second equation to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.

From 1st equation, we get,

x = (1.3- 0.3y)/0.2 _________________(1)

Now, substitute the value of x in the given second equation to get,

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

Now, substitute the value of y in equation (1), we get,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

(v) Given,

√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.

From 1st equation, we get,

x = – (√3/√2)y __________________(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0

⇒ y = 0

Now, substitute the value of y in equation (1), we get,

x = 0

Therefore, x = 0 and y = 0.

(vi) Given,

(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.

From 1st equation, we get,

(3/2)x = -2 + (5y/3)

⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)

Putting the value of x in the given second equation to get,

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

Now, substitute the value of y in equation (1), we get,

(3x/2) – 5(3)/3 = -2

⇒ (3x/2) – 5 = -2

⇒ x = 2

Therefore, x = 2 and y = 3.

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II), we get

x = (11-3y)/2 ………………….(III)

Substituting the value of x in equation (II), we get

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24

-7y = -35

y = 5……………………………………..(IV)

Putting the value of y in equation (III), we get

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2, y = 5

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore the value of m is -1.

3. 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.

Solution:

Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26

x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39.

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

Solution:

Let the larger angle by xo and smaller angle be yo.

We know that the sum of two supplementary pair of angles is always 180o.

According to the question,

x + y = 180o……………. (1)

x – y = 18o ……………..(2)

From (1), we get x = 180o – y …………. (3)

Substituting (3) in (2), we get

180o – y – y =18o

162o = 2y

y = 81o ………….. (4)

Using the value of y in (3), we get

x = 180o – 81o

= 99o

Hence, the angles are 99o and 81o.

(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.

Solution:

Let the cost a bat be x and cost of a ball be y.

According to the question,

7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (II)

From (I), we get

y = (3800-7x)/6………………..(III)

Substituting (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800-7 ×500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

(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 charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

According to the question,

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 × 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:

Let the fraction be x/y.

According to the question,

(x+2) /(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

(x+3) /(y+3) = 5/6

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get x = (-4+9y)/11 …………….. (3)

Substituting the value of x in (2), we get

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x = (-4+9×9 )/11 = 7

Hence the fraction is 7/9.

(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?

Solutions:

Let the age of Jacob and his son be x and y respectively.

According to the question,

(x + 5) = 3(y + 5)

x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5)

x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)

Substituting the value of x in (2), we get

3y + 10 – 7y = -30

-4y = -40

y = 10 ………………… (4)

Substituting the value of y in (3), we get

x = 3 x 10 + 10 = 40

Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

NCERT Solutions For Class 10 Chapter 1 Real Number Exercise 1.4

noreply@blogger.com (Unknown) — Tue, 26 Apr 2022 20:16:00 +0530

Exercise 1.4 Page: 17

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2352) (vii) 129/(225775) (viii) 6/15 (ix) 35/50 (x) 77/210

Solutions:

Note: If the denominator has only factors of 2 and 5 or in the form of 2m ×5n then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

(i) 13/3125

Factorizing the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Since, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

(ii) 17/8

Factorizing the denominator, we get,

8 = 2×2×2 = 23

Since, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

(iii) 64/455

Factorizing the denominator, we get,

455 = 5×7×13

Since, the denominator is not in the form of 2m × 5n, thus 64/455 has a non-terminating decimal expansion.

(iv) 15/ 1600

Factorizing the denominator, we get,

1600 = 26×52

Since, the denominator is in the form of 2m × 5n, thus 15/1600 has a terminating decimal expansion.

(v) 29/343

Factorizing the denominator, we get,

343 = 7×7×7 = 73 Since, the denominator is not in the form of 2m × 5n thus 29/343 has a non-terminating decimal expansion.

(vi)23/(2352)

Clearly, the denominator is in the form of 2m × 5n.

Hence, 23/ (2352) has a terminating decimal expansion.

(vii) 129/(225775)

As you can see, the denominator is not in the form of 2m × 5n.

Hence, 129/ (225775) has a non-terminating decimal expansion.

(viii) 6/15

6/15 = 2/5

Since, the denominator has only 5 as its factor, thus, 6/15 has a terminating decimal expansion.

(ix) 35/50

35/50 = 7/10

Factorising the denominator, we get,

10 = 2 × 5

Since, the denominator is in the form of 2m × 5n thus, 35/50 has a terminating decimal expansion.

(x) 77/210

77/210 = (7× 11)/ (30 × 7) = 11/30

Factorising the denominator, we get,

30 = 2 × 3 × 5

As you can see, the denominator is not in the form of 2m × 5n .Hence, 77/210 has a non-terminating decimal expansion.

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solutions:

(i) 13/3125

13/3125 = 0.00416

(ii) 17/8

17/8 = 2.125

(iii) 64/455 has a Non terminating decimal expansion

(iv)15/ 1600

15/1600 = 0.009375

(v) 29/ 343 has a Non terminating decimal expansion

(vi)23/ (2352) = 23/(8×25)= 23/200

23/ (2352) = 0.115

(vii) 129/ (225775) has a Non terminating decimal expansion

(viii) 6/15 = 2/5

(ix) 35/50 = 7/10

35/50 = 0.7

(x) 77/210 has a non-terminating decimal expansion.

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

Solutions:

(i) 43.123456789

Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

(ii) 0.120120012000120000. . .

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5.

Class 10 NCERT Solution Math Chapter 15 – Probability Exercise: 15.2

noreply@blogger.com (Unknown) — Mon, 4 Apr 2022 09:24:00 +0530

Exercise: 15.2 (Page No: 311)

1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days?

Solution:

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5×5 = 25

(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) = 5/25 = ⅕

(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

So, P(both visiting on the consecutive days) = 8/25

(iii) P (both visiting on the different days) = 1-P (both visiting on the same day)

So, P (both visiting on the different days) = 1-(⅕) = ⅘

2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is

(i) even?

(ii) 6?

(iii) at least 6?

Solution:

The table will be as follows:

+	1	2	2	3	3	6
1	2	3	3	4	4	7
2	3	4	4	5	5	8
2	3	4	4	5	5	8
3	4	5	5	6	6	9
3	4	5	5	6	6	9
6	7	8	8	9	9	12

So, the total number of outcomes = 6×6 = 36

(i) E (Even) = 18

P (Even) = 18/36 = ½

(ii) E (sum is 6) = 4

P (sum is 6) = 4/36 = 1/9

(iii) E (sum is atleast 6) = 15

P (sum is atleast 6) = 15/36 = 5/12

3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball

is double that of a red ball, determine the number of blue balls in the bag.

Solution:

It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x+5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ P (drawing a blue ball) = [x/(x+5)] ——–(i)

Similarly,

P (drawing a red ball) = [5/(x+5)] ——–(i)

From equation (i) and (ii)

x = 10

So, the total number of blue balls = 10

4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the

box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now

double of what it was before. Find x

Solution:

Total number of black balls = x

Total number of balls = 12

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (getting black balls) = x/12 ——————-(i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x+6

Now, P (getting black balls) = (x+6)/18 ——————-(ii)

It’s given that, the probability of drawing a black ball now is double of what it was before

(ii) = 2 × (i)

(x+6)/18 = 2 × (x/12)

x + 6 = 3x

2x = 6

∴ x = 3

5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at

random from the jar, the probability that it is green is ⅔. Find the number of blue balls

in the jar.

Solution:

Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24-x

P(getting green marble) = x/24

From the question, x/24 = ⅔

So, the total green marbles = 16

And, the total blue marbles = 24-16 = 8

Class 10 NCERT Solution Math Chapter 15 – Probability Exercise: 15.1

noreply@blogger.com (Unknown) — Mon, 4 Apr 2022 09:23:00 +0530

Exercise: 15.1 (Page No: 308)

1. Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = ___________.

(ii) The probability of an event that cannot happen is __________. Such an event is called ________.

(iii) The probability of an event that is certain to happen is _________. Such an event is called _________.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.

(v) The probability of an event is greater than or equal to ___ and less than or equal to __________.

Solution:

(i) Probability of an event E + Probability of the event ‘not E’ = 1.

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to Solution: a true-false question. The Solution: is right or wrong.

(iv) A baby is born. It is a boy or a girl.

Solution:

(i) This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.

(ii) Even this statement does not have equally likely outcomes as the player may shoot or miss the shot.

(iii) This statement has equally likely outcomes as it is known that the solution is either right or wrong.

(iv) This statement also has equally likely outcomes as it is known that the newly born baby can either be a boy or a girl.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution:

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

4. Which of the following cannot be the probability of an event?

(A) 2/3 (B) -1.5 (C) 15% (D) 0.7

Solution:

The probability of any event (E) always lies between 0 and 1 i.e. 0 ≤ P(E) ≤ 1. So, from the above options, option (B) -1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Solution:

We know that,

P(E)+P(not E) = 1

It is given that, P(E) = 0.05

So, P(not E) = 1-P(E)

Or, P(not E) = 1-0.05

∴ P(not E) = 0.95

6. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavored candy?

(ii) a lemon flavored candy?

Solution:

(i) We know that the bag only contains lemon-flavored candies.

So, The no. of orange flavored candies = 0

∴ The probability of taking out orange flavored candies = 0/1 = 0

(ii) As there are only lemon flavored candies, P(lemon flavored candies) = 1 (or 100%)

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

Let the event wherein 2 students having the same birthday be E

Given, P(E) = 0.992

We know,

P(E)+P(not E) = 1

Or, P(not E) = 1–0.992 = 0.008

∴ The probability that the 2 students have the same birthday is 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red?

(ii) not red?

Solution:

The total number of balls = No. of red balls + No. of black balls

So, the total no. of balls = 5+3 = 8

We know that the probability of an event is the ratio between the no. of favourable outcomes and the total number of outcomes.

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Probability of drawing red balls = P (red balls) = (no. of red balls/total no. of balls) = 3/8

(ii) Probability of drawing black balls = P (black balls) = (no. of black balls/total no. of balls) = 5/8

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

(i) red?

(ii) white?

(iii) not green?

Solution:

The Total no. of balls = 5+8+4 = 17

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of red balls = 5

P (red ball) = 5/17 = 0.29

(ii) Total number of white balls = 8

P (white ball) = 8/17 = 0.47

(iii) Total number of green balls = 4

P (green ball) = 4/17 = 0.23

∴ P (not green) = 1-P(green ball) = 1-(4/7) = 0.77

10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a 50 p coin?

(ii) will not be a ₹5 coin?

Solution:

Total no. of coins = 100+50+20+10 = 180

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of 50 p coin = 100

P (50 p coin) = 100/180 = 5/9 = 0.55

(ii) Total number of ₹5 coin = 10

P (₹5 coin) = 10/180 = 1/18 = 0.055

∴ P (not ₹5 coin) = 1-P (₹5 coin) = 1-0.055 = 0.945

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

Solution:

The total number of fish in the tank = 5+8 = 13

Total number of male fish = 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (male fish) = 5/13 = 0.38

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Solution:

Total number of possible outcomes = 8

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of favourable events (i.e. 8) = 1

∴ P (pointing at 8) = ⅛ = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = ½ = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 1

13. A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Solution:

Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½ = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = ½ = 0.5

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Solution:

Total number of possible outcomes = 52

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total numbers of king of red colour = 2

P (getting a king of red colour) = 2/52 = 1/26 = 0.038

(ii) Total numbers of face cards = 12

P (getting a face card) = 12/52 = 3/13 = 0.23

(iii) Total numbers of red face cards = 6

P (getting a king of red colour) = 6/52 = 3/26 = 0.11

(iv) Total numbers of jack of hearts = 1

P (getting a king of red colour) = 1/52 = 0.019

(v) Total numbers of king of spade = 13

P (getting a king of red colour) = 13/52 = ¼ = 0.25

(vi) Total numbers of queen of diamonds = 1

P (getting a king of red colour) = 1/52 = 0.019

15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Solution:

Total numbers of cards = 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Numbers of queen = 1

P (picking a queen) = ⅕ = 0.2

(ii) If the queen is drawn and put aside, the total numbers of cards left is (5-4) = 4

(a) Total numbers of ace = 1

P (picking an ace) = ¼ = 0.25

(b) Total numbers of queen = 0

P (picking a queen) = 0/4 = 0

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution:

Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132+12 = 144 pens

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P(picking a good pen) = 132/144 = 11/12 = 0.916

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution:

(i) Number of defective bulbs = 4

The total number of bulbs = 20

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = ⅕ = 0.2

(ii) Since 1 non-defective bulb is drawn, then the total numbers of bulbs left are 19

So, the total number of events (or outcomes) = 19

Number of non-defective bulbs = 19-4 = 15

So, the probability that the bulb is not defective = 15/19 = 0.789

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

Solution:

The total number of discs = 90

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of discs having two digit numbers = 81

(Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90-9 = 81)

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = 18/90 = ⅕ = 0.2

19. A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting

(i) A?

(ii) D?

Solution:

The total number of possible outcomes (or events) = 6

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) The total number of faces having A on it = 2

P (getting A) = 2/6 = ⅓ = 0.33

(ii) The total number of faces having D on it = 1

P (getting D) = ⅙ = 0.166

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

Solution:

First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome and the area of the circle will be the favourable outcome.

So, the area of the rectangle = (3×2) m2 = 6 m2

and,

The area of the circle = πr2 = π(½)2 m2 = π/4 m2 = 0.78

∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

Solution:

The total numbers of outcomes i.e. pens = 144

Given, numbers of defective pens = 20

∴ The numbers of non defective pens = 144-20 = 124

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total numbers events in which she will buy them = 124

So, P (buying) = 124/144 = 31/36 = 0.86

(ii) Total numbers events in which she will not buy them = 20

So, P (not buying) = 20/144 = 5/36 = 0.138

22. Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your Solution:.

Solution:

If 2 dices are thrown, the possible events are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

So, the total numbers of events: 6×6 = 36

(i) It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1)

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)

So, P(sum 3) = 2/36

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

So, P (sum 4) = 3/36

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

So, P (sum 5) = 4/36

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

So, P (sum 6) = 5/36

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

So, P (sum 7) = 6/36

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

So, P (sum 8) = 5/36

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

So, P (sum 9) = 4/36

E (sum 10) = (4,6), (6,4), and (5,5)

So, P (sum 10) = 3/36

E (sum 11) = (5,6), and (6,5)

So, P (sum 11) = 2/36

E (sum 12) = (6,6)

So, P (sum 12) = 1/36

So, the table will be as:

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.

23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution:

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) = 6/8 = ¾ = 0.75

24. A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution:

Outcomes are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

So, the total number of outcome = 6×6 = 36

(i) Method 1:

Consider the following events.

A = 5 comes in first throw,

B = 5 comes in second throw

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

So, P(not A) = 1-(6/36) = 5/6

∴ The required probability = (5/6)×(5/6) = 25/36

Method 2:

Let E be the event in which 5 does not come up either time.

So, the favourable outcomes are [36–(5+6)] = 25

∴ P(E) = 25/36

(ii) Number of events when 5 comes at least once = 11(5+6)

∴ The required probability = 11/36

25. Which of the following arguments are correct and which are not correct? Give reasons for your Solution:.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

Solution:

(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P (getting two heads) = ¼

and, P (getting one of the each) = 2/4 = ½

∴ This statement is incorrect.

(ii) Since the two outcomes are equally likely, this statement is correct.

Class 10 NCERT Solution Math Chapter 14 – Statistics Exercise 14.4

noreply@blogger.com (Unknown) — Sun, 3 Apr 2022 21:42:00 +0530

Exercise 14.4 Page: 293

1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Daily income in Rupees	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Solution

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

Daily income	Frequency	Cumulative Frequency
Less than 120	12	12
Less than 140	14	26
Less than 160	8	34
Less than 180	6	40
Less than 200	10	50

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

2.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:

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.

Class interval	Number of students(Frequency)	Cumulative Frequency
Less than 38	0	0
Less than 40	3-0=3	3
Less than 42	5-3=2	5
Less than 44	9-5=4	9
Less than 46	14-9=5	14
Less than 48	28-14=14	28
Less than 50	32-28=4	32
Less than 52	35-22=3	35

The class 46 – 48 has the maximum frequency, therefore, this is modal class

Here, l = 46, h = 2, f1= 14, f0= 5 and f2 = 4

The mode formula is given as:

Now, Mode =

= 46 + 0.95 = 46.95

Thus, mode is verified.

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.

Production Yield	50-55	55-60	60-65	65-70	70-75	75-80
Number of farms	2	8	12	24	38	16

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

Solution:

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha)	Number of farms
More than or equal to 50	100
More than or equal to 55	100-2 = 98
More than or equal to 60	98-8= 90
More than or equal to 65	90-12=78
More than or equal to 70	78-24=54
More than or equal to 75	54-38 =16

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on

this graph paper. The graph obtained is known as more than type ogive curve.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the CBSE Term II exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4( it could vary as per question).
Topics covered in Chapter 14, Statistics are;
Mean of Grouped Data
Mode of Grouped Data
Median of Grouped Data
Graphical Representation of Cumulative Frequency Distribution
List of Exercises in class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Question ( 9 long)
Exercise 14.2 Solutions 6 Question ( 6 long)
Exercise 14.3 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)
NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.
The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.
You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form or in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE Term II exam.

Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics

The solutions for the statistics chapter works as a reference for the students.
It will help students to score marks against the questions asked from the statistics chapter.
Students can prepare and do the revision for chapter 14 with this source.
The questions of statistics are solved by subject experts.
The content of the material is as per the term-wise CBSE Syllabus (2021-22) and guidelines.
Statistics can also be understood in a much better way by using the other solutions which are provided at RAJ EDUCATION HUB. The solutions are prepared to help students perform well in the CBSE exams.

Class 10 NCERT Solution Math Chapter 14 – Statistics Exercise 14.3

noreply@blogger.com (Unknown) — Sun, 3 Apr 2022 21:31:00 +0530

Exercise 14.3 Page: 287

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

Monthly consumption(in units)	No. of customers
65-85	4
85-105	5
105-125	13
125-145	20
145-165	14
165-185	8
185-205	4

Solution:

Find the cumulative frequency of the given data as follows:

Class Interval	Frequency	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	8	64
185-205	4	68
	N=68

From the table, it is observed that, n = 68 and hence n/2=34

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, n = 68, Cf = 22, f = 20, h = 20

Median is calculated as follows:

=125+((34−22)/20) × 20

=125+12 = 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

f1=20, f0=13, f2=14 & h = 20

Mode formula:

Mode = l+ [(f1-f0)/(2f1-f0-f2)]×h

Mode = 125 + ((20-13)/(40-13-14))×20

=125+(140/13)

=125+10.77

=135.77

Therefore, mode = 135.77

Calculate the Mean:

Class Interval	fi	xi	di=xi-a	ui=di/h	fiui
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135	0	0	0
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
	Sum fi= 68				Sum fiui= 7

x̄ =a+h ∑fiui/∑fi =135+20(7/68)

Mean=137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of x & 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:

Given data, n = 60

Median of the given data = 28.5

Where, n/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25+x

Lower limit of median class, l = 20,

Cf = 5+x,

f = 20 & h = 10

Substitute the values

28.5=20+((30−5−x)/20) × 10

8.5 = (25 – x)/2

17 = 25-x

Therefore, x =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

Since,

60=5+20+15+5+x+y

Now, substitute the value of x, to find y

60 = 5+20+15+5+8+y

y = 60-53

y = 7

Therefore, the value of x = 8 and y = 7.

3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Age (in years)	Number of policy holder
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:

Class interval	Frequency	Cumulative frequency
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

Given data: n = 100 and n/2 = 50

Median class = 35-45

Then, l = 35, cf = 45, f = 33 & h = 5

Median = 35+((50-45)/33) × 5

= 35 + (5/33)5

= 35.75

Therefore, the median age = 35.75 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as 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 leaves.

Solution:

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

Class Interval	Frequency	Cumulative frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	9	17
144.5-153.5	12	29
153.5-162.5	5	34
162.5-171.5	4	38
171.5-180.5	2	40

So, the data obtained are:

n = 40 and n/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Median = 144.5+((20-17)/12)×9

= 144.5+(9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a life time of 400 neon lamps.

Lifetime (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 lifetime of a lamp.

Solution:

Class Interval	Frequency	Cumulative
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

Data:

n = 400 &n/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, Cf = 130,

f = 86 & h = 500

Median = 3000 + ((200-130)/86) × 500

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

Therefore, the median life time of the lamps = 3406.97 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in 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 number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

To calculate median:

Class Interval	Frequency	Cumulative Frequency
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

Given:

n = 100 &n/2 = 50

Median class = 7-10

Therefore, l = 7, Cf = 36, f = 40 & h = 3

Median = 7+((50-36)/40) × 3

Median = 7+42/40

Median=8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode = 7+((40-30)/(2×40-30-16)) × 3

= 7+(30/34)

= 7.88

Therefore mode = 7.88

Calculate the Mean:

Class Interval	fi	xi	fixi
1-4	6	2.5	15
4-7	30	5.5	165
7-10	40	8.5	340
10-13	16	11.5	184
13-16	4	14.5	51
16-19	4	17.5	70
	Sum fi = 100		Sum fixi = 825

Mean = x̄ = ∑fi xi /∑fi

Mean = 825/100 = 8.25

Therefore, mean = 8.25

7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

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:

Class Interval	Frequency	Cumulative frequency
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

Given: n = 30 and n/2= 15

Median class = 55-60

l = 55, Cf = 13, f = 6 & h = 5

Median = 55+((15-13)/6)×5

Median=55 + (10/6) = 55+1.666

Median =56.67

Therefore, the median weight of the students = 56.67

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the CBSE Term II exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4( it could vary as per question).

Topics covered in Chapter 14, Statistics are;

Mean of Grouped Data
Mode of Grouped Data
Median of Grouped Data
Graphical Representation of Cumulative Frequency Distribution

List of Exercises in class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Question ( 9 long)
Exercise 14.2 Solutions 6 Question ( 6 long)
Exercise 14.3 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.

The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.

You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form or in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE Term II exam.

Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics

The solutions for the statistics chapter works as a reference for the students.
It will help students to score marks against the questions asked from the statistics chapter.
Students can prepare and do the revision for chapter 14 with this source.
The questions of statistics are solved by subject experts.
The content of the material is as per the term-wise CBSE Syllabus (2021-22) and guidelines.

Statistics can also be understood in a much better way by using the other solutions which are provided at RAJ EDUCATION HUB. The solutions are prepared to help students perform well in the CBSE exams.

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−4298 as a rational number with denominator 7

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प्रधानमंत्री की शक्तियां

प्रधानमंत्री की नियुक्ति, शक्तियां एवं कार्य

प्रधानमंत्री का कार्यकाल

प्रधानमंत्री की न्युक्ति कौन करता है?

प्रधानमंत्री की नियुक्ति की प्रक्रिया

प्रधानमंत्री की शक्तियां एवं कार्य

प्रधानमंत्री और मंत्रीपरिषद के बीच संबंध

प्रधानमंत्री व संसद के बीच संबंध

निष्कर्ष (Conclusion) :-

GK QUIZ

राष्ट्रपति की शक्तियां

राष्ट्रपति की सैलरी व भत्ते

राष्ट्रपति की शक्तियां एवं अधिकार

कार्यपालिका शक्ति

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Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.7

Exercise 3.7 Page: 68

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.6

Exercise 3.6 Page: 67

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.5

Exercise 3.5 Page: 62

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.4

Exercise 3.4 Page: 56

Class 10 NCERT Solution Math Chapter 3 – Pair of Linear Equations in Two Variables Exercise - 3.3

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics

Class 10 NCERT Solution Math Chapter 14 – Statistics Exercise 14.3

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics

$\frac{- 42}{98}$ as a rational number with denominator $7$