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NCERT Solutions for Class 4 पर्यावरण अध्ययन Chapter 27 कोशिश हुई कामयाब

एन.सी.ई.आर.टी. पाठ्यपुस्तक (पृष्ठ संख्या 212)
प्रश्न 1.
तुम स्कूल कैसे जाते हो?
उत्तर:
में साइकिल से स्कूल जाता हूँ।

प्रश्न 2.
पता करो, लद्दाख कहाँ है। वह कैसा इलाका है?
उत्तर:
लद्दाख कश्मीर में है। यह भारत की सबसे ऊँची पठारी है। यह समुद्र तल से 3000 मीटर की ऊँचाई पर है। यह एक बहुत ही ठंढ़ा इलाका है।

एन.सी.ई.आर.टी. पाठ्यपुस्तक (पृष्ठ संख्या 213)
प्रश्न 1.
तुम्हें स्कूल में क्या-क्या करना अच्छा लगता है?
उत्तर:
मुझे स्कूल में पढ़ना, अच्छी बातें सीखना तथा दोस्तों के साथ खेलना अच्छा लगता है।

प्रश्न 2.
तुम्हें क्या स्कूल जाना अच्छा लगता है?
उत्तर:
हाँ, मुझे स्कूल जाना बहुत अच्छा लगता है।

प्रश्न 3.
यदि तुम कभी स्कूल नहीं जा पाते, तो तुम्हें कैसा लगता?
उत्तर:
यदि मैं कभी स्कूल नहीं जा पाता हूँ, तो मुझे अपने दोस्तों की याद आती है तथा बहुत बुरा लगता है।

एन.सी.ई.आर.टी. पाठ्यपुस्तक (पृष्ठ संख्या 215)
बताओ
प्रश्न 1.
चुसकिट किस-किस की मदद से स्कूल पहुँच पाई?
उत्तर:
चुसकिट अब्दुल, स्कूल के बच्चों तथा स्कूल के शिक्षकों की मदद से स्कूल पहुँच पाई।

प्रश्न 2.
अगर तुम अब्दुल होते, तो तुम क्या-क्या करते?
उत्तर:
यदि मैं अब्दुल होता तो मैं भी चुसकिट को स्कूल पहुँचाने में अब्दुल की तरह ही मदद करता।

प्रश्न 3.
चुसकिट स्कूल तो पहुँच गई। पर स्कूल के अंदर उसे कुछ परेशानियाँ हो सकती हैं? कौन-कौन सी?
उत्तर:
चुसकिट को स्कूल में घूमने में, एक क्लास से दूसरे क्लास तक जाने में तथा खेलने में परेशानी हो सकती है।

प्रश्न 4.
अगर तुम चुसकिट के दोस्त होते, तो उसकी मदद कैसे-कैसे करते?
उत्तर:
मैं चुसकिट को उसके व्हील चेयर पर स्कूल में घुमाता, उसके साथ स्कूल के मैदान में खेलता तथा खेलने में उसकी मदद करता।

प्रश्न 5.
क्या तुम्हारे स्कूल में पहियों वाली कुर्सी के लिए रैंप बने हैं?,
उत्तर:
हाँ, मेरे स्कूल में पहियों वाली कुर्सी के लिए रैंप बने हैं।

प्रश्न 6.
क्या तुम्हारे घर के आस-पास कोई ऐसा बच्चा रहता है, जिसे किन्हीं कारणों से स्कूल जाने में परेशानी हो रही हो?
उत्तर:
हाँ, मेरे घर के पास एक बच्चा है, जो स्कूल नहीं जाता है क्योंकि वह अंधा है।

प्रश्न 7.
क्या तुम उस बच्चे की मदद करना चाहोगे? कैसे?
उत्तर:
मैं उसके माता-पिता को समझाऊँगा कि उसे स्कूल भेजें। मैं उसे स्कूल जाने में मदद करूंगा।

प्रश्न 8.
अपने घर के आस-पास की इमारतों को देखो। क्या उनमें पहियों वाली कुर्सी अंदर ले जाने की सुविधा है?
उत्तर:
नहीं, उन घरों में पहिये वाली कुर्सी अंदर ले जाने की सुविधा नहीं है।

आओ, बना कर देखें
प्रश्न 1.
रैंप और पहियों वाली कुर्सी का चित्र कॉपी में बनाओ।
उत्तर:
रैंप और पहियों वाली कुर्सी
NCERT Solutions for Class 4 पर्यावरण अध्ययन Chapter 27 कोशिश हुई कामयाब 1

प्रश्न 2.
तुम भी अपना एक पुल बनाओ। इसके लिए सामान तुम्हें अपने आस-पास ही मिल सकता है, जैसे-आइसक्रीम की इंडियाँ, प्लास्टिक के चम्मच, छोटी डंडियाँ, रस्सी, सुतली आदि। अपने सारे दोस्तों को भी पुल बनाने के लिए कहो।
उत्तर:
NCERT Solutions for Class 4 पर्यावरण अध्ययन Chapter 27 कोशिश हुई कामयाब 2

प्रश्न 3.
अब समूह के साथ मिलकर एक मॉडल बनाओ। मॉडल में खेत, नदियाँ, पर्वत, सड़क और रेल की पटरियाँ बनाओ। इसके लिए तुम चिकनी मिट्टी, रेत, कंकड़-पत्थर के टुकड़े, टहनी आदि काम में ले सकते हो। अब इस मॉडल में अलग-अलग जगहों पर पुल रखो।
उत्तर:
स्वयं करो।

प्रश्न 4.
चुसकिट को स्कूल पहुँचाओ
NCERT Solutions for Class 4 पर्यावरण अध्ययन Chapter 27 कोशिश हुई कामयाब 3
उत्तर:

NCERT Solutions for Class 4 पर्यावरण अध्ययन

The post NCERT Solutions for Class 4 पर्यावरण अध्ययन Chapter 27 कोशिश हुई कामयाब appeared first on Learn CBSE.

Understanding Partition Politics, Memories, Experiences – CBSE Notes for Class 12 History

• The British policy of Divide and Rule played a key role in spreading of communalism.
• Earlier the British attitude towards the Muslim was not favourable, they think that they were responsible for the revolt of 1857.
• But soon they felt that due to their behaviour Hindus grew stronger, so they reversed their policy.
• Now, they began to take side with the Muslims and turned against Hindus.
• Bengal was partitioned in 1905 by Lord Curzen. He said Bengal was partitioned due to administrative problems.
• The real objective of British behind the partition of Bengal was to sow the seed of disunity between the Hindus and the Muslims.
• By the act of 1909 British government gave the Muslims the right of separate electorate.
• In, 1916 Lucknow Pact was signed between Congress and the Muslim League. It was an important landmark step forward in achieving Hindu-Muslim unity. But it was really an agreement for cooperation in the political field on the basis of common programme.
• In February 1937, elections to the provincial assembly were held, in which only few had the right to vote.
• To solve the political crisis of India, Lord Attlee sent Cabinet Mission to India.
• The Muslim League, on 6th June 1946 accepted the Cabinet Mission Plan as the foundation of Pakistan was inherent in it, but Congress opposed it.
• To solve the political tangle of India Lord Mount Batten arrived India. He proposed his plan on 3 June 1947, in which he stated that country would be divided into two Dominions,
i. e. India and Pakistan. It was accepted by both Congress and Muslim League.
1. Unionist Party: This party stood for the interests of all landlords in Punjab. It was founded in the year 1923.
2. Confederation: It refers to a union of fairly autonomous and sovereign states with a central government.
3. Arya Samaj: Founded by Swami Dayanand Saraswati in the year 1875.
4. Muslim League: Founded in the year 1906.
5. Hindu Mahasabha: Founded in the year 1915.
6. Federal Union: Autonomous federation given autonomous status. Its Central Government has some limited powers.
7. Lucknow Pact: A pact between Congress and Muslim League signed in the year 1916.
8. Pakistan: Came into existence after partition of India. Choudhary Rehmat Ali, a Punjabi- Muslim student at Cambridge, coined the name ‘Pakistan’ for the first time in the year 1933.
9. Muslim League’s Resolution at Lahore: In 1940, the Muslim League moved a resolution at Lahore demanding a measure of autonomy for the Muslim-majority areas.
10. General Elections of 1946: In the year 1946, last general elections were held of pre-independence. Major political parties of India contested the election. The Indian Congress won massively in the general elections of constituencies. The Muslim also won in Muslim constituencies.
11. Cabinet Mission: A three-member Cabinet Mission arrived in India in March 1946.
Time line:
1905 Partitioned on Bengal
1906 Muslim League was formed
1916 Lucknow Pact was signed
22 Dec, 1939 Deliverance day, observed by Muslim League
16 Aug, 1946 Direct Action Day plan celebrated by Muslim League
3 June, 1947 Declaration and acceptance
15 August, 1947 India became free and emerged as an independent nation.

The post Understanding Partition Politics, Memories, Experiences – CBSE Notes for Class 12 History appeared first on Learn CBSE.

UK Board 10th Result 2019: The Uttarakhand Board of School Education (UBSE) will announce the Uttarakhand Board Result 2019 for Class 10 and 12 examination Results by the last week of May. UBSE released results for class 10 today i.e 30 May 2019 after the press conference at 10: 30 AM. The UBSE will also announce the Uttarakhand Class 10 Board Result 2019 and Uttarakhand Class 12 Students will be able to check their respective results from the official websites ubse.uk.gov.in and uaresults.nic.in,ukresults.nic.in..

Direct Link to Check UK Board Class 10 Results 2019

State Board Results

The general trend of last few years and even academic experts have backed the reports suggesting that UBSE will declare the UK Board Result 2019 as per their academic calendar of the Board. After the declaration, a live link to official websites for the Uttarakhand Board Result 2019 will also be available on this page

The result will also be published on the official website: ubse.uk.gov.in

UK Board Class 10 Results Latest Updates.

30th May 2019: UK board 10th result to be released today at 10:30 Am. Stay tuned for more updates.

UK Board Results 2019

28th May 2019: As per the latest update on the official website of UBSE the Uttarakhand Board SSLC and HS result will be released on the official website on 30th May 2019 by 10:30 am. Click Here to read the official notification on the Uk board result date 2019.

22nd May 2019: UBSE has officially released the Uttarakhand Board 10th and 12th result date. As per the official notification, the UBSE result will be released on 30th May 2019 on the official website.

17th May 2019: According to the media reports the Uttarakhand Board results will be released by 31st May 2019. It is expected that the board will be releasing the UK board result through a press conference.

11th May 2019: According to NDTV, UK Board 10th Result will be announced by last week of May 2019. Uttarakhand Board Secretary Nita Tiwari said that Uttarakhand Board Class 10 Result will either be released this month end or if not in the 1st week of June 2019.

Apart from this few news reports say that the UBSE 10th result will be released by 3rd week of May 2019. However, there is no official declaration of the Uttarakhand Board Result date and candidates can expect the confirmation of the dates shortly.

10th May 2019: According to Times Now News, the Uttarakhand Board of School Education confirmed the Uk Board 10th Result Date. UBS will release the UBSE 10th Board Result on 31st May 2019.

Uttarakhand Board Result 2019 Overview

The Uttarakhand Board of School Education has released the timetable for the upcoming Class 10 and Class 12 Examinations 2019. However, the board is rather tight-lipped on the date of declaration of the results. Taking a cue from the trend of the previous year, the board is expected to announce the results by the last week of May

Name of the Exam	Uttarakhand Class 12 Board Exam
Conducting Body	Uttarakhand Board of School Education (UBSE)
Exam Mode	Offline
Exam Start Date	1st March 2019
Exam End Date	26th March 2019
AHSEC Result	Last Week of May 2019 (Expected)
Official Website	ubse.uk.gov.in

Uttarakhand Board 10th Result 2019 – How to Check?

Uttarakhand Board of School Education will be releasing the UK Board result for Class 10 in the online mode and students will be able to check their results from the official website – ubse.uk.gov.in. Let’s go through the step-by-step process to check and download your UK Board 10th result 2019

1st Step: Visit the UK Board official website – ubse.uk.gov.in

2nd Step: Click on “Results/Admit Card” from the left-hand side of the page.

3rd Step: A new page opens. Now click on the link provided to check your UK Board Class 10 result.

4th Step: Enter your Roll Number which is mentioned on the UK Board Class 10 admit card.

5th Step: The UK Board 10th Result will appear on the screen. Download the result and take printouts for future reference.

Uttarakhand Board 10th Result – Previous Year Statistics

Every year, a large number of students participate in the UBSE 10th Main/final Examination of UK Board. All students can check their Uttarakhand 10th timetable 2019 on the official website. It is generally being seen that the students go through tremendous pressure and stress during results time. The only way to ease the tension is by looking at the key facts and figures from the last year’s result. This will help the students to set realistic expectations from the results. For your benefit, we have provided the same below.

The Board had conducted the Class 10 Examinations from March 6 to March 24, 2018. The Uttarakhand Class 10 and Class 12 Results 2018 were released by state School Education Minister Arvind Pandey. According to reports, around 74.57 percent students were declared successful in Class 10. The girls performed better in the exams with a pass percentage at 80 percent compared to the boys at 68.96 percent.

Previous Year Statistics	2018
Total number of students appeared	146,166
Overall percentage	74.57%
The pass percentage of boys	68.96%
The pass percentage of girls	80%

Details Mentioned on Uttarakhand Result 2019

Uttarakhand class 10 result will contain the following details mentioned below

Roll Number
Student’s Name
Date of Birth
Mother’s Name
Father’s Name
School Details
Subjects Details
Marks Obtained in Each Subject
Total Marks Obtained
Percentage
Qualifying Status

Uttarakhand Board 10th Result 2019 – Re-evaluation/Rechecking

Students who are unsatisfied with their scores obtained in Uttarakhand Board 10th result 2019 will have the option to get their answer scripts rechecked/re-evaluated. Students will be able to avail this facility by filling in the requisite forms and paying the fee prescribed by the board. Any changes in the final mark tally will be updated in the mark sheet of the students.

Uttarakhand Board 10th Result 2019 – Compartmental Exam

In case the students have flunked in any of the examination paper, they should not feel dejected. The students will have another opportunity to improve their scores by appearing in the compartment examination. To apply for the compartment examination, they should contact their respective schools and fill out the prescribed forms.

About Uttarakhand Board of School Education

The Uttarakhand Board of School Education (UBSE) was established in the year 2001. It was established to regulate the secondary level education in the state. Before the formulation of the Uttarakhand Board of School Education (UBSE), the board level examinations were conducted under Uttaranchal Shiksha Evam Pariksha Parishad which was later renamed Uttarakhand Board of School Education (UBSE) in the year 2008. The board has made tremendous growth in the very short span of time and currently, has more than 10000 affiliated schools and every year 3 lakh plus students participate in the UK Board 10th Examination and UK Board 12th Examination.

We hope the detailed article on UK Board 10th Result 2019 is helpful. If you have any doubt regarding this article or UK Board 10th Result 2019, drop your comments in the comment section below and we will get back to you as soon as possible.

The post UK Board 10th Result 2019 Released Today | Check UBSE Board 10th Result 2019 @ ubse.uk.gov.in appeared first on Learn CBSE.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1
Question 1.
Find the range of heights of any ten students of your class.
Solution:
Let us have the heights of 10 students are as follows:
140 cm, 141.5 cm, 138 cm, 150 cm, 161 cm,
138 cm, 140.5 cm, 135.5 cm, 160 cm, 158 cm
Here, minimum height = 135.5 cm
Maximum height =161 cm
∴ Range = Maximum height – Minimum height
= 161 cm — 135.5 cm = 25.5 cm
Hence, the required range = 25.5 cm.

Question 2.
Organise the following marks in a class assessment in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Solution:
Let us form a frequency distribution table:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling 1

(i) 9 is the highest marks.
(ii) 1 is the lowest marks.
(iii) Range = Max. marks – Min. marks
= 9 – 1 = 8
(iv) Arithmetic mean = $\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{100}{20}=5$

Question 3.
Find the mean of first five whole numbers.
Solution:
First 5 whole numbers are 0, 1, 2, 3, 4
∴ Mean $=\frac{0+1+2+3+4}{5}=\frac{10}{5}=2$
Hence, the required mean = 2.

Question 4.
A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100
Find the mean score.
Solution:
Following are the scores of the runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100
∴ Mean = $=\frac{\text { Sum of all runs }}{\text { Number of innings }}$
$\begin{aligned} &=\frac{58+76+40+35+46+45+0+100}{8} \\ &=\frac{400}{8}=50 \end{aligned}$
Hence, the required mean = 50.

Question 5.
Following table shows the points of each player scored in four games:

Player	Game 1	Game 2	Game 3	Game 4
A	14	16	10	10
B	0	8	6	4
C	8	11	Did not play	13

Now answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Solution:
(i) Number of points scored by A in all games are
Game 1 – 14, Game 2 = 16, Game 3 = 10, Game 4 = 10
∴ Average score = $=\frac{14+16+10+10}{4}$
$=\frac{50}{4}=12.5$
(ii) Since, C did not play Game 3, he played only 3 games. So, the total will be divided by 3.
(iii) Number of points scored by B in all the games are Game 1 = 0, Game 2 = 8, Game 3 = 6, Game 4 = 4
∴ Average score = $\frac{0+8+6+4}{4}=\frac{18}{4}=4.5$
(iv) Mean score of C
$=\frac{8+11+13}{3}=\frac{32}{3}=10.67$
Mean score of C = 10.67
While mean score of A = 12.5
Clearly, A is the best performer.

Question 6.
The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the
(i) highest and the lowest marks obtained by the students.
(ii) range of the marks obtained.
(iii) mean marks obtained by the group.
Solution:
Marks obtained are:
85, 76, 90, 85, 39, 48, 56, 95, 81 and 75
(i) Highest marks = 95
Lowest marks = 39
(ii) Range of the marks
= Highest marks – Lowest marks
= 95 – 39 = 56
(iii) Mean marks
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling 3

Question 7.
The enrolment in a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820
Find the mean enrolment of the school for this period.
Solution:
Mean enrolment
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling 4
Thus, the required mean = 2058.

Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day	Rainfall (in mm)
Monday	0.0
Tuesday	12.2
Wednesday	2.1
Thursday	0.0
Friday	20.5
Saturday	5.5
Sunday	1.0

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution:
(i) Maximum rainfall = 20.5 mm
Minimum rainfall = 0.0 mm
∴ Range = Maximum rainfall – Minimum rainfall
= 20.5 mm – 0.0 mm = 20.5 mm

(ii) Mean rainfall
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling 5

(iii) Number of days on which the rainfall was less than the mean rainfall = Monday, Wednesday, Thursday, Saturday, Sunday = 5 days.

Question 9.
The heights of 10 girls were measured in cm and the results are as follows:
135, 150, 139, 128, 151, 132, 146, 149, 143, 141
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(Hi) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?
Solution:
(i) Height of the tallest girl = 151 cm.
(ii) Height of the shortest girl = 128 cm.
(iii) Range = Height of tallest girl – Height of the shortest girl
= 151 cm – 128 cm = 23 cm.
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling 6
(v) Number of girls having more height than the mean height
= 150, 151, 146, 149 and 143 = 5 girls

NCERT-Solutions-For-Class-7-Maths-Data-Handling-Exercise-3.1-01

NCERT Solutions Maths Science Social English Sanskrit Hindi RD Sharma

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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.2
Question 1.
The scores in mathematics test (out of 25) of 15 students is as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Solution:
Given data:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Let us arrange the given data in increasing order
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 1
Since 20 occurs 4 times (highest)
∴ Mode = 20
n = 15 (odd)
∴ Median = $\frac{n+1}{2} \text { th term }=\frac{15+1}{2}$
= 8th term = 20
Thus, median = 20 and mode = 20
∴ Mode and median are same.

Question 2.
The runs scored in a cricket match by 11 players is as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
Find the mean, mode and median of this data. Are the three same?
Solution:
Given data:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 2
Arranging the given data in increasing order, we get

Here, 15 occurs 3 times (highest)
∴ Mode = 15
n = 11 (odd)
∴ Median = $\left(\frac{11+1}{2}\right)^{t h}$ term = 6th term = 15
Thus mean = 39, mode = 15 and median = 15
No, they are not same.

Question 3.
The weights (in kg) of 15 students of a class are:
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Solution:
Given data: 38, 42, 35, 37,45, 50, 32,43, 43,40, 36, 38, 43, 38, 47
Arranging in increasing order, we get
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 4
(i) Here, 38 and 42 occur 3 times (highest)
Thus mode = 38 and 43
n = 15(odd)
Median = $\left(\frac{n+1}{2}\right)^{\mathrm{th}} \text { term }=\left(\frac{15+1}{2}\right)^{\mathrm{th}}$ term
= 8th term = 40
Thus mode 38 and 43 and median = 40
(ii) Yes, the given data has two modes i.e. 38 and 43.

Question 4.
Find the mode and median of the data:
13, 16, 12, 14, 19, 12, 14, 13, 14
Solution:
Arranging the given data in increasing order, we get
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 5
Here, 14 occur 3 times (highest)
Thus, mode = 14
n = 9(odd)
∴ Median = $\left(\frac{n+1}{2}\right)^{\text { th }} \text { term }=\left(\frac{9+1}{2}\right)^{\text { th }}$ term
= 5th term = 14
Hence, mode = 14 and median = 14.

Question 5.
Tell whether the statement is true or false.
(i) The mode is always one of the number in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
Solution:
(i) True
(ii) False
(iii) True
(iv) False

NCERT-Solutions-For-Class-7-Maths-Data-Handling-Exercise-3.2-01

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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.3
Question 1.
Use the bar graph to answer the following questions.
(a) Which is the most popular pet?
(b) How many students have dog as a pet?
Solution:
From the given bar graph in figure, we have
(a) Cats are the most popular pet among the students.
(b) 8 students have dog as a pet animal.
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 1

Question 2.
Read the bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:
(i) About how many books were sold in 1989, 1990, 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which year were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 2
Solution:
From the given bar graph, we have
(i) Number of books sold in the year 1989 is about 180, in 1990 is about 490 and in 1992 is about 295.
(ii) About 475 books were sold in 1990. About 225 books were sold in the year 1992.
(iii) Fewer than 250 books were sold in the years 1989 and 1992.
(iv) On y-axis, the line is divided into 10 small parts of 10 books each. So, we can estimate the number of books sold in 1989 is about 180.

Question 3.
Number of children in six different classes are given below. Represent the data on a bar graph.

Class	Number of children
Fifth	135
Sixth	120
Seventh	95
Eighth	100
Ninth	90
Tenth	80

(a) How would you choose a scale?
(b) Answer the following questions:

Which class has the maximum number of children? And the minimum?
Find the ratio of students of class sixth to £he students of class eighth.

Solution:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 3
(a) Scale on y-axis is 1 cm = 10 students
(b)

Fifth class has the maximum number of children i.e., 135.
Tenth class has the minimum number of children i.e., 80.
Number of children in class eight = 100
∴ Ratio of class sixth to the students of class

Question 4.
The performance of a student in 1st term and 2nd term is given. Draw a double bar graph choosing appropriate scale and answer the following:

Subject	1st term (M.M. 100)	2nd term (M.M. 100)
English	67	70
Hindi	72	65
Math	88	95
Science	81	85
S. Science	73	75

(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Solution:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 5
(i) In Math, the performance of the students improved the most.
(ii) In social science, the performance of the students improved the least.
(iii) Yes, in Hindi the performance of the students has gone down.

Question 5.
Consider this data collected from survey of a colony.

Favourite sport	Watching	Participating
Cricket	1240	620
Basket ball	470	320
Swimming	510	320
Hockey	430	250
Athletics	250	105

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Solution:
(i) The above bar graph depicts the number of people who are watching and who are participating in sports.
(ii) Cricket is the most popular sport.
(iii) Watching the sports is more preferred by the people.
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 6

Question 6.
Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this chapter. Plot a double bar graph using the data and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and maximum temperature.

Temperature of cities as on 20.6.2006
City	Max.	Min.
Ahmedabad	38°C	29°C
Amritsar	37°C	26°C
Bengaluru	28°C	21°C
Chennai	36°C	27°C
Delhi	38°C	28°C
Jaipur	39°C	29°C
Jammu	41°C	26°C
Mumbai	32°C	27°C

Solution:
Double bar graph:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 7
(i) Jammu has the largest difference between the maximum and minimum temperature i.e. 41°C – 26°C = 15°C
(ii) Hottest city is Jammu with 41°C temperature and coldest city is Bengaluru with 21°C temperature.
(iii)

Bengaluru having its maximum temperature 28°C is less than the minimum temperature 29°C in Ahmedabad.
Bengaluru having its maximum tem-perature 28°C is less than the maximum temperature 29°C in Jaipur.

(iv) Mumbai has the least difference between its minimum and maximum temperatures i.e. 32°C – 27°C = 5°C

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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.4
Question 1.
Tell whether the following situations are certain to happen, impossible to happen, can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A dice when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
Solution:

Event	Chance
(i) You are older today than yesterday.	Certain to happen.
(ii) A tossed coin will land heads up.	Can happen but not certain.
(iii) A dice when tossed shall land up 8 on the top.	Impossible.
(iv) The next traffic light seem will be green.	Can happen but not certain.
(v) Tomorrow will be cloudy day.	Can happen but not certain.

Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with number 5?
Solution:
(i) Total number of marbles marked with the number from 1 to 6 = 6
∴ n(S) = 6
Number of marble marked with 2=1
∴ n(E) = 1
∴ Required probability = $\frac{n(\mathrm{E})}{n(\mathrm{S})}=\frac{1}{6}$

(ii) Number of marble marked with 5 = 1
∴ n(E) = 1
∴ Required probability = $\frac{n(\mathrm{E})}{n(\mathrm{S})}=\frac{1}{6}$

Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:
Coin has 2 faces—Head (H) and Tail (T)
∴ Sample space S(n) = 2
Number of successful event n(E) = 1
∴ Required probability = $\frac{n(\mathrm{E})}{n(\mathrm{S})}=\frac{1}{2}$

Sample Space: The sample space of an experiment is the number of all possible outcomes of that experiment.

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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1
Question 1.
Complete the given column of the table:

Solution:

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19; (n = 1)
(b) 7n + 5 = 19; in – -2)
(c) 7n + 5 = 19; (n = 2)
(d) 4p – 3 = 13; (p = 1)
(e) 4p – 3 = 13; (p = -4)
(f) 4p-3 = 13; (p = 0)
Solution:
(a) n + 5 = 19 (n = 1)
Put n = 1 in LHS
1 + 5 = 6 ≠ 19 (RHS)
Since LHS ≠ RHS
Thus n = 1 is not the solution of the given equation.

(b) 7n + 5 = 19; (n = -2)
Put n = – 2 in LHS
7 × (-2) + 5 = -14 + 5 = -9 ≠ 19 (RHS)
Since LHS ≠ RHS
Thus, n = -2 is not the solution of the given equation.

(c) 7n+ 5 = 19; (n = 2)
Put n = 2 in LHS
7 × 2 + 5 = 14 + 5 = 19 = 19 (RHS)
Since LHS = RHS
Thus, n – 2 is the solution of the given equation.

(d) 4p – 3 = 13; (p = 1)
Put p = 1 in LHS
4 × 1 – 3 = 4 – 3 = 1 ≠ 13 (RHS)
Since LHS ≠ RHS
Thus, p = 1 is not the solution of the given equation.

(e) 4p – 3 = 13; (p = -A)
Put p = -4 in LHS
4 × (-4) – 3 = -16 – 3 = -19 ≠ 13 (RHS)
Since LHS ≠ RHS
Thus p = -4 is not the solution of the given equation.

(f) 4p – 3 = 13; (p = 0)
Put p = 0 in LHS
4 × (0) – 3 = 0 – 3 = -3 ≠ 13 (RHS)
Since LHS ≠ RHS
Thus p – 0 is not the solution of the given equation.

Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Solution:
(i) 5p + 2 = 17
For p = 1, LHS
= 5 × 1 + 2 = 5 + 2 = 7 ≠ 17 (RHS)
For p = 2, LHS = 5 × 2 + 2
= 10 + 2 = 12 ≠ 17 (RHS)
For p = 3, LHS = 5 × 3 + 2
= 15 + 2 = 17 = 17 (RHS)
Since the given equation is satisfied for p = 3 Thus, p = 3 is the required solution.

(ii) 3m – 14 = 4
For m = 1, LHS = 3 × 1 – 14
= 3 – 14 = -11 ≠ 4 (RHS)
For m = 2, LHS = 3 × 2 – 14 = 6 – 14
= -8 ≠ 4 (RHS)
For m = 3, LHS = 3 × 3 – 14 = 9 – 14
= -5 ≠ 4 (RHS)
Form m = 4, LHS = 3 × 4 – 14
= 12 – 14 = -2 ≠ 4 (RHS)
For m = 5, LHS = 3 × 5 – 14
= 15 – 14 = -1 ≠ 4 (RHS)
For m = 6, LHS = 3 × 6 – 14
= 18 – 14 = 4 (=) 4 (RHS) .
Since, the given equation is satisfied for m = 6.
Thus, m = 6 is the required solution.

Question 4.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 4

Question 5.
Write the following equations in statement forms.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 7
Solution:

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 1)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Solution:
(i) Let m be the Parmit’s marbles.
∴ Irfan’s marble = 5m + 7
Total number of Irfan’s marble is given by 37.
Thus, the required equation is 5m + 7 = 37

(ii) Let Laxmi’s age bey years.
∴ Laxmi’s father’s age = 3y + 4
But the Laxmi’s father age is given by 49
Thus the required equation is 3y + 4 = 49

(iii) Let the lowest score be l.
∴ The highest score = 2l + 1
But the highest score is given by 87.
Thus, the required equation is 2l + 1 = 87

(iv) Let each base angle be ‘b’ degrees.
∴ Vertex angle of the triangle = 2b
Sum of the angles of a triangle = 180°
∴ Required equation is b + b + 2b = 180° or 4b = 180°

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1

NCERT CBSE Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT Solutions for Class 7 Chapter 4 Simple Equations Exercise 4.1
NCERT Solutions for Maths Chapter 4 Simple Equations Exercise 4.1
NCERT Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.1

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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2
Question 1.
Given first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = -7
(f) y -4 = 4
(g) y + 4 = 4
(h) y + 4 = -4
Solution:
(a) x – 1 = 0
Adding 1 to both sides, we get
x – 1 + l = 0 + 1 ⇒ x = 1
Thus, x = 1 is the required solutions.
Check: Put x = 1 in the given equations
x – 1 = 0
1 – 1 = 0
0 = 0
LHS = RHS
Thus x = 1 is the correct solution.

(b) x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = 0 – 1 ⇒ x = -1
Thus x = -1 is the required solution.
Check: Put x = -1 in the given equation
-1 + 1 = 0
0 = 0
LHS = RHS
Thus x = -1 is the correct solution.

(c) x – 1 = 5
Adding 1 to both sides, we get
x – 1 + 1 = 5 + 1 ⇒ x = 6
Thus x = 6 is the required solution.
Check: x – 1 = 5
Putting x = 6 in the given equation
6 – 1 = 5 ⇒ 5 = 5
LHS = RHS
Thus, x = 6 is the correct solution.

(d) x + 6 = 2
Subtracting 6 from both sides, we get
x + 6 – 6 = 2 – 6 ⇒ x = -4
Thus, x = -4 is the required solution.
Check: x + 6 = 2
Putting x = -4, we get
-4 + 6 = 2 ⇒ 2 = 2 LHS = RHS
Thus x = -4 is the correct solution.

(e) y – 4 = -7
Adding 4 to both sides, we get
y – 4 + 4 = -7 + 4 ⇒ y = -3
Thus, y = -3 is the required solution.
Check: y – 4 = -7
Putting y = -3, we get
-3 – 4 = -7 ⇒ -7 = -7
LHS = RHS
Thus, y = -3 is the correct solution.

(f) y – 4 = 4
Adding 4 to both sides, we get
y – 4 + 4 = 4 + 4 ⇒ y = 8
Thus, y = 8 is the required solution.
Check: y – 4 = 4
Putting y = 8, we get
8 – 4 = 4 ⇒ 4 = 4
LHS = RHS
Thus y = 8 is the correct solution.

(g) y + 4 = 4
Subtracting 4 from both sides, we get
y + 4 – 4 = 4 – 4 ⇒ y = 0
Thus y = 0 is the required solution.
Check: y + 4 = 4
Putting y = 0, we get
0 + 4 = 4 ⇒ 4 = 4
LHS = RHS
Thus y = 0 is the correct solution.

(h) y + 4 =-4
Subtracting 4 from both sides, we get
y + 4 – 4 = -4 – 4 ⇒ y = -8
Thus, y = -8 is the required solution.
Check: y + 4 = -4
Putting y = -8, we get
-8 + 4 = -4 ⇒ -4 = -4
LHS = RHS
Thus, y = -8 is the correct solution.

Question 2.
Give first the step you will use to separate the variable and then solve the following equation:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 1
Solution:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 4

Question 3.
Give the steps you will use to separate the variables and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) $\frac{20 p}{3}=40$
(d) $\frac{3 p}{10}=6$
Solution:
(a) 3n – 2 = 46
⇒ 3n- 2 + 2 = 46+ 2 (adding 2 to both sides)
⇒ 3n = 48
⇒ 3n + 3 = 48 ÷ 3
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 5

(b) 5m + 7 = 17
⇒ 5m+ 7 – 7 = 17 – 7 (Subtracting 7 from both sides)
⇒ 5m = 10
⇒ 5m + 5 = 10 ÷ 5 (Dividing both sides by 5)
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 6

(d) $\frac{3 p}{10}=6$
⇒ $\frac{3 p}{10} \times 10=6 \times 10$ (Multiplying both sides by 10)
⇒ 3p = 60
⇒ 3p ÷ 3 = 60 ÷ 3 (Dividing both sides by 3)
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 8

Question 4.
Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) $\frac{p}{4}=5$
(d) $\frac{-p}{3}=5$
(e) $\frac{3 p}{4}=6$
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:
(a) 10p = 100
⇒ 10p ÷ 10 = 100 ÷ 10 (Dividing both sides by 10)
$p=\frac{100}{10}=10$
Thus p= 10

(b) 10p + 10 = 100
⇒ 10p + 10 – 10 = 100 -10 (Subtracting 10 from both sides)
⇒ 10p = 90
⇒ 10p ÷ 10 = 90 ÷ 10 (Dividing both side by 10)
$p=\frac{90}{10}=9$
Thus p = 9

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 9

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 10

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 11

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 14

(l) 2q + 6 = 12
⇒ 2q + 6 – 6 = 12 – 6 ( Subtracting 6 from both sides)
⇒ 2q = 6
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 13

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2
NCERT Solutions for Class 7 CBSE Maths Chapter 4 Simple Equations Exercise 4.2
NCERT CBSE Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2

NCERT Solutions for Class 7 Chapter 4 Simple Equations Exercise 4.2

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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3
Question 1.
Solve the following equations:

Solution:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 6

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 7

Question 2.
Solve the following equations:
(а) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = -21
(d) -4(2 + x) = 8
(e) 4(2 – x) = 8
Solution:
(a) 2(x + 4) = 12
⇒ $\frac{2(x+4)}{2}=\frac{12}{2}$ (Dividing both sides by 2)
⇒ x + 4 = 6
⇒ x = 6 – 4 (Transposing 4 to RHS)
⇒ x – 2
Check: Put x = 2 in LHS
2(2 + 4) = 2 × 6 = 12 RHS as required

(b) 3(n – 5) = 21
⇒ $\frac{3(n-5)}{3}=\frac{21}{3}$ (Dividing both sides by 3)
⇒ n – 5 = 7
⇒ n = 7 + 5 (Transposing 5 to RHS)
n = 12
Check: Put n = 12 in LHS
3(12 – 5) = 3 × 7 = 21 RHS as required

(c) 3(n – 5) = -21
⇒ $\frac{3(n-5)}{3}=\frac{-21}{3}$ (Dividing both sides by 3)
⇒ n – 5 = -7
⇒ n = -7 + 5 (Transposing 5 to RHS)
⇒ n = -2
Check: Put n = -2 in LHS
3(-2 – 5) = 3 × -7
= -21 RHS as required

(d) -4(2 + x) = 8
⇒ $\frac{-4(2+x)}{-4}=\frac{8}{-4}$ (Dividing both sides by -4)
⇒ 2 + x = -2
⇒ x = -2 – 2 (Transposing 2 to RHS)
⇒ x = —4
Check: Put x = -4 in LHS
-4(2 – 4) = -4 × -2 = 8 RHS as required

(e) 4(2-x) = 8
⇒ $\frac{4(2-x)}{4}=\frac{8}{4}$ (Dividing both sides by 4)
⇒ 2 – x = 2 – 2 (Transposing 2 to RHS)
⇒ -x = 0
∴ x = 0 (Multiplying both sides by -1)
Check: Put x = 0 in LHS
4(2 – 0) = 4 × 2 = 8 RHS as required

Question 3.
Solve the following equations:
(a) 4 = 5(p- 2)
(b) -4 = 5(p – 2)
(c) 16 = 4 + 3 (t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)
Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 8

(c) 16 = 4 + 3 (t + 2)
⇒ 16 – 4 = 3(t + 2) (Transposing 4 to LHS)
⇒ 12 = 3 (t + 2)
⇒ $\frac{12}{3}=\frac{3(t+2)}{3}$ (Dividing both sides by 3)
⇒ 4 = t + 2
⇒ 4 – 2 = t (Transposing 2 to LHS)
⇒ 2 = t or t = 2
Check: Put t = 2 in RHS
4 + 3(2 + 2) = 4 + 3 × 4 = 4 + 12
= 16 LHS as required

(d) 4 + 5(p – 1) = 34
⇒ 5(p – 1) = 34 – 4(Transposing 4 to RHS)
⇒ 5(p – 1) = 30
⇒ $\frac{5(p-1)}{5}=\frac{30}{5}$ (Dividing both sides by 5)
⇒ p – 1 = 6
⇒ P = 7
Check: Put p = 7 in LHS
4 + 5(7 – 1) = 4 + 5 × 6
= 4 + 30 = 34 RHS as required

(e) 0 = 16 + 4(m – 6)
⇒ 0 — 16 = 4(m – 6) (Transposing 16 to LHS)
⇒ -16 = 4(m – 6)
⇒ $-\frac{16}{4}=\frac{4(m-6)}{4}$ (Dividing both sides by 4)
⇒ -4 = m – 6
⇒ -4 + 6 = m (Transposing 6 to LHS)
⇒ 2 = m
or m = 2
Check: Put m = 2 in RHS
16 + 4(2 – 6) = 16 + 4 × (-4) = 16 – 16 = 0 LHS as required

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x – -2.
Solution:
(a) Possible equations are:
10x + 2 = 22; $\frac{x}{5}=\frac{2}{5}$ ; 5x – 3 = 7
(b) Possible equations are:
3x= -6; 3x + 7 = 1; 3x + 10 = 4

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3
NCERT CBSE Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3

NCERT Solutions for Class 7 Chapter 4 Simple Equations Exercise 4.3

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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4
Question 1.
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from $\frac{5}{2}$ of the numbers, the result is 23.
Solution:
(a) Let the required number be x.
Step I: 8x + 4
Step II: 8x + 4 = 60 is the required equation
Solving the equation, we have
8x + 4 = 60
⇒ 8x = 60 – 4 (Transposing 4 to RHS)
⇒ 8x = 56
⇒ $\frac{8 x}{8}=\frac{56}{8}$ (Dividing both sides by 8)
⇒ x = 7
Thus, x – 7 is the required unknown number.

(b) Let the required number be x.
Step I: $\frac{1}{5}$ x – 4
Step II: $\frac{1}{5}$ x – 4 = 3 is the required equation. 5
Solving the equation, we get
$\frac{1}{5}$ x – 4 = 3
⇒ $\frac{1}{5}$ x =4 + 3 (Transposing 4 to RHS)
⇒ $\frac{1}{5} x=7$
⇒ $\frac{1}{5} x \times 5=7 \times 5$ (Multiplying both sides by 5)
⇒ x = 35 is the required unknown number,

(c) Let the required number be x.
Step I: $\frac{3}{4}$ x + 3
Step II: $\frac{3}{4}$ x + 3 = 21 is the required equation.
Solving the equation, we have
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 1
⇒ x = 24 is the required unknown number.

(d) Let the required unknown number be x.
Step I: 2x – 11
Step II: 2x -11 = 15 is the required equations.
Solving the equation, we have
2x – 11= 15
⇒ 2x = 15 + 13 (Transposing 11 to RHS)
⇒ 2x = 28
⇒ $\frac{2 x}{2}=\frac{28}{2}$ (Dividing both sides by 2)
⇒ x = 14 is the required unknown number,

(e) Let the required number be x.
Step I: 50 – 3x
Step II: 50 – 3x = 8 is the required equations.
Solving the equation, we have
50 – 3x = 8
⇒ -3x = 8 – 50 (Transposing 50 to RHS)
⇒ -3x = -42
⇒ $\frac{-3 x}{-3}=\frac{-42}{-3}$ (Dividing both sides by -3)
⇒ x = 14 is the required unknown number.

(f) Let the required number be x.
Step I: x + 19
Step II: $\frac{x+19}{5}$
Step III: $\frac{x+19}{5}$ = 8 is the required equation.
Solving the equation, we have
$\frac{x+19}{5}$ = 8
⇒ $\frac{x+19}{5}$ × 5 = 8 × 5(Multiplying both sides by 5)
⇒ x + 19 = 40
⇒ x = 40 – 19 (Transposing 19 to RHS)
∴ x = 21 is the required unknown number.

(g) Let the required number be x.
Step I: $\frac{5}{2}$ x – 7
Step II: $\frac{5}{5}$ – 7 = 23 is the required equation.
Solving the equation, we have
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 3
⇒ x = 12 is the required unknown number.

Question 2.
Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angle are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°?)
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:
(a) Let the lowest score be x.
Step I: Highest marks obtained = 2x + 7
Step II: 2x + 7 = 87 is the required equation. Solving the equation, we have
2x + 7 = 87
⇒ 2x = 87 – 7 (Transposing 7 to RHS)
⇒ 2x = 80
⇒ $\frac{2 x}{2}=\frac{80}{2}$ (Dividing both sides by 2)
⇒ x = 40 is the required lowest marks.

(b) Let each base angle be x degrees.
Step I: Sum of all angles of the triangle (x + x + 40) degrees.
Step II: x + x + 40 = 180°
⇒ 2x + 40° = 180°
Solving the equation, we have
2x + 40° = 180°
2x = 180° – 40° (Transposing 40° to RHS)
2x = 140°
⇒ $\frac{2 x}{2}=\frac{140^{\circ}}{2}$ (Dividing both sides by 2)
⇒ x = 70°
Thus the required each base angle = 70°

(c) Let the runs scored by Rahul = x
Runs scored by Sachin = 2x
Step I: x + 2x = 3x
Step II: 3x + 2 = 200
Solving the equation, we have
3x + 2 = 200
⇒ 3x = 200 – 2 (Transposing 2 to RHS)
⇒ 3x = 198
⇒ $\frac{3 x}{3}=\frac{198}{3}$ (Dividing both sides by 3)
⇒ x = 66
Thus, the runs scored by Rahul is 66 and the runs scored by Sachin = 2 × 66 = 132

Question 3.
Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(iii) People of Sundargram planted trees in a village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Solution:
(i) Let the number of marbles with Parmit be
Step I: Number of marbles that Irfan has = 5x + 7
Step II: 5x + 7 = 37 Solving the equation, we have 5x + 7 = 37
⇒ 5x = 37 – 7 (Transposing 7 to RHS)
⇒ 5x = 30
⇒ $\frac{5 x}{5}=\frac{30}{5}$ (Dividing both sides by 5)
⇒ x = 6
Thus, the required number of marbles = 6.

(ii) Let Laxmi’s age be x years.
Step I: Father’s age = 3x + 4
Step II: 3x + 4 = 49
Solving the equation, we get
3x + 4-= 49
⇒ 3x = 49 – 4 (Transposing to RHS)
⇒ 3x = 45
⇒ $\frac{3 x}{3}=\frac{45}{3}$ (Dividing both sides by 3)
⇒ x = 15
Thus, the age of Laxmi = 15 years

(iii) Let the number of planted fruit tree be x.
Step I: Number of non-fruit trees = 3x + 2
Step II: 3x + 2 = 77
Solving the equation, we have
3x + 2 = 77
⇒ 3x = 77 – 2 (Transposing 2 to RHS)
⇒ 3x = 75
⇒ $\frac{3 x}{3}=\frac{75}{3}$ (Dividing both sides by 3)
⇒ x = 25
Thus, the required number of fruit tree planted = 25

Question 4.
Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Solution:
Suppose my identity number is x.
Step I: 7 + 50
Step II: 7x + 50 + 40 = 300
or 7x + 90 = 300
Solving the equation, we have
7x + 90 = 300
⇒ 7x = 300 – 90 (Transforming 90 to RHS)
⇒ 7x = 210
⇒ $\frac{7 x}{7}=\frac{210}{7}$ (Dividing both sides by 7)
⇒ x = 30
Thus, my identity is 30.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4

NCERT CBSE Solutions for Class 7 Simple Equations Exercise 4.4

NCERT Solutions Maths Science Social English Sanskrit Hindi RD Sharma

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Character Sketch of Eppie in Silas Marner

Character Sketch of Eppie: Eppie is Silas Marner’s adopted daughter who fills the void left in his heart after the betrayal of his closest friend and, later, the theft of his gold. She is the biological daughter of Molly Farren and Godfrey Cass, but raised as Silas Marner’s daughter. She enters Silas’ life when she follows a bright light to the door of his cottage and goes to sleep straight in front of the fireplace. Silas and the people of Raveloe think she has been sent to Silas from above. Silas becomes possessive of her and wishes to learn to do everything himself, so that the little girl will be attached to him from the start.

Young Eppie is endlessly curious and demanding. Her desires are infectious, and as she hungrily explores the world around her, so does Silas. By the time Eppie is three, she shows signs of mischievousness, escaping from the cottage and going missing for a while, though she is soon found. When she is punished for her act by being locked in the coal-hole, she takes a liking to the place and, as a result, is reared without punishment. She becomes an object of fascination and affection among the people of Raveloe.

She is Silas Marner’s adopted daughter who fills the void left in his heart after the betrayal of his closest friend and, later, the theft of his gold. She is the child of Godfrey Cass and his secret wife, Molly Farren. Eppie is beautiful and lively and loves Silas unconditionally. When, towards the end of the narrative, she discovers her real father, she chooses to stay with Silas for he is the only family she has known since she was two years old.

She is the biological daughter of Molly Farren and Godfrey Cass, but raised as Silas Marner’s adopted daughter. She brings Silas closer to the other villagers. Unlike his gold, which exacerbated his isolation and did not respond to his attentions, young Eppie is endlessly curious and demanding. Her desires are infectious, and as she hungrily explores the world around her, so does Silas. Eppie tempts Silas away from his work to play outside.

By the time Eppie is three, she shows signs of mischievousness and escapes from the cottage and goes missing for a while, though she is soon found. When punished for her act by being locked in the coal-hole, she enjoys it and thus Silas rears her without punishment.

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Character Sketch of Dunstan Cass in Silas Marner

Character Sketch of Dunstan Cass: The Squire’s younger son, Dunstan Cass, more commonly called by the nickname Dunsey, is a sneering and unpleasant young man with a taste for gambling and drinking. He is a reckless and manipulative man who will do anything or say anything to get what he wants. He is attracted to greed and wealth, and has no conscience whatsoever.

Dunstan Cass is shown as a contrast to Godfrey. While Godfrey is merely weak, Dunstan Cass is literally a bad man. He is yainraffogant, and selfish as well as dishonest. Like Godfrey, he is primarily interested in what he himself wants, but he lacks any virtues. He refuses to pay the debt money, tells Godfrey to arrange the money on his own and blackmails him with the secret of Godfrey’s marriage to the opium addict Molly Farren. He even suggests and offers to sell Godfrey’s prized horse, Wildfire. However, he accidentally kills the horse while jumping a fence at the hunt.

Dunstan Cass is a cruel, lazy, manipulative and greedy person, ready to do or say anything to get what he wants. He blackmails his brother, Godfrey, by threatening to spill the beans about the latter’s secret marriage. He convinces his brother to sell off his favourite horse, Wildfire. Dunstan Cass strikes a good bargain for the animal with Bryce, but his foolish greed gets the animal killed. While returning home, he stops by at Silas’s cottage and steals his money. He disappears after the act and his dead body is found years later when Stone-pits is drained.

He steals poor Silas Marner’s money. He is thought to have disappeared somewhere, but his dead body’s skeleton is found in the stone-pits when they were drained after sixteen years. Dunsey is found with Silas’ money. He, thus, serves as a contrast to Godfrey, as a means of relieving Silas of his gold, and as a reminder to Godfrey that truth will eventually reveal itself.

Dunstan Cass, the squire’s younger son, is an evil man with a taste for gambling and drinking. He is a reckless and manipulative man who will do anything or say anything to get what he wants. He is attracted to greed and wealth, and has no conscience whatsoever. He refuses to pay the debt money and tells Godfrey to arrange the money on his own. simultaneously blackmailing him with the secret of Godfrey’s marriage to the drunken opium addict Molly Farren. He even suggests and offers to sell Godfrey’s prized horse, Wildfire. However, he inadvertently kills the horse by jumping one fence too many at the hunt. Later on. he steals Silas Marner’s gold, but drowns in the stone pit by falling into it in the darkness. Thus he meets a just end.

Godfrey, is looked upon by the people of Raveloe as a decent, serious and potentially prosperous young man. Dunstan on the other hand, is a sneering and unpleasant young man with a taste for gambling and drinking. He is a reckless and manipulative man who gets Godfrey to do just about anything he wants, through whatever means are convenient.

Dunstan is a direct contrast to Godfrey. Where Godfrey is merely weak, Dunstan is completely evil He is vain, arrogant and selfish, as well as dishonest. Godfrey is weak-willed and usually unable to think much beyond his immediate material comfort.

In other words, Godfrey was no different than his brother in terms of immorality; he had indeed married the low-life woman, he was in debt, he was drinking more and more in the company of Dunstan, and he was too weak of character to face up to his mistakes.

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Character Sketch of Dolly Winthrop in Silas Marner

Character Sketch of Dolly Winthrop: Content, simple, and not extremely academic, we find in the character of Dolly Winthrop the quintessential example of the classical country woman: the matron of all, who is always there to provide comfort and advice when most needed. Known for being the wheelwright’s wife, and for her famous lard cakes, Dolly epitomises comfort, devotion and joy. She has an instinctive faith that contrasts with Silas’ initial distrust of God. She represents the best of Raveloe, the community spirit, and has real interest and concern for others. It is ultimately she who will serve as the backbone of support in Silas’ raising of Eppie. She is a fervent believer in the customs and traditions of Raveloe, especially the religious aspect of celebrations.

Dolly Winthrop and her lard cakes come to the rescue to try and cheer up the very mortified Silas, who is under a deep depression that arises as a result of having had all of his gold stolen. Aside from the comfort food, Dolly Winthrop brings with her some advice: Silas should start going to church because it will do him more good than he thinks. When Sills starts caring for Eppie, Dolly advises him how to care for a child. She is particularly helpful, offering advice, giving him clothing outgrown by her own child, and helping to bathe and care for the girl. She persuades Silas to have the child christened. She insists that Silas should not spoil Eppie and he should punish her either by spanking her or by putting her in the coal-hole to frighten her. Later, she is Eppie’s godmother and Silas’ trusted adviser on religion and life. Silas goes to seek her advice whenever he has a problem, whether it concerns Eppie’s welfare or his past. Dolly makes him realise that he should trust the world.

Dolly Winthrop is the wife of the wheelwright, Ben Winthrop, and the mother of Aaron. Dolly Winthrop takes it upon herself to help Silas raise Eppie. She persuades Silas to trust in God always. Dolly Winthrop later becomes Eppie’s godmother and mother-in-law. She is kind and patient, and is devoutly religious. She becomes Eppie’s godmother and later, her mother-in-law.

Dolly Winthrop is the most lovable character in ‘Silas Marner’ because she is a typical classical country woman, content, simple and not academically minded. She is always there to provide comfort and advice when most needed. Dolly epitomises comfort, devotion and joy. She has an instinctive faith that contrasts with Silas’ initial distrust of God. She represents the best of Raveloe, having the community sprit with real interest and concern for others. She is a fervent believer in the customs and traditions of Raveloe, especially the religious aspect of celebrations. When Silas starts caring for Eppie, Dolly Winthrop advises him how to care for a child. She is particularly helpful in helping Silas look after Eppie. Later, she is Eppie’s godmother and Silas’ trusted adviser in religion and life. Silas goes to seek her advice whenever he faces a problem, whether it concerns Eppie’s welfare or his past.

Dolly Winthrop was the wheelwright’s wife. She was “a woman of scrupulous conscience”and was eager to fulfil her duties. She had a mellow and patient character and was not at all quarrelsome. She was the quintessential example of the classical country woman, the matron of all who was selfless and was always there to provide comfort and advice when most needed.

Dolly Winthrop was content and simple. Known for her famous lard cakes, Dolly epitomised comfort, devotion and joy. She represents the best of Raveloe, the community spirit and real interest and concern for others. She is a fervent believer in the customs and traditions of Raveloe, especially the religious aspect of celebration. For Dolly, faith in God provides not only an incentive to do good work herself, but also a trust that others in the community will do their part.

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Character Sketch of Molly Farren in Silas Marner

Character Sketch of Molly Farren: Molly Farren is Godfrey’s secret wife, whom Godfrey has told that he would rather die than acknowledge as his wife. She knows there is a dance being held at the Red House and plans to crash the party in order to take revenge on Godfrey. She decided to go in her dingy rags, with her faded face and her little child and disclose herself to the Squire as his elder son’s wife.

Molly Farren is addicted to opium and knows that this, not Godfrey, is the primary reason for her troubles. However she resents Godfrey’s wealth and comfort and believes that he should support her. As she has been walking from the morning, she is tired, and to comfort herself, consumes the last portion of opium she has. The drug makes her drowsy and she becomes unconscious.

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Character Sketch of Priscilla Lammeter in Silas Marner

Character Sketch of Priscilla Lammeter : The introduction of Priscilla Lammeter uncovers the hypocrisy and pettiness of the assembled rich folk. She provides something of a feminist point of view, thanking her stars that there are pretty woman like her sister Nancy Lammeter to keep the men away from her. She is blunt in her remarks, which offends the Miss Gunns, but she does not seem to care. She continues to offer her strong opinions while they finish dressing, speaking on men and marriage. She likes to see the men mastered’ and as for marriage, she says that “Mr Have-your-own-way is the best husband.”

Although plain and certainly the uglier, Priscilla Lammeter is more worldly-wise than Nancy. She is described as ‘square-shouldered’, ‘clumsy’ and ‘high-featured’. However, Priscilla Lammeter is also skilled in many trades, is no-nonsense and actually has more common sense than Nancy in many ways.

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Character Sketch of Squire Cass in Silas Marner

Character Sketch of Squire Cass: The Squire Cass is a tall, stout man of sixty, with a face in which the knit brow and rather hard glance seemed contradicted by the slack and feeble mouth. His appearance is slovenly and disgusting, further exaggerated by his behaviour. However, he has ‘self-possession and authoritativeness of voice and carriage’ that distinguishes him from ordinary people. He speaks in a ‘ponderous coughing fashion’ and lives an idle life, failing to heed his troubles until they are impossible to ignore. He only reaches decisions in fits of anger, making violent rash decisions that he refuses to revoke even when his head has cooled.

Squire Cass has a very sharp tongue and he doesn’t mince his words. He banishes Dunstan by saying “Let him turn ostler and keep himself. He shan’t hang on me any more.”

Squire Cass is the richest man in Raveloe. The Squire is described as lazy, complacent, selfish and short-tempered. He is the father of Godfrey and Dunstan. However, he seems to care more for his money than his sons. He allows them to do whatever they please as long as they do not involve his tenants in any way.

The relationship between Squire Cass and his sons can be characterised as the relationship an administrator has with two bad, lazy, ineffective and difficult employees who also happen to be related by blood to him. Hence, it is twice more difficult to fire an employee who is related to you by family, because the drama increases and more bridges are burned in the end. It seems he has spoiteld his sons, not out of affection, but simply out of neglect. There are no courtesies in conversation, no shared mealtimes and Squire Cass does not trust either of his sons to let them administer the Red House. He treats his children more like tenants than like family, to be dismissed or ordered about or evicted. Hence, it is the relationship of master and servant, or boss and employee, between the father and his sons.

Squire Cass is acknowledged as the most important and greatest man in Raveloe. The winter feasts in Raveloe are times of great merrymaking. Although the finest of these may be at Mr Osgood’s, possibly the greatest abundance is to be found at the Red House, home of Squire Cass. The Squire’s wife is long dead, and his sons have turned out to be vagabonds. The younger son, Dunstan, more commonly called by the nickname Dunsey, is “a spiteful jeering fellow”. He is a sneering and unpleasant young man with a taste for gambling and drinking and other unsavoury activities.

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NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.1
Question 1.
Find the ratio of:
(a) ₹ 5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hours
Solution:
(a) ₹ 5 to 50 paise
Converting the given quantities into same units, we have
₹ 5 = 5 × 100 = 500 paise
∴ ₹ 5 : 50 paise
= 500 paise : 50 paise [∵ ₹ 1 = 100 paise]
= 10 : 1
So, required ratio is 10 : 1.

(b) 15 kg to 210 g
Converting the given quantities into same units, we have
15 kg = 15 × 1000
= 15000 g [∵ 1 kg = 1000 g]
∴ 15 kg : 210 g = 15000 g : 210 g
= 1500 : 21
= 500 : 7
So, the required ratio is 500 : 7.

(c) 9 m to 27 cm
Converting the given quantities into same units, we have
9 m = 9 × 100 = 900 cm
∴ 9m: 27 cm = 900 cm : 27 cm [∵ 1 m = 100 cm]
= 100 : 3
So, the required ratio is 100 : 3.

(d) 30 days to 36 hours
Converting the given quantities into same
units, we have
30 days = 30 × 24 hours [ ∵ 1 day = 24 hours]
= 720 hours
∴ 30 days : 36 hours
= 720 hours : 36 hours = 20:1
So, the required ratio is 20 : 1.

Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:
Using Unitary Method, we have
6 students require 3 computers
∴ 1 student will require = $\frac{3}{6}$ computers
∴ 24 students will require = $\frac{3}{6}$ x 24 computers
= 3 × 4 computers = 12 computers
Hence the number of computers required = 12.

Question 3.
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs.
Area of Rajasthan = 3 lakh km2 and area of UP = 2 lakh km2.
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
Solution:
Given:
Population of Rajasthan = 570 lakhs
Population of UP = 1660 lakhs
Area of Rajasthan = 3 lakh km²
Area of UP = 2 lakh km²
(i) Number of people per km² of Rajasthan 5
$=\frac{570 \text { lakhs }}{3 \text { lakh } \mathrm{km}^{2}}$
= 190 per km²
Number of people in UP = 1660 lakhs
Area of UP = 2 lakh km²
Number of people per km² of UP
= $=\frac{1660 \text { lakhs }}{2 \text { lakh } \mathrm{km}^{2}}=830 \text { per } \mathrm{km}^{2}$
Since 190 per km²< 830 per km²
(ii) Rajasthan is less populated state.

NCERT Solutions for Class 7th Maths Chapter 8 Comparing Quantities Exercise 8.1
NCERT Solutions Comparing Quantities Exercise 8.1

NCERT Solutions Maths Science Social English Sanskrit Hindi RD Sharma

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NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.2
Question 1.
Convert the given fractional numbers to per cents:

Solution:

Question 2.
Convert the given decimal fractions to per cents:
(а) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Solution:
(a) 0.65 = $\frac{0.65 \times 100}{100}$ = 0.65 × 100% = 65%
(b) 2.1 = $\frac{2.1 \times 100}{100}$ = 2.1 ×100% = 210%
(c) 0.02 = $\frac{0.02 \times 100}{100}$ = 0.02 × 100% = 2%
(d) 12.35 = $\frac{12.35 \times 100}{100}$ = 12.35 × 100% = 1235%

Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 5
Solution:
(i) Fraction of coloured part = $\frac{1}{4}$
∴ Percentage of coloured parts 100

(ii) Fraction of coloured part = $\frac{3}{6}$
∴ Percentage of coloured parts
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 7

(iii) Fraction of coloured part = $\frac{6}{8}$
∴ Percentage of coloured parts
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 8

Question 4.
Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg
Solution:
(a) 15% of 250 = $\frac{15}{100} \times 250=\frac{75}{2}=37.5$
(b) 1% of 1 hour = 1% of 60 minutes [∵ 1 h = 60 min.]
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 9

Question 5.
Find the whole quantity if
(a) 5% of it is 600
(b) 12% of it is? 1080
(c) 40% of it is 500 km
(d) 70% of it is 14 minutes
(e) 8% of it is 40 litres
Solution:
Let the required whole quantity be x.
(a) 5% of x = 600
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 10
Thus the required whole quantity is 12,000.

(b) 12% of x = ₹ 1080
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 11
Thus, the required quantity is ₹ 9,000.

(c) 40% of x = 500 km
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 12
Thus, the required quantity = 1250 km.

(d) 70% of x = 14 minutes
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 13
Thus, the required quantity = 20 minutes,

(e) 8% of x = 40 litre 8
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 14
Thus, the required quantity = 500 litres

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Solution:
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 16

Question 7.
In a city, 30% are females, 40% are males and remaining are children. What per cent are children?
Solution:
Given: 30% are females
40% are males
Total Percentage of females and males
= 30% + 40% = 70%
∴ Percentage of children
= (100 – 70)% = 30%

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the Percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Total number of voters = 15,000
Percentage of the voters who voted = 60%
∴ Percentage of the voters who did not vote
= (100 – 60)% = 40%
Actual number of voters who did not vote
= 40% of 15,000
$=\frac{40}{100} \times 15,000=6,000$

Question 9.
Meena saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let Meena’s salary by ₹ x.
∴ 10% of x = ₹ 400
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 15
Thus, her salary is ₹ 4000.

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solution:
Number of matches played by the cricket team = 20
Percentage of the matches won by them = 25%
i.e. $\frac{25}{100} \times 20=5$ matches
Thus, the number of matches won by them = 5

NCERT Solutions for Class 7th Maths Chapter 8 Comparing Quantities Exercise 8.2
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.2

NCERT Solutions Maths Science Social English Sanskrit Hindi RD Sharma

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NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3
Question 1.
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:
(a) Here, CP = ₹ 250
SP = ₹ 325
Since SP > CP
∴ Profit = SP – CP
= ₹ 325 – ₹ 250 = ₹ 75
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 1
Hence, the required profit = ₹ 75
and Profit per cent = 30%

(b) Here, CP = ₹ 12,000
SP = ₹ 13,500
Since SP > CP
∴ Profit = SP – CP
= ₹ 13,500 – ₹ 12,000 = ₹ 1,500
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 2
Hence, the required profit = ₹ 1500 × 100
profit % = $12 \frac{1}{2}$ %

(c) Here, CP = ₹ 2500
SP = ₹ 3000
Since SP > CP
∴ Profit = SP – CP
= ₹ 3000 – ₹ 2500 = ₹ 500
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 3
Hence, the required profit = ₹ 500 and profit% = 20%

(d) Here, CP = ₹ 250
SP = ₹ 150
Here CP > SP
∴ Loss = CP – SP
= ₹ 250 – ₹ 150 = ₹ 100
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 4
Hence, the required loss = ₹ 100 and loss% = 40%

Question 2.
Convert each part of the ratio to Percentage:
(a) 3:1
(b) 2:3:5
(c) 1 : 4
(d) 1:2:5
Solution:
(a) 3 : 1
Sum of the ratio parts = 3 + 1 = 4
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 5

(b) 2 : 3 : 5
Sum of the ratio parts = 2 + 3 + 5 = 10
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 6

(d) 1 : 2 : 5
Sum of the ratio parts = 1 + 2 + 5 = 8
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 8

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the Percentage decrease.
Solution:
Initial population = 25,000
Decreased population = 24,500
Decrease in population
= 25,000 – 24,500 = 500
Percentage of decrease = $\frac{500 \times 100}{25000}$ = 2%
Hence the Percentage of decrease in population = 2%.

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the Percentage of price increase?
Solution:
Original price of the car = ₹ 3,50,000
Price increased next year = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000
∴ Percentage of the increase in the price
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 9
Hence, the Percentage of increase in price = $5 \frac{5}{7} \%$

Question 5.
I buy a TV for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
Here, CP = ₹ 10,000
Profit = 20%
SP = ?
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 10
Hence, the required money got by me = ₹ 12,000.

Question 6.
Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
SP of the washing machine = ₹ 13,500
Loss = 20%
CP = ?
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 11
Hence, the cost price of the machine = ₹ 16875.

Question 7.
(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the Percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Solution:
(i) Sum of the ratio parts = 10 + 3 + 12 = 25
∴ Percentage of carbon in chalk
= $=\frac{3}{25} \times 100 \%=12 \%$
Hence, the Percentage of carbon in chalk = 12%

(ii) Weight of carbon = 3 g
∴ Weight of chalk = $=\frac{3}{3}$ × 25 g = 25 g
Hence, the weight of chalk = 25 g

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
CP of book = ₹ 275
Loss = 15%
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 12
Hence, the required selling price = ₹ 233.75

Question 9.
Find the amount to be paid at the end of 3 years in each case.
(a) Principal = ₹ 1200 at 12% p.a.
(b) Principal = ₹ 7500 at 5% p.a.
Solution:
(a) Given: Principal = ₹ 1200
Rate of interest = 12% p.a., T = 3 years
∴ Interest = $\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{1200 \times 12 \times 3}{100}$
Amount = Principal + Interest
= ₹ 1200 + ₹ 432 = ₹ 1632
Hence, the required amount = ₹ 1632

(b) Given: Principal = ₹ 7500
Rate = 5% p.a.
Time = 3 years
∴ Interest = $\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{7500 \times 5 \times 3}{100}$
= ₹1125
Amount = Principal + Interest
= ₹ 7500 + 11125 = ₹ 8625
Hence, the required amount = ₹ 8625.

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:
Given: Principal = ₹ 56,000
Interest = ₹280
Time = 2 years
Rate = ?
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 13
Hence, the required rate = 0.25%

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:
Given: Interest = ₹ 45
Time = 1 year
Rate = 9% p.a.
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 14
Hence, the required sum = ₹ 500.

NCERT Solutions for Class 7th Maths Chapter 8 Comparing Quantities Exercise 8.3

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