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# NCERT Solutions Class 7 Maths Chapter 7 Comparing Quantities

Last updated date: 12th Sep 2024
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## NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities - FREE PDF Download

NCERT Solutions for Class 7 Maths Chapter 7 PDF solutions are created to help students apply the formulas to compute ratios and compare quantiles to solve various problems. In NCERT Solutions for Class 7 Maths for Chapter 7 Comparing Quantities, students will learn the concepts of ratio and proportion. They will also learn the concepts of unitary method, percentage, and simple interest and how they can apply them practically. The conversion of fractions and decimals into percentages and vice-versa is also covered in Class 7 Maths Comparing Quantities.

Table of Content
1. NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities - FREE PDF Download
2. Glance on Maths Chapter 7 Class 7 - Comparing Quantities
3. Access Exercise wise NCERT Solutions for Chapter 7 Maths Class 7
4. Exercises Under NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities
5. Access NCERT Solutions for Class 7 MathsChapter 7 – Comparing Quantities
5.1Exercise 7.1
5.2Exercise 7.2
6. Overview of Deleted Syllabus for CBSE Class 7 MathsComparing Quantities
7. Class 7 Maths Chapter 7: Exercises Breakdown
8. Other Study Material for CBSE Class 7 Maths Chapter 7
9. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs

## Glance on Maths Chapter 7 Class 7 - Comparing Quantities

• NCERT Solution for Class 7 Maths Chapter 7 Comparing Quantities consists of topics like  Percentage – another way of comparing quantities, Converting decimals to percentages, Use of percentages, Ratios to percents, Prices related to an item or buying and selling, Charges given on borrowed money or simple interest

• From Decimals to Percentages unlocks the secret, showing you how to convert decimals into percentages with ease.

• Learn how percentages are used and how percentages are a special way of expressing ratios, making comparisons between quantities even more intuitive.

• Calculate discounts using percentages, markups, and other pricing strategies for buying and selling.

• Learn how to calculate the charge on borrowed money expressed as a percentage of the principal amount.

• This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 7 - Comparing Quantities, which you can download as PDFs.

• There are two exercises (21 fully solved questions) in class 7th maths chapter 7 Comparing Quantities.

## Access Exercise wise NCERT Solutions for Chapter 7 Maths Class 7

 Current Syllabus Exercises of Class 7 Maths Chapter 7 NCERT Solutions of Class 10 Maths Comparing Quantities Exercise 7.1 NCERT Solutions of Class 10 Maths Comparing Quantities Exercise 7.2

## Exercises Under NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities

### Exercise 7.1: Percentage Increase and Decrease

• Calculating percentage increase or decrease in various contexts.

• Applying percentage changes to real-life problems.

• Using percentage changes to compare different quantities.

### Exercise 7.2: Profit and Loss

• Understanding profit and loss in terms of percentages.

• Solving problems related to profit and loss.

• Calculating profit and loss in various scenarios.

• Introduction to the concept of simple interest.

• Using the formula for simple interest to solve problems.

• Understanding the relationship between principal, rate, and time.

## Access NCERT Solutions for Class 7 MathsChapter 7 – Comparing Quantities

### Exercise 7.1

1. Convert the given fractional numbers to percent:

(a) $\frac{\text{1}}{\text{8}}$

Ans: A fractional number is given, $\frac{1}{8}$

$\frac{1}{8} =\frac{1}{8} \times 100 \%$

$=\frac{25}{2} \%=12.5 \%$

$\Rightarrow \frac{1}{8}=12.5 \%$

(b) $\frac{\text{5}}{\text{4}}$

Ans: A fractional number is given, $\frac{\text{5}}{\text{4}}$

$\frac{5}{4}=\frac{5}{4} \times 100 \%$

$=5 \times 25 \%$

$=125 \%$

$\Rightarrow \frac{5}{4}=125 \%$

(c) $\frac{\text{3}}{\text{40}}$

Ans: A fractional number is given, $\frac{\text{3}}{\text{40}}$

$\frac{3}{40}=\frac{3}{40} \times 100 \%$

$\quad=\frac{15}{2} \%$

$=7.5 \%$

$\Rightarrow \frac{3}{40}=7.5 \%$

(d) $\frac{\text{2}}{\text{7}}$

Ans: A fractional number is given, $\frac{\text{2}}{\text{7}}$

$\frac{2}{7}=\frac{2}{7} \times 100 \%$

$=\frac{200}{7} \%$

$=28 \frac{4}{7} \%$

$\Rightarrow \frac{2}{7}=28 \frac{4}{7} \%$

2. Convert the given decimal fractions to percents:

(a) $\text{0}\text{.65}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 0.65=\frac{65}{100}$

$\Rightarrow 0.65=65 \%$

(b) $\text{2}\text{.1}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 2.1=\frac{21}{10} \times 100 \%$

$\Rightarrow 2.1=210 \%$

(c) $\text{0}\text{.02}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 0.02=\frac{2}{100}$

$\Rightarrow 0.02=2 \%$

(d) $\text{12}\text{.35}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 12.35=\frac{1235}{100} \times 100 \%$

$\Rightarrow 12.35=1235 \%$

3. Estimate what part of the figure is colored and hence find the percent which is colored:

(i)

Ans: It is clear from the figure that the colored part is $\text{= }\frac{\text{1}}{\text{4}}$

Therefore, the required percentage of colored parts,

is $=\frac{1}{4} \times \frac{25}{25}$

$=\frac{25}{100}$

$=25 \%$

(ii)

Ans: It is clear from the figure that the colored part is $\text{= }\frac{\text{3}}{\text{5}}$

Therefore, the required percentage of colored parts,

is $\text{= }\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{20}}{\text{20}}$

$\text{= }\frac{\text{60}}{\text{100}}$

$\text{= 60 }\%$

(iii)

Ans: It is clear from the figure that the colored part is $\text{= }\frac{\text{3}}{\text{8}}$

Therefore, the required percentage of colored parts,

is $\text{= }\frac{\text{3}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{100}}{\text{100}}$

$\text{= }\frac{\text{300}}{\text{8}}\text{ }\%$

$\text{= 37}\text{.5 }\%$

4. Find:

(a) $\text{15 }\%$ of $\text{250}$

Ans: Needed to find the required percentage of the given number $\text{250}$ i.e.,

$\text{15 }\%\text{ }$ of $\text{250}$ $\text{= }\frac{\text{15}}{\text{100}}\text{ }\!\!\times\!\!\text{ 250}$

$\text{= 15 }\!\!\times\!\!\text{ 2}\text{.5}$

$\text{= 37}\text{.5}$

$\Rightarrow \text{ 15 }\%\text{ }$ of $\text{250}$$\text{= 37}\text{.5} (b). \text{1 }\% of \text{1} hour Ans: Needed to find the required percentage i.e., \text{1 }\% of \text{1} hour \text{= 1 }\% of \text{60} minutes \text{= 1 }\% of \text{60 }\!\!\times\!\!\text{ 60} seconds \text{= }\frac{\text{1}}{\text{100}}\text{ }\!\!\times\!\!\text{ 60 }\!\!\times\!\!\text{ 60} seconds \text{= 36} seconds \Rightarrow \text{1 }\% of \text{1} hour \text{= 36} seconds (c). \text{20 }\% of Rs\text{2500} Ans: Needed to find the required percentage of the given currency Rs\text{2500} i.e., \text{20 }\% of Rs\text{2500} \text{= }\frac{\text{20}}{\text{100}}\text{ }\!\!\times\!\!\text{ 2500} \text{= 20 }\!\!\times\!\!\text{ 25} \text{=} Rs \text{500} \Rightarrow \text{20 }\% of Rs\text{2500}$$\text{=}$Rs$\text{500}$

(d). $\text{75 }\%$ of $\text{1}$ kg

Ans: Needed to find the required percentage of the given quantity $\text{1}$ kg i.e.,

$\text{75 }\%$ of $\text{1}$ kg $\text{= }\frac{75}{\text{100}}\text{ }\!\!\times\!\!\text{ 1}$ kg

$\text{= 0}\text{.75}$kg

$\Rightarrow \text{75 }\%$ of $\text{1}$ kg $\text{= 0}\text{.75}$kg

5. Find the whole quantity if:

(a). $\text{5 }\%$ of it is $\text{600}$

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{5 }\%$ of $\text{x}$ $\text{= 600}$

$\Rightarrow \text{ }\frac{\text{5}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 600}$

$\Rightarrow \text{ x = }\frac{\text{600 }\!\!\times\!\!\text{ 100}}{\text{5}}$

$\Rightarrow \text{ x = 12000}$

(b). $\text{12 }\%$ of it is Rs$\text{1080}$

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{12 }\%$ of $\text{x}$ $\text{= Rs 1080}$

$\Rightarrow \text{ }\frac{\text{12}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 1080}$

$\Rightarrow \text{ x = }\frac{\text{1080 }\!\!\times\!\!\text{ 100}}{\text{12}}$

$\Rightarrow \text{ x = Rs 9000}$

(c). $\text{40 }\%$ of it is $\text{500}$ km

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{40 }\%$ of $\text{x}$ $\text{= 500}$km

$\Rightarrow \text{ }\frac{\text{40}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 500}$

$\Rightarrow \text{ x = }\frac{\text{500 }\!\!\times\!\!\text{ 100}}{\text{40}}$

$\Rightarrow \text{ x = 1250}$km

(d). $\text{70 }\%$ of it is $\text{14}$ minutes

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{70 }\%$ of $\text{x}$ $\text{= 14}$minutes

$\Rightarrow \text{ }\frac{\text{70}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 14}$

$\Rightarrow \text{ x = }\frac{\text{14 }\!\!\times\!\!\text{ 100}}{\text{70}}$

$\Rightarrow \text{ x = 20}$ minutes

(e). $\text{8 }\%$ of it is $\text{40}$ liters

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{8 }\%$ of $\text{x}$ $\text{= 40}$liters

$\Rightarrow \text{ }\frac{\text{8}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 40}$

$\Rightarrow \text{ x = }\frac{\text{40 }\!\!\times\!\!\text{ 100}}{\text{8}}$

$\Rightarrow \text{ x = 500}$ liters

6. Convert given percents to decimal fraction and also fraction to simplest form:

(a). $\text{25 }\%$

Ans: We have given a percent $\text{25 }\%$

Fraction form$\text{= }\frac{\text{25}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{1}}{\text{4}}$

Decimal form $\text{= 0}\text{.25}$

(b). $\text{150 }\%$

Ans: We have given a percent $\text{150 }\%$

Fraction form $\text{= }\frac{\text{150}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{3}}{\text{2}}$

Decimal form $\text{= 1}\text{.5}$

(c). $\text{20 }\%$

Ans: We have given a percent $\text{20 }\%$

Fraction form $\text{= }\frac{\text{20}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{1}}{\text{5}}$

Decimal form $\text{= 0}\text{.2}$

(d). $\text{5 }\%$

Ans: We have given a percent $\text{5 }\%$

Fraction form $\text{= }\frac{\text{5}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{1}}{\text{20}}$

Decimal form $\text{= 0}\text{.05}$

7. In a city, $\text{30 }\%$ are females, $\text{40 }\%$ are males and remaining are children. What percent are children?

Ans:  Let the percentage of children be $\text{x }\%$

It is given that the percentage of females and males are $\text{30 }\%$ and $\text{40 }\%$ respectively.

And, the total percentage $\text{= 100 }\%\text{ =}$ Percentage of males and Percentage of females and Percentage of children

$\Rightarrow \text{ 100 }\%\text{ = 30 }\%\text{ + 40 }\%\text{ + x }\%$

$\Rightarrow \text{ 100 }\%\text{ = 70 }\%\text{ + x }\%$

$\Rightarrow \text{ x }\%\text{ = 100 }\%\text{ - 70 }\%\text{ }$

$\Rightarrow \text{ x }\%\text{ = 30 }\%$

Thus $\text{30 }\%\text{ }$is the population of children in the city.

8. Out of $\text{15,000}$ voters in a constituency, $\text{60 }\%$ voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

Ans: The total number of voters $\text{= 15,000}$

The percentage of people who voted $\text{= 60 }\%$

So, the percentage of people who didn’t vote $\text{= 100 }\%\text{ - 60 }\%$

$\text{= 40 }\%$

And, the number of actual candidates, who didn’t vote $=40 \%$ of $\text{15000}$

$\text{= 6000}$

Thus, $\text{6,000}$ people out of $\text{15,000}$ did not vote.

9. Meeta saves Rs $\text{400}$ from her salary. If this is $\text{10 }\%$ of her salary. What is her salary?

Ans:  Let $\text{x}$ be the salary of Meeta.

Since $\text{10 }\%$ of her salary $\text{= Rs 400}$

$\Rightarrow \text{ 10 }\%\text{ of x = 400}$

$\Rightarrow \text{ 10 }\%\text{ }\!\!\times\!\!\text{ x = 400}$

$\Rightarrow \text{ }\frac{\text{10}}{\text{100}}\text{x = 400}$

$\Rightarrow \text{ x = Rs 4,000}$

Therefore, the salary of Meeta is $\text{Rs 4,000}$.

10. A local cricket team played $\text{20}$ matches in one season. It won $\text{25 }\%$ of them. How many matches did they win?

Ans: The local cricket team played $\text{20}$ matches.

They won $\text{25 }\%$ of matches out of $\text{20}$

Therefore, the number of matches the cricket team won $\text{= 25 }\%\text{ of 20}$

$\text{= }\frac{\text{25}}{\text{100}}\text{ }\!\!\times\!\!\text{ 20}$

$\text{= 5}$ matches

So, the local cricket team won $\text{5}$ matches out of $\text{20}$.

### Exercise 7.2

1. Tell what is the profit or loss in the following transactions. Also, find the profit per cent or loss per cent in each case.

(a). Gardening shears bought for $\text{Rs 250}$ and sold for $\text{Rs 325}$.

Ans: Here, the Cost price of the gardening shears $\text{= Rs 250}$

Also, the Selling price of gardening shears $\text{= Rs 325}$

Since, the Selling Price (S.P.) is greater than Cost Price (C.P.)

Therefore, here we have is the profit.

$\therefore$ Profit $\text{=}$ S.P. $-$ C.P. $\text{= 325-250}$

$\Rightarrow$ Profit $\text{=}$ $\text{Rs 75}$

So, the profit percentage $\text{= }\frac{\text{Profit}}{\text{C}\text{.P}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Profit$\text{ }\%\text{ = }\frac{\text{75}}{\text{250}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Profit $\text{ }\%\text{ = 30 }\%$

(b). A refrigerator bought for $\text{Rs 12,000}$ and sold at $\text{Rs 13,500}$.

Ans: Here, the Cost price of the refrigerator $\text{= Rs 12,000}$

Also, the Selling price of the refrigerator $\text{= Rs 13,500}$

Since, the Selling Price (S.P.) is greater than Cost Price (C.P.)

Therefore, here we have is the profit.

$\therefore$ Profit $\text{=}$ S.P. $-$ C.P. $\text{= 13,500 - 12,000}$

$\Rightarrow$ Profit $\text{=}$ $\text{Rs 1,500}$

So, the profit percentage $\text{= }\frac{\text{Profit}}{\text{C}\text{.P}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Profit $\text{ }\%\text{ = }\frac{\text{1500}}{12000}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Profit $\text{ }\%\text{ = 12}\text{.5 }\%$

(c). A cupboard bought for $\text{Rs 2,500}$ and sold at $\text{Rs 3,000}$.

Ans: Here, the Cost price of the cupboard $\text{= Rs 2,500}$

Also, the Selling price of the cupboard $\text{= Rs 3,000}$

Since, Selling Price (S.P.) is greater than Cost Price (C.P.)

Therefore, here we have is the profit.

$\therefore$ Profit $\text{=}$ S.P. $-$ C.P. $\text{= 3,000 - 2,500 }$

$\Rightarrow$ Profit $\text{=}$ $\text{Rs 500}$

So, the profit percentage $\text{= }\frac{\text{Profit}}{\text{C}\text{.P}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Profit $\text{ }\%\text{ = }\frac{\text{500}}{\text{2500}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Profit $\text{ }\%\text{ = 20 }\%$

(d). A skirt bought for $\text{Rs 250}$ and sold at $\text{Rs 150}$.

Ans: Here, the Cost price of the skirt $\text{= Rs 250}$

Also, the Selling price of the skirt $\text{= Rs 150}$

Since, Selling Price (S.P.) is lower than Cost Price (C.P.)

Therefore, here we have is the loss.

$\therefore$ Loss $\text{=}$ C.P. $\text{-}$ S.P.$\text{= 250 - 150}$

$\Rightarrow$ Loss $\text{=}$ $\text{Rs 100}$

So, the loss percentage $\text{= }\frac{\text{Loss}}{\text{C}\text{.P}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Loss$\text{ }\%\text{ = }\frac{\text{100}}{\text{250}}\text{ }\!\!\times\!\!\text{ 100}$

$\Rightarrow$ Loss$\text{ }\%\text{ = 40 }\%$

2. Convert each part of the ratio to percentage:

(a). $\text{3 : 1}$

Ans: The given ratio is $\text{3 : 1}$

The total of the parts of the given ratio is $\text{3 + 1 = 4}$

So, the fractional part of the given ratio is $\frac{\text{3}}{\text{4}}\text{ : }\frac{\text{1}}{\text{4}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ 100 : }\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ 100}$

$\text{= 75 }\%\text{ : 25 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 75 }\%\text{ : 25 }\%$

(b). $\text{2 : 3 : 5}$

Ans: The given ratio is $\text{2 : 3 : 5}$

The total of the parts of the given ratio is $\text{2 + 3 + 5 = 10 }$

So, the fractional part of the given ratio is $\frac{\text{2}}{\text{10}}\text{ : }\frac{\text{3}}{\text{10}}\text{ : }\frac{\text{5}}{\text{10}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{2}}{\text{10}}\text{ }\!\!\times\!\!\text{ 100 : }\frac{\text{3}}{\text{10}}\text{ }\!\!\times\!\!\text{ 100 : }\frac{\text{5}}{\text{10}}\text{ }\!\!\times\!\!\text{ 100}$

$\text{= 20 }\%\text{ : 30 }\%\text{ : 50 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 20 }\%\text{ : 30 }\%\text{ : 50 }\%$

(c). $\text{1 : 4}$

Ans: The given ratio is $\text{1 : 4}$

The total of the parts of the given ratio is $\text{1 + 4 = 5}$

So, the fractional part of the given ratio is $\frac{\text{1}}{\text{5}}\text{ : }\frac{\text{4}}{\text{5}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{1}}{\text{5}}\text{ }\!\!\times\!\!\text{ 100 : }\frac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ 100}$

$\text{= 20 }\%\text{ : 80 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 20 }\%\text{ : 80 }\%$

(d). $\text{1 : 2 : 5}$

Ans: The given ratio is $\text{1 : 2 : 5}$

The total of the parts of the given ratio is $\text{1 + 2 + 5 = 8}$

So, the fractional part of the given ratio is $\frac{\text{1}}{\text{8}}\text{ : }\frac{\text{2}}{\text{8}}\text{ : }\frac{\text{5}}{\text{8}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{1}}{\text{8}}\text{ }\!\!\times\!\!\text{ 100 : }\frac{\text{2}}{\text{8}}\text{ }\!\!\times\!\!\text{ 100 : }\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ 100}$

$\text{= 12}\text{.5 }\%\text{ : 25 }\%\text{ : 62}\text{.5 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 12}\text{.5 }\%\text{ : 25 }\%\text{ : 62}\text{.5 }\%$

3. The population of a city decreased from $\text{25,000}$ to $\text{24,500}$. Find the percentage decrease.

Ans: The population of a city got decreased from $\text{25,000}$ to $24,500$

The decreased population $\text{= 25,000 - 24,500 = 500}$

Decreased Percentage $\text{= }\frac{\text{Decreased Population}}{\text{Original Population }}\text{ }\!\!\times\!\!\text{ 100}$

$\text{= }\frac{500}{25000}\text{ }\times \text{ 100}$

$\text{= 2 }\%$

Thus, the required percentage decrease is $2%$.

4. Arun bought a car for $\text{Rs 3,50,000}$. The next year, the price went up to $\text{Rs 3,70,000}$. What was the percentage of price increase?

Ans: The price of the car got increased from $\text{Rs 3,50,000}$ to $\text{Rs 3,70,000}$

So, the change in the price of the car $\text{= 3,70,000 - 3,50,000}$ $\text{= Rs 20,000}$

Therefore, Increased percentage $\text{= }\frac{\text{Change in Amount}}{\text{Original Amount}}\text{ }\!\!\times\!\!\text{ 100}$

$\text{= }\frac{20000}{350000}\text{ }\times \text{ 100}$

$\text{= 5}\frac{\text{5}}{\text{7}}\text{ }\%$

Thus, the percentage of the increased price is $\text{ 5}\frac{\text{5}}{\text{7}}\text{ }\%$

5. I buy a T.V. for $\text{Rs 10,000}$ and sell it at a profit of $\text{20 }\%$. How much money do I get for it?

Ans: The money I got by selling the T.V. is known as the selling price.

Since, I bought it at $\text{Rs 10,000}$, therefore

The Cost Price of T.V. $\text{= Rs 10,000}$

We know that, S.P. = C.P. $\text{+}$ Profit           …… (1)

As, I sell it at $\text{20 }\%$ profit.

Then, the Profit $\text{= 20 }\%$ of C.P.

$\Rightarrow$ Profit $\text{= }\frac{\text{20}}{\text{100}}\text{ }\!\!\times\!\!\text{ 10000}$

$\Rightarrow$ Profit $\text{= Rs 2,000}$

Now, from equation (1) we get,

S.P. $\text{=}$ C.P. $\text{+}$ Profit

$\Rightarrow \text{ S}\text{.P}\text{. = 10,000 + 2,000}$

$\Rightarrow \text{ S}\text{.P}\text{. = Rs 12,000}$

Hence, I sold the T.V. at a price of $\text{Rs 12,000}$

6. Juhi sells a washing machine for $\text{Rs 13,500}$. She loses $\text{20 }\%$ in the bargain. What was the price at which she bought it?

Ans: Juhi sells the washing machine for $\text{Rs 13,500}% Therefore, the selling price of the washing machine$\text{= Rs 13,500}$As, she loses$\text{20 }\%$in the bargain, Therefore, the loss percent$\text{= 20 }\%$Loss$\text{= 20 }\%\text{  of C}\text{.P}\text{.}\Rightarrow $Loss$\text{= }\frac{\text{20}}{\text{100}}\text{  }\!\!\times\!\!\text{  C}\text{.P}\text{.}\Rightarrow $Loss$\text{= }\frac{\text{C}\text{.P}\text{.}}{\text{5}}$We know that, S.P. = C.P.$-$Loss$\Rightarrow \text{ S}\text{.P}\text{. = C}\text{.P}\text{. - }\frac{\text{C}\text{.P}\text{.}}{\text{5}}\Rightarrow \text{ 13500 = }\frac{\text{4 C}\text{.P}\text{.}}{\text{5}}\Rightarrow$C.P.$\text{= }\frac{\text{13500  }\!\!\times\!\!\text{ 5}}{\text{4}}\Rightarrow \text{ C}\text{.P}\text{. = Rs 16,875}$Hence, the price at which Juhi bought the washing machine is$\text{Rs 16,875}$. 7. (i) Chalk contains Calcium, Carbon, and Oxygen in the ratio$\text{10 : 3 : 12}$. Find the percentage of Carbon in chalk. Ans: We have given the ratio of the component present in a Chalk, i.e., the ratio of Calcium, Carbon, and Oxygen as$\text{10 : 3 : 12}$Therefore, the total parts$\text{10 + 3 +12 = 25}$Thus, the part of the carbon$\text{= }\frac{\text{3}}{\text{25}}\Rightarrow $The percentage of Carbon in the chalk is$\text{= }\frac{\text{3}}{\text{25}}\text{  }\!\!\times\!\!\text{  100  = 12 }\%$(ii) If in a stick of chalk, Carbon is$\text{3}$g, what is the weight of the chalk stick? Ans:$\text{3}$g Carbon is present in the chalk. Let the weight of the chalk is$\text{x}$g. Then from part (i)$\text{12 }\%$of$\text{x = 3}\Rightarrow \text{ }\frac{\text{12}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 3}\Rightarrow \text{ x = }\frac{\text{3  }\!\!\times\!\!\text{  100}}{\text{12}}\Rightarrow \text{ x = 25}$g Thus, the weight of the chalk is$\text{25 g}$. 8. Amina buys a book for$\text{Rs 275}$and sells it at a loss of$\text{15 }\%$. How much does she sell it for? Ans: Let the Selling Price (S.P.) of the book is$\text{Rs x}$. Since, Amina bought the book for$\text{Rs 275}$Therefore, the Cost Price (C.P.) of the book is$\text{Rs 275}$She losses$\text{15 }\%$while selling the book Lost Percent$\text{= 15 }\%$and,$\Rightarrow \text{ }$Loss$\text{= 15 }\%$of C.P$\Rightarrow \text{ }$Loss$\text{= 15 }\%$of$\text{Rs 275}\Rightarrow \text{ }$Loss$\text{= }\frac{\text{15}}{\text{100}}\text{  }\!\!\times\!\!\text{  275}\Rightarrow \text{ }$Loss$\text{= Rs 41}\text{.25}$We know that, S.P. = C.P.$-$Loss$\Rightarrow $S.P.$\text{= 275 - 41}\text{.25}\Rightarrow $S.P.$\text{= Rs 233}\text{.75}$Thus, Amina sells her book at$\text{Rs 233}\text{.75}$9. Find the amount to be paid at the end of$\text{3}$years in each case: (a) Principal$\text{= Rs 1,200}$at$\text{12 }\%$p.a. Ans: We know that Amount$\text{=}$Principal$\text{+}$Simple Interest (S.I.) And, S.I.$\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}$Here, P$\text{=}$Principal$\text{= Rs 1,200}$R$\text{=}$Rate$\text{= 12 }\%\text{  }$and, T$\text{=}$Time$\text{= 3}$years Therefore, S.I.$\text{= }\frac{\text{1200 }\!\!\times\!\!\text{ 12 }\!\!\times\!\!\text{ 3}}{\text{100}}\Rightarrow \text{ S}\text{.I}\text{. = Rs 432}$So, the Amount$\text{= 1200 + 432}\Rightarrow $Amount$\text{= Rs 1,632}$(b) Principal$\text{= Rs 7,500}$at$5%$p.a. Ans: We know that Amount$\text{=}$Principal$\text{+}$Simple Interest (S.I.) And, S.I.$\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}$Here, P$\text{=}$Principal$\text{= Rs 7,500}$R$\text{=}$Rate$\text{= 5 }\%\text{  }$and, T$\text{=}$Time$\text{= 3}$years Therefore, S.I.$\text{= }\frac{\text{7500 }\!\!\times\!\!\text{ 5 }\!\!\times\!\!\text{ 3}}{\text{100}}\Rightarrow \text{ S}\text{.I}\text{. = Rs 1,125}$So, the Amount$\text{= 7,500 + 1,125}\Rightarrow $Amount$\text{= Rs 8,625}$10. What rate gives$\text{Rs 280}$as interest on a sum of$\text{Rs 56,000}$in$\text{2}$years? Ans: Given, Simple Interest (S.I.)$\text{= Rs 280}$Principal$\text{= Rs 56,000}$and, Time$\text{= 2}$years Let$\text{R}$be the required rate of interest. As, S.I.$\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}\Rightarrow \text{280 = }\frac{\text{56000 }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ 2}}{\text{100}}\Rightarrow \text{R = }\frac{\text{280 }\!\!\times\!\!\text{ 100}}{\text{56000 }\!\!\times\!\!\text{ 2}}\Rightarrow $$R = 0.25\%$ Thus, the required rate of interest is, $R = 0.25\%$ 11. If Meena gives an interest of$\text{Rs 45}$for one year at$\text{9 }\%$rate p.a. What is the sum she has borrowed? Ans: Let the sum / Principal borrowed by Meena is P Given, Simple Interest, S.I.$\text{= Rs 45}$Rate of Interest, R$\text{= 9 }\%\text{  p}\text{.a}\text{.}$and, Time,$\text{T = 1}$year As, S.I.$\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}\Rightarrow\text{45 = }\frac{\text{P }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 1}}{\text{100}}\Rightarrow\text{P = }\frac{\text{45 }\!\!\times\!\!\text{ 100}}{\text{9 }\!\!\times\!\!\text{ 1}}\Rightarrow\text{P = Rs 500}$Thus, the sum Meena borrowed is,$\text{Rs 500}\$.

## Overview of Deleted Syllabus for CBSE Class 7 MathsComparing Quantities

 Chapter Dropped Topics Comparing Quantities 7.1 - Introduction 7.2 - Equivalent ratios.

## Class 7 Maths Chapter 7: Exercises Breakdown

 Exercise Number of Questions Exercise 7.1 10 Questions & Solutions Exercise 7.2 11 Questions & Solutions

## Important Formulas

• Converting Fractions/Decimals to Percent: Percent (%) = (fraction or decimal) x 100

• Converting Percent to Fraction/Decimal: Fraction/Decimal = Percent (%) / 100

• Percentage: Percentages are defined as the ratios which are expressed as a fraction of 100. “%” is used to represent percentage.

Percentage (%) = (fraction or decimal) x 100

• Buying and Selling

(i) Selling price (SP): Selling price is the price at which an item or product is sold out.

(ii) Cost price (CP): Cost price is the buying price of an item or product.

Profit = Selling price – Cost price

Loss = Cost price – Selling price

• If Selling Price > Cost Price, then it is profit.

• If Seling Price = Cost Price, then it is neither profit nor loss.

• If Cost Price > Selling Price, then it is a loss.

• Profit and Loss

Profit Percentage= $\frac{profit}{cost\,price}\times 100$

Loss Percentage= $\frac{loss}{cost\,price}\times 100$

## Facts

• Various quantities (of the same kind) are compared using their ratios.

• If two fractions are equal, their ratios are equivalent.

• When two ratios are equal then the four quantities are in proportion.

• Percentages are numerators of fractions with denominator 100. The percentage is also a way of comparing quantities.

• To convert a percent into decimal, drop the sign of percent and then shift the decimal point two places to the left.

• To convert a fraction into percent, multiply the fraction by 100, and write % sign on the right of the number.

• Profit = SP – CP (when SP > CP).

• Loss = CP – SP (when CP > SP).

• Percent gain or loss is always calculated on CP.

• Money borrowed is called the principal.

• Simple interest = (P x R x T)/100

• Amount = Principal + Interest

## Ratio

Comparison means to know “ how many times one quantity is of the other”, or to know – what part one quantity is of the other. It is called the ratio of quantities of the same kind and in the same units.

The ratio of two numbers ‘a’ and ‘b’ (b ≠ 0) is a/b and it is denoted by a:b. A ratio in the simplest form is also called the ratio in the lowest terms.

Note: To compare two quantities or to find the ratio of two quantities, their units must be the same.

Different ratios can also be compared by writing them as ‘like-fractions’.

Let us go through some examples to get a better understanding. Also, you can download and refer to the Exercise 8.1 in NCERT Solutions for class 7 chapter 11 from Vedantu for free, and prepare for your exams.

Example 1: Find the ratio of 5 km and 400 m.

Solution: The units of the two distances in the question are different.

First we need to convert the distances to the same unit.

5 km = 5 x 1000 m = 5000 m

Now, 5 km : 400 m = 5000 m : 400 m

= 5000 : 400 = 5000/200 : 400/200

= 25 : 2 (Since, the HCF of 5000 and 400 is 200)

Thus, the required ratio is 25:2.

Example 2: Find the ratio of 30 days and 36 hours.

Solution: Since the given durations have different units, we need to first convert them to the same unit.

30 days = 30 x 24 hours = 720 hours

Now, 30 days : 36 hours = 720 hours : 36 hours

= 720/36 : 36/36 (Since, the HCF of 720 and 36 is 36)

=  20 : 1

Thus, the required ratio is 20:1.

### Equivalent Ratios

We can compare various ratios by converting them to like fractions. If the like fractions are equal, then the given ratios are said to be equivalent.

Example: Are the ratios 2 : 3 and 3: 4 equivalent?

Solution: The corresponding fractions of the ratios (2:3 and 3:4) are 2/3 and 3/4.

Now, 2/3 = 2/3 x 4/4 = 8/12 (∴ LCM of 3 and 4 is 12)

and, 3/3 = 3/4 x 3/4 = 9/12

since, 8 < 9

or, 8/12 < 9/12 or 2/3 < 3/4

Thus, 2:3 and 3:4 are not equivalent.

Referring to the Exercise 8.2 in NCERT Solutions of Chapter 11 for class 7, you need to learn what percentage is and the conversion of fractional numbers to Percentage.

### Percentage

Percentages are numerators of fractions having denominators as 100. Percentages are used in comparing results. As mentioned before, the percent is represented by the symbol % and it is expressed as parts of hundredths.

Example: 9 % means 9 out of 100. It is written as 9% = 9/100 = 0.09.

Percentage when Total is not Hundred.

In case, the total is not a hundred, then we need to convert the fraction into an equivalent fraction with denominator 100. For this purpose, we can use any method. Some methods are given below:

Method I: We multiply the given fraction by 100/100. [ ∵ 100/100 = 1], which does not change the value of the fraction.

Example: 12/20 = 12/20 x 100/100 = 3/5 x 100/100 = 60/100 = 60%

Method II: We can use the unitary method.

Example: For 20, the corresponding value is 8

For 1, the corresponding value is 8/20.

For 100, the corresponding value is (8/20  x 100) % = 40%

Method III: We multiply and divide the given fraction by a number such that the denominator becomes 100.

Example: 12/20 = 12/20 x 5/5 = 60/100 = 60%

### Ratios to Percent

Sometimes, the parts of a whole quantity are given in the form of ratios. We can convert them into percentages. Let us look into an example to understand this well.

Example: The population of a state increased from 5,50,5000 to 6,05,000. What is the percentage increase in the population?

Solution:  Initial population = 5,50,000

Increased population = 6,05,000

∴  Increase in population = 6,05,000 – 5,50,000

= 55,000

∴ Percent increase = 55,000/5,50,000  x 100% = 10%

Note:  We convert the increase or decrease in a quantity as a percentage of the initial amount.

### Profit or Loss as a Percentage

We know that the buying price of any item is called its cost price (CP) and the price at which it is sold is called its selling price (SP).

Note

1. If CP < SP, then there is a profit in the transaction, and Profit = SP – CP.

2. If CP > SP, then there is a loss in the transaction and Loss = CP – SP.

3. Percent profit or percent loss is always calculated on the CP.

### Simple Interest

The money borrowed is called the Principal (or sum borrowed). For using a bank's money for some time, the borrower has to pay some extra money to the bank. This extra money is called Interest. The total money paid back with interest is called Amount.

Amount = Principal + Interest

Interest is generally given in percent for a period of one year. Generally, the rate of interest is written in percent per year or per annum.

So, Principal is denoted by P, Rate of interest by R, and Time by T.

Now,  simple interest, or Interest = (Principal x Rate x Time)/100, or I = PRT/100

## Conclusion

Chapter 7, "Comparing Quantities," provides a comprehensive understanding of ratios, percentages, profit and loss, and simple interest. These concepts are essential for solving real-life problems and are foundational for higher-level mathematics. In previous years' exams, questions from this chapter have been frequently asked, reflecting its importance in the curriculum. Typically, 5-7 questions are asked annually, covering various topics within the Class 7 Maths Chapter Comparing Quantities. These questions range from simple calculations of percentages and ratios to more complex problems involving profit, loss, and simple interest.

## Other Study Material for CBSE Class 7 Maths Chapter 7

 S. No Important Links for Chapter 7 Comparing Quantities 1 Class 7 Comparing Quantities Important Questions 2 Class 7 Comparing Quantities Revision Notes 3 Class 7 Comparing Quantities NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 7 Maths Chapter 7 Comparing Quantities

1. What is the main focus of Chapter 8 - Comparing Quantities?

The main focus of the NCERT Class 7 Maths Chapter 7 is to help students understand and work with ratios, percentages, profit and loss, and simple interest. These concepts are essential for comparing different quantities in various contexts.

2. How do you convert a ratio into a percentage in Comparing Quantities?

To convert a ratio into a percentage, divide the first part of the ratio by the second part, multiply the result by 100, and add the percentage symbol (%).

3. What is the formula for calculating simple interest in comparing quantities?

SI = (P × R × T) / 100

Here's what each variable represents:

• SI: Simple Interest (the amount of interest earned or paid)

• P: Principal amount (the initial amount borrowed or invested)

• R: Rate of interest (expressed as a percentage per year - R%)

• T: Time period (usually in years)

4. How many questions are typically asked from comparing quantities in previous years' exams?

Typically, 5-7 questions are asked annually from class 7 math chapter 7 in exams, highlighting its significance in the curriculum.

6. How many wins can Rs 4 change for Rs 400?

To get the answer to this question you have to divide 400 by 4. Hence, the answer to this question will be 100. For a more detailed explanation, visit Vedantu where you will get the step by step solution to this question and many other questions.