NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities

Comparing Quantities is a chapter that introduces how to express comparisons through ratios and percentages, and how to calculate increase or decrease in quantities. The chapter also covers concepts such as profit and loss, simple interest, and conversion between different forms of comparison like fractions, decimals, and percentages. These NCERT Solutions offer step-by-step explanations that help students develop clarity and confidence in solving problems related to comparing quantities using appropriate methods and formulas.

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

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1. NCERT Solutions for Maths Chapter 7 Comparing Quantities Class 7- Important Points
2. NCERT Solutions for Maths Chapter 7 Comparing Quantities Class 7
3. NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities (Exercise)
4. Comparing Quantities Maths Class 7 Chapter 7-Topics
5. NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities - Points to Remember
6. NCERT Solutions for Class 7 Maths Chapter-Wise
7. NCERT Solutions for Class 7 Subject Wise

These solutions are solved by subject matter experts at Careers360 making it an important and reliable study resource that is of great help during exam preparation. The NCERT Solutions for Class 7 Maths helps students understand fundamental topics of each chapter of Class 7 with step-by-step detailed solutions.

NCERT Solutions for Maths Chapter 7 Comparing Quantities Class 7- Important Points

Percent Calculation: Percent represents a part of a whole, expressed as a fraction of 100.

Percent = $\frac{\text{Part}}{\text{Whole}}\times 100$

Part (How Many) = $\frac{\text{Percent}}{100}$ × Whole

Conversion: You can convert fractions or decimals to percentages by multiplying them by 100. Similarly, to convert percentages to fractions or decimals, you divide the percentage by 100.

Fraction or Decimal to Percent = (Fraction or Decimal) × 100

Percent to Fraction or Decimal = $\frac{\text{Percent}}{100}$

Increase or Decrease as Percent: You can find the percentage increase or decrease by comparing the change in value to the original value.

Increase or Decrease as Percent: Percentage Increase = $\frac{\text{Amount of Change}}{\text{Original Amount}}\times 100$

Cost Price (CP) and Selling Price (SP): Cost Price (CP) is the price you pay for an item, and Selling Price (SP) is the price at which you sell it. If SP is higher than CP, you make a profit. If SP is lower, you incur a loss.

If $\mathrm{CP}<\mathrm{SP}\to$ Profit $=\mathrm{SP}-\mathrm{CP}$

If $\mathrm{CP}>\mathrm{SP}\to$ Loss $=\mathrm{CP}-\mathrm{SP}$

If $\mathrm{CP}=\mathrm{SP}\to$ No profit, No loss

Profit percent = $\frac{\text{Profit}}{\text{CP}}\times 100$

Loss Percent = $\frac{\text{Loss}}{\text{CP}}\times 100$

Simple Interest (S.I.): Simple Interest is a way to calculate the extra amount you earn or pay on a sum of money over time. It depends on the principal amount, the interest rate, and the time period.

S.I. = $\frac{\text{Principal}\times \text{Rate}\times \text{Time}}{100}$

Amount = Principal + Interest

NCERT Solutions for Maths Chapter 7 Comparing Quantities Class 7

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NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities (Exercise)

Question: 1 Convert the given fractional numbers to per cents.

$(a)\frac{1}{8}$
$(b)\frac{5}{4}$
$(c)\frac{3}{40}$
$(d)\frac{2}{7}$

Answer: To convert the given fractional into the per cent we have to do multiplication in numerator and denominator by 100
Now,
$(a)\frac{1}{8}$
$\Rightarrow \frac{1}{8}\times 100=\frac{100}{8}\mathrm{\%}=12.5\mathrm{\%}$

$(b)\frac{5}{4}$
$\Rightarrow \frac{5}{4}\times 100=\frac{500}{4}\mathrm{\%}=125\mathrm{\%}$

$(c)\frac{3}{40}$
$\Rightarrow \frac{3}{40}\times 100=\frac{300}{40}\mathrm{\%}=7.5\mathrm{\%}$

$(d)\frac{2}{7}$
$\Rightarrow \frac{2}{7}\times 100=\frac{200}{7}\mathrm{\%}=28.57\mathrm{\%}$

Question: 2 Convert the given decimal fractions to per cents.

(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35

Answer: To convert the given fractional decimal into a per cent, multiply the denominator and numerator by 100.
So,
(a) 0.65
$0.65\times 100=65\mathrm{\%}$

(b) 2.1
$2.1\times 100=210\mathrm{\%}$

(c) 0.02
$0.02\times 100=2\mathrm{\%}$

(d) 12.35
$12.35\times 100=1235\mathrm{\%}$

Question: 3(i) Estimate what part of the figure is coloured and hence find the percent which is coloured.

Answer: We have,
The fraction of the coloured part = $\frac{1}{4}$
Therefore,
$\frac{1}{4}\times \frac{100}{100}=\frac{100}{4}\mathrm{\%}=25\mathrm{\%}$

Question: 3(ii) Estimate what part of the figure is coloured and hence find the percent which is coloured.

Answer: In the given figure, the fraction of the coloured part is $3/6$
Therefore, Percentage of the coloured part
$=\frac{3\times 100}{5\times 100}=\frac{300}{5}\mathrm{\%}=60\mathrm{\%}$

Question: 3(iii) Estimate what part of the figures is coloured and hence find the percent which is coloured.

Answer: In the given figure, the fraction of the coloured part is $3/8$
Therefore, Percentage of the coloured part
$=\frac{3\times 100}{8\times 100}=\frac{300}{8}\mathrm{\%}=37.5\mathrm{\%}$

Question:4 Find: (a) 15 $\mathrm{\%}$ of 250
(b) 1 $\mathrm{\%}$ of 1 hour
(c) 20 $\mathrm{\%}$ of Rs 2500
(d) 75 $\mathrm{\%}$ of 1 kg

Answer: Here per cent implies for ( $\mathrm{\%}\to \frac{1}{100}$ )

Therefore,
(a) 15% of 250
$15\times \frac{1}{100}\times 250=\frac{75}{2}$
= 37.5

(b) 1 $\mathrm{\%}$ of 1 hour
= 1 $\mathrm{\%}$ of 60 minutes
$=\frac{1}{100}\times 60=\frac{3}{5}min$
$=\frac{3}{5}\times 60=36sec$

(c) 20 $\mathrm{\%}$ of Rs 2500
$=\frac{20}{100}\times 2500=Rs500$

(d) 75 $\mathrm{\%}$ of 1 kg
We know that 1kg = 1000 gm, therefore,
$\frac{75}{100}\times 1000g=750g$
=0.75 kg

Question: 5 Find the whole quantity if

(a) 5% of it is 600.
(b) 12% of it is Rs 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 litres.

Answer: (a) Let the whole quantity be X
Therefore, 5 % of X = 600
$$\begin{gathered} \Rightarrow \frac{5}{100}\times X=600 \\ \Rightarrow X=\frac{600\times 100}{5}=12000 \end{gathered}$$
Thus the required whole quantity is 12000

(b) 12 % of X = Rs 1080
$$\begin{gathered} \Rightarrow \frac{12}{100}\times X=1080 \\ \Rightarrow X=\frac{1080\times 100}{12}=9000 \end{gathered}$$

(c) 40 % of X = 500 km
$$\begin{gathered} \Rightarrow \frac{40}{100}\times X=500 \\ \Rightarrow X=\frac{50000\times 100}{40}=1250km \end{gathered}$$

(d) 70% of X = 14 minutes
$$\begin{gathered} \Rightarrow \frac{70}{100}\times X=14min \\ \Rightarrow X=\frac{1400\times 100}{70}=20min \end{gathered}$$

(e) 8 % of X = 40 litre
$$\begin{gathered} \Rightarrow \frac{8}{100}\times X=40L \\ \Rightarrow X=\frac{4000}{8}=500L \end{gathered}$$

Question: 6 Convert given per cent to decimal fraction and also to fraction in simplest forms:

(a) 25%
(b) 150%
(c) 20%
(d) 5%

Answer: (a) 25 %
$\Rightarrow \frac{25}{100}=0.25=\frac{1}{4}$

(b) 150%
$\Rightarrow \frac{150}{100}=1.5=\frac{3}{2}$

(c) 20%
$\Rightarrow \frac{20}{100}=0.2=\frac{1}{5}$

(d) 5%
$\Rightarrow \frac{5}{100}=0.05=\frac{1}{20}$

Question: 7 In a city, 30% are females, 40% are males and the remaining are children. What per cent are children?

Answer: Given that,
30 % are female and
40 % are males
Total percentage of females and males
= 30 % + 40% = 70%
Therefore, 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 out how many actually did not vote?

Answer: Given that,
Total number of voters = 15000
Percentage of voters who voted = 60%
Therefore, the remaining 40% of voters didn't vote

Thus, the number of people who did not vote
= $40\mathrm{\%}of15000$
$=\frac{40}{100}\times 15000$
$=6000$

Question: 9 Meeta saves Rs 4000 from her salary. If this is 10% of her salary. What is her salary?

Answer: Given that,
10 % of her salary = 4000
Let her total salary be X
Therefore,
10% of X = 4000
$\Rightarrow \frac{10}{100}\times X=4000$
$X=\frac{4000}{10}\times 100$
X = Rs 40,000

Question: 10 A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

Answer: Given that,
Total matches played by the cricket team = 20
and, a match won = 25%

According to the question,
Number of winning match = 25 % of 20
$\Rightarrow \frac{25}{100}\times 20=5$

Hence they won a total of 5 matches out of 20 matches.

Question: 1(a) Tell what is the profit or loss in the following transaction. Also, find profit per cent or loss per cent in each case.

Gardening shears were bought for Rs 250 and sold for Rs 325.

Answer: Given that,
Selling price = Rs 325
Cost price = Rs 250
Since SP > CP
Therefore, profit = SP-CP = Rs 75

And, Profit %
$\Rightarrow \frac{75}{250}\times 100=30\mathrm{\%}$

Question: 1(b) Tell what is the profit or loss in the following transaction. Also, find profit per cent or loss per cent in each case.

A refrigerator was bought for Rs 12,000 and sold at Rs 13,500.

Answer: Given that,
CP = Rs 12000
SP = Rs 13500

Since SP > CP
Therefore, Profit = SP - CP = 1500

Profit %$=\frac{1500}{12000}\times 100=12.5\mathrm{\%}$

Question: 1(c) Tell what is the profit or loss in the following transactions. Also, find profit per cent or loss per cent in each case.

A cupboard bought for Rs 2,500 and sold at Rs 3,000.

Answer: We have,
CP = Rs 2500
SP = Rs 3000

Since SP > CP
Therefore, Profit = SP - CP = 500

Profit % $=\Rightarrow \frac{500}{2500}\times 100=20\mathrm{\%}$

Question: 1(d) Tell what is the profit or loss in the following transaction. Also, find profit per cent or loss per cent in each case.

A skirt was bought for Rs 250 and sold at Rs 150.

Answer: We have,
SP = Rs 150
CP = Rs 250

Since CP > SP
Therefore, Loss = CP-SP = Rs 100

Loss% $=\frac{100}{250}\times 100=40\mathrm{\%}$

Question: 2(a) Convert each part of the ratio to a percentage:

3 : 1

Answer: (a) 3 : 1
The total sum of the ratio is 3 + 1 = 4
Therefore, the percentage of the first part
$\frac{3}{4}\times 100=75\mathrm{\%}$

Percentage of the second part
$=\frac{1}{4}\times 100=25\mathrm{\%}$

Question: 2(b) Convert each part of the ratio to a percentage:

2 : 3: 5

Answer: (b) 2 : 3: 5
Sum of the ratio part = 2 + 3 + 5 = 10
Therefore, the percentage of the first part
$=\frac{2}{10}\times 100=20\mathrm{\%}$

The percentage of the second part
$=\frac{3}{10}\times 100=30\mathrm{\%}$

The percentage of the third part
$=\frac{5}{10}\times 100=50\mathrm{\%}$

Question: 2(c) Convert each part of the ratio to a percentage:

1:4

Answer: (c) 1:4
Sum of the ratio part = 4 + 1 = 5
therefore, the percentage of the first part
$=\frac{1}{5}\times 100=20\mathrm{\%}$

the percentage of the second part
$=\frac{4}{5}\times 100=80\mathrm{\%}$

Question: 2(d) Convert each part of the ratio to a percentage:

1 : 2 : 5

Answer: (d) 1 : 2 : 5

Sum of the ratio part = 1 + 2 + 5 = 8
therefore, the percentage of the first part
$=\frac{1}{8}\times 100=12.5\mathrm{\%}$

The percentage of the second part
$=\frac{2}{8}\times 100=25\mathrm{\%}$

The percentage of the third part
$=\frac{5}{8}\times 100=62.5\mathrm{\%}$

Question: 3 The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Answer: Given that,
Initial population = 25,000
Final population = 24,500

Total decrement = 1000 (25000 - 24500)
Therefore, the percentage in decrease
$\Rightarrow \frac{500}{25000}\times 100=2\mathrm{\%}$

Question: 4 Arun bought a car for Rs 3,50,000. The next year, the price went up to Rs 3,70,000. What was the Percentage of price increase?

Answer: Given that,
Original price (OP) = 3,50,000
Increased price (IP) = 3,70,000

Therefore, increase in price = OP - IP = 20,000

Thus, the percentage of the increased price
$=\frac{20,000}{3,50,000}\times 100=5.71\mathrm{\%}$

Question: 5 I buy a T.V. for Rs 10,000 and sell it at a profit of 20%. How much money do I get for it?

Answer: Here we have,
CP = Rs 10,000
Profit = $20\mathrm{\%}$
SP =?

We know that,
$SP=CP(1+\frac{profit}{100})$ ...........(i)

By substituting the values in eq (i), we get

$SP=10,000(1+\frac{20}{100})$
$=10,000\times (\frac{6}{5})$
= Rs 12,000

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

Answer: We have,
Sp of the washing m/c = Rs 13,500
Loss(%) = 20%
CP =?

We know that,
$SP=CP(1-\frac{Loss}{100})$
Putting the values in the above equation we get;

$$\begin{gathered} 13,500=CP(1-\frac{20}{100}) \\ 13500=CP(1-\frac{1}{5})=CP\times (\frac{4}{5}) \end{gathered}$$
Therefore,
$CP=13,500\times \frac{5}{4}=Rs16875$

Question: 7(i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.

Answer: We have,
The ratio of calcium, carbon and oxygen in the chalk = 10 : 3: 12
Now, Sum of the ratio = 10 + 3 + 12 = 25

Therefore, the percentage of carbon in the chalk
$\Rightarrow \frac{3}{25}\times 100=12\mathrm{\%}$

Question: 7(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?

Answer: We have the weight of the carbon = 3g
Therefore, the weight of the chalk
$\Rightarrow \frac{3}{3}\times 25g=25g$

Hence the weight of the chalk is 25g

Question: 8 Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?

Answer: Here we have,
CP of the book = Rs 275
Loss = 15%

We know that,
$SP=CP(1-\frac{Loss}{100})=275(1-\frac{15}{100})$
$=275\times (\frac{85}{100})$
$=Rs233.75$

Hence the required selling price is Rs 233.75

Question: 9(a) Find the amount to be paid at the end of 3 years in this case:

Principal = Rs 1,200 at 12% p.a.

Answer: Given that,
Principal (PA)= Rs 1,200 at 12% p.a. (Rate of interest)
and T = 3 year
We know that,
$Interest=\frac{P\times R\times T}{100}$
Now, putting the values in the above equation, we get
$\Rightarrow \frac{1200\times 12\times 3}{100}=Rs432$

So, Amount = PA + PI = 1200 + 432 = Rs 1632

Question: 9(b) Find the amount to be paid at the end of 3 years in this case:

(b) Principal = Rs 7,500 at 5% p.a.

Answer: Given that,
Principal (PA)= Rs 7,500 at 5% p.a. (Rate of interest)
and T = 3 year
We know that,
$Interest=\frac{P\times R\times T}{100}$
Now, putting the values in the above equation, we get
$\Rightarrow \frac{7,500\times 5\times 3}{100}=Rs1125$

So, Amount = PA + PI = 7,500 + 1125 = Rs 8625

Question: 10 What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?

Answer: Given that,
Principal = Rs 56,000
Interest = Rs 280
Time = 2 year

Rate= ?

Therefore,
$Rate=\frac{100\times I}{P\times T}$
$=\frac{100\times 280}{56,000\times 2}=0.25\mathrm{\%}$

Question: 11 If Meena gives an interest of Rs 45 for one year at a 9% rate p.a. What is the sum she has borrowed?

Answer: Given that,
Principal =?
Interest = Rs 45
Time = 1 year

Rate= 9% p.a

Therefore,
$Principal=\frac{100\times I}{R\times T}$
$=\frac{100\times 45}{9\times 1}=Rs500$

Hence she borrowed a total of Rs 500

Comparing Quantities Maths Class 7 Chapter 7-Topics

8.1 Introduction

8.2 Equivalent Ratios

8.3 Percentage – Another Way Of Comparing Quantities

8.3.1 Meaning Of Percentage

8.3.2 Converting Fractional Numbers To Percentage

8.3.3 Converting Decimals To Percentage

8.3.4 Converting Percentages To Fractions Or Decimals

8.3.5 Fun With Estimation

8.4 Use Of Percentages

8.4.1 Interpreting Percentages

8.4.2 Converting Percentages To “How Many”

8.4.3 Ratios To Percents

8.4.4 Increase Or Decrease As a percent

8.5 Prices Related To An Item Or Buying And Selling

8.5.1 Profit Or Loss As A Percentage

8.6 Charge Given On Borrowed Money Or Simple Interest

8.6.1 Interest For Multiple Years

NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities - Points to Remember

Percentage:

Percent = $\frac{\text{Part}}{\text{Whole}}\times 100$

Part (How Many) = $\frac{\text{Percent}}{100}$ × Whole

Conversion:

Fraction or Decimal to Percent = (Fraction or Decimal) × 100

Percent to Fraction or Decimal = $\frac{\text{Percent}}{100}$

Increase or Decrease as Percent:

Percentage Increase or decrease = $\frac{\text{Amount of Change}}{\text{Original Amount}}\times 100$

Cost Price (CP) and Selling Price (SP):

If $\mathrm{CP}<\mathrm{SP}\to$ Profit $=\mathrm{SP}-\mathrm{CP}$

If $\mathrm{CP}>\mathrm{SP}\to$ Loss $=\mathrm{CP}-\mathrm{SP}$

If $\mathrm{CP}=\mathrm{SP}\to$ No profit, No loss

Profit percent = $\frac{\text{Profit}}{\text{CP}}\times 100$

Loss Percent = $\frac{\text{Loss}}{\text{CP}}\times 100$

Simple Interest (S.I.):

S.I. = $\frac{\text{Principal times Rate times Time}}{100}$

Amount = Principal + Interest

NCERT Solutions for Class 7 Maths Chapter-Wise

NCERT Solutions for Class 7 Subject Wise

Students can access the subject-wise solutions for Class 7 through the link below.

• NCERT Solutions for Class 7 Maths
• NCERT Solutions for Cass 7 Science

Students can also check NCERT Books and NCERT Syllabus here:

• NCERT Syllabus Class 7 Maths
• NCERT Books Class 7
• NCERT Syllabus Class 7