NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

You must have come across many real-life situations involving comparisons such as discounts during shopping, profit or loss in business, interest on savings, or percentage increases in prices. In mathematics, these comparisons are made using quantities like ratios, percentages, and proportions. For example, when prices increase by a certain percentage, the overall cost of a product also changes. Similarly, calculating interest helps us understand how much extra we earn or pay over time. These NCERT Solutions are created by the expert team at craeers360, keeping the latest syllabus and pattern of CBSE 2025-26.

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This Story also Contains

1. Comparing Quantities Class 8 Questions And Answers PDF Free Download
2. Comparing Quantities Class 8 Solutions - Important Formulae
3. Comparing Quantities Class 8 Maths NCERT Solutions (Exercise)
4. NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities - Topics
5. NCERT Solutions for Class 8 Maths: Chapter Wise
6. Importance of Solving NCERT Questions of Class 8 Maths Chapter 7
7. NCERT Solutions for Class 8 - Subject Wise
8. NCERT Books and NCERT Syllabus

In Comparing Quantities Class 8, you will learn to deal with practical problems involving percentages, profit and loss, discount, simple interest, and compound interest. All these concepts are explained in the NCERT textbook using solved examples, real-life word problems, and easy-to-follow steps. Through these NCERT solutions for Class 8 Maths Chapter 7 Comparing Quantities, students can cover all the problems related to ratios, percentages, discounts, profit-loss, and simple compound interest. Here you will get the detailed NCERT Solutions for Class 8 Maths.

Comparing Quantities Class 8 Questions And Answers PDF Free Download

Comparing Quantities Class 8 Solutions - Important Formulae

• Profit: Profit = Selling price - Cost price

• Loss: Loss = Cost price - Selling price

• Profit or Loss Determination:

• If SP > CP, then it is a profit.

• If SP = CP, then it is neither profit nor loss.

• If CP > SP, then it is a loss.

• Discount: Discount = Marked Price - Sale Price

• Discount Percentage: Discount % = (Discount × 100) / Marked Price

• Profit Percentage: Profit Percentage = (Profit / Cost Price) × 100

• Loss Percentage: Loss Percentage = (Loss / Cost Price) × 100

• Percentage Increase: Percentage Increase = (Change in Value / Original Value) × 100

• Simple Interest: Simple Interest = (Principal × Rate × Time) / 100

• Compound Interest Formula: Compound Interest = Amount - Principal

• Sales Tax or VAT: Sales tax or VAT = (Cost Price × Rate of Sales Tax) / 100

• Billing Amount: Billing Amount = Selling price + VAT

Comparing Quantities Class 8 Maths NCERT Solutions (Exercise)

Question 1(I): In a primary school, the parents were asked the number of hours they spend per in helping their children to do homework. There were 90 parents who helped $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours. The distribution of parents according to the time for which they said they helped is given in the adjoining figure; 20% helped for more than $1\frac{1}{2}$ hours; 30% helped for $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours; 50% did not help at all.

Using this, answer the following :

(i) How many parents were surveyed?

Answer:

Let the number of parents surveyed be X.

30% of them helped for $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours.

In total, there were 90 such parents.

$\therefore$ 30% of X = 90

$$\begin{gathered} \Rightarrow \frac{30}{100}\times X=90 \\ \Rightarrow X=90\times 100/30=300 \end{gathered}$$

Therefore, there were a total of 300 parents who were surveyed.

Question 1(ii): In a primary school the parents were asked about the number of hours they spend per day helping their children to do homework. There were 90 parents who helped for $\frac{1}{2}$ to $1\frac{1}{2}$ hours. The distribution of parents according to the time for which they said they helped is given in the adjoining figure; 20% helped for more than $1\frac{1}{2}$ hours; 30% helped for $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours; 50% did not help at all.

Using this, answer the following :

(ii) How many said that they did not help?

Answer: From the previous question number of parents surveyed X = 300.

Given that, 50% did not help.

Therefore, the number of parents that did not help = 50% of 300 = 150.

Question (iii): In a primary school, the parents were asked about the number of hours they spend pery in helping their children to do homework.There were 90 parents who helped for $\frac{1}{2}$ hour

to $1\frac{1}{2}$ hours. The distribution of parents according to the time for which they said they helped is given in the adjoining figure; 20% helped for more than $1\frac{1}{2}$ hours; 30% helped for $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours; 50 % did not help at all.

Using this, answer the following :

(iii) How many said that they helped for more than $1\frac{1}{2}$ hours?

Answer:

From the first question number of parents surveyed X = 300

Given, 20% helped for more than $1\frac{1}{2}$ hours.

Therefore, number of such parents = 20% of 300 = 2 x 30 = 60.

Question 1(a): A shop gives 20% discount. What would the sale price of each of these be?

(a) A dress marked at Rs 120.

Answer:

The marked price of dress = Rs120

Discount rate = 20%

$$\begin{gathered} Totaldiscount=MarkedPrice\times \frac{Discountrate}{100} \\ Totaldiscount=120\times \frac{20}{100} \\ Totaldiscount=Rs.24 \end{gathered}$$

Selling price (SP) = Marked price - Total discount = Rs (120- 24) = Rs 96.

Question 1(b): A shop gives a 20% discount. What would the sale price of each of these be?

(b) A pair of shoes marked at Rs 750

Answer:

The marked price of shoes=Rs 750

Discount rate = 20%

$$\begin{gathered} Totaldiscount=MarkedPrice\times \frac{Discountrate}{100} \\ Totaldiscount=750\times \frac{20}{100} \\ Totaldiscount=Rs.150 \end{gathered}$$

Selling price= Marked price- Total discount =Rs (750- 150) = Rs 600.

Question 1(c): A shop gia ves 20% discount. What would the sale price of each of these be?

(c) A bag marked at Rs 250

Answer:

The marked price of the bag = Rs 250

Discount rate=20%

$$\begin{gathered} Totaldiscount=MarkedPrice\times \frac{Discountrate}{100} \\ Totaldiscount=250\times \frac{20}{100} \\ Totaldiscount=Rs.50 \end{gathered}$$

Selling price = Marked price - Total discount = Rs (250- 50) = Rs 200.

Question 2: A table market maratd ar Rs 15,000 is available for Rs 14,400. Find the discount given and the discount.

Answer:

Marked price = Rs 15000

Selling Price = Rs 14400

We know that discount amount = Marked price - Selling price

Discount amount = Rs 15000-Rs 14400 = Rs 600

$$\begin{gathered} Discountpercent=\frac{Discount}{MarkedPrice}\times 100 \\ \therefore Discountpercent=\frac{600}{15000}\times 100=4\mathrm{\%} \end{gathered}$$

Therefore, Discount percentage = 4%

Question 3: An almirah is sold at Rs 5,225 after allowing a discount of 5%. Find its marked price.

Answer:

Selling price = Rs 5225

Discount percent = 5%

$$\begin{gathered} Sellingprice=Markedprice-DiscountAmount \\ Sellingprice=Markedprice-Markedprice\times \frac{5}{100} \\ Sellingprice=Markedprice\times \frac{95}{100} \\ Markedprice=Sellingprice\times \frac{100}{95} \\ Markedprice=\frac{100}{95}\times 5225 \end{gathered}$$

Therefore, the marked price = Rs 5500

Question 1(a): Find the ratio of the following.

The speed of a cycle is 15 km per hour to the speed of a scooter is 30 km per hour.

Answer:

The ratio of the speed of the cycle to the speed of the scooter =

$\frac{15km/hour}{30km/hour}=\frac{15}{30}=\frac{1}{2}$

= 1:2

Question 1(b): Find the ratio of the following.

5m to 10km

Answer:

To find the ratio, we need to make both quantities of the same unit.

We know, 1 km = 1000 m.

$\therefore$ 10 km = 10x1000 m = 10000 m

Therefore the required ratio =

$\frac{5m}{10\times 1000m}=\frac{5}{10000}=\frac{1}{2000}$

= 1 : 2000

Question 1(c): Find the ratio of the following.

50 paise to Rs. 5

Answer:

To find the ratio, we first make the quantities of the same unit.

We know, that Rs. 1 = 100 paisa. $\therefore$ Rs. 5 = 5x100 = 500 paisa

Therefore, the required ratio =

$\frac{50paisa}{Rs.5}=\frac{50paisa}{5\times 100paisa}=\frac{1}{10}$

= 1:10

Question 2 (a): Convert the following ratios into percentages.

(a) 3:4

Answer:

To convert a ratio to a percentage, we multiply the ratio by 100%.

$\therefore$ The required percentage of the ratio 3:4 =

$\frac{3}{4}\times 100\mathrm{\%}=75\mathrm{\%}$

Question 2(b): Convert the following ratios to percentages.

(b) $2:3$

Answer:

To convert a ratio to a percentage, we multiply the ratio by 100%.

$\therefore$ The required percentage of ratio 2:3 =

$\frac{2}{3}\times 100\mathrm{\%}=66.67\mathrm{\%}$

Question 3: 72% of 25 students are interested in mathematics. How many are not interested in mathematics?

Answer:

Given;

72% of 25 students are interested in mathematics

$\therefore$ Percentage of students not interested in mathematics = (100 - 72)% = 28 %

$\therefore$ The number of students not interested in mathematics = 28% of 25

$=\frac{28}{100}\times 25=\frac{28}{4}$

$=7$

Therefore, 7 students (out of 25) are not interested in mathematics.

Question 4: A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Answer:

Given,

Win percentage of the team = 40%

This means that they won 40 matches out of 100 matches played.

$\therefore$ They won 10 matches out of $\frac{100}{40}\times 10$ = 25 matches played.

Therefore, they played a total of 25 matches in all.

Question 5: If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning?

Answer:

Let the amount of money Chameli had in the beginning = Rs. X

She spends 75% of the money.

$\therefore$ Percentage of money left = (100 - 75)% = 25%

Since she has Rs. 600 left.

$\therefore$ 25% of X = 600

$\Rightarrow \frac{25}{100}\times X=600$

$\Rightarrow \frac{X}{4}=600$

$\Rightarrow X=600\times 4$

X =2400

Therefore, Chameli had Rs. 2400 with her in the beginning.

Question 6: If 60% of people in the city like cricket, 30% like football, and the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

Answer:

Given,

Total number of people = 50 lakhs

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

Since remaining people like other games,

$\therefore$ Percentage of people who like other games = {100 - (60 + 30) } = (100 - 90) = 10%

10% of the people like other games.

Now,

$\therefore$ Number of people who like cricket = 60% of 50 lakhs

$=\frac{60}{100}\times 50=30lakhs$

$\therefore$ Number of people who like football = 30% of 50 lakhs

$=\frac{30}{100}\times 50=15lakhs$

$\therefore$ Number of people who like other games = 10% of 50 lakhs

$=\frac{10}{100}\times 50=5lakhs$

Question 1: During a sale, a shop offered a discount of 10% on the market prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

Answer:

Since the 10% discount is on all the items, we can calculate the Selling price by totaling the Cost price of all items bought.

Now,

Total Cost price(CP) of the items he bought = CP of a pair of jeans + CP of two shirts = Rs.(1450 + 850 + 850) = Rs. 3150

$\therefore$ The selling price of these items = (100- 10)% of Rs. 3150 = 90% x Rs.3150 = Rs. 2835

$\therefore$ The customer has to pay Rs. 2835.

Question 2: The price of a TV is Rs 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Answer:

Given,

Cost price of the TV = Rs. 13000

Sales tax at the rate of 12%

$\therefore$ Selling price = CP + Sales Tax = CP + 12% of CP

= 112% of CP =

$\frac{112}{100}\times 13000=Rs.14560$

$\therefore$ Vinod will have to pay Rs. 14,560.

Question 3: Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs 1,600, find the marked price.

Answer:

The discount given was 20% which means if CP is Rs. 100 then the SP is Rs. 80

$\therefore$ If SP is Rs. 80 then CP is Rs. 100

For SP of Rs. 1600, CP

$=\frac{100}{80}\times 1600=Rs.2000$

$\therefore$ The marked price is Rs. 2000.

Question 4: I purchased a hair dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.

Answer:

Given,

VAT = 8%

Let the original price be Rs. 100

Original price + VAT = Rs. 100 + Rs. $\left(8\mathrm{\%}of100\right)$

Original price + VAT = Rs. 100 + Rs. 8 = Rs. 108

$\therefore$ If the price after VAT is Rs.5400, then the price before VAT is $Rs.\frac{100}{108}\times 5400=Rs.5000$

$\therefore$ The price before VAT was added is Rs. 5000.

Question 5: An article was purchased for Rs 1,239 including a GST of 18%. Find the price of the article before GST was added.

Answer:

Given,

GST = 18%

Cost with GST included = Rs. 1239

Cost without GST = x Rs.

$x+(\frac{18}{100}\times x)=1239$

Cost before GST+ GST = cost with GST

$x+(\frac{9x}{50})=1239$

x = 1050

Price before GST = 1050 rupees

Question 1(i): The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) Find the population in 2001.

Answer:

Let the population in 2001 be P

Compound rate of increase = 5% p.a.

The population in 2003 will be more than in 2001

Time period, n = 2 years (2001 to 2003)

$\therefore$ $$\begin{gathered} 54000=P(1+\frac{5}{100}{)}^2 \\ ⟹P=\frac{54000\times 100\times 100}{105\times 105}\approx 48980 \end{gathered}$$

Therefore, the population in 2001 was 48980 (approx)

Question 1(ii): The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(ii) what would be its population in 2005?

Answer:

Let the population in 2001 be P'

Compound rate of increase = 5% p.a.

The population in 2005 will be more than in 2003

Time period, n = 2 years (2003 to 2005)

$\therefore$ $$\begin{gathered} {P}^{\prime }=54000(1+\frac{5}{100}{)}^2 \\ ⟹P^{\prime }=\frac{54000\times 105\times 105}{100\times 100}\approx 59535 \end{gathered}$$

Therefore, the population in 2005 will be 59535 (approx)

Question 2: In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Answer:

Given,

The initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours = $P(1+\frac{R}{100}{)}^n$

$=506000(1+\frac{2.5}{100}{)}^2\approx 531616$

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

Question 3: A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Answer:

Given,

Principal = Rs 42,000

Rate of depreciation = 8% p.a

$\therefore$ Reduction = 8% of Rs 42000 per year

$=\frac{42000\times 8\times 1}{100}=Rs3360$

$\therefore$ Value at the end of 1 year = Rs (42000 – 3360) = Rs 38,640

NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities - Topics

• Recalling Ratios and Percentages
• Finding the Increase or Decrease Percent
• Finding Discounts
• Prices Related to Buying and Selling (Profit and Loss)
• Sales Tax/Value Added Tax/Goods and Services Tax
• Compound Interest
• Deducing a Formula for Compound Interest
• Rate Compounded Annually or Half Yearly (Semi-Annually)
• Applications of Compound Interest Formula

NCERT Solutions for Class 8 Maths: Chapter Wise

Importance of Solving NCERT Questions of Class 8 Maths Chapter 7

• 
  
• This chapter deals with percentages, profit & loss, discounts, and simple and compound interest – all of which are used in daily life, shopping, banking, and finance.
• 
  
• Solving NCERT questions helps strengthen core concepts like ratio, proportion, percentage increase/decrease, and interest calculation.
• 
  
• Regular practice boosts speed and accuracy in solving numerical problems, which is useful for competitive exams later on.
• 
  
• Detailed, step-by-step explanations for each problem make it easy for students to understand and apply mathematical concepts related to ratios, percentages, and discounts.
• 
  
• It lays a strong base for commercial mathematics topics in higher classes, especially classes 9 and 10.

NCERT Solutions for Class 8 - Subject Wise

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

NCERT Books and NCERT Syllabus

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8