case study based on comparing quantities class 8

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Comparing Quantities

Quantity is the extent, size or sum of countable or measurable discrete events, objects or phenomena. One can compare quantities if one has knowledge of ratios, discounts, and percentages. Let’s us learn more about them in detail in the upcoming sections.

Ratios and Percentages

• In the ratio, a:b, a is called antecedent and b is called consequent.
• The ratio of two numbers is always expressed in the lowest form. For example, If the ratio of two quantities is 17:34. Then, reduce it to 1:2 [17 X 2 = 34]
• The order of the ratio is important. If we interchange the antecedent and the consequent then we get a different ratio. For example, 1:2 ≠ 2:1

The antecedent and consequent in a ratio are always whole numbers. When they are not, convert them into whole numbers.

• Multiple percentages are not additive in nature.

EXAMPLE 1: Two numbers are in the ratio 1:23. On adding 4 to the first and 5 to the second, their ratios become 1:13. Find the numbers.

SOLUTION:  Let the numbers be 1x and 23x.

Adding 4 to the first number, we get= 1x + 4

Adding 5 to the second number, we get= 23x + 5

The ratio of new numbers are in the ratio (1x + 4) : (23x + 5) which is equal to 1:13

Thus, (1x + 4) : (23x + 5) = 1:13

(1x + 4)/(23x + 5) = 1/13

13(1x + 4) = 1(23x + 5)             [Doing cross multiplication]

13x + 52 = 23x + 5

x = 4.7 Hence, the numbers are 4.7 and 23X4.7= 108.1 [ANS]

EXAMPLE 2:  Out of the salary of Rs.123456, I kept 1/2 as savings. Out of the remaining money, I spend 47% on food and 18% on rent. How much do I spend on food and house rent?

SOLUTION:  My salary is Rs. 123456

My savings s = 1/2  x  123456 = Rs. 0.61728

Remaining Money= Rs. (123456 - 61728) = Rs. 61728

Amount of money I spend on food = 47% x 61728 = (47/100) x 61728 = 0.47 x 61728

                                                                                                       = Rs. 29012.16

Amount of money I spend on rent = 18% x 61728 = 18/100 x 61728 = 0.18 x 61728

                                                                                                     = Rs. 11111.04

Hence, I spend Rs.29012.16 on food and Rs.11111.04 on rent. 

Increase or Decrease Percent

EXAMPLE 1: The population of a town in a particular year increased by 10%. Next year it decreased by 10%. Find the net increase or decrease percent in the initial population.

SOLUTION: Let the initial population of the town be x. Population increased by= 10% X x= 0.1 X x =0.1x Increase Population= (x+0.1x)= 1.1x Next year, population decreased by= 10% X 1.1x= 0.1 X 1.1x= 0.11x Decreased population= 1.1x- 0.11x= 0.99x 0.99x<x, clearly the population decreased.

Decrease percent = x - 0.99x/x x 100 = 1%

Hence, the decrease in population is 1%. 

EXAMPLE 2: The price of onions increased by 12%. By how much percent should you reduce your consumption in the house so that your expenditure on onions does not increase?

SOLUTION:  Decrease in consumption= {[%Price Increase]/[100 + % Price Increase]} X 100

                                                     = 12/[100 + 12] x 100

                                                     = (12/112) x 100 = 10.7%

 Hence, I should reduce my consumption by 10.7% if the price of onions has increased by 12%.

Hence, I should reduce my consumption by 10.7% if the price of onions has increased by 12%.

Finding Discounts

• Subtract the discount from the original price For example, Discounted price for the shoes = Rs.(3456 - 414.72)                                              = Rs. 3041.28

EXAMPLE 1: An item marked at Rs.1234 is sold for Rs.567. What is the discount and discount%?

SOLUTION:  Discount Amount= Rs.(1234 – 567)= Rs. 667 Percentage Discount = (Discount Amount)/(Marked Price) X 100                               = 667/1234 X 100                               = 54.05% Hence, Discount= Rs.667; Percentage Discount= 54.05% 

EXAMPLE 2: The list price of a dress is Rs.8901. A discount of 23% is announced on sales. What are the amount of discount on it and its sales price? SOLUTION: Amount of discount= 23% X Rs.8901= Rs.2047.23 Sales Price= Rs.[8901- 2047.23] =Rs.6853.77 Hence, discount amount= Rs2047.23; sales price=Rs.6853.77. [Ans]

Profit & Loss

Cost Price: The price at which an article is purchased is called its cost price. It is abbreviated as CP. 

Selling Price: The price at which an article is sold is called its selling price. It is abbreviated as SP.

If the selling price is more than the cost price, the seller is said to have a profit or gain.

If the selling price is less than the cost price, the seller is said to have incurred a loss.

EXAMPLE 1:  Arya sold two digital cameras at Rs. 30,000 each. On one she gains 12% and on the other, she loses 12%. What percent does she gain/loss on the transaction? SOLUTION:  For the first digital camera:

 Gain= 12%

Let cost price(CP) = Rs. 100

Therefore, Selling Price(SP) = Rs. [100 + 12%/100] X 100= Rs.112

When the selling price is Rs.112, Cost Price is Rs. 100.

When the selling price is Re. 1, Cost Price = Rs. 100/112

When the selling price is Rs.30, 000, Cost Price = Rs. (100/112) x 30000 = Rs. 26785.71

For the second digital camera:

Let cost price is Rs. 100.

Therefore, Selling Price-Rs. (100 - 12) = Rs. 88

When the selling price is Rs.88, cost price is Rs. 100.

When the selling price is Re. 1, cost price is Rs. 100/88.

When the selling price is Rs.30000, Cost Price is Rs. (100/88) X 30000 = Rs.34090.90

The total cost price of two digital cameras = Rs (26785.71+34090.90) = Rs.60876.61

The total selling price of two digital cameras = Rs. 2 X 30000= Rs.60,000

Thus, cost price > selling price, so there is a loss.

Loss CP - SP = Rs. (60876.61 - 60000) Rs. 876.61 

Loss Percentage= Loss/CP X 100 = 876.61/60876.61 X 100 = 1.4% 

Hence, Arya loses 1.4% on the whole transaction. Alternative Method: If two commodities are sold at the same price and the seller gets the same amount of gain & loss on each commodity then the seller always incurs a loss on the whole transaction.

Loss% = [Common Loss or Gain%/10] 2  = [12/10]² = 1.44%. 

EXAMPLE 2:  Krishna bought a laptop for Rs. 123456 and spent Rs.7890 on its spares. He later sold the laptop for Rs. 234567. Find her gain or loss percent.

  SOLUTION:  Cost price includes the overhead expenses also.

Therefore, cost price = Rs. (123456 + 7890) = Rs. 131346

And SP = Rs. 234567

Since Selling Price > Cost Price, there is a profit.

Profit = SP - CP = Rs. [234567 - 131346] = Rs.103221

Profit% = Profit/CP X 100 = 103221/131346 x 100 = 78.5%

Hence, her total profit is 78.5%. 

Sales Tax / Value-Added Tax

Sales Tax is calculated on the sales price.

• Value Added Tax (VAT) is a tax on the value added at each transfer of goods, from the original manufacturer to the retailer. For example, Assuming the rate of tax is 12% and a person purchases an article for Rs.345. The tax he pays= Rs.(12% X 345) = Rs. 41.4 Now, if he sells the same article for Rs. 678 The tax he recovers= Rs.(12% X 678) = Rs. 81.36 VAT = Tax recovered on the sale – Tax he paid on the purchase        = Rs.(81.36 – 41.4) = Rs.39.96

EXAMPLE 1: Swathi bought the following articles from a departmental store

Calculate the total bill paid, including sales tax paid by Swathi to the departmental store.

SOLUTION: Cost price of 3 dress= Rs.1234 X 3 = Rs.3702 Sales tax on dresses = Rs. (12% X 3702) = Rs. 444.24 Amount paid for 3 dresses = Rs(3702+444.24) = Rs 4146.24

Cost price of 2 bags= Rs 567 X 2 = Rs 1134 Sales Tax on 2 bags = Rs (3% X 1134) = Rs 34.02 Amount paid for 2 bags = Rs(34.02+1134) = Rs1168.02

Cost Price of 1 Laptop = Rs89012 Sales Tax on Laptop = Rs. (4% X 89012) = Rs.3560.48 Amount paid for 1 laptop = Rs. (89012 + 3560.48) = Rs.92572.48

Cost Price of 2 dinner set = Rs. (2 X 345) = Rs. 690 Sales Tax on 2 dinner sets= Rs(5% X 690) = Rs.34.5 Amount paid for 2 dinner sets = Rs.(34.5+690) = 724.5

Hence, total amount of bill = Rs.[4146.24+1168.02+92572.48+724.5) = Rs.98611.24

EXAMPLE 2: A shopkeeper bought an AC at a discount of 12% from the wholesaler, the printed of the AC is Rs.34567. The shopkeeper sells it to a consumer at a discount of 8% on the printed price. If the rate of sales tax is 9%, find i) the VAT paid by the shopkeeper. ii) the total amount that the consumer pays for the AC.

SOLUTION: M.P. = Rs.34567 Discount= 12% for shop-owner Discount = 8% for consumer Sales Tax = 9%

Discount for shopkeeper = Rs.(12% X 34567) = Rs. 4148.04 Cost for Shopkeeper = Rs(34567 – 4148.04) = Rs. 30418.96 Cost for Consumer = Rs(34567 – (8% X 34567)) = Rs. 31801.64

• Tax charged by the shopkeeper= Rs.(9% X 31801.64) = Rs. 2862.1476 Tax paid by the shopkeeper = Rs.(9% X 30418.96) = Rs.2737.7064 Hence, VAT paid by the shopkeeper = Rs.(2862.1476 – 2737.7064) = Rs. 124.44
• Amount paid by the consumer = Rs.(31801.64 +2862.1476) = Rs. 34663.78[Ans]

Compound Interest

• Compound Interest is the interest calculated on the principal amount and also on the accumulated interest of previous periods.

When interest is compounded annually:

When interest is compounded half-yearly:

When interest is compounded quarterly:

When interest is compounded annually but the time is in fraction, say 1 2/3 years

When rates are different for different years say R1%, R2%, and R3% for three consecutive years, then:

EXAMPLE 1:  Calculate the amount and compound interest on Rs.12, 345 for 6 7/8 years at 9% per annum compounded annually.

Amount = Rs. 12345 X [1 + 9/100] 6  x [1 + {(7 x 9/8)/100}] = Rs. 22334.22 

Compound Interest = Amount - P = Rs.(22334.22 - 12345) = Rs. 9989.22 

EXAMPLE 2:  Sonal borrowed Rs.23456 from a Bank to buy a bike at a rate of 7% p.a. compounded yearly. What amount will she pay at the end of 8 years and 9 months to clear the loan?

SOLUTION: 9 months = 9/12 years = 3/4 years

Total years = 8 ¾  years

Amount = Rs.23456 x [1 + 7/100] 8  × [1 +3 x 7/4/100] = 42,417.61 

Practice these questions

Q1) The ratio of the number of boys and a number of girls in a school of 1234 students is 5:6. If 78 new boys are admitted, find how many new girls should be admitted to make the ratio 1:2.

Q2) Rachel gets 91 marks in her exam. These are 23% of the total marks. Find the Maximum number of marks.

Q3) Increase 123 by 4% and then decrease it by 5%. Find the final number.

Q4) The marked price of a water cooler is Rs. 14650. The shopkeeper offers an off-season discount of 8% on it. Find its selling price.

Q5) Monica bought 2 dozen eggs for Rs. 156. Since 7 of them broke, he incurred a loss of Rs.120 on selling them. What was the selling price of one egg?

Q6) A manufacturer printed the price of his goods as Rs.1120 per article. He allowed a discount of 4% to the wholesaler who in his turn allowed a discount of 3% on the printed price to the retailer. If the prescribed rate of sales tax on the goods is 5% and the retailer sells it to the consumer at the printed price then find the value-added tax paid by the wholesaler and the retailer.

Q7) Vivan took a loan of Rs. 654987 from the bank. If the company charges interest at 15% per annum, compounded quarterly, what amount will discharge his debt after one year?

• The percentage is not cumulative in nature.
• Sales tax is are collected by the retailer when the final sale in the supply chain is reached via a sale to the end consumer.
• Compound Interest is the interest calculated on the initial principal and also on the accumulated interest of previous periods of a deposit or loan.

Quiz for Comparing Quantities

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NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 8 are in accordance with the latest CBSE guidelines and marking schemes.

Class 8 Maths Chapter 8 Exercise 8.1 Solutions

Class 8 Maths Chapter 8 Exercise 8.2 Solutions

Class 8 Maths Chapter 8 Exercise 8.3 Solutions

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• NCERT Solutions for Class 8 Maths Chapter 8 - Comparing Quantities
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NCERT Solutions for CBSE Class 8 Maths Chapter 8: Comparing Quantities Free PDF Download

NCERT Solutions for Chapter 8 Class 8 Maths are available here in a PDF format. Subject experts with years of experience have prepared these solutions as per the updated CBSE Class 8 guidelines. The PDF covers all the exercise questions of Class 8 Maths Chapter 8 that will aid in your revision and help you perform well in your upcoming Maths exams. 

These NCERT Solutions for Class 8 are provided in simple steps to assist students in solving problems in the best way possible. One of the benefits of referring to the NCERT Solutions for Class 8 Maths Chapter 8 is that students will get a conceptual understanding of the topic. Class 8 students can download NCERT Solutions for Maths Chapter 8 PDF for free from Vedantu.

What Does Comparing Quantities Mean in Maths?

In Maths, comparing quantities means examining the differences between numbers, quantities or values to decide whether the given quantity is greater, smaller, or equal to another quantity.

Topics Covered in Class 8 Maths Chapter 8

8.1 Introduction

8.2: Equivalent Ratio

8.3: Percentage: Another Way of Comparing Quantities

8.3.1: Meaning of Percentage

8.3.2: Converting Fractional Number To Percentage

8.3.3: Converting Decimal To Percentage

8.3.4 Converting Percentage To Fraction or Decimals

8.3.5: Fun With Estimation

8.4: Use of Percentage

8.4.1: Interpreting Percentage

8.4.2: Converting Percentage To How Many

8.4.3: Ratios To Percents

8.4.4: Increase or Decrease As Percents

8.5: Price Related To an Item or Buying and Selling

8.5.1: Profit or Loss as Percentage

8.6: Charges that are given on borrowed money or simple interest

8.6.1: Interest for Multiple years

Access NCERT Solutions for Class-8 Maths Chapter 8 – Comparing Quantities

Exercise: 8.1

1. Find the ratio of the following:

(a) Speed of a cycle $15$km per hour to the speed of scooter $30$km per hour.

Ans: Speed of a cycle $ = $ $15$ km

Speed of scooter $ = $$30$ km

Ratio of the speed of a cycle to the speed of scooter 

$ = \dfrac{{15}}{{30}}$

$ = \dfrac{1}{2}$  

The required ratio is $1:2$ .

(b) $5$m to $10$km.

Ans:  $5$ m to $10$ km.

Since 1km $ = $$1000$ m

$ \Rightarrow $$\dfrac{{5\;{\text{m}}}}{{10\;{\text{km}}}}\; = \,\dfrac{5}{{10}} \times \dfrac{1}{{1000}}$

$ = \dfrac{1}{{2000}}$

$ \Rightarrow 1:2000$

The required ratio is $1:2000$ .

(c) $50$ paise to Rs $5$.

Ans:   $50$ paise to Rs $5$

Since Rs 1 $ = $ $100$ paise      

$ \Rightarrow \dfrac{{50paise}}{{Rs5}}$                      

$ = \dfrac{{50}}{5} \times \dfrac{1}{{100}}$                

$ = \dfrac{1}{{10}}$                     

$ \Rightarrow 1:10$  

The required ratio is $1:10$ .

2. Convert the following ratios to percentages.

Ans: $3:4$ $ = $ $\dfrac{3}{4}$

$ \Rightarrow $$\dfrac{3}{4} \times \dfrac{{100}}{{100}}$                         

$ = $$0.75 \times 100\% $  

The required ratio to percentage is $75\% $

Ans: $2:3$ $ = $ $\dfrac{2}{3}$

$ = \dfrac{2}{3} \times \dfrac{{100}}{{100}}$

$ = \dfrac{{200}}{3}\% $  

$ = 66\dfrac{2}{3}\% $

The required ratio to percentage is $66\dfrac{2}{3}\% $

3. $72\% $ of  $25$ students are good in mathematics. How many are not good in mathematics?

Ans: Total number of students $ = $ 25 .

Percentage of students are good in mathematics  $ = $$72\% $

Percentage of students who are not good in mathematics  $ = $ $(100 - 72)\% $

$ \Rightarrow 28\% $

$\therefore $ Number of students who are not good in mathematics $ = $$28\%  \times 25$

$ \Rightarrow \dfrac{{28}}{{100}} \times 25$

$ \Rightarrow \dfrac{{28}}{4}$

$ \Rightarrow 7$

Students are not good in mathematics $ = 7$

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 ?

Ans: The total number of matches won by the football team $ = 10$ .

Percentage of team $ = 40\% $

The total number of matches played by the team $ = ?$

The total number of matches played by the team 

$ \Rightarrow 40\%  \times x = 10$

$ \Rightarrow \dfrac{{40}}{{100}} \times x = 10$

$ \Rightarrow x = 10 \times \dfrac{{100}}{{40}}$

$ \Rightarrow x = \dfrac{{100}}{4}$

$ \Rightarrow x = 25$

The total number of matches played by the team $ = 25$ .

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

Ans: Chameli’s money after spend $ = 600$

Percentage of money after spend $ = 75\% $

Beginning amount of chameli $ = ?$

Percentage of beginning amount 

$ = \left( {100{\text{ }} - {\text{ }}75} \right)\% \qquad $

Beginning amount of chameli

$ \Rightarrow 25\%  \times x = 600$

$ \Rightarrow \left( {\dfrac{{25}}{{100}}} \right) \times x = 600$

$ \Rightarrow \left( {\dfrac{1}{4}} \right) \times x = 600$

$ \Rightarrow x = 600 \times 4$

$ \Rightarrow x = 2400$

Beginning amount of chameli $ = 2400$ .

6.  If $60\% $ people in city like cricket,$30\% $like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are $50$ lakh, find the exact number who like each type of game.

Ans: Total number of people in city $ = 50$ lakh

Percentage of people like cricket $ = 60\% $

Percentage of people like football $ = 30\% $

Percentage of people like other games $ = ?$

Number of people like each type of game $ = ?$

Percentage of people like other games 

$ = \left( {100 - 60 - 30} \right)\% $  

$ = \left( {100 - 90} \right)\% $

Percentage of people like other games $ = 10\% $

Number of people like cricket 

$ = \left( {60\%  \times 50} \right)$

$ = \left( {\dfrac{{60}}{{100}} \times 50} \right)$

$ = \dfrac{{60}}{2}$

$ = 30$ lakh

Number of people like cricket $ = 30$ lakh

Number of people like football

$ = \left( {30\%  \times 50} \right)$

$ = \left( {\dfrac{{30}}{{100}} \times 50} \right)$

$ = \dfrac{{30}}{2}$

$ = 15$ lakh

Number of people like football $ = 15$ lakh

Number of people like other games

$ = \left( {10\%  \times 50} \right)$

$ = \left( {\dfrac{{10}}{{100}} \times 50} \right)$

$ = \dfrac{{10}}{2}$

$ = 5$ lakh.

Number of people like other games $ = 5$ lakh.

Exercise 8.2

1. A man got a $10\% $ increase in his salary. If his new salary is Rs $1,54,000$ . Find his original salary.

Ans: Percentage of increment in salary $ = 10\% $

New salary $ = 1,54,000$

Original salary $ = x$

Original salary $ + $ Increment $ = $ New salary. 

$ \Rightarrow x + 10\%  \times x = 1,54,000$

$ \Rightarrow x\left( {1 + 10\% } \right) = 1,54,000$

$ \Rightarrow x\left( {1 + \dfrac{{10}}{{100}}} \right) = 1,54,000$

$ \Rightarrow x\left( {\dfrac{{100 + 10}}{{100}}} \right) = 1,54,000$

$ \Rightarrow x\left( {\dfrac{{110}}{{100}}} \right) = 1,54,000$

$ \Rightarrow x = 1,54,000 \times \left( {\dfrac{{100}}{{110}}} \right)$

$ \Rightarrow x = 1400 \times 100$

$ \Rightarrow x = 140000$

The original salary $ = 1,40,000$ .

2. On Sunday $845$ people went to the Zoo. On Monday only $169$ people went. What is the per cent decrease in the people visiting the zoo on Monday?

Ans: People went to zoo in Sunday $ = 845$

People went to zoo in Monday $ = 169$

Percentage decrease in number of people visit in Monday $ = ?$

Percentage decrease in number of people visit in Monday $ = \left( {\dfrac{{Decrease{\text{ }}in{\text{ }}number{\text{ }}of{\text{ }}people{\text{ }}visit{\text{ }}in{\text{ }}Monday}}{{People{\text{ }}went{\text{ }}to{\text{ }}zoo{\text{ }}in{\text{ }}Sunday}}} \right) \times 100\% $

Decrease in number of people visit in Monday

$ = 845 - 169$

Decrease in number of people visit in Monday $ = 676$ .

Percentage decrease in number of people visit in Monday

$ = \left( {\dfrac{{676}}{{845}} \times 100} \right)\% $

$ = 0.8 \times 100\% $

Percentage decrease in number of people visit in Monday $ = 80\% $

3. A shopkeeper buys $80$ articles for Rs $2,400$ and sells them for a profit of $16\% $. Find the selling price of one article.

Ans: Total number of articles $ = 80$

Articles price $ = 2400$

Percentage of profit $ = 16\% $

Selling price of one article $ = ?$

Profit percent $ = \left( {\dfrac{{profit}}{{C.P}}} \right) \times 100$

Selling price of one article $ = $ C.P. $ + $ Profit

Cost of one article 

$ = \left( {\dfrac{{Articles{\text{ }}price}}{{Total{\text{ }}number{\text{ }}of{\text{ }}articles}}} \right)$

$ = \left( {\dfrac{{2400}}{{80}}} \right)$

Cost of one article $ = 30$

$16 = \dfrac{{Profit}}{{30}} \times 100$ Profit $ = \dfrac{{16 \times 30}}{{100}}$

Profit $ = 0.16 \times 30$

Profit $ = 4.80$

$ = $ Rs $\left( {30 + 4.80} \right)$

$ = $ Rs $34.80$

4. The cost of an article was Rs $15,500$. Rs $450$ were spent on its repairs. If it is sold for a profit of $15\% $, find the selling price of the article.

Ans:  Cost of an article \[ = \] Rs. \[15,500\]

Spent for repair = Rs. $450$     

Profit precentag  $ = 15\% $   

Selling price of one article  $ = ?$

Total cost of an article $ = $ Cost of an article $ + $ Spent for repair

$ = $ Rs. $15500$ $ + $ Rs. $450$

$ = $ Rs $15950$  

$15 = \left( {\dfrac{{profit}}{{15950}}} \right) \times 100$

Profit $ = \dfrac{{15 \times 15950}}{{100}}$

Profit $ = 15 \times 159.5$

Profit $ = 2392.5$

$ = $ Rs $2392.5$

$ = $ Rs $18342.5$

Selling price of one article $ = $ Rs $18342.5$

5. A VCR and TV were bought for Rs $8,000$ each. The shopkeeper made a loss of $4\% $ on the VCR and a profit of $8\% $ on the TV. Find the gain or loss percent on the whole transaction.

Ans: Cost of VCR and TV $ = 8,000$

Loss Percentage in VCR $ = 4\% $

Profit percentage in TV $ = 8\% $

Whole gain and loss $ = ?$

Loss percentage  $ = 4\% $

if C.P. is Rs. $100$ , then S.P. is Rs. $96$ .

when C.P. is Rs. $8000$

S.P $ = \left( {\dfrac{{96}}{{100}} \times 8000} \right)$

S.P $ = 96 \times 80$

S.P $ = $ Rs. $7680$

Profit percentage $ = 8\% $

if C.P. is Rs. $100$ , then S.P. is Rs. $108$ .

S.P $ = \left( {\dfrac{{108}}{{100}} \times 8000} \right)$

S.P $ = 108 \times 80$

Selling price $ = $ Rs. $8640$

Total selling price $ = $ Rs. $7680 + $ Rs $8640$

$ = $ Rs. $16320$

Total cost price $ = $ Rs. $8000$$ + $ Rs. $8000$

$ = $ Rs. $16000$

Since total selling price greater than total cost price .

Profit $ = $ Rs. $16320$$ - $ Rs. $16000$

$ = $ Rs. $320$

$ = \left( {\dfrac{{320}}{{16000}}} \right) \times 100$

$ = \dfrac{{320}}{{160}}$

The gain percentage of shopkeeper in whole transaction $ = 2\% $

6. During a sale, a shop offered a discount of $10\% $ on the marked 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?

Ans: Discount percentage $ = 10\% $

Price of pair jeans $ = $ Rs. $1450$

Price of shirt $ = $ Rs. $850$

Discount $ = $ Marked price $ - $ Sale price.

One shirt price $ = $ Rs. $850$

Two shirt price = $2 \times $ Rs. $850$

$ \Rightarrow $ Rs. $1,700$

Total marked price = Rs. $\left( {1,450 + 1,700} \right)$

= Rs. $3,150$

Discount = Rs. $\left( {10\%  \times 3150} \right)$

= Rs. $\left( {\dfrac{{10}}{{100}} \times 3150} \right)$

Discount = Rs. $315$ .

Discount $ = $ Total Marked price $ - $ Sale price

$ \Rightarrow $ Rs $315$ = Rs $3150$ − Sale price

$ \Rightarrow $ Sale price $ = $ Rs $\left( {3150 - 315} \right)$

$ \Rightarrow $ Sale price $ = $ Rs $2835$

Customer paid $ = $ Rs $2835$ .

7. A milkman sold two of his buffaloes for Rs $20,000$ each. On one he made a gain of $5\% $ and on the other a loss of $10\% $. Find his overall gain or loss.

(Hint: Find CP of each).

Ans: Buffaloes $ = $ Rs. $20,000$

Gain percentage $ = 5\% $

Loss percentage $ = 10\% $

Overall gain or loss $ = ?$

Gain  percentage $ = 5\% $

if C.P. is Rs. $100$ , then S.P. is Rs. $105$ .

when C.P. is Rs. $20,000$

S.P $ = \left( {\dfrac{{100}}{{105}} \times 20,000} \right)$

S.P $ = 100 \times 190.47$

Selling price $ = $ Rs. $19047$

if C.P. is Rs. $100$ , then S.P. is Rs. $90$ .

S.P $ = \left( {\dfrac{{100}}{{90}} \times 20,000} \right)$

S.P $ = 100 \times 222.222$

Selling price $ = $ Rs. $22222.22$ .

Total selling price $ = $ Rs. $20,000 + $ Rs $20,000$

$ = $ Rs. $40,000$

Total cost price $ = $ Rs. $19047.62$$ + $ Rs. $22222.22$

$ = $ Rs. $41269.84$

Since total selling price less than total cost price .

Loss $ = $ Rs. $41269.84$$ - $ Rs. $40,000$

$ = $ Rs. $1269.84$

The overall loss of milkman $ = $ Rs. $1269.84$ .

8. 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.

Ans: Price of TV $ = $ Rs. $13,000$ .

Sales tax percentage $ = 12\% $

Vinod have to pay $ = ?$

if  Rs. $100$ , then Tax to be pay is Rs. $12$ .

when  Rs. $13,000$

Tax to be pay $ = \left( {\dfrac{{12}}{{100}} \times 13,000} \right)$

Tax to be pay $ = 12 \times 130$

Tax to be pay $ = $ Rs. $1,560$ .

Vinod have to pay $ = $ price of TV $ + $ Tax to be pay

$ = $ Rs. $13,000$$ + $ Rs. $1560$

$ = $ Rs. $14560$

Vinod have to pay $ = $ Rs. $14560$ .

9. 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.

Ans: Discount in skates $ = 20\% $

Total amount $ = 1,600$

Marked price $ = x$

Discount percent \[\]$ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$  

Discount percent $ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$

$ \Rightarrow 20 = \left( {\dfrac{{Discount}}{x}} \right) \times 100$

Discount $ = \dfrac{{20 \times x}}{{100}}$

Discount $ = \dfrac{{1 \times x}}{5}$

Discount $ = $ Marked price $ - $ Total amount

$ \Rightarrow \dfrac{{1 \times x}}{5}$$ = x - 1600$

$ \Rightarrow 1600 = x - \dfrac{1}{5}x$

$ \Rightarrow 1600 = \dfrac{{5x - x}}{5}$

$ \Rightarrow 1600 = \dfrac{{4x}}{5}$

$ \Rightarrow \dfrac{{1600 \times 5}}{4} = x$

$ \Rightarrow x = 400 \times 5$

$ \Rightarrow x = 2000$

Marked price $ = 2000$ .

10. I purchased a hair-dryer for Rs $5,400$ including $8\% $ VAT. Find the price before VAT was added.

Ans: Hair-dryer rate include VAT $ = 5,400$

Tax percentage $ = 8\% $

Rate before VAT $ = ?$

VAT $ = 8\% $

If VAT without Rs. $100$ , then price is Rs. $108$

when  Rs. $5400$

Rate before VAT $ = \left( {\dfrac{{100}}{{108}} \times 5400} \right)$

Rate before VAT $ = 100 \times 50$

Rate before VAT $ = $ Rs. $5000$ .

Exercise 8.3

1. Calculate the amount and compound interest on

(a) Rs $10800$ for $3$ years at \[12\dfrac{1}{2}\% \] per annum compounded annually.

(b) Rs \[18000\] for\[2\dfrac{1}{2}\] years at \[10\% \]  per annum compounded annually.

(c) Rs \[62500\] for \[1\dfrac{1}{2}\] years at \[8\% \]  per annum compounded half yearly.

(d) Rs \[8000\] for \[1\] year at \[9\% \]  per annum compound half yearly.

(You could use the year by year calculation using SI formula to verify)

(e) Rs \[10000\] for \[1\] year at \[8\% \] per annum compounded half yearly.

(a) Principal (P) \[ = \] Rs. \[10,800\] .

Rate (R) = \[12\dfrac{1}{2}\% \]

\[ \Rightarrow \dfrac{{24 + 1}}{2}\% \]

\[ \Rightarrow \dfrac{{25}}{2}\% \] (annual)

Number of years n \[ = 3\]

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

A \[ = 10,800{\left( {1 + \dfrac{{25}}{{2 \times 100}}} \right)^3}\]

A \[ = 10,800{\left( {1 + \dfrac{{25}}{{200}}} \right)^3}\]

A  \[ = 10,800{\left( {\dfrac{{200 + 25}}{{200}}} \right)^3}\]

A \[ = 10,800{\left( {\dfrac{{225}}{{200}}} \right)^3}\]

A \[ = 10,800{\left( {\dfrac{9}{8}} \right)^3}\]

A \[ = 10800{\left( {1.125} \right)^3}\]

A \[ = 10800 \times 1.423828\]

A \[ = 15377.34\] (approximately)

\[ \Rightarrow \] Rs. \[\left( {15377.34 - 10800} \right)\]

\[ \Rightarrow \] Rs. \[4577.34\] .

(b) Principal (P) \[ = \] Rs. \[18,000\] .

Rate (R) = \[10\% \] (annual)

Number of years n \[ = 2\dfrac{1}{2}\] Years (2 Years and 6 month)

The amount for 2 years and 6 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years.

Firstly, the amount for 2 years has to be calculated.

A \[ = 18000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\]

A \[ = 18,000{\left( {1 + \dfrac{1}{{10}}} \right)^2}\]

A  \[ = 18,000{\left( {\dfrac{{10 + 1}}{{10}}} \right)^2}\]

A \[ = 18,000{\left( {\dfrac{{11}}{{10}}} \right)^2}\]

A \[ = 18,000{\left( {1.1} \right)^2}\]

A \[ = 18,000 \times 1.21\]

A \[ = 21780\] (approximately)

By taking Rs. \[21,780\] as principal.

The S.I for next 6 month will be calculated.

i.e \[\dfrac{1}{2}\] year

S.I \[ = \] Rs \[\left( {\dfrac{{21,780 \times \dfrac{1}{2} \times 10}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{21,780 \times 5}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{21,78}}{2}} \right)\]

S.I \[ = \] Rs \[1089\] .

Interest for first two years \[ = \] Rs. \[\left( {21780 - 18000} \right)\]

\[ = \] Rs. \[3780\] .

And interest for next half years \[ = \] Rs. \[1089\] .

Total compound interest \[ = \] Rs. \[3780\]\[ + \]\[1089\]

\[ = \] Rs. \[4869\] .

A \[ = \] P \[ + \] Compound interest

\[ \Rightarrow \] Rs. \[\left( {18000 + 4869} \right)\]

\[ \Rightarrow \] Rs. \[22,869\] .

Amount \[ = \] Rs. \[22,869\] .

(c) Principal (P) \[ = \] Rs. \[62,500\] .

Rate (R) = \[8\% \] per annum or \[4\% \] per half year.

Number of years n \[ = 1\dfrac{1}{2}\] years

A \[ = 62,500{\left( {1 + \dfrac{4}{{100}}} \right)^3}\]

A \[ = 62,500{\left( {1 + \dfrac{1}{{25}}} \right)^3}\]

A  \[ = 62,500{\left( {\dfrac{{25 + 1}}{{25}}} \right)^3}\]

A \[ = 62,500{\left( {\dfrac{{26}}{{25}}} \right)^3}\]

A \[ = 62,500{\left( {1.04} \right)^3}\]

A \[ = 62,500\left( {1.1248} \right)\]

A \[ = 70,304\] (approximately)

\[ \Rightarrow \] Rs. \[\left( {70304 - 62500} \right)\]

\[ \Rightarrow \] Rs. \[7,804\] .

(d) Principal (P) \[ = \] Rs. \[8,000\] .

Rate (R) = \[9\% \] (annual) or \[\dfrac{9}{2}\% \] (half year)

Number of years n \[ = 1\]

There will be 2 half years in 1 year.

A \[ = 8,000{\left( {1 + \dfrac{9}{{200}}} \right)^2}\]

A \[ = 8,000{\left( {1 + 0.045} \right)^2}\]

A  \[ = 8,000{\left( {1.045} \right)^2}\]

A \[ = 8,000 \times 1.092\]

A \[ = 8,736.2\] (approximately)

\[ \Rightarrow \] Rs. \[\left( {8736.2 - 8000} \right)\]

\[ \Rightarrow \] Rs. \[736.20\] .

(e) Principal (P) \[ = \] Rs. \[10,000\] .

Rate (R) = \[8\% \] per annum or 4% per half year.

Number of years n \[ = 1\] year

A \[ = 10,000{\left( {1 + \dfrac{4}{{100}}} \right)^2}\]

A \[ = 10,000{\left( {1 + \dfrac{1}{{25}}} \right)^2}\]

A  \[ = 10,000{\left( {\dfrac{{25 + 1}}{{25}}} \right)^2}\]

A \[ = 10,000{\left( {\dfrac{{26}}{{25}}} \right)^2}\]

A \[ = 10,000{\left( {1.04} \right)^2}\]

A \[ = 10,000 \times 1.0816\]

A \[ = 10,816\] (approximately)

\[ \Rightarrow \] Rs. \[\left( {10816 - 10000} \right)\]

\[ \Rightarrow \] Rs. \[816\] .

2. Kamala borrowed Rs \[26400\] from a Bank to buy a scooter at a rate of \[15\% \]p.a. compounded yearly. What amount will she pay at the end of \[2\] years and \[4\] months  to clear the loan?

(Hint: Find A for \[2\] years with interest is compounded yearly and then find SI on 

the\[2\] nd year amount for \[\dfrac{4}{{12}}\]years.)

Ans: Principal (P) \[ = \] Rs. \[26,400\]

Rate (R) = \[15\% \] per annum 

Number of years n  =  \[2\dfrac{4}{{12}}\] year

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years.

A \[ = 26,400{\left( {1 + \dfrac{{15}}{{100}}} \right)^2}\]

A \[ = 26,400{\left( {1 + \dfrac{3}{{20}}} \right)^2}\]

A  \[ = 26,400{\left( {\dfrac{{20 + 3}}{{20}}} \right)^2}\]

A \[ = 26,400{\left( {\dfrac{{23}}{{20}}} \right)^2}\]

A \[ = 26,400{\left( {1.15} \right)^2}\]

A \[ = 26,400 \times 1.3225\]

A \[ = 34,914\] (approximately)

By taking Rs. \[34,914\] as principal.

The S.I for next \[\dfrac{1}{3}\] years will be calculated.

S.I \[ = \] Rs \[\left( {\dfrac{{34914 \times \dfrac{1}{3} \times 15}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{34914 \times 0.333 \times 15}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{174569.999}}{{100}}} \right)\]

S.I \[ = \] Rs \[1745.70\] .

Interest for first two years \[ = \] Rs. \[\left( {34914 - 26400} \right)\]

\[ = \] Rs. \[8514\] .

And interest for next \[\dfrac{1}{3}\] years \[ = \] Rs. \[1,745.70\] .

Compound interest \[ \= \] Rs. \[\left( {8514 + 1745.70} \right)\]

\[ \Rightarrow \] Rs. \[10,259.70\] .

Amount \[ = \] P \[ + \] C.I

\[ \Rightarrow \] Rs. \[\left( {26,400 + 10,259.70} \right)\]

\[ \Rightarrow \] Rs. \[36,659.70\] .

3. Fabina borrows Rs $12,500$ at $12\% $ per annum for $3$years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Ans: Fabina and Radha borrowed money (P) $ = 12,500$

Percentage per annum for Fabina (R) $ = 12\% $

Percentage per annum for Radha (R) $ = 10\% $

Years (T) $ = 3$

Interest paid by Fabina $ = \dfrac{{P \times R \times T}}{{100}}$

$ = $ Rs. $\left( {\dfrac{{12500 \times 12 \times 3}}{{100}}} \right)$

$ = $ Rs. $125 \times 12 \times 3$

$ = $ Rs. $4,500$

Amount paid by Radha at the end of 3 years

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 12500{\left( {1 + \dfrac{{10}}{{100}}} \right)^3}$

A $ = 12500{\left( {1 + \dfrac{1}{{10}}} \right)^3}$

A $ = 12500{\left( {1 + 0.1} \right)^3}$

A $ = 12500{\left( {1.1} \right)^3}$

A = $12500 \times 1.331$

A = $16637.5$

\[ \Rightarrow \] Rs. \[\left( {16637.50 - 12500} \right)\]

\[ \Rightarrow \] Rs. \[4,137.50\] .

The interest paid by Fabina is Rs. $4,500$ and by Radha is Rs. \[4,137.50\] .

$\therefore $ Fabina pays more interest.

\[ \Rightarrow \] Rs. $4,500 - $ Rs. \[4,137.50\] .

\[ \Rightarrow \] Rs. $362.50$

$\therefore $ Fabina will have to pay  Rs. $362.50$ more.

4. I borrowed Rs $12,000$ from Jamshed at $6\% $ per annum simple interest for $2$ years. Had I borrowed this sum at $6\% $ per annum compound interest, what extra amount would I have to pay?

Ans: Principle (P) $ = $ Rs $12,000$

Rate $ = $ $6\% $ per annum

Time (T) $ = $$2$ years.

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

$ = $ Rs. $\left( {\dfrac{{12,000 \times 6 \times 2}}{{100}}} \right)$

$ = $ Rs. $120 \times 6 \times 2$

$ = $ Rs. $1,440$

Find amount A,

A $ = 12,000{\left( {1 + \dfrac{6}{{100}}} \right)^2}$

A $ = 12,000{\left( {1 + \dfrac{3}{{50}}} \right)^2}$

A $ = 12,000{\left( {1 + 0.06} \right)^2}$

A $ = 12,000{\left( {1.06} \right)^2}$

A = $12,000 \times 1.1236$

A = $13,483.20$

\[ \Rightarrow \] Rs. \[\left( {13483.20 - 12000} \right)\]

\[ \Rightarrow \] Rs. \[1,483.20\] .

Difference in interests = C.I. $ - $ S.I

\[ \Rightarrow \] Rs. \[\left( {1,483.20 - 1,440} \right)\]

\[ \Rightarrow \] Rs. \[43.20\] .

The extra amount to be paid $ = $ Rs. \[43.20\] .

5. Vasudevan invested Rs $60,000$ at an interest rate of $12\% $ per annum compounded half yearly. What amount would he get

(i) after $6$ months?

(ii) after $1$ year?

Ans: (i)Given :

Principal (P) $ = $ Rs. $60,000$

Rate $ = $$12\% $ per annum or $6\% $ per half year.

Number n $ = $ $6$ months or $1$ half year.

A $ = 60,000{\left( {1 + \dfrac{6}{{100}}} \right)^1}$

A $ = 60,000{\left( {\dfrac{{100 + 6}}{{100}}} \right)^{}}$

A $ = 60,000{\left( {\dfrac{{106}}{{100}}} \right)^{}}$

A $ = 60,000{\left( {1.06} \right)^{}}$

A = $63,600$

(ii) There are 2 half years in 1 year.

Number n $ = 2$  

A $ = 60,000{\left( {1 + \dfrac{6}{{100}}} \right)^2}$

A $ = 60,000{\left( {\dfrac{{100 + 6}}{{100}}} \right)^2}$

A $ = 60,000{\left( {\dfrac{{106}}{{100}}} \right)^2}$

A $ = 60,000{\left( {1.06} \right)^2}$

A $ = 60,000 \times 1.1236$

A = $67416$ .

6. Arif took a loan of Rs $80,000$ from a bank. If the rate of interest is $10\% $ per annum, find the difference in amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is

(i) Compounded annually.

(ii) Compounded half yearly.

(i) Principal (P) $ = $ Rs. $80,000$

Rate $ = $ $10\% $ per annum.

Number n $ = $$1\dfrac{1}{2}$ years.

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.Firstly, the amount or 1 year has to be calculated.

A $ = 80,000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$

A $ = 80,000\left( {\dfrac{{100 + 10}}{{100}}} \right)$

A $ = 80,000\left( {\dfrac{{110}}{{100}}} \right)$

A $ = 80,000\left( {1.1} \right)$

A = $88,000$

By taking Rs. \[88,000\] as principal.

The S.I for next \[\dfrac{1}{2}\] years will be calculated.

S.I \[ = \] Rs \[\left( {\dfrac{{88000 \times 10 \times \dfrac{1}{2}}}{{100}}} \right)\]

S.I \[ = \] Rs \[8800 \times 1 \times 0.5\]

S.I \[ = \] Rs \[4,400\] .

Interest for first years \[ = \] Rs. \[\left( {88000 - 80000} \right)\]

\[ = \] Rs. \[8000\] .

And interest for next \[\dfrac{1}{2}\] years \[ = \] Rs. \[4,400\] .

Total Compound interest 

\[ \Rightarrow \] Rs. \[\left( {8000 + 4400} \right)\]

\[ \Rightarrow \] Rs. \[12400\] .

\[ \Rightarrow \] Rs. \[\left( {80000 + 12400} \right)\]

\[ \Rightarrow \] Rs. \[92,400\] .

(ii) Given :

The interest is compounded half yearly.

Rate $ = 10\% $ per annum or $5\% $ per half year.

Number n  $ = 1\dfrac{1}{2}$   years

A $ = 80,000{\left( {1 + \dfrac{5}{{100}}} \right)^3}$

A $ = 80,000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^3}$

A $ = 80,000{\left( {\dfrac{{105}}{{100}}} \right)^3}$

A $ = 80,000{\left( {1.05} \right)^3}$

A = $80,000 \times 1.157625$

A = $92,610$ .

Difference in amount 

= Rs. $\left( {92,610 - 92,400} \right)$

= Rs. $210$ .

7. Maria invested Rs \[8,000\] in a business. She would be paid interest at $5\% $ per annum compounded annually. Find.

(i) The amount credited against her name at the end of the second year

The interest for the $3$ rd year.

(i) Priniciple (p) $ = $ Rs $8000$

Rate(R)             $ = $$5\% $ per annum

Number(n)      $ = $$2$ years

A $ = 8,000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

A $ = 8,000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

A $ = 8,000{\left( {\dfrac{{105}}{{100}}} \right)^2}$

A $ = 8,000{\left( {1.05} \right)^2}$

A = $8,000 \times 1.1025$

(ii) The interest for the next one year, i.e. the third year, has to be calculated. By taking Rs $8820$ as principal, the S.I. for the next year will be calculated

S.I $ = $ Rs $\left( {\dfrac{{8820 \times 5 \times 1}}{{100}}} \right)$

S.I $ = $ Rs $\left( {\dfrac{{44100}}{{100}}} \right)$

S.I $ = $ Rs $441$

8. Find the amount and the compound interest on Rs $10,000$ for years at $10\% $ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Ans: Principle (p) $ = $ Rs $10,000$

Rate(R)             $ = $ $10\% $ per annum or $5\% $ per half year.

Number(n)      $ = $ $1\dfrac{1}{2}$ years.

There will be $3$ half years in $1\dfrac{1}{2}$ years.

A $ = 10,000{\left( {1 + \dfrac{5}{{100}}} \right)^3}$

A $ = 10,000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^3}$

A $ = 10,000{\left( {\dfrac{{105}}{{100}}} \right)^3}$

A $ = 10,000{\left( {1.05} \right)^3}$

A = $10,000 \times 1.157625$

A = $11576.25$

\[ \Rightarrow \] Rs. \[\left( {131576.25 - 10000} \right)\]

\[ \Rightarrow \] Rs. \[1,576.25\] .

The amount for 1 year and $6$ months can be calculated by first calculating the amount for  1 year using the compound interest formula, and then calculating the simple interest for $6$ months on the amount obtained at the end of 1  year.The amount for the first year has to be calculated first.

A $ = 10,000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$

A $ = 10,000\left( {\dfrac{{100 + 10}}{{100}}} \right)$

A $ = 10,000\left( {\dfrac{{110}}{{100}}} \right)$

A $ = 10,000\left( {1.1} \right)$

A = $11,000$

By taking Rs. \[11,000\] as principal.

S.I \[ = \] Rs \[\left( {\dfrac{{11000 \times 10 \times \dfrac{1}{2}}}{{100}}} \right)\]

S.I \[ = \] Rs \[1100 \times 1 \times 0.5\]

S.I \[ = \] Rs \[550\] .

Interest for first years \[ = \] Rs. \[\left( {11000 - 10000} \right)\]

\[ = \] Rs. \[1000\] .

\[ \Rightarrow \] Rs. \[\left( {1000 + 550} \right)\]

\[ \Rightarrow \] Rs. \[1550\] .

Therefore, the interest would be more when compounded half yearly than the interest when compounded annually.

9. Find the amount which Ram will get on Rs $4096$, he gave it for $18$ months at $12\dfrac{1}{2}\% $ per annum, interest being compounded half yearly.

Ans:  Principle (p) $ = $ Rs $10,000$

Rate(R)             $ = $ $12\dfrac{1}{2}\% $ per annum or $\dfrac{{25}}{4}\% $ per half year.

Number(n)      $ = $ $18$ months.

There will be $3$ half years in $18$ months.

A $ = 4096{\left( {1 + \dfrac{{25}}{{400}}} \right)^3}$

A $ = 4096{\left( {1 + \dfrac{1}{{16}}} \right)^3}$

A $ = 4096{\left( {\dfrac{{16 + 1}}{{16}}} \right)^3}$

A $ = 4096{\left( {\dfrac{{17}}{{16}}} \right)^3}$

A $ = 4096{\left( {1.0625} \right)^3}$

 A = $4096 \times 1.1994628$

 A = $4913$

Amount = $4913$ .

10. The population of a place increased to $54000$ in $2003$ at a rate of $5\% $ per annum

(i) find the population in $2001$

(ii) what would be its population in $2005$?

Ans:  Population in the year $2003$$ = 54,000$

(i) $54000$ $ = $ population in 2001 $ \times $ ${\left( {1 + \dfrac{5}{{100}}} \right)^2}$

$54000$$ = $ population in 2001 $ \times $${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

$54000$$ = $ population in 2001 $ \times $${\left( {\dfrac{{105}}{{100}}} \right)^2}$

$54000$$ = $ population in 2001 $ \times $${\left( {1.05} \right)^2}$

$54000$$ = $ population in 2001 $ \times $$1.1025$

$\dfrac{{54000}}{{1.1025}}$$ = $ population in 2001 

population in 2001 $ = 48979.591$

(ii) population in 2005

population in 2001 $ = $$54000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

population in 2001 $ = 54000$${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

population in 2001  $ = 54000{\left( {\dfrac{{105}}{{100}}} \right)^2}$

population in 2001 $ = 5400{\left( {1.05} \right)^2}$

population in 2001 $ = 54000 \times 1.1025$

population in 2001 $ = 59.535$

11. 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 $5,06,000$.

Ans:  Initial count of bacteria $ = 5,06,000$

Bacteria at the end of $2$ hours

$ = 506000{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}$

$ = 506000{\left( {1 + 0.025} \right)^2}$

$ = 506000{\left( {1.025} \right)^2}$

$ = 506000\left( {1.050625} \right)$

$ = 531616.25$ (approximately)

The count of bacteria at the end of 2 hours $ = 531616.25$ .

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

Ans:  Principal (P) $ = $ Rs $42,000$

Rate(R)             $ = $ $8\% $ per annum .

Number(n)      $ = $ $1$ year.

S.I \[ = \] Rs \[\left( {\dfrac{{42000 \times 8 \times 1}}{{100}}} \right)\]

S.I \[ = \] Rs \[8 \times 420\]

S.I \[ = \] Rs \[3360\] .

Value after $1$ year \[ = \] Rs. \[\left( {42000 - 3360} \right)\]

\[ = \] Rs. \[38,640\] .

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The solutions to all the exercise questions in the chapter Comparing Quantities are given in the PDF. So students can refer to these solutions to understand how to present the answers in the exam in a step-wise method.

As these NCERT Solutions for Class 8 Comparing Quantities are available on Vedantu mobile app as well, students can refer to these solutions at any time and anywhere. Thus, studying these solutions on Vedantu will be highly time-saving for all students.

Our NCERT Solutions for Class 8 Comparing Quantities are provided in a single place. Therefore, revising the revision chapter before the exams become convenient as the questions for all the exercise is available in one place.

NCERT Solutions for Class 8 Maths Chapter 8  Comparing Quantities are available for free to download.

Important Questions for Practice

Express 10& and 35% as decimals.

Calculate the ratio of 7 m to 49 m.

5. If 45% of 15 students like English, find out the number of students who dislike English.

A shopkeeper bought two watches for Rs. 8,000 each. After selling the watches, there was a loss of 4% on the 1st watch while a profit of 8% on the 2nd watch. Calculate the overall gain or loss per cent on the whole transaction made by the shopkeeper.

In Class 8, there are 120 girls. Calculate the total number of students if 12% of the total students are boys.

NCERT Solutions for Class 8 Maths Chapter 8 on Comparing Quantities by Vedantu are a helpful resource for students. This chapter likely covers important concepts related to comparing quantities, such as understanding ratios, percentages, and their applications in real-life situations. The solutions provided by Vedantu aim to simplify complex mathematical ideas, making them more accessible to students. One crucial section could be the practical application of comparing quantities, where students learn how these mathematical concepts are used in everyday scenarios. Overall, these solutions serve as a valuable tool for students to grasp and apply mathematical concepts effectively.

FAQs on NCERT Solutions for Class 8 Maths Chapter 8 - Comparing Quantities

1. Define ratio.

A ratio means comparing two quantities of similar units. A ratio of two quantities of the same unit is derived by dividing one quantity by the other.

2. Define interest.

Interest is the extra money paid by the post offices or banks on the money deposited with them. Interest is also paid by the people when they borrow money.

3. What is known as the conversion period?

The time period after which the interest is added each time to form a new principal is known as the conversion period.

4. What is known as overhead expenses?

Overhead expenses are the additional expenses made after buying an article and are included in the cost price of the article.

5. What is the full form of GST?

GST stands for Goods and Services Tax and is imposed on the supply of goods or services or both.

7. Where can I find NCERT Solutions for Chapter 8 “Comparing Quantities” of Class 8 Maths?

Students can easily find NCERT Solutions for Chapter 8 “Comparing Quantities” of Class 8 Maths on the internet. NCERT Solutions for Class 8 maths are available at free of cost on the Vedantu website and on the Vedantu app. Students can download NCERT Solutions free of cost. All Solutions are explained properly. They can click on this NCERT Solutions for Chapter 8 of Class 8 Maths to download solutions. NCERT Solutions can help students to prepare well for their exams. They can understand the concepts given in Chapter 8 easily. 

NCERT Solutions for Class 8 Maths

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Comparing Quantities

Ncert solutions for chapter 8 comparing quantities class 8 maths.

Find the ratio of the following:

(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour. (b) 5 m to 10 km (c) 50 paise to Rs. 5

(a)  Speed of cycle = 15 km/hr

Speed of scooter = 30 km/hr

Hence ratio of speed of cycle to that of scooter = 15 : 30 = 15/30 = 1/2= 1 : 2

(b)   ∵  1 km = 1000 m

∴  10 km = 10 ×1000  = 10000 m

∴  Ratio = 5m/10000m = 1/2000  = 1 : 2000

(c)   ∵  Rs 1 = 100 paise

∴  Rs 5 = 5 × 100  = 500 paise

Hence Ratio = 50 paise/ 500 paise = 1/10 = 1 : 10

Convert the following ratios to percentages:

(a) 3 : 4 (b) 2 : 3

(a)  Percentage of 3 : 4 = 3/4  ×100 % = 75%  

(b)  Percentage of 2 : 3 = 2/3  × 100% = 66.2/3%

Total number of students = 25

Number of good students in mathematics  = 72% of 25 = 72/100  ×25 = 18  

Number of students not good in mathematics  = 25 – 18 = 7

Hence percentage of students not good in mathematics = 7/25  × 100  = 28%

Let total number of matches be x.

According to question,

40% of total matches = 10

⇒  40% of x = 10  

⇒  40/100  × x = 10

⇒  x = (10 ×100)/40 = 25

Hence total number of matches are 25.

Total percentage of money she didn't spent =  100% - 75%=25%

According to question,   

⇒ 25% = 600

⇒  1% = 600/25

⇒ 100% = 600/ 25  × 100 

Hence the money in the beginning was Rs 2,400.

Number of people who like cricket = 60%

Number of people who like football = 30%

Number of people who like other games = 100% – (60% + 30%) = 10%

Now Number of people who like cricket = 60% of 50,00,000 = 60/100  × 50,00,000 = 30,00,000  

And Number of people who like football

= 30% of 50,00,000

=  30/100  × 50,00,000  = 15,00,000

∴  Number of people who like other games = 10% of 50,00,000

= 10/100  × 50,00,000  = 5,00,000

Hence, number of people who like other games are 5 lakh.

Let original salary be Rs.100.

Therefore New salary i.e., 10% increase

= 100 + 10 = Rs.110

∵  New salary is Rs.110, when original salary = Rs.100

∴  New salary is Rs.1, when original salary = 100/110 

∴  New salary is Rs.1,54,000, when original salary = 100/110  × 154000  = Rs.1,40,000

Hence original salary is Rs. 1,40,000.

On Sunday, people went to the Zoo  = 845

On Monday, people went to the Zoo = 169

Number of decrease in the people = 845 – 169 = 676

Decrease percent = 676/845  × 100  = 80%

Hence decrease in the people visiting the Zoo is 80%.

No. of articles = 80

Cost Price of articles = Rs. 2,400

And Profit = 16%

∵  Cost price of articles is Rs.100, then selling price = 100 + 16 = Rs.116

∴  Cost price of articles is Rs.1, then selling price = 116/100

∴  Cost price of articles is Rs.2400, then selling price = 116/100  × 2400   = Rs.2784

Hence, Selling Price of 80 articles = Rs.2784

Therefore Selling Price of 1 article

= 2784/80 = Rs.34.80

Here, C.P. = Rs.15,500 and Repair cost = Rs.450

Therefore Total Cost Price = 15500 + 450 = Rs.15,950

Let C.P. be Rs.100, then S.P. = 100 + 15 = Rs.115

∵  When C.P. is Rs.100, then S.P. = Rs.115

∴   When C.P. is Rs.1, then S.P. = 115/100

∴  When C.P. is Rs.15950, then S.P.

=115/100  × 15950   = Rs.18,342.50

Cost price of VCR = Rs.8000 and Cost price of TV = Rs.8000

Total Cost Price of both articles

= Rs.8000 + Rs.8000 = Rs. 16,000

Now VCR is sold at 4% loss.

Let C.P. of each article be Rs.100, then S.P. of VCR = 100 – 4 = Rs.96

∴  When C.P. is Rs.100, then S.P. = Rs.96

∴  When C.P. is Rs.1, then S.P. = 96/100

∴  When C.P. is Rs.8000, then S.P.

= 96/100  ×8000  = Rs.7,680

And TV is sold at 8% profit, then S.P. of TV = 100 + 8 = Rs.108

∵  When C.P. is Rs.100, then S.P. = Rs.108

∴  When C.P. is Rs.1, then S.P. = 108/100

= 108/100  × 8000  = Rs.8,640

Then, Total S.P.

= Rs.7,680 + Rs.8,640 = Rs. 16,320

Since S.P. >C.P.,

Therefore Profit = S.P. – C.P.

= 16320 – 16000 = Rs.320

And Profit% = Profit/ cost price  × 100  

= 320/16000  × 100  = 2%

Therefore,the shopkeeper had a gain of 2% on the whole transaction.

Rate of discount on all items = 10%

Marked Price of a pair of jeans = Rs.1450 and Marked Price of a shirt = Rs.850

Discount on a pair of jeans

= (Rate  × M.P)/100 = (10 ×1450)/100  = Rs.145

∴  S.P. of a pair of jeans = Rs.1450 – Rs.145 = Rs.1305

Marked Price of two shirts = 2 × 850  = Rs.1700

Discount on two shirts =  (Rate  × M.P)/100 = (10   × 1700)/100   = Rs.170

∴  S.P. of two shirts = Rs.1700 – Rs.170 = Rs.1530

Therefore the customer had to pay = 1305 + 1530

= Discount on a pair of jeans

= (Rate  × M.P)/100 = (10 ×1450)/100  

∴  S.P. of a pair of jeans

= Rs.1450 – Rs.145 = Rs.2,835

Thus,the customer will have to pay Rs.2,835

S.P. of each buffalo = Rs.20,000

S.P. of two buffaloes = 20,000  × 2  = Rs.40,000

One buffalo is sold at 5% gain.

Let C.P. be Rs.100, then S.P. = 100 + 5 = Rs.105

∵  When S.P. is Rs.105, then C.P. = Rs.100

∴  When S.P. is Rs.1, then C.P. = 100/105

∴  When S.P. is Rs.20,000, then C.P.

= 100/105  × 20000   = Rs.19,047.62

Another buffalo is sold at 10% loss.

Let C.P. be Rs.100, then S.P. = 100 – 10 = Rs.90

∵  When S.P. is Rs.90, then C.P. = Rs.100

∴  When S.P. is Rs.1, then C.P. = 100/90

= 100/90  × 20000  = Rs.22,222.22

Total C.P. = Rs.19,047.62 + Rs.22,222.22

= Rs.41,269.84

Since C.P. >S.P.

Therefore here it is loss.

Loss = C.P. – S.P.

= Rs.41,269.84 – Rs. 40,000.00 = Rs.1,269.84

The overall loss of milkman was Rs.1269.84

C.P. = Rs.13,000 and S.T. rate = 12%

Let C.P. be Rs.100, then S.P. for purchaser

 = 100 + 12 = Rs.112

∵  When C.P. is Rs.100, then S.P. = Rs.112

∴  When C.P. is Rs.1, then S.P. = 112/100

∴  When C.P. is Rs.13,000, then S.P.

= 112/100  × 13000  = Rs.14,560

He will have to pay Rs.14,560

S.P. = Rs.1,600 and Rate of discount = 20%

Let M.P. be Rs.100, then S.P. for customer = 100 – 20 = Rs.80

∵  When S.P. is Rs.80, then M.P. = Rs.100

∴  When S.P. is Rs.1, then M.P. = 100/80

∴  When S.P. is Rs.1600, then M.P.

= 100/ 80  × 1600   = Rs.2,000

Thus, the marked price was Rs. 2,000

C.P. = Rs.5,400 and Rate of VAT = 8%

Let C.P. without VAT is Rs. 100, then price including VAT = 100 + 8 = Rs.108

∵  When price including VAT is Rs.108, then original price = Rs.100

∴  When price including VAT is Rs.1, then original price = 100/108

∴  When price including VAT is Rs.5400, then original price = 100/108  × 5400   = Rs.5000 

Thus, the price of Hair Dryer before the addition of VAT was Rs 5000

Calculate the amount and compound interest on: (a) Rs.10,800 for 3 years at 12.1/2%   per annum compounded annually. (b) Rs.18,000 for 2.1/2   years at 10% per annum compounded annually. (c) Rs.62,500 for 1.1/2   years at 8% per annum compounded annually.

(d) Rs.8,000 for 1  years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).

(e) Rs.10,000 for 1  years at 8% per annum compounded half yearly.

(a)  Here, Principal (P) = Rs. 10800, Time(n) = 3 years,

Rate of interest (R) = 12.1/2% = 25/2 %

= 10800  × 9/8  × 9/8  × 9/8  

   = Rs. 15,377.34 (approx.)

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

(b)  Here, Principal (P) = Rs. 18,000, Time (n) = 2.1/2  years, Rate of interest (R)

= 18000(11/10) 2  = 18000  × 11/10  ×11/10   

= Rs. 21,780

Interest for 1/2 years on Rs. 21,780 at rate of 10% = (21780 ×10 ×1)/100  = Rs. 1,089

Total amount for 2.1/2 years

= Rs. 21,780 + Rs. 1089 = Rs. 22,869

= Rs. 22869 – Rs. 18000 = Rs. 4,869

(c)  Here, Principal (P) = Rs. 62500, Time (n) = 1.1/2 = 3/2 years = 3 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

= 62500 (26/25) 3  

= 62500  × 26/25  × 26/25  × 26/25  

= Rs. 70,304

= Rs. 70304 – Rs. 62500 = Rs. 7,804

(d)  Here, Principal (P) = Rs. 8000, Time (n) = 1 years = 2 years(compounded half yearly)

Rate of interest (R) = 9% = 9/2 % (compounded half yearly)

= 8000 (209/200) 2

= 8000  × 209/200  × 209/200  

= Rs. 8,736.20

= Rs. 8736.20 – Rs. 8000

 = Rs. 736.20

(e)  Here, Principal (P) = Rs. 10,000, Time (n) = 1 years = 2 years (compounded half yearly)

= 10000 (26/25) 2  

= 10000  × 26/26  × 26/25  

= Rs. 10,816

= Rs. 10,816 – Rs. 10,000 = Rs. 816

Here, Principal (P) = Rs. 26,400, Time(n) = 2 years 4 months, Rate of interest (R) = 15% p.a.

= 26400(23/20) 2   = 26400  × 23/20  × 23/20  

= Rs. 34,914

Interest for 4 months = 4/12 = 1/3  years at the rate of 15% = (34914  × 15 ×1)/100   

= Rs. 1745.70

= Rs. 36,659.70

Here, Principal (P) = Rs.12,500, Time (T) = 3 years, Rate of interest (R)

= Rs. 16,637.50

= Rs. 16,637.50 – Rs. 12,500 = Rs. 4,137.50

Thus, Fabina pays more interest

= Rs. 4,500 – Rs. 4,137.50 = Rs. 362.50

Here, Principal (P) = Rs.12,000, Time (T) = 2 years, Rate of interest (R) = 6% p.a.

Simple Interest = (P ×R ×T)/100  

= (12000  × 6 × 2)/100  = Rs. 1,440

Had he borrowed this sum at 6% p.a., then

= 12000(53/50) 2   - 12000 

= 12000  × 53/50  × 53/50 - 12000  

= Rs. 13,483.20 – Rs. 12,000

= Rs. 1,483.20

Difference in both interests

= Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

Thus ,the extra amount to be paid is Rs.43.20

(i)  Here, Principal (P) = Rs. 60,000,

Time (n)= 6 months = 1 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

= 60000 (53/50) 1  

= 60000  × 53/50  

= Rs. 63,600

  After 6 months Vasudevan would get amount Rs. 63,600.

(ii)  Here, Principal (P) = Rs. 60,000,

Time (n) = 1 year = 2 year(compounded half yearly)

= 60000(53/50) 2  

= 60000  × 53/50  × 53/50  

= Rs. 67,416

After 1 year Vasudevan would get amount Rs. 67,416.

(i)  Here, Principal (P) = Rs. 80,000, Time (n)=  1.1/2 years, Rate of interest (R) = 10%

= 80000 (11/10) 1

= Rs. 88,000

  Interest for 1/2 year = (88000 ×10 ×1)/(100 ×2)  

= Rs. 4,400

  Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

(ii)  Here, Principal (P) = Rs.80,000,

Time (n) =  1.1/2 year = 3/2 years (compounded half yearly)

Rate of interest (R) = 10% = 5% (compounded half yearly)

= 80000(21/20) 3

= 80000  × 21/20  × 21/20  × 21/20  

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

(i)  Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time (n) = 2 years

= 8000(21/20) 2  

= 8000  × 21/20  × 21/20  

= Rs. 8,820

(ii)  Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time (n) = 3 years

= 8000(21/20) 3  

= 8000  × 21/20  × 21/20  × 21/20  

= Rs. 9,261

Interest for 3 rd  year = A – P

= Rs. 9,261 – Rs. 8,820 = Rs. 441

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time (n) = 1.1/2  years = 3 years (compounded half yearly)

= 10000 (21/20) 3  

= 10000  × 21/20  × 21/20  × 21/20  

= Rs. 11,576.25

= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time (n) =  1.1/2years

= 10000(11/10) 1  

= 10000  × 11/10

= Rs. 11,000

Interest for 1/2 year = (11000  × 1  × 10)/(2 ×100)   = Rs. 550

∴  Total amount = Rs. 11,000 + Rs. 550

= Rs. 11,550

Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000

= Rs. 1,550

Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

Here, Principal (P) = Rs. 4096,

Rate of Interest (R) = 12.1/2 = 25/2%  

= 25/4%  (compounded half yearly)

Time (n)= 18 months = 1.1/2  years = 3 years (compounded half yearly)

= 4096(17/16) 3  

 = 4096  × 17/16  × 17/16  × 17/16  

= Rs. 4,913

(i)  Here,  A 2003  = Rs. 54,000, R = 5%, n = 2 years

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

⇒  54000 =  P 2001  (21/20) 2  

⇒  54000 =  P 2001  ×21/20  × 21/20  

⇒ P 2001  = (54000  × 20  × 20)/(21  × 21)

               =48,979.5

⇒      P 2001  = 48,980 (approx.)

(ii)  According to question, population is increasing. Therefore population in 2005,

= 54000 (1 + 5/100) 2  

= 54000(1+1/20) 2   

= 54000(21/20) 2  

= 54000  × 21/20  × 21/20  

Hence population in 2005 would be 59,535.

Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours

After 2 hours, number of bacteria,

= 506000(1+ 2.5/100) 2   

= 506000(1+25/1000) 2

= 506000(1+1/40) 2   

= 506000(41/40) 2

= 506000  × 41/40  × 41/40  

= 5,31,616.25

  Hence, number of bacteria after two hours are 531616 (approx.).

Here, Principal (P) = Rs. 42,000, Rate of Interest (R) = 8%, Time (n) = 1 years

Amount (A)  = P ( 1 - R/100) n   

= 42000(1 - 8/100) 1

= 42000(1+ 2/25) 1  

= 42000 (27/25) 1   

= 42000  × 27/25  

= Rs. 38,640

Hence, the value of scooter after one year is Rs. 38,640.

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Related chapters.

• Rational Numbers
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CBSE Class 8th Maths Value Based Questions Chapter 8 Comparing Quantities PDF Download

CBSE Class 8th Maths Value Based Questions Chapter 8 Comparing Quantities are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 8th are available here at Free of cost. These questions are expected to be asked in the Class 8th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

CBSE Class 8th Maths Value Based Questions Chapter 8 Comparing Quantities

Most of these Maths Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Maths Value Based Questions for board examinations. These Maths Value Based Questions are prepared by Directorate of Education, Delhi.

CBSE Maths Value Based Questions Class 8th Chapter 8 Comparing Quantities PDF

The purpose of the Maths Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Maths Value Based Questions for Class 8th Chapter 8 Comparing Quantities download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 8th exam are collected from various sources. It covers questions asked in previous year examinations.

CBSE Maths Value Based Questions for Class 8th Chapter 8 Comparing Quantities Free PDF

Class 8th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Maths Value Based Questions Chapter 8 Comparing Quantities certainly help students to obtain good marks in the examinations. 

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Important Questions for CBSE Class 8 Maths Chapter 8 Comparing Quantities

Home » CBSE » Important Questions for CBSE Class 8 Maths Chapter 8 Comparing Quantities

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Chapter 8 of Class Maths introduces students to Comparing Quantities. The concept of comparing quantities sets up the basis for many higher Maths topics and is a practically essential skill to have to solve real-life problems. Hence, enough practice is required to build quick mental Maths skills, which will help students in the examinations, enabling them to save enough time on calculations.

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Extramarks is the best study adviser for students and helps them with comprehensive online study solutions from Class 1 to Class 12. Our  Maths experts have prepared various NCERT solutions to help students in their studies and exam preparation. Students can refer to our Important Questions Class 8 Maths Chapter 8 to practise exam-oriented questions. We have collated questions from various sources such as NCERT textbooks and exemplars, CBSE sample papers , CBSE past year question papers , etc. Students can prepare well for their exams and tests by solving various chapter questions from our comparing quantities class 8 extra questions .

To get maximum benefit during exam preparations, students can register on the Extramarks website and get full access to Important Questions Class 8 Maths Chapter 8 and other study materials, including NCERT solutions, CBSE revision notes, etc.

Get Access to CBSE Class 8 Maths Extra Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

Comparing Quantities Class 8 Extra Questions with Solutions

Our in-house Maths faculty experts have collected an entire list of Important Questions Class 8 Maths Chapter 8 by referring to various sources. For each question, the team experts have  prepared a step-by-step explanation that will help students understand the concepts used in each question. Also, the questions are chosen in a way that would cover full chapter topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak points. And improvise by further focusing on weaker sections of the chapter.

Given below are a few of the questions and answers from our question bank of Maths Class 8 Chapter 8 Important Questions:

Question 1: Find the rate of discount given on a shirt whose selling price is ₹1092 after deducting a discount of ₹208 on its marked price.

Answer 1: we know that the SP = ₹1092

The discount = ₹208

By using the formula market price = SP + Discount

= 1092 + 208 = ₹1300

∴ The discount% = (discount/market price) × 100

= (208/1300) × 100 = 16%

Question 2: A scooter was bought at the cost of ₹ 42,000. Its value drops down to the rate of 8% per annum. Find its value after one year.

Answer 2: Principal = Cost price = ₹ 42,000

Depreciation rate = 8% of ₹ 42,000 per year

                                = ( Principal x Rate x Time period)/100

                                = (42000 x 8 x 1)/100

                                = ₹ 3360

Therefore , the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.

Question 3: A particular football team won 10 matches out of all the total number of matches they played. If their winning percentage was 40 %, how many matches did they play?

Answer 3: Let the total number of matches played be x.

The team won around 10 matches, and the team’s winning percentage was 40%.

40/100 × x = 10

40x = 10 × 100

x = 1000/40

Hence, the team played 25 matches.

Question 4:In a particular school, there are 456 girls. Compute the total number of students if 24% of the total students are boys.

Answer 4: The total number of students is 100.

It is given that 24% are boys.

So, total number of boys = 24% of 100 = 24

Therefore, the total number of girls will be = (100-24)%, i.e. 76%.

So, the total number of girls = 76% of 100 = 76

But, it is given that there are 456 girls.

Now, for 76 girls, the total number of students is 100

For 1 girl, the total number of students will be 100/76

Therefore, for 456 girls, the total number of students will be = 100/76 × 456 = 45600/76 = 600

Therefore, the total number of students = 600.

Question 5: I borrowed an amount ₹ 12000 from Jamshed at a rate of 6% per annum simple interest for 2 years. Had I borrowed this sum at a rate of 6% per annum compound interest, what is the extra amount I would have to pay?

Answer 5: Principle = ₹ 12000

Rate = 6% per annum

Time period = 2 years

Simple Interest = (P x R x T)/100

= (12000 x 6 x 2)/100

To find the compound interest,

the amount (A) has to be calculated

A = P(1 + R/100)n

     = 12000(1 + 6/100)2

     = 12000(106/100)2

     = 12000(53/50)2

     = ₹ 13483.20

∴ Compound Interest = A − P

           = ₹ 13483.20 − ₹ 12000

           = ₹ 1,483.20

Compound Interest − Simple Interest = ₹ 1,483.20 − ₹ 1,440

                  = ₹ 43.20

Hence, the extra amount to be paid is ₹ 43.20.

Question 6: Find the cost price when:

SP = ₹34.40 and Gain = 7 ½ %

Answer 6: Cost price = (100 / (100+ Gain %)) × SP

= (100/ (100+ (15/2))) × 34.40

= (100/ (215/2)) × 34.40

= (200/215) × 34.40

∴ the cost price is ₹ 32

Question 7:₹ 62500 for 1½ years at 8% per annum compounded half yearly.

Answer 7: Principal (P) = ₹ 62,500

Rate = 8% per annum or 4% per half-year

Number of years = 1½

There will be 3 and half years in 1½ years

Amount, A = Principle (1 + Rate/100)time period

= 62500(1 + 4/100)3

= 62500(104/100)3

= 62500(26/25)3

Compound Interest = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804

Question 8:In a certain laboratory, the bacteria count in a particular experiment increased at 2.5% per hour. Find the bacteria at the end of exactly 2 hours if the count was initially 5,06,000.

Answer 8: The initial count of bacteria = 5,06,000

Bacteria at the end of exactly 2 hours = 506000(1 + 2.5/100)2

= 506000(1 + 1/40)2

= 506000(41/40)2

= 531616.25

Therefore, the bacteria count at the end of 2 hours will be 5,31,616 (approx.).

Question 9:A person shops and spends 75% of his money. If he is now left with Rs. 600, find out how much he had in the beginning.

Answer 9: Let “x” be the initial amount he had initially.

As per the given question, the person spent 75% of Rs.x and is left with Rs. 600.

So, the amount he spent = x – 600

=> 75% of x = x – 600

=> (75/100) × x = x – 600

=> 75x = 100x – 60,000

=>25x = 60,000

Or, x = 2400.

Thus, the person had Rs. 2400 initially.

Question 10:A student scored 150 out of 200 in the subject maths and got 120 marks out of 180 in the subject science. In which subject did the student perform better?

Answer 10: Express the following marks in the form of ratios.

For the subject maths,the ratio = 150/200 = 3/4

For the subject science, the ratio = 120/180= 2/3

Here, the ratio 3/4 shows 75% (3/4× 100 = 75) and the ratio 2/3 shows 66.6% (2/3× 100 = 66.6).

Therefore, the student performed better in the maths exam than in his science exam.

Question 11: The population of a particular place increased to 54000 in 2003 at 5% per annum

(i) find the population of the place in the year 2001

(ii) what would be the population of the place in the year 2005?

Answer 11: (i) Population in the year 2003 = 54,000

54,000 = (Population in the year 2001) (1 + 5/100)2

54,000 = (Population in the year 2001) (105/100)2

Population in the year 2001 = 54000 x (100/105)2

Thus, the population of a place in the year 2001 was approximately 48,980

(ii) Population in the year 2005 = 54000(1 + 5/100)2

= 54000(105/100)2

= 54000(21/20)2

Hence, the population of a place in the year 2005 would be 59,535.

Question 12:Compute the amount and compound interest on the principal amount ₹ 10,800 for 3 years at 12½ % per annum compounded annually.

Answer 12 :Principal (P) = ₹ 10,800

Rate (R) = 12½ % = 25/2 % (annual)

Number of years (n) = 3

Amount (A) = P(1 + R/100)n

= 10800(1 + 25/200)3

= 10800(225/200)3

= 15377.34375

= ₹ 15377.34 (approximately)

Compound interest = A – P 

                                = ₹ (15377.34 – 10800) = ₹ 4,577.34

Question 13:Evaluate the ratio of 5 m to 20 km.

Answer 13: First, make both the units the same.

So, we must convert 20 km to the equivalent metre, i.e. “m”.

=> 20km = (20 × 1000)m

Therefore, the ratio of 5 m to 20 km = 5/20000 = 1:4000

Question 14: The marked price of a particular water cooler is ₹4650. The shopkeeper proposes an off-season discount of 18% on it. Find its selling price.

Answer 14: we know that the marked price = ₹4650

The discount = 18%

And the Discount in amount = 18% of the market price

= (18/100) × 4650

By using the formula, SP = marked price – Discount

= 4650 – 837 = ₹ 3,813

Question 15: Arun bought a new pair of skates at a sale where the Discount is given 20%. If the amount Arun pays is ₹ 1,600, find out the marked price.

Answer 15: Let the marked price be x

Discount percent = Discount/Marked Price x 100

20 = Discount/x × 100

Discount = 20/100 × x

Discount = Marked price – Sale price

x/5 = x – ₹ 1600

x – x/5 = 1600

4x/5 = 1600

x = 1600 x 5/4

Therefore, the marked price was ₹ 2000.

Question 16: How much did she have in the beginning if Chameli had ₹600 left after spending 75% of her money?

Answer 16: Let the amount of money which Chameli had, in the beginning, be x

After spending amount 75% of ₹x, she was left with ₹600

So, (100 – 75)% of x = ₹600

Or, 25% of x = ₹600

25/100 × x = ₹600

x = ₹600 × 4

Hence, Chameli had ₹2,400 in the beginning.

Question 17:Express 25% and 12% as decimals.

Answer 17: 25% = 25/100 = 0.25

                      12% = 12/100 = 0.12

Question 18: A man got a 10% increase in his salary. If his new salary amounted to ₹1,54,000, what was his original salary?

Answer 18: Let the original salary be x

The new salary is ₹1,54,000

Original salary + Increment = New Salary

The increment is of 10% of the original salary

So, (x + 10/100 × x) = 154000

x + x/10 = 154000

11x/10 = 154000

x = 154000 × 10/11

Thus, the original salary was ₹1,40,000.

Question 19: A particular cell phone was marked at 40% above the cost price, and a discount of 30% was given on its marked price. Find out the gain or loss percent made by the shopkeeper.

Answer 19: Let us consider the CP of goods as x

The market price of the goods when goods marked above the 40% CP is

Market price = x + (40x/100) = 140x/100 = 1.4x

So the Discount = 30%

Discount amount = 30% of 1.40x = 0.42x

∴ The SP = Market price – Discount

= 1.4x – 0.42x= 0.98x

Since SP is less than CP, it’s a loss

We know that Loss = CP – SP

= x – 0.98x = 0.02x

∴ Loss % = (Loss × 100)/ CP

= (0.02x × 100)/ x

Question 20: Express 45% and 78% as a fraction.

Answer 20: 45% = 45/100 = 9/20

                      78% = 78/100 = 39/50

Question 21: Obtain the amount Ram will get on ₹ 4,096, and he gave it for 18 months at 12½ per annum, interest being compounded half-yearly.

Answer 21: Principal = ₹ 4,096

Rate = 12½ per annum = 25/2 per annum = 25/4 per half-year

Time period = 18 months

There will be exactly 18 months in 3 half years 

Thus, amount A = P(1 + R/100)n

= 4096(1 + 25/(4 x 100))3

= 4096 x (1 + 1/16)3

= 4096 x (17/16)3

Hence, the required amount is ₹ 4,913.

Question 22: A bookseller sells a book for ₹100, gaining ₹20. His gain percentage is

• d) None of these

Answer 22 : we know that the SP = ₹100

By using the formula CP = SP – Gain

∴ Gain% = (Gain × 100) /CP

= (20 × 100) / 80

Question 23: Find out the amount and the compound interest on ₹ 10,000 for 1½ years at the rate of 10% per annum, compounded half yearly. Could this interest be more than the interest he would get if it were compounded annually?

Answer 23: P = ₹ 10,000

Rate = 10% per annum = 5% per half-year

Time period = 1½ years

There will 3 half years in 1½ years

Amount, A = P(1 + R/100)n

= 10000(1 + 5/100)3

= 10000(105/100)3

= ₹ 11576.25

Compound interest = Amount − Principal 

= ₹ 11576.25 − ₹ 10000

= ₹ 1,576.25

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

= 10000(1 + 10/100)1

= 10000(110/100)

By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

= (11000 x 10 x ½)/100

So, the interest for particularly the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000

Therefore, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550

So the actual difference between two interests = 1576.25 – 1550 = 26.25

Thus, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.

Question 24: Maria invested ₹ 8,000 in a particular business. She would eventually be paid interest at 5% per annum compounded annually. Find out

(i) The amount credited against her name, particular at the end of the second year

(ii) The interest for the third year

Answer 24: (i) P = ₹ 8,000

R = 5% per annum

n = 2 years

= 8000(1 + 5/100)2

= 8000(105/100)2

(ii) The interest for the next year, i.e. the third year, has to be calculated. By taking ₹

8,820 as principal, the Simple Interest for the next year will be calculated.

= (8820 x 5 x 1)/100

Question 31: Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get?

(i) after 6 months?

(ii) after 1

Answer 31:(i) P = ₹ 60,000

Rate = 12% per annum = 6% per half-year

n = 6 months = 1 half-year

Amount, A = Principle(1 + Rate/100)time period

= 60000(1 + 6/100)1

= 60000(106/100)

= 60000(53/50)

(ii) There are almost 2 and half years in 1 year

So, time period = 2

= 60000(1 + 6/100)2

= 60000(106/100)2

= 60000(53/50)2

Question 25:Fabina borrows ₹ 12,500 at the rate of 12% per annum for 3 years at simple interest, and Radha borrows the same amount for the same period at 10% per annum, compounded annually. Who pays more interest, Fabina and Radha and by how much?

Answer 25: Interest paid by Fabina = (Principle x Rate x Time period)/100

= (12500 x 12 x 3)/100

Amount paid by Radha at the end of 3 years = A = Principle(1 + Rate/100)time period

Amount = 12500(1 + 10/100)3

= 12500(110/100)3

= ₹ 16637.50

Compound Interest = Amount – Principal  = ₹ 16637.50 – ₹ 12500 

                                                                      = ₹ 4,137.50

The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50

Therefore, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Therefore, Fabina will have to pay ₹ 362.50 more than Radha.

Question 26:Kamala borrowed ₹ 26400 from a particular Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to pay off the loan?

(Hint: Find A for 2 years with interest compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)

Answer 26: Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2

years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years

First, the amount for 2 years has to be calculated

Amount, A = Principal (1 + Rate/100)time period

= 26400(1 + 15/100)2

= 26400(1 + 3/20)2

= 26400(23/20)2

By taking ₹ 34,914 as the principal amount, the S.I. for the next 1/3 years can be calculated.

Simple Interest = (34914 × 1/3 x 15)/100 

                            = ₹ 1745.70

Interest for the first two years

                            = ₹ (34914 – 26400) 

                            = ₹ 8,514

And interest for the next 1/3 year = ₹ 1,745.70

Total Compound Interest = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70

Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70

Question 27: Oranges are purchased at 5 for ₹10 and sold at 6 for ₹15. His gain percentage is

Answer 27: we know that the CP of 1 orange = ₹10/5 = ₹2

SP of 1 orange = ₹15/6 = ₹2.5

Since SP is more than CP, it’s a Gain

Gain = SP – CP

=2.5 – 2

= ((0.5) × 100) / 2

Benefits Of Solving Important Questions Class 8 Maths Chapter 8

Practice is the key to scoring 100%  in Maths. The foundation of fundamental apprehension is the Maths taught in  Classes 8, 9, and 10 and it’s important to step up their learning experience and eliminate “maths phobia” among students. . We recommend students access  Extramarks to obtain important questions in Class 8 Maths Chapter 8. By systematically solving questions and going through the required solutions, students will get  the confidence to solve any tough  questions in the given chapter ”Comparing Quantities”.

Below are a few benefits of frequently solving comparing quantities class 8 important questions:

•  Our  Maths  subject experts have carefully assembled the most important questions in  Class 8 Maths Chapter 8 by inspecting many past years’  question papers.
• The questions and solutions provided are based on the latest CBSE syllabus and as per CBSE guidelines. So the students can completely count on it.
• The questions covered in our set of important questions in Class 8 Maths Chapter 8 are entirely based on several topics covered in the chapter ”Comparing Quantities”. It is suggested that students revise and clear all their doubts before solving all these important questions. 
• By practicing class 8 maths chapter 8 extra questions students will get a general idea of how the paper will be prepared. Practising questions similar to the exam questions would help the students gain confidence, score 100%  marks in their examinations, and set their own benchmark. .

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Q.1 A washing machine was sold for 5760 after giving successive discounts of 15% and 10% respectively. What was the marked price

Marks: 4 Ans

Selling price of washing machine = 5760 Two successive discounts are 15% and 10%. Let marked price of washing machine = x S . P . of washing machine after first discount = x 100 ˆ’ 15 100 = 85 x 100 = 17 x 20 S . P . of washing machine after second discount = 17 x 20 100 ˆ’ 10 100 = 17 x 20 — 90 100 = 153 x 200 Then , according to condition 153 x 200 = Rs . 5760 x = 5760 — 200 153 x = 7529 .40 approx Thus , the marked price of washing machine is 7529 .40.

Q.2 Find the compound interest on 1,60,000 for 2 years at 10% per annum when compounded semi-annually.

Marks: 2 Ans

Principal = 160000, rate = 10% per annum = 5% per half year Time = 2 years = 4 half years. Amount = 160000 — 1 + 5 100 4 = 160000 — 21 20 — 21 20 — 21 20 — 21 20 = 1 , 94 , 481 Compound Interest = 1 , 94 , 481 ˆ’ 1 , 60 , 000 = 34,481

Q.3 Find the amount on a principal of 2000 for 2 years at 10% per annum compounded annually. Also find the compound interest.

Marks: 1 Ans

A = P 1 + r 100 n = 2000 1 + 10 100 2 = 2000 — 110 100 — 110 100 = 2420 Amount = 2420 Compound Interest = Amount – Principal = 2420 – 2000 = 420

Q.4 When 5% sale tax is added on the purchase of a bedsheet of 300, find the buying price or the cost price of the bedsheet.

5 % of 300 is 5 100 — 300 = 15 buying price of Bedsheet is 300 + 15 = 315

Q.5 Rakesh bought a watch for 800 and sold it for 1000. Mukesh bought a car for 4,00,000 and sold it for 4,20,000. Who made a better sale, Rakesh or Mukesh

C.P. of watch for Rakesh = 800 S.P. of watch for Rakesh = 1000 Profit on watch to Rakesh = 1000 ˆ’ 800 = 200 Rate of Profit = 200 1000 — 100 = 20 % C.P. of car for Mukesh = 4,00,000 S.P. of car for Mukesh = 4,20,000 Profit on car for Mukesh = 4,20,000 ˆ’ 4,00,000 Profit on car for Mukesh = 20 , 000 Rate of Profit = 20 , 000 4,00,000 — 100 = 5 % So , Rakesh made a better sale.

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Important questions for class 8 maths, chapter 1 - rational numbers.

Chapter 2 - Linear Equation in One Variable

Chapter 3 - understanding quadrilaterals, chapter 4 - practical geometry, chapter 5 - data handling, chapter 6 - squares and square roots, chapter 7 - cubes and cube roots, chapter 9 - algebraic expressions and identities, chapter 10 - visualising solid shapes, chapter 11 - mensuration, chapter 12 - exponents and powers, chapter 13 - direct and inverse proportions, chapter 14 - factorisation, chapter 15 - introduction to graphs, chapter 16 - playing with numbers, faqs (frequently asked questions), 1. what are the formulas in ”comparing quantities” class 8.

Following are the Important formulas for comparing Quantities Class 8 Maths:

• Cost Price= Selling Price – Profit
• Cost Price= Selling Price + Loss
• Selling Price=Cost Price + Profit
• Selling Price=Cost Price – Loss
• Profit= Selling Price – Cost Price
• Loss= Cost Price – Selling Price
• Profit %= Profit/ cost Price x 100
• Loss %= Loss/ Cost Price x 100

2. How can I score well in Class 8 Maths examinations?

Maths is a subject which requires a lot of practice. To score well in Maths, one must have a strong conceptual understanding of the chapter, be good in calculations, practice questions regularly, give mock tests from time to time, get feedback and avoid silly mistakes. Regular practice with discipline, working diligently and conscientiously towards your goal, will definitely ensure a 100% in your exams.  

3. What can I get from the Extramarks website?

Extramarks is one of the best educational platforms and it has its own archive of educational resources, which assists students in acing their exams. You can get all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT-based mock tests, CBSE revision notes, and  Class 8 Maths Chapter 8 important questions on the Extramarks website. Apart from this, you can get comprehensive guidance from our subject experts and doubt-clearing sessions once you sign up on our official website for any study resources.     

4. How many total chapters will students study in Class 8 Maths?

There are 16 chapters in Class 8 Maths. The list of chapters is given below:

• Chapter 1- Rational Numbers
• Chapter 2 – Linear Equations in One Variable
• Chapter 3 – Understanding Quadrilaterals
• Chapter 4 – Practical Geometry
• Chapter 5 – Data Handling
• Chapter 6 – Square and Square Roots
• Chapter 7 – Cube and Cube Roots
• Chapter 8 – Comparing Quantities
• Chapter 9 – Algebraic Expressions and Identities
• Chapter 10 – Visualising Solid Shapes
• Chapter 11- Mensuration
• Chapter 12 – Exponents and Powers
• Chapter 13 – Direct and Inverse Proportions
• Chapter 14 – Factorisation
• Chapter 15 – Introduction to Graphs
• Chapter 16 – Playing with Numbers

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Unit 7: Comparing quantities

Percent change word problems.

• Percent word problem: penguins (Opens a modal)
• Percent word problem: magic club (Opens a modal)
• Percent word problems Get 5 of 7 questions to level up!
• Percentage change word problems Get 3 of 4 questions to level up!

Profit and loss

• Intro to profit and loss (Opens a modal)
• Finding profit percent (Opens a modal)
• Profit and loss percent problems Get 3 of 4 questions to level up!
• Profit and loss word problems Get 3 of 4 questions to level up!

Tax and discount

• Percent word problems: tax and discount (Opens a modal)
• Tax and tip word problems Get 3 of 4 questions to level up!
• Discount, markup, and commission word problems Get 3 of 4 questions to level up!

Compound interest

• Intro to compound interest (Opens a modal)
• Solved example: compound interest (Opens a modal)
• Find compound interest Get 3 of 4 questions to level up!
• Word problems on compound interest Get 3 of 4 questions to level up!

Comparing Quantities Class 8 Formulas

Comparing quantities class 8 formulas are fundamental to understanding the theory of comparing mathematical quantities and solving numerical problems based on them. For students who find the concept of comparing quantities difficult, it is important to gain complete knowledge of all basic concepts and formulas. This article provides a list of all comparing quantities class 8 formulas in brief. It also covers solved examples and other details that will benefit students in handling everyday problems.

List of Comparing Quantities Class 8 Formulas

Students will get a better understanding of comparing Quantities Class 8 Formulas by referring to the list provided below:

• Discount = Marked Price – Sale Price
• Discount percentage = ( Discount/ Marked Price ) x 100
• Overhead expenses are those additional expenses that are made after buying an article and are included in the cost price.
• Cost Price = Buying price + Overhead expenses
• Sales tax = Tax percentage of Bill Amount
• Simple Interest = ( Principal x Rate x Time )/100
• Compound Interest Formula = Amount - Principal
• Amount = Principal ( 1 + Rate/100)n

Applications of Comparing Quantities Class 8 Formulas

In our everyday life, we come across terms like discount, profit, loss, interest. Let us see how comparing quantities class 8 formulas play an important role in real life.

• Understanding the shopping transactions is an important skill to have. The profit, loss and discount formulas of comparing quantities are extremely helpful in understanding the same.
• The time value of money is the essence of modern financial systems and comparing quantities class 8 formulas help in the evaluation of the amount of money over a period of time-based on different parameters.

Tips to Memorize Comparing Quantities Class 8 Formulas

Comparing quantities class 8 formulas can be a bit confusing to remember if the terms associated with them are not clear to the students. Hence, below are some of the tips that can help students memorize these formulas with much ease:

• Comparing quantities class 8 formulas has terms like cost price, market price, principal, amount. The students must make sure to understand the meaning of these terms first before moving ahead to memorize the formulas. They can get help from teachers or their friends to get clarity on the same.
• Once they are clear with the terms, they can try creating a story around the formula, or weave the initials of the formula into some word of their choice so as to quickly memorize them.
• Students can put the formula images as wallpaper on their gadgets which will enable them to do a quick revision whenever they use their mobile or laptop.
• Practice solved examples given in the textbook. This will give maximum coverage of the usage of formulas in different contexts.

Comparing Quantities Class 8 Formulas Examples

Example 1: Arvind opened a cracker shop and bought crackers in wholesale worth 1000 INR at a 10% discount. Arvind then sold the crackers in his shop and paid 5% tax on the bill amount. What is the net profit of Arvind from his cracker shop?

Solution: Total cost price for Arvind = 1000 - ( 1000* 10/100)

= 1000 - 100

Net selling price for Arvind = 1000 - (1000 * 5/100)

= 1000 - 50

Net Profit for Arvind = Selling Price - Cost Price

= 950 -900 = 50 INR.

Example 2: Kangna took a loan of 40,000 INR at a simple interest rate of 10% per annum and spent 20,000 INR. The remaining 20,000 INR Kangana loaned it to her best friend Dimple at 5 % compound interest. After five years, what is the net asset and liability of Kangana?

Solution: Total amount to be paid by Kangana = Principal + Simple Interest

= 40000 + (40000 * 10 * 5)/100

= 40000 + 20000

Thus, 60000 INR is the liability to be paid to the bank by Kangana

Total amount received by Kangana = Principal + Compound Interest

= 20000 ( 1+ 5/20)5

= 20000 ( 1+.25)5

= 20000 * 3.05

= 61000 INR is the asset to be received from Dimple.

Students can download the printable Maths Formulas C lass 8 sheet from below.

FAQs On Comparing Quantities Class 8 Formulas

What are the Important Formulas Covered Under Comparing Quantities Class 8 Formulas?

The important formulas covered under comparing quantities class 8 formulas are as below:

What are the Basic Formulas in Comparing Quantities Class 8 Formulas?

The basic formulas in comparing quantities class 8 formulas will help students understand the basics of transactions and the calculation of interest on money over a period of time. Also, these formulas will help explain the relationship between the selling price, marked price, cost price, discount, interest, amount, etc.

What are the Important Concepts Covered in Comparing Quantities Class 8 Formulas?

The important concepts covered in comparing quantities class 8 formulas are how to find the discount and sales tax, calculating the value of money over a particular period of time using formulas of simple interest and compound interest.

How Many Formulas are There in Comparing Quantities Class 8 Formulas?

There are around 8 formulas in comparing quantities class 8 formulas that can be remembered easily if the students follow the tips mentioned in this article on a consistent basis. The main formulas include calculating discounts, sales tax, simple interest and compound interest.

How to Memorize Comparing Quantities Class 8 Formulas?

Due to the introduction of a lot of new terms like selling price, marked price, interest, etc., the students often get confused while memorizing the comparing quantities class 8 formulas. This article will help them with the tips to overcome this hurdle.

• Before memorizing the formulas, students must ensure that they have understood the meaning of all the terms involved in the formulas. They can seek help from professors or friends to gain a better understanding of concepts.
• Once students understand the logical details, they can revise the formulas through formula sheets or explanatory video resources. Also, the students can set formula images as wallpaper on their gadgets.
• Students must ensure to practice all of the textbook's solved examples. This will provide the most comprehensive coverage on formula usage in various contexts.

• School Guide
• CBSE Notes for Class 8
• CBSE Notes for Class 9
• CBSE Notes for Class 10
• CBSE Notes for Class 11
• CBSE Notes for Class 12
• NCERT Solutions
• English Grammar
• Basic Maths Formulas

NCERT Class 8 Maths : Latest Syllabus based Study Material

Learn ncert class 8 maths with our easy-to-follow ncert-based study material. we've compiled everything a student needs to learn class 8 maths, equipping you with the core concepts to top your exams. we'll guide you through the essential concepts from all chapters in a clear, engaging, and easy-to-understand way, so you can learn all concepts of class 8 maths and unlock your potential in this field to score 95+ marks in all your exams..

Maths Solutions

Important Formulas

Discover a smoother learning journey through our effortless roadmap

Chapter 1: Rational Numbers

Rational Number: Definition, Examples, Worksheet

Natural Numbers | Definition, Examples, Properties

Whole Numbers - Definition, Properties and Examples

Integers - Definition, Properties and Worksheet

Properties of Rational Numbers

Closure Property

Commutative Property - Definition | Commutative Law and Examples

Associative Property

Distributive Property

Chapter 2: Relations and Functions

Linear Equations in One Variable

Algebraic Expressions in Maths

Equation in Maths

Linear Equations in One Variable - Solving Equations which have Linear Expressions on one Side and Numbers on the other Side | Class 8 Maths

Solve Linear Equations with Variable on both Sides

Reducing Equations to Simpler Form | Class 8 Maths

Chapter 3: Understanding Quadrilaterals

16 articles

Polygon - Shape, Formula, Types, and Examples

Convex Polygon

Regular Polygon

Types of Polygons

Triangles in Geometry

Quadrilateral - Definition, Properties, Types, Formulas, Examples

Area of Pentagon

Understanding Quadrilaterals - Measures of the Exterior Angles of a Polygon

Types of Quadrilaterals and Their Properties

Kite - Quadrilaterals

Introduction to Parallelogram: Properties, Types, and Theorem

Properties of Parallelograms

Diagonal of Parallelogram Formula

Rhombus: Definition, Properties, Formula, Examples

Rectangle | Definition, Properties, Formulas

Chapter 4: Data Handling

Data Handling

Pictograph in Statistics, Creating, Reading, Examples, Advantages/Disadvantages, Practice Questions

Bar Graph | Examples & How To Draw a Bar Graph

Probability in Maths

Random Experiment - Probability

Chapter 5: Squares and Square Roots

Squares and Square Roots

Square Numbers

Squares and Cubes

Pythagorean Triples

Square Root

How to calculate a square root?

Square root of a number by Repeated Subtraction method

Long Division Method to find Square root with Examples

Square Root of Decimals

Chapter 6: Cube and Cube Roots

Cubes and Cube Roots

Cube Root 1 to 30

Chapter 7: Comparing Quantities

Ratios and Percentages

Fractions - Definition, Properties, Types, Operations & Examples

Ratio Formula

Discount Formula

Sales Tax, Value Added Tax, and Goods and Services Tax - Comparing Quantities | Class 8 Maths

Compound Interest | Class 8 Maths

Compound Interest Formula

Chapter 8: Algebraic Expressions and Identities

Addition of Algebraic Expressions

Subtraction of Algebraic Expressions

Multiplication of Algebraic Expression

Multiplying Polynomials

Chapter 9: Mensuration

Area of Polygons

Area of Triangle | Formula and Examples

Area of Quadrilateral

Visualizing Solid Shapes

Cube - Definition, Shape & Formula

Cuboid - Shape and Properties

What is a Cone?

Cylinder | Shape, Formula and Examples

Surface Area of Cube

Surface Area of Cuboid

Surface Area of Cylinder

Volume of a Cube

Volume of Cuboid | Formula and Examples

Volume of Cylinder

Volume and Capacity - Mensuration | Class 8 Maths

Chapter 10: Exponents and Power

Negative Exponents

Laws of Exponents

Laws of Exponents & Use of Exponents to Express Small Numbers in Standard Form - Exponents and Powers | Class 8 Maths

Chapter 11: Direct and Inverse Proportions

Direct Proportion in Mathematics

Direct and Inverse Proportions

Chapter 12: Factorisation

Factors of a Number

Factorization of Polynomial

How to find Common Factors?

Division of Algebraic Expressions

Chapter 13: Introduction to Graphs

Introduction to Graphs | Class 8 Maths

Maths Solution

RD Sharma Class 8 Solutions for Maths: Chapter Wise PDF

Maths Formulas

CBSE Class 8 Maths Formulas

This NCERT Class 8 Maths Study Material is designed by expert teachers to help you learn the theory and Applications of Class 8 mathematics. Follow along with the interactive tutorials that demonstrate how to solve problems and apply mathematical concepts. In addition, GeeksforGeeks has also provided detailed answers to previous year's questions, so students can practice and learn how to answer them correctly. This helps students to understand the concepts behind the questions and to improve their problem-solving skills.

What do we offer?

GeeksforGeeks has designed this study material to make an interactive and enjoyable experience for learning class 8 Maths. Our expert teachers provide clear and concise explanations of difficult concepts to keep you motivated and focused. We've also incorporated practice problems and quizzes with solutions to help you test your knowledge and score 95+ Marks in your Maths exam.

Key Highlights of this Tutorial:

• Based on the Latest NCERT Board Class 8 Syllabus
• Focus on Application: Bridge the gap between theory and application. We focus on how Mathematical concepts are utilized in real-world situations, making learning easy and engaging.
• Highlights All formulas: Important Maths formulas are covered throughout the lessons, equipping you with the necessary formulas to score 95+ marks in your class 8 Maths exam with confidence.

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Loved the Mathematics lessons on GeeksforGeeks! Easy to understand content and paper set made studying not just simple but interesting too. A big thumbs up from me!

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GeeksforGeeks made Mathematics so clear! The lessons are short and sweet, and I can learn anytime. It's like having a teacher always ready to help.

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Mathematics is fun with GeeksforGeeks! The explanations are super simple, and I love doing the quizzes. It's my go-to for studying now.

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Q.1 How to study for class 8 maths?

1. Focus on NCERT: Understand the book, and solve problems yourself. 2. Practice regularly: NCERT problems, worksheets, and past exams. 3. Plan and be consistent: Schedule study time, and stick to it. 4. Get help: Ask your teacher, tutor, or classmate.

Q2. Which math book is best for class 8 NCERT?

1. NCERT (Essential): Core resource, that covers all concepts. 2. Reference Books (Practice): RD Sharma, RS Aggarwal for extra practice. 3. Past Papers (Exams): Practice for format, questions, and timing.

Q3. How to get 100 in Maths class 8?

To score full marks in math, you need to have a clear understanding of the concepts, practice regularly, focus on accuracy, manage your time effectively, use shortcuts and tricks, and seek help when needed.

Q4. Should we use Study Materials for NCERT Class 8 exam preparation?

For the students of Class 8, the important benefit of using the NCERT Study Materials is for Class 8 exam preparation. This Study material focuses deeply on the fundamentals of every concept according to the latest NCERT syllabus and curriculum.

Q5. How useful are NCERT Class 8 Maths notes?

Revising from notes can be very useful and fruitful as it helps in preparation just before the exam.

Q6. Which is the toughest chapter in math class 8?

While difficulty can vary, some chapters are mentioned more often as challenging: 1. Comparing Quantities (ratio, proportion, percentage) 2. Algebraic Expressions (variables, expressions) 3. Mensuration (area, volume, perimeter)

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• Class 8 Maths MCQs
• Chapter 8 Comparing Quantities

Class 8 Maths Chapter 8 Comparing Quantities MCQs

Class 8 Maths Chapter 8 Comparing Quantities MCQs (Questions and Answers) are provided online for free. These multiple-choice questions are prepared by our subject experts according to the  CBSE syllabus (2022-2023 ) and NCERT guidelines. The objective questions are also available chapter-wise at BYJU’S to make students learn each concept thoroughly and help them to score better marks in exams. Also, learn important questions for class 8 Maths here.

Practice more and test your skills on Class 8 Maths Chapter 8 Comparing Quantities MCQs with the given PDF here.

MCQs Questions on Class 8 Chapter 8 Comparing Quantities

Multiple choice questions (MCQs) are available for Class 8 Chapter 8 Comparing Quantities chapter. All the problems have four multiple options, in which one is the right answer. Students have to solve each question and choose the correct answer.

1. The ratio of speed of cycle 12 km per hour to the speed of scooter 36 km per hour is

D. None of the above

Explanation: Speed of cycle/Speed of scooter = 12/36 = ⅓

2. The ratio of 10m to 10 km is:

Explanation: 10m/10km = 10m/10000m = 1/1000

3. The percentage of 1:4 is:

Explanation: 1:4 = 1/4

(1×100)/(4×100) = 1/4 x 100% = 0.25 x 100% = 25%

4. The percentage of 2:5

Explanation: 2:5 = ⅖ = ⅖ x 100% = 0.4 x 100% = 40%

5. If 50% of students are good at science out of 20 students. Then the number of students good at science is:

Explanation: 50% of students out of 20 students = 50% of 20

= (50/100) x 20

6. The price of a motorcycle was Rs. 34,000 last year. It has increased by 20% this year. The price of motorcycle now is:

A. Rs. 36,000

B. Rs. 38,800

C. Rs. 40,800

D. Rs. 32,000

Explanation: 20% of Rs.34,000 = 20/100 x 34,000 = Rs.6800

New price = Rs. 34,000+Rs.6800

= Rs. 40,800

7. An item marked at Rs. 840 is sold for Rs. 714. The discount % is:

Explanation: Discount = Marked Price – Sale Price

= 840 – 714

Discount % = (126/840) x 100% = 15%

8. A person got a 10% increase in his salary. If his salary was Rs. 50000, then the new salary is:

A. Rs. 55000

B. Rs. 60000

C. Rs. 45000

D. Rs. 65000

Explanation: Previous salary = Rs. 50000

10% of Rs.50000 = (10/100) x 50000 = Rs. 5000

New salary = Rs. 50000 + Rs. 5000

= Rs. 55000/-

9. The cost of the article was Rs. 15500 and Rs. 500 was spent on its repairing. If it is sold for a profit of 15%. The selling price of the article is:

A. Rs.16400

B. Rs.17400

C. Rs.18400

D. Rs.19400

Explanation: Total cost = 15500+500 = 16000

Profit % = (Profit/Cost price) x 100

Profit = (Profit% x Cost price)/100

P = (15×16000)/100 = 2400

Selling Price = Profit + cost price = 2400 + 16000 = Rs.18400

10. Waheeda bought an air cooler for Rs. 3300 including a tax of 10%. The price of the air cooler before VAT was added is:

A. Rs. 2000

B. Rs. 3000

C. Rs. 2500

D. Rs. 2800

Explanation:10% VAT on Rs.100 will make it Rs.110

So, for price including VAT Rs.110, the original price is Rs.100

Then, Price including VAT Rs. 3300, the original price = Rs. (100/110) x 3300 = Rs. 3000.

11. The ratio of 50 paise to Rs. 1 is:

Rs. 1 = 100 paise

50 paise : 100 paise

12. The Ratio of 10m to 1 Km is Equal To:

Explanation: 10m/1km = 10m/1000m = 1/100

13. The ratio of 1m and 100 cm is:

Answer: D. 1:1

Explanation: One meter is equal to hundred centimeters.

1m = 100 cm

Thus, their ratio will be:

⇒ 100cm:100cm

14. The percentage of ratio 2:3 is:

Explanation: 2:3 = ⅔

Converting into percentage, we have to multiply the ration by 100.

⅔ x 100% = 200/3 = 66.66..%

15. The ratio 4 : 25 converted to percentage is:

Answer: C. 16%

4 : 25 =4/25 x 100% = 16%

16. Convert the fraction ⅛ into a percentage.

Answer: A. 12.5%

Explanation: Multiply the given fraction by 100 to get the percentage value.

⅛ x 100 = 12.5%

17. The ratio of speed of a motorbike 50 km per hour and speed of cycle 20 km per hour is:

Answer: C. 5:2

Explanation: Speed of motorbike is 50km/hr

Speed of cycle is 20km/hr

Ratio of their speeds are = 50/20 = 5/2 = 5:2

18. If 40% of students passed in the Mathematics test, out of 40 students in the class. What is the number of students who have passed?

Explanation: 40% of students out of 40 students will be:

= (40/100) x 40

19. 25% of 60 students of a class likes to play football. How many students does not like to play football?

Answer: A. 45

Explanation: Number of students = 60

25% of 60 students = 25/100 x 60 = 15

Number of students who does not play football = 60 – 15 = 45

20. Out of 40 students, 25% passed in a class. How many students did actually pass?

Explanation:

25% of 40 = 25/100 x 40 = 10

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MCQ Questions for Class 8 Maths Chapter 8 Comparing Quantities with Answers

We have compiled the NCERT MCQ Questions for Class 8 Maths Chapter 8 Comparing Quantities with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 8 Maths with Answers on a daily basis and score well in exams. Refer to the Comparing Quantities Class 8 MCQs Questions with Answers here along with a detailed explanation.

Comparing Quantities Class 8 MCQs Questions with Answers

Choose the correct option.

Question 1. The ratio of 5 km per hour to 500 m per hour is (a) 1 : 10 (b) 10 : 1 (c) 1 : 5 (d) 5 : 1

Answer: (a) 1 : 10

Question 2. The fraction \(\frac {3}{5}\) converted to percentage is (a) 20% (b) 30% (c) 40% (d) 60%

Answer: (d) 60%

Question 3. Out of 60 students in a class, 25% passed. How many students passed? (a) 10 (b) 15 (c) 30 (d) 40

Answer: (b) 15

Question 4. VAT is always calculated on which of the following? (a) Selling price (b) Cost price (c) Marked price (d) Profit or loss

Answer: (a) Selling price

Question 5. The marked price of a book is Rs 200. The shopkeeper gives 25% discount on it. What is the selling price of the book? (a) Rs 50 (b) Rs 150 (c) Rs 175 (d) Rs 125

Answer: (b) Rs 150

Question 6. The simple interest on Rs 4000 for 2 years at 8% per annum is (a) Rs 180 (b) Rs 640 (c) Rs 1200 (d) Rs 1600

Answer: (b) Rs 640

Question 7. A sum of Rs 1600 at 5% p.a. compound interest amount to Rs 1764 in (a) 2 years (b) 10 years (c) 8 years (d) 4 years

Answer: (a) 2 years

Question 8. On what principal, simple interest of 5 years at the rate of 6% is Rs 90? (a) Rs 250 (b) Rs 300 (c) Rs 400 (d) Rs 450

Answer: (b) Rs 300

Question 9. If the interest is compound quarterly, then rate will be (a) half of the annual rate (b) double the annual rate (c) four times the annual rate (d) one-fourth of the annual rate

Answer: (c) four times the annual rate

Question 10. The compound interest on Rs 4000 for one year at 5% p.a payable half-yearly is: (a) Rs 400 (b) Rs 405 (c) Rs 200 (d) Rs 202.50

Answer: (d) Rs 202.50

Question 11. On which of the following profit percent or loss percent is calculated? (a) SP (b) CP (c) marked price (d) none of these.

Answer: (b) CP

Question 12. The discount is always calculated on which of the following? (a) SP (b) CP (c) marked price (d) none of these.

Answer: (c) marked price

Fill in the blanks

Question 1. …………………… is always calculated on the marked price.

Answer: Discount

Question 2. Cost price + Profit gives ……………………

Answer: Selling price

Question 3. Comparison of two quantities by division is called ……………………

Answer: Ratio and Proportion

Question 4. Difference between simple interest and compound interest for one year on the same sum and rate is equal to ……………………

Question 5. The difference between C.I. and S.I on Rs 50,000 at 5% p.a for one year is ……………………

Hope the information shed above regarding NCERT MCQ Questions for Class 8 Maths Chapter 8 Comparing Quantities with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 8 Maths Comparing Quantities MCQs Multiple Choice Questions with Answers, feel free to reach us so that we can revert back to us at the earliest possible.