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**NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities** are developed by experts to ensure the requirements of a CBSE Class 8 student. These class 8 NCERT Solutions are important and will not confuse students; they instead help them in reaching a conclusion using the ideal problem-solving method. Solving practice questions from NCERT textbook gives students the advantage of adequate preparation, well before the CBSE final exams.

Knowledge Glow NCERT Solutions for Class 8 Maths aims to bring out the best in students by further skill-building exercises that are apt to their class, strengths and interests. Referring to these solutions will increase the confidence of students in CBSE Class 8 Mathematics exams. NCERT Solutions for Class 8 is an invaluable help to students in their exam preparation.

### NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.1

**Question 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.**

(b) 5 m to 10 km

(c) 50 paise to ₹ 5

**Solution:**

**(a) Speed of cycle : Speed of Scooter = 15 km per hour : 30 km per hour**= 1530 = 12

Hence, the ratio = 1 : 2

**(b) 5 m to 10 km**= 5 m : 10 × 1000 m [∵ 1 km = 1000 m]

= 5 m : 10000 m

= 1 : 2000

Hence, the ratio = 1 : 2000

**(c) 50 paise to ₹ 5**

= 50 paise : 5 × 100 paise

= 50 paise : 500 paise

ratio = 1 : 10

**Question 2. Convert the following ratios to percentages:**(a) 3 : 4

(b) 2 : 3

**Solution:**

a) 3:4 = ¾ = ¾ x 100% = 0.75 x 100% = 75%

b) 2:3 = 2/3 = 2/3 x 100% = 0.666 x 100% = 66.66% = 66⅔%

**Question 3. 72% of 25 students are good in mathematics. How many are not good in mathematics?**

**Solution:**

Number of students who are good in mathematics = 72% of 25

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

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

**Solution:**

Let Assume the total number of matches played by the team be x.

Considering that the team won 10 matches and the winning percentage of the team was 0%.

⇒ 40/100 × x = 10

40x = 10 × 100

40x = 1000

x = 1000/40

= 100/4

= 25

Therefore, the team played 25 matches.

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

**Solution:**

Let the amount of money which Chameli had in the beginning be x

Considering that, after spending 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

= ₹2400

Therefore, Chameli had ₹2400 is the beginning.

**Question 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.**

**Solution:**

**Total number of people = 50,00,000**

Percentage of people who like other games = (100 – 60 – 30)%

= (100 – 90)%

= 10%

Total number of people = 50 lakhs

*So,*

Number of people who like cricket = 60/100 x 50 = 30 lakhs

Number of people who like football = 30/100 x 50 = 15 lakhs

Number of people who like other games = 10/100 x 50 = 5 lakhs

### NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.2

**Question 1. A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.**

**Solution:**

Let Original Salary x

Considering that the new salary is ₹ 1,54,000

Original salary + Increment = New salary

Considering that the increment is 10% of the original salary

So, (x + 10/100 × x) = 154000

x + x/10 = 154000

11x/10 = 154000

x = 154000 × 10/11

= 140000

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

**Question 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?**

**Solution:**

Consider that on Sunday, 845 people went to the zoo and on Monday, 169 people went to the zoo

Decrease in the number of people = 845 – 169 = 676

Thus,

Percentage decrease = (Decrease in the number of people/Number of people who went to zoo on Sunday) x 100%

= (676/845 x 100)%

= 80%

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

**Solution:**

Given that the shopkeeper buys 80 articles for ₹ 2,400

Cost of one article = 2400/80 = ₹ 30

Profit percentage = 16%

Profit percentage = Profit/C.P x 100

16 = Profit/30 x 100

Profit = (16 x 30)/100

= ₹ 4.8

Therefore, selling price of one article = C.P + Profit

= ₹ (30 + 4.80)

= ₹ 34.80

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

**Solution:**

Total cost of an article = Cost + Overhead expenses

= ₹15500 + ₹450 = ₹15950

Profit percentage = 15%

Profit percentage = Profit/C.P x 100

15 = Profit/15950 x 100

Profit = (15 x 15950)/100 = 2392.50

Therefore, Selling price of the article = C.P + Profit

= ₹(15950 + 2392.50) **= ₹18342.50**

**Question 5. A VCR and TV were bought for ₹ 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.**

**Solution:**

C.P. of a VCR = ₹ 8000

The shopkeeper made a loss of 4 % on VCR

This means if C.P. is ₹ 100, then S.P. is ₹ 96. When C.P. is ₹ 8000

S.P. = (96/100 x 8000) = ₹ 7680

C.P. of a TV = ₹ 8000

The shopkeeper made a profit of 8 % on TV.

This means that if C.P. is ₹ 100, then S.P. is ₹ 108.

When C.P. is ₹ 8000,

S.P. = (108/100 x 8000) = ₹ 8640

Total S.P. = ₹ 7680 + ₹ 8640 = ₹ 16320

Total C.P. = ₹ 8000 + ₹ 8000 = ₹ 16000

Since, total S.P.> total C.P. ⇒ profit

Profit = ₹ 16320 − ₹ 16000 = ₹ 320

Profit % on the whole transaction = Profit/Total CP x 100

= 320/16000 x 100 = 2%

**So the shopkeeper got 2% profit on the whole transaction.**

**Question 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 ₹ 1450 and two shirts marked at ₹ 850 each?**

**Solution:**

Total marked price = ₹ (1,450 + 2 × 850)

= ₹ (1,450 +1,700)

= ₹ 3,150

Given that the discount rate = 10%

Discount = ₹ (10/100 x 3150)

= ₹ 315

Also, Discount = Marked price − Sale price

₹ 315 = ₹ 3150 − Sale price

∴ Sale price = ₹ (3150 − 315)

= ₹ 2835

So customer has to pay ₹ 2835.

**Question 7. A milkman sold two of his buffaloes for ₹ 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.**

**Solution:**

S.P. of each buffalo = ₹ 20,000

A milkman made a gain of 5% while selling one buffalo

This means if C.P. is ₹ 100, then S.P. is ₹ 105.

C.P. of one buffalo = 100/105 × 20000

= ₹ 19,047.62

Other 2nd buffalo was also sold at a loss of 10%

This means that if C.P. is ₹ 100 then S.P. is ₹90

∴ C.P. of other buffalo = 100/90 × 20000

= ₹ 22222.22

Total C.P. = ₹ 19047.62 + ₹ 22222.22 = ₹ 41269.84

Total S.P. = ₹ 20000 + ₹ 20000 = ₹ 40000

Loss = ₹ 41269.84 − ₹ 40000 = ₹ 1269.84

Hence, the total loss to the milkman was ₹ 1,269.84

**Question 8. The price of a TV is ₹ 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,**

**Solution:**

On ₹100, the tax to be paid = ₹12

Here, on ₹13000, tax to be paid will be = 12/100 × 13000 = ₹1560

Required amount = Cost + Sales Tax

= ₹ 13000 + ₹ 1560 = ₹ 14560

Hence, Vinod has to pay ₹ 14,560 for the T.V.

**Question 9. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.**

**Solution:**

Let the marked price be x

Discount percent = Discount/Marked Price x 100

20 = Discount/x × 100

Discount = 20/100 × x

= x/5

Discount = Marked price – Sale price

x/5 = x – ₹ 1600

x – x/5 = 1600

4x/5 = 1600

x = 1600 x 5/4 **= 2000**

Hence, the marked price was ₹2000.

**Question 10. I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.**

**Solution:**

Price includes VAT

So, 8% VAT means that if the price without VAT is ₹100,

then the price including VAT is ₹108

if the price including of VAT is ₹108, original price = ₹ 100

When the price including of VAT is ₹5400, original price = ₹ (100/108 × 5400)

= ₹ 5000

Hence the price of the hair dryer before the addition of VAT was ₹ 5000.

### NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3

**Question 1. Calculate the amount and compound interest on**

**(a) ₹ 10,800 for 3 years at 12½ % per annum compounded annually.**

**Solution:**

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)

C.I. = A – P = ₹ (15377.34 – 10800) = ₹ 4,577.34

**(b) ₹ 18000 for 2½ years at 10% per annum compounded annually.**

**Solution:**

Principal (P) = ₹ 18,000

Rate (R) = 10% annual

Number of years (n) = 2½

The amount of 2 years and 6 months can be calculated by calculating the amount for 2 years.

By using the compound interest formula and then calculating the simple interest of 6 months from the amount obtained at the end of 2 years

First, the amount for 2 years must be calculated

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

= 18000(1 + 1/10)^{2}

= 18000(11/10)^{2} = ₹ 21780

By taking the principal as ₹ 21,780, calculate the S.I. for the next ½ year.

S.I. = (21780 x ½ x 10)/100

= ₹ 1089

Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780

And, interest for the next ½ year = ₹ 1089

Total C.I. = ₹ 3780 + ₹ 1089

= ₹ 4,869

Therefore,

Amount, A = P + C.I.

= ₹ 18000 + ₹ 4869

= ₹ 22,869

**(c) ₹ 62500 for 1½ years at 8% per annum compounded half yearly.**

**Solution:**

Principal (P) = ₹ 62,500

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

Number of years = 1½

It will be 3 half years in 1½ years

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

= 62500(1 + 4/100)^{3}

= 62500(104/100)^{3}

= 62500(26/25)^{3} **= ₹ 70304**

C.I. = A–P

= ₹ 70304 – ₹ 62500 **= ₹ 7,804**

**(d) ₹ 8000 for 1 year at 9% per annum compound half yearly.**

**(You could use the year by year calculation using SI formula to check it)**

**Solution:**

Principal (P) = ₹ 8000

Rate of interest = 9% per annum or 9/2% per half year

Number of years = 1 year

There will be 2 half years in 1 year

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

= 8000(1 + 9/200)^{2}

= 8000(209/200)^{2} = 8736.20

C.I. = A – P

= ₹ 8736.20 – ₹ 8000 = ₹ 736.20

**(e) ₹ 10000 for 1 year at 8% per annum compounded half yearly.**

**Solution:**

Principal (P) = ₹ 10,000

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

Number of years = 1 year

There are 2 half years in 1 year

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

= 10000(1 + 4/100)^{2} = 10000(1 + 1/25)^{2} = 10000(26/25)^{2}

= ₹ 10816

C.I. = A – P = ₹ 10816 – ₹ 10000 = ₹ 816

**2. Kamala borrowed ₹ 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 S.I. on the 2nd year amount for 4/12 years.)**

**Solution:**

Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

years using the compound interest formula and then calculating months of simple interest on the amount received at the end of 2 years

The amount for 2 years should to be calculated

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

= 26400(1 + 15/100)^{2}

= 26400(1 + 3/20)^{2}

= 26400(23/20)^{2} **= ₹34914**

By taking ₹ 34,914 as principal, calculate the S.I. for the next 1/3 years

S.I. = (34914 × 1/3 x 15)/100 = ₹ 1745.70

Interest for first two years = ₹ (34914 – 26400) = ₹ 8,514

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

Total C.I. = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70

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

**3. Fabina borrows ₹ 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?**

**Solution:**

Interest paid by Fabina = (P x R x T)/100

= (12500 x 12 x 3)/100

= 4500

Amount paid by Radha at the end of 3 years = A = P(1 + R/100)^{n}

A = 12500(1 + 10/100)^{3}

= 12500(110/100)^{3}

= ₹ 16637.50

C.I. = A – P = ₹ 16637.50 – ₹ 12500 = ₹ 4,137.50

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

Thus, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Hence, Fabina will have to pay ₹ 362.50 more.

**4. I borrowed ₹ 12000 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?**

**Solution:**

P = ₹ 12000

R = 6% per annum

T = 2 years

S.I. = (P x R x T)/100

= (12000 x 6 x 2)/100

= ₹ 1440

To find the compound interest, the amount (A) will be calculated

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

= 12000(1 + 6/100)^{2}

= 12000(106/100)^{2}

= 12000(53/50)^{2}

= ₹ 13483.20

∴ C.I. = A − P

= ₹ 13483.20 − ₹ 12000

= ₹ 1,483.20

C.I. − S.I. = ₹ 1,483.20 − ₹ 1,440

= ₹ 43.20

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

**5. 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**

**Solution:**

(i) P = ₹ 60,000

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

n = 6 months = 1 half year

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

= 60000(1 + 6/100)^{1}

= 60000(106/100)

= 60000(53/50)

= ₹ 63600

(ii) There are 2 half years in 1 year

So, n = 2

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

= 60000(1 + 6/100)^{2}

= 60000(106/100)^{2}

= 60000(53/50)^{2} = **₹67416**

**NCERT Solutions for Class 8 Maths Chapter 8 – Comparing Quantities**

**NCERT Solutions for Class 8 Maths** is very important to study because this chapter gives basic information about taxes, interest, profit, loss and much more. The concepts covered in this chapter are:

- Discount is a reduction given on marked price. The Discount = Marked Price – Sale Price.
- The discount can be calculated if the discount rate is given. Discount = Discount % of the Marked Price.
- Additional costs incurred after the purchase of the product are included in the cost price and are called overhead costs. CP = purchase price + Overhead expenses
- The state collects sales tax from the sale of goods, which is added to the amount of the invoice. Sales tax = tax % of the invoice amount.
- GST stands for Goods and Services Tax and is levied on the supply of goods or services or both.
- Compound interest is the interest calculated on the previous year’s amount (A = P + I)