# NCERT Solution for class 8 maths chapter 8 - Comparing Quantities

The **NCERT solutions for class 8 maths** chapter 8 Comparing Quantities is prepared by academic team of Physics Wallah. We have prepared **NCERT solutions** for all exercise of chapter 8. Given below is step by step solutions of all questions given in NCERT textbook for chapter 8. Read chapter 8 theory make sure you have gone through the theory part of chapter 8 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions for class 8 maths with **revision notes** of all **maths formula** of class 8 maths. Do read these contents before moving to solve the exercise of NCERT chapter 8

### NCERT Solutions for Class 8 Maths 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) 50 m to 10 km

(c) 50 paise to Rs.5

**Solution :**

(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 = = = 1 : 2000

(c) `1 = 100 paise

∴`5 = 5 × 100= 500 paise

∴Hence Ratio = = = 1 : 10

**Question 2.**

Convert the following ratios to percentages:

(a) 3 : 4

(b) 2 : 3

**Solution :**

(a) Percentage of 3 : 4 =

= 75%

(b) Percentage of 2 : 3 =

=

**Question 3.**

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

**Solution :**

Total number of students = 25

Number of good students in mathematics = 72% of 25 = = 18

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

Hence percentage of students not good in mathematics = = 28%

**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 total number of matches be** x,**

According to question,

Hence total number of matches are 25.

**Question 5.**

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

**Solution :**

Let her money in the beginning be `x.

According to question,

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

**Question 6.**

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

**Solution :**

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

= = 30,00,000

And Number of people who like football

= 30% of 50,00,000

= 15,00,000

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

= = 5,00,000

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

### NCERT Solutions for Class 8 Maths Exercise 8.2

**Question 1.**

A man got 10% increase in his salary. If his new salary is Rs.1,54,000, find his original salary.

**Solution :**

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 =

∴ New salary is Rs.1,54,000, when original salary = = Rs.1,40,000

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

**Question 2.**

On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday?

**Solution :**

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 = = 80%

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

**Question 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.

**Solution :**

No. of articles = 80

∵Cost Price of articles = Rs. 2,400And 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 =

Cost price of articles is Rs.2400, then selling price = Rs.2784

Hence, Selling Price of 80 articles = Rs.2784

Therefore Selling Price of 1 article

= Rs.34.80

**Question 4.**

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

**Solution :**

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. =

When C.P. is Rs.15950, then S.P.

= 15950 = Rs.18,342.50

**Question 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.

**Solution :**

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. =

∴ When C.P. is Rs.8000, then S.P.

= 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. =

∴When C.P. is Rs.8000, then S.P.

= = 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% =

== 2%

**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 Rs.1450and two shirts marked at Rs.850 each?

**Solution :**

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

= = 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 =

= 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

=

= Rs.145

∴ S.P. of a pair of jeans

= Rs.1450 – Rs.145 = Rs.2,835

**Question 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)

**Solution :**

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. =

∴ When S.P. is Rs.20,000, then C.P.

= = 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. =

∴When S.P. is Rs.20,000, then C.P.

== 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

**Question 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.

**Solution :**

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. =

∴ When C.P. is Rs.13,000, then S.P.

== Rs.14,560

**Question 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.

**Solution :**

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. =

∴ When S.P. is Rs.1600, then M.P.

= = Rs.2,000

**Question 10.**

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

**Solution :**

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 =

∴ When price including VAT is Rs.5400, then original price = = Rs.5000

### NCERT Solutions for Class 8 Maths Exercise 8.3

**Question 1.**

Calculate the amount and compound interest on:

(a) Rs.10,800 for 3 years at per annum compounded annually.

(b) Rs.18,000 foryears at 10% per annum compounded annually.

(c) Rs.62,500 for years at 8% per annum compounded annually.

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

**Solution :**

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

Rate of interest (R) =

= Rs. 15,377.34

Compound Interest (C.I.) = A – P

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

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

= 10% p.a.

Amount (A) =

=

Interest for years on Rs. 21,780 at rate of 10% = = Rs. 1,089

Total amount for years

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

Compound Interest (C.I.) = A – P

= Rs. 22869 – Rs. 18000 = Rs. 4,869

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

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

= Rs. 70,304

Compound Interest (C.I.) = A – P

= 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% = (compounded half yearly)

Amount (A) =

= Rs. 8,736.20

Compound Interest (C.I.) = A – P

= Rs. 8736.20 – Rs. 8000

= Rs. 736.20

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

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

Amount (A) =

= Rs. 10,816

Compound Interest (C.I.) = A – P

= Rs. 10,816 – Rs. 10,000 = Rs. 816

**Question 2.**

Kamala borrowed Rs.26,400 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 **4/12** years).

**Solution :**

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

Amount for 2 years (A) =

Interest for 4 months = years at the rate of 15% =

= Rs. 1745.70

Total amount = Rs. 34,914 + Rs. 1,745.70

= Rs. 36,659.70

**Question 3.**

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

**Solution :**

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

= 12% p.a.

Simple Interest for Fabina =

= Rs. 16,637.50

C.I. for Radha = A – P

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

Here, Fabina pays more interest

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

**Question 4.**

I borrowsRs.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?

**Solution :**

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

Simple Interest =

= 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

**Question 5.**

Vasudevan invested Rs.60,000 at an interest rate of 12% per ann^um compounded half yearly. What amount would he get:

(i) after 6 months?

(ii) after 1 year?

**Solution :**

(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)

= 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)

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

Amount (A) =

= Rs. 67,416

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

**Question 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

**Solution :** years if the interest is:

(i) compounded annually.

(ii) compounded half yearly.

**Solution :**

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

Amount for 1 year (A)

= Rs. 88,000

Interest for

= Rs. 4,400

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

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

Time **(n) ** = year = 3year (compounded half yearly)

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

Amount (A) =

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

**Question 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.

(ii) The interest for the third year.

**Solution :**

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

Amount (A)

= Rs. 8,820

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

Amount (A)

= id="docs-internal-guid-2c688946-7fff-2a9f-114e-3f51e432463a" >

= Rs. 9,261

Interest for 3^{rd} year = A – P

= Rs. 9,261 – Rs. 8,820 = Rs. 441

**Question 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?

**Solution :**

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

Time == 3 years (compounded half yearly)

Amount (A)

= Rs. 11,576.25

Compound Interest (C.I.) = A – P

= 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) ** = years

Amount (A) for 1 year

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.

**Question 9.**

Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months at per annum, interest being compounded half yearly.

**Solution :**

Here, Principal (P) = Rs. 4096,

Rate of Interest (R) =

= (compounded half yearly)

Time** (n) **= 18 months = years = 3 years (compounded half yearly)

Amount (A)

= Rs. 4,913

**Question 10.**

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(i) Find the population in 2001.

(ii) What would be its population in 2005?

**Solution :**

(i) Here, A_{2003} = Rs. 54,000, R = 5%, = 2 years

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

Here population is increasing.

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

= 59,535

Hence population in 2005 would be 59,535.

**Question 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.

**Solution :**

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

After 2 hours, number of bacteria,

= 5,31,616.25

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

**Question 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.

**Solution :**

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

= Rs. 38,640

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

### Notes,worksheet and solved question for Maths class 8

- class 8 maths notes on chapter Liner equation in one variable
- class 8 maths notes on chapter algebric expression
- class 8 maths notes on chapter Mensuration
- class 8 maths notes on chapter Square and square roots
- class 8 maths notes on chapter statistice
- class 8 maths notes on chapter practical Geometry
- class 8 maths notes on chapter commericial maths
- class 8 maths notes on chapter solid shape
- class 8 maths notes on chapter quadrilaterals
- class 8 maths notes on chapter exponents
- class 8 maths notes on chapter factorisation
- class 8 maths notes on chapter inverse proporation
- class 8 maths notes on chapter cube and cube roots

Check your marks in a chapter which you have complited in school from Physics Wallah chapter wise online test just click on the link given below

chapter wise online test for class 8 maths