NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

NCERT
SOLUTIONS
CLASS - 8TH

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Book : Mathematics Ncert Solutions | Chapter-8 Maths

Class : 8th
Subject : Maths
Chapter : 8
Chapter Name : Comparing Quantities

Exercise 8.1

Q1 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

Answer. (a) Ratio of the speed of cycle to the speed of scooter = 15

30
= 1 : 2

(b) Since 1 km = 1000 m,
Required ratio = =
10km
5m 5m

10×1000m
= 1 : 2000

(c) Since Re 1 = 100 paise,
Required ratio = =
50 paise 50 paise
= = 1 : 10
Rs5 500 paise

Page : 119 , Block Name : Exercise 8.1

Q2 Convert the following ratios to percentages.
(a) 3 : 4
(b) 2 : 3

3 3 100 3
3 : 4 = = × = × 100% = 75%
4 4 100 4

2 2 100 2 200

Answer. 2 : 3 = 3
=
3
×
100
=
3
× 100% =
3
%

66×3+2 2
= ( ) % = 66 %
3 3

Page : 119 , Block Name : Exercise 8.1

Q3 72% of 25 students are interested in mathematics. How many are not interested in
mathematics?

Answer. It is given that 72% of 25 students are good in mathematics.
Percentage of students who are not good in mathematics = (100 — 72)% = 28%
.•.Number Of students who are not good in mathematics = × 25 = 7
28

100

Thus, 7 students are not good in mathematics.

Page 1 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
Page : 119 , Block Name : Exercise 8.1

Q4 A football team won 10 matches out of the total number of matches they played. If their win
percentage was 40, then how many matches did they play in all?

Answer. Let the total number of matches played by the team be x.
It is given that the team won 10 matches and the winning percentage of the team was 40%.
Therefore,
40
× x = 10
100

100
x = 10 ×
40

x = 25

Thus, the team played 25 matches.

Page : 119 , Block Name : Exercise 8.1

Q5 If Chameli had Rs.600 left after spending 75% of her money, how much did she have in the
beginning?

Answer. Let the amount of money which Chameli had in the beginning be x.
It is given that after spending 75% of Rs x, she was left with Rs 600.
Therefore,
(100 − 75)% of x = Rs600

Or, 25% of x = Rs600

25
× x = Rs 600
100

100
x = Rs(600 × ) = Rs 2400
25

Page : 119 , Block Name : Exercise 8.1

Q6 If 60% people in a 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 is 50 lakh, nd the
exact number who like each type of game.

Answer. Percentage of people who like other games — (100 - 60 - 30 )%
= ( 100 - 90 )% = 10%
Total number of people = 50 lakh
Therefore, number Of people who like cricket = ( lakh = 30 lakh
60
× 50)
100

Number of people who like football = ( 30

100
× 50) lakh = 15 lakh

Number Of people who like other games = ( 10

100
× 50) lakh = 5 lakh

Page : 120 , Block Name : Exercise 8.1

Exercise 8.2
Page 2 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths

Q1 A man got a 10% increase in his salary. If his new salary is Rs.1,54,000, nd his original salary.

Answer. Let the original salary be x. It is given that the new salary is Rs 1,54,000.
Original salary + Increment = New salary
However, it is given that the increment is 10% of the original salary.
Therefore,
10
x + × x = 154000
100

110x
= 154000
100

100
x = (154000 × )
110

x = 140000

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

Page : 125 , Block Name : Exercise 8.2

Q2 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?

Answer. It is given that on Sunday, 845 people went to the zoo and on Monday, 169 people went.
Decrease in the number of people = 845 — 169 = 676
Percentage decrease = (
Decrease in the number of people ×100
)%
Number of people who went to zoo on sunday

676
= ( × 100) %
845

= 80%

Page : 125 , Block Name : Exercise 8.2

Q3 A shopkeeper buys 80 articles for Rs.2,400 and sells them for a pro t of 16%. Find the selling
price of one article.

Answer. It is given that the shopkeeper buys 80 articles for Rs 2,400.
Cost of one article = Rs
2400
= Rs 30
80

Pro t percent = 16
Profit
Profit Percent = × 100
C.P.

Profit
16 = × 100
Rs 30

16×30
Profit = Rs( ) = Rs 4.80
100

Selling price of one article = C.P. + Pro t = Rs (30 + 4.80) = Rs 34.80

Page : 125 , Block Name : Exercise 8.2

Q4 The cost of an article was Rs.15,500. Rs.450 were spent on its repairs. If it is sold for a pro t of
15%, nd the selling price of the article.

Page 3 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
Answer. Total cost of an article = = Cost + Overhead expenses
= Rs 15500 + Rs 450
= Rs 15950
Profit
Profit % = × 100
C.P.

Profit
15 = × 100
Rs15950

15950×15
Profit = Rs( ) = Rs2392.50
100

.•.Selling price of the article = C.P. + Pro t = Rs.(15950 + 2392.50)
= Rs 18342.50

Page : 125 , Block Name : Exercise 8.2

Q5 A VCR and TV were bought for Rs.8,000 each. The shopkeeper made a loss of 4% on the VCR
and a pro t of 8% on the TV. Find the gain or loss percent on the whole transaction.

Answer. C.P. of a VCR = Rs 8000
The shopkeeper made a loss of 4% on VCR.
This means if C.P. is Rs 100, then S.P. is Rs 96.
When C.P. is Rs 8000, S.P. = Rs( 108

100
× 8000) = Rs 8640

C.P. of a TV = Rs 8000
The shopkeeper made a pro t of 8 % on TV.
This means that if C.P. is Rs 100, then S.P. is Rs 108.
When C.P. is Rs 8000, S.P. = Rs( 108

100
× 8000) = Rs 8640

Total S.P. = Rs.7680 + Rs 8640 = Rs 16320
Total C.P. = Rs 8000 + Rs 8000 = Rs 16000
Since total S.P.> total C.P., there a pro t.
Pro t = Rs.16320 - Rs.16000 = Rs 320
Profit
Profit % = × 100
C. P.

320
= × 100 = 2%
16000

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

Page : 125 , Block Name : Exercise 8.2

Q6 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?

Answer. Total marked price = Rs.(1,450 + 2 x 850) = Rs.(1,450 +1,700) = Rs.3,150
Given that, discount % = 10 %
Discount = /(\operatorname{Rs}\left(\frac{10}{100} \times 3150\right)=\operatorname{Rs} 315\)
Also, Discount = Marked price — Sale price
Rs 315 = Rs 3150 - Sale price
Therefore, Sale price = Rs (3150 - 315) = Rs. 2835

Page 4 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
Thus, the customer will have to pay Rs 2,835.

Page : 125 , Block Name : Exercise 8.2

Q7 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)

Answer. S.P. of each buffalo = Rs 20000
The milkman made a gain of 5% while selling one buffalo.
This means if C.P. is Rs 100, then S.p. is Rs 105.
C.P. of one buffalo = Rs(20000 × 100

105
) = Rs19, 047.62

Also, the second buffalo was sold at a loss of 10%.
This means if C.P. is Rs 100, then S.P. is Rs 90.
Therefore, C.P. Of other buffalo = Rs (20000 ×
100
) 22222.22
90
=Rs

Total C.P. = Rs 19047.62 + Rs 22222.22 = Rs 41269.84
Total s.P. = Rs 20000 + Rs 20000 = Rs 40000
Loss = Rs 41269.84 - Rs 40000 = Rs 1269.84
Thus, the overall loss Of milkman was Rs 1,269.84.

Page : 125 , Block Name : Exercise 8.2

Q8 The price of a TV is Rs.13,000. The sales tax charged on it is at the rate of 12%. Find the
amount that Vinod will have to pay if he buys it.

Answer. On Rs 100, the tax to be paid = Rs.12
On Rs 13000, the tax to be paid will be ? = Rs( = Rs 1560
12
× 13000)
100

Required amount = Cost + Sales Tax = Rs 13000 + Rs 1560 = Rs 14560
Thus, Vinod will have to pay Rs 14,560 for the T.V.

Page : 125 , Block Name : Exercise 8.2

Q9 Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays
is Rs.1,600, nd the marked price.

Answer. Let the marked price be x,
Discount
Discount percent = × 100
Marked price

Discount
20 = × 100
x

20 1
Discount = × x = x
100 5

Also,
Discount = Marked Price - Sale Price

Page 5 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
1
x = x − Rs 1600
5

1
x − x = Rs 1600
5

4
x = Rs 1600
5

5
x = Rs(1600 × ) = Rs 2000
4

Thus, the marked price was Rs 2000.

Page : 125 , Block Name : Exercise 8.2

Q10 I purchased a hair-dryer for Rs.5,400 including 8% VAT. Find the price before VAT was added.

Answer. The price includes VAT.
Thus, 8% VAT means that if the price without VAT is Rs 100, then price including VAT will be Rs.
108.
When price including VAT is Rs 108, original price = Rs 100
100
= Rs( × 5400)
When price including VAT is Rs 5400. original price 108

= Rs5000

Thus, the price of the hair-dryer before the addition of VAT was Rs 5,000.

Page : 125 , Block Name : Exercise 8.2

Q11 An article was purchased for Rs.1239 including GST of 18%. Find the price of the article before
GST was added?

Answer. 1016

Page : 125 , Block Name : Exercise 8.2

Exercise 8.3

Q1 Calculate the amount and compound interest on
(a) Rs.10,800 for 3 years at 12 % per annum compounded annually.
1

2

(b) Rs.18,000 for 2 1

2
years at 10% per annum compounded annually.
(c) Rs.62,500 for 1 years at 8% per annum compounded half yearly.
1

2

(d) Rs.8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year
calculation using SI formula to verify).
(e) Rs.10,000 for 1 year at 8% per annum compounded half yearly.

Answer. (a) Principal (P) = Rs 10, 800
1

Rate (R) = 12
2
%
=
25

2
% (annual)
Number of years (n) = 3

Page 6 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
n
R
P(1 + )
100

3
25
= Rs[10800(1 + ) ]
200

3
225
= Rs[10800( ) ]
200

225 225 225
= Rs(10800 × × × )
200 200 200

= Rs. 15377.34375
= Rs. 15377.34 (approximately)
C.I. = A - P = Rs. (15377.34 - 10800) = Rs. 4,577.34

(b) Principal (P ) = Rs18, 000

Rate (R) = 10% annual

1
Number of years (n) = 2 years
2

The amount for 2 years and 6 months can be calculated by rst 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.
2
1 11 11
A = Rs [18000(1 + ) ] = Rs (18000 × × ) = Rs21780
10 10 10

By taking Rs 21780 as principal, the S. l. for the next half year will be calculated.
1
21780× ×10
2
= Rs( ) = Rs 1089
100

Therefore, Interest for the rst 2 years = Rs (21780 - 18000) = Rs 3780
And interest for the next half year = Rs 1089
Therefore, Total C.I. = Rs 3780 + Rs 1089 = Rs4,869
A = P + C.I. = Rs 18000 + Rs 4869 = Rs 22,869

(c) Principal (P) = Rs 62,500
Rate = 8% per annum or 4% per half year
Number of years = 1 1

2

There Will be 3 half years in 1 1

2
years.
n 3
R 4
A = P(1 + ) = Rs [62500(1 + ) ]
100 100

26 26 26
= Rs(62500 × × × )
25 25 25

= Rs70304

C.I. = A - P = Rs 70304 - Rs 62500 = Rs. 7,804

(d) Principal (P) = Rs. 8000
Rate of interest = 9% per annum or % per half year
9

2

Number of years = 1 year
Page 7 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
There will be 2 half years in 1 year.
n
R
A = P(1 + )
100

2
9
= Rs[8000(1 + ) ]
200

2
209
=Rs[8000( ) ] = Rs8, 736.20
200

C.I. = A - P = Rs 8736.20 - Rs. 8000 = Rs.736.20

(e) Principal (P) = Rs 10,000
Rate = 8% per annum or 4% per half year
Number of years = 1 year
There are 2 half years in 1 year.
n
R
A = P(1 + )
100

2 2
4 1
= Rs[10000(1 + ) ] = Rs [10000(1 + ) ]
100 25

26 26
= Rs(10000 × × ) = Rs10, 816
25 25

C.I. = A - P = Rs 10816 - Rs. 10000 = Rs. 816

Page : 133 , Block Name : Exercise 8.3

Q2 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 nd SI on the 2nd year amount for
4

12

years).

Answer. Principal (P) = Rs 26,400
Rate (R) = 15% per annum
Number of years (n) = 2 years 4

12

The amount for 2 years and 4 months can be calculated by rst 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.
Firstly, the amount for 2 years has to be calculated.
2 2
15 3
A = Rs[26400(1 + ) ] = Rs[26400(1 + ) ]
100 20

23 23
= Rs(26400 × × ) = Rs 34, 914
20 20

By taking Rs 34,914 as principal, the S.I. for the next 1

3
years will be calculated.
1
34914× ×15

S.I. = Rs(
3
) = Rs1, 745.70
100

Interest for the rst two years Rs.(34914 — 26400) = Rs 8,514
Page 8 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
And interest for the next 1

3
year = Rs 1,745.70
Total C.I. = RS (8514 + RS 1745.70) = Rs 10,259.70
Amount = P + C.I. = Rs 26400 + Rs 10259.70 = Rs 36,659.70

Page : 133 , Block Name : Exercise 8.3

Q3 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?

Answer. Interest paid by Fabina =
P ×R×T

100

12500×12×3
= Rs( ) = Rs 4, 500
100

n

Amount paid by Radha at the end of 3 years = A = P(1 + R

100
)

3
10
A = Rs[12500(1 + ) ]
100

110 110 110
= Rs(12500 × × × ) = Rs 16, 637.50
100 100 100

C.I. = A - P = Rs 16637.50 - Rs 12500 = Rs 4,137.50
The interest paid by Fabina is Rs 4,500 and by Radha is Rs 4,137.50.
Thus, Fabina pays more interest.
Rs 4500 - Rs 4137.50 = Rs 362.50
Hence, Fabina will have to pay Rs 362.50 more.

Page : 134 , Block Name : Exercise 8.3

Q4 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?

Answer. P = Rs 12000
R = 6% per annum
T = 2 years
P×R×T 12000×6×2
SI. = = Rs ( ) = Rs1, 440
100 100

To nd the compound interest, the amount (A) has to be calculated.
′′ 2
R 6
A = P(1 + ) = Rs [12000(1 + ) ]
100 100

2
3 53 53
= Rs[12000(1 + ) ] = Rs (12000 × × )
50 50 50

= Rs13, 483.20

Therefore, C.I. = A - P = Rs 13483.20 - Rs 12000 = Rs 1,483.20
C.I. - S.l. = Rs 1,483.20 - Rs 1,440 = Rs 43.20
Thus, the extra amount to be paid is Rs. 43.20

Page 9 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
Page : 134 , Block Name : Exercise 8.3

Q5 Vasudevan invested 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?

Answer. (i) P = Rs.60,000
Rate = 12% per annum = 6% per half year
n = 6 months = 1 half year
n
R
A = P(1 + )
100

1
6 106
= Rs[60000(1 + ) ] = Rs(60000 × ) = Rs63, 600
100 100

(ii) There are 2 half years in 1 year.
n=2
2
6 106 106
A = Rs [60000(1 + ) ] = Rs (60000 × × ) = Rs67, 416
100 100 100

Page : 134 , Block Name : Exercise 8.3

Q6 Arif took a loan of Rs.80,000 from a bank. If the rate of interest is 10% per annum, nd the
difference in amounts he would be paying after 1 years if the interest is
1

2

(i) compounded annually.
(ii) compounded half yearly.

Answer. (i) P = Rs 80,000
R = 10% per annum
n = 1 years
1

2

The amount for 1 year and 6 months can be calculated by rst 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 for 1 year has to be calculated.
1
10
A = Rs[80000(1 + ) ]
100

1 11
= Rs[80000 (1 + )] = Rs(80000 × ) = Rs88, 000
10 10

By taking Rs 88,000 as principal, the Sl for the next year will be calculated.
1

2
1
88000×10×
P×R×T 2
S. L. = = Rs ( ) = Rs4, 400
100 100

Interest for the rst year = Rs 88000 — Rs 80000 = Rs 8000
And interest for the next year = Rs 4,400
1

2

Total C.I. = Rs 8000 + Rs 4,400 = Rs 1,2400

Page 10 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
A = P + C.I. = Rs (80000 + 12400) = Rs 92,400

(ii) The interest is compounded half yearly.
Rate = 10% per annum = 5% per half year
There will be three half years in 1 years.
1

2

3 3
5 1
A = Rs [80000(1 + ) ] = Rs [80000(1 + ) ]
100 20

21 21 21
= Rs (80000 × × × ) = Rs92, 610
20 20 20

Difference between the amounts = Rs 92,610 - Rs 92,400 = Rs 210

Page : 134 , Block Name : Exercise 8.3

Q7 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 3rd year.

Answer. (i) P = Rs 8,000
R = 5% per annum
n = 2 years
2 2
5 1
A = Rs [8000(1 + ) ] = Rs (8000(1 + ) )
100 20

21 21
= Rs (8000 × × ) = Rs8, 820
20 20

(ii) The interest for the next one year, i.e. the third year, has to be calculated.
By taking Rs 8,820 as principal, the S.I. for the next year will be calculated.
8820×5×1
= Rs( ) = Rs 441
100

Page : 134 , Block Name : Exercise 8.3

Q8 Find the amount and the compound interest on Rs.10,000 for 1 1

2
years at 10% per annum,
compounded half yearly. Would this interest be more than the interest he would get if it was
compounded annually?

Answer. P = Rs 10,000
Rate = 10% per annum = 5% per half year
n = 1 years
1

2

There will be 3 half years in 1 1

2
years.

Page 11 of 13 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-8 Maths
3 3 3
5 1 1
A = Rs [10000(1 + ) ] = Rs [10000(1 + ) ]] = Rs [10000(1 + ) ]
100 20 20

21 21 21
= Rs (10000 × × × ) = Rs11, 576.25
20 20 20

C.I. = A - P
= Rs 11576.25 - Rs 10000 = Rs 1,576.25
The amount for 1 year and 6 months can be calculated by rst 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 rst year has to be calculated rst.
1
10 1
A = Rs [10000(1 + ) ] = Rs [10000 (1 + )]
100 10

11
= Rs(10000 × ) = Rs11, 000
10

By taking Rs 11,000 as the principal, the S.I. for the next half year will be calculated.
1
11000×10×
2
= Rs( ) = Rs 550
100

Therefore, Interest for the rst year = Rs 11000 — Rs 10000 = Rs 1,000
Therefore, Total compound interest = Rs 1000 + Rs 550 = Rs 1,550
Therefore, the interest would be more when compounded half yearly than the interest when
compounded annually.

Page : 134 , Block Name : Exercise 8.3

Q9 Find the amount which Ram will get on Rs.4096, if he gave it for 18 months at 12 % 2 per
1

2

annum, interest being compounded half yearly.

P = Rs4, 096

1
= 12 %
2
Answer.
25
R = 18 per annum = %
4

n = 18 months

There will be 3 half years in 18 months.
Therefore,
3 3
25 1
A = Rs [4096(1 + ) ] = Rs[4096(1 + ) ]
400 16

17 17 17
= Rs(4096 × × × ) = Rs 4, 913
16 16 16

Thus, the required amount is Rs. 4,913.

Page : 134 , Block Name : Exercise 8.3

Q10 The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) nd the population in 2001.

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Book : Mathematics Ncert Solutions | Chapter-8 Maths
(ii) what would be its population in 2005?

Answer. (i) It is given that, population in the year 2003 = 54,000
Therefore,
2
5
(1+ )
100
54000 = ( Population in 2001)

Population in 2001 = 54000 × 20

21
×
20

21
= 48979.59

Thus, the population in the year 2001 was approximately 48,980.

2

(ii) Population in 2005 = 54000(1 + 5

100
)

2
1 21 21
= 54000(1 + ) = 54000 × × = 59, 535
20 20 20

Thus, the population in the year 2005 would be 59,535.

Page : 134 , Block Name : Exercise 8.3

Q11 In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of
2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Answer. The initial count of bacteria is given as 5,06,000.
2

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

2
1 41 41
= 506000(1 + ) = 506000 × ×
40 40 40

= 531616.25 = 5, 31, 616( approx. )

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

Page : 134 , Block Name : Exercise 8.3

Q12 A scooter was bought at Rs.42,000. Its value depreciated at the rate of 8% per annum. Find its
value after one year.

Answer. Principal = Cost price of the scooter = Rs 42,000
Depreciation = 8% Of Rs 42,000 per year
42000×8×1
= Rs( )
100

= Rs 3, 360

Value after 1 year = Rs 42000 - Rs 3360 = Rs 38,640

Page : 134 , Block Name : Exercise 8.3

Page 13 of 13 Aglasem Schools