 Lesson: 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) 5m to 10 km
(c) 50 paise to Rs. 5
(a) 1:2 (b) 1:2000 (c) 1:10
Solution:
(a) Ratio of the speed of cycle to the speed of scooter = 15: 30 = 1:2
(b) Ratio of 5 m to 10 km = Ratio of 5m to 10000m = 5: 10000 = 1:2000
(c) Ratio of 50 paise to Rs. 5 = Ratio of 50 paise to 500 paise = 50: 500 = 1: 10
Question 2
Convert the following ratios to percentages:
(a) 3: 4 (b) 2: 3
(a) 75% (b)
2
66 %
3
Solution:
(a) Ratio = 3: 4
Fraction =
3
4
Percentage =
3 25 75
100
4 25
= 75%
(b) Ratio = 2:3
Fraction =
2
3 Percentage =
2 100 200 1 2
66 %
3 100 3 100 3
Question 3
72% of 25 students are good in mathematics.
How many are not good in mathematics?
7 students
Solution:
Percentage of students not good in Mathematics = (100 72) % = 28%
So, number of students not good in Mathematics = 28% of 25 =
28
25
100
= 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?
25
Solution:
Let the total number of matches played by the football team be x.
Their win percentage was 40%.
So, 40% of x = 10 or,
40
100
× x = 10 or, x =
10 100
40
= 25.
Question 5
If Chameli had Rs. 600 left after spending 75% of her money, how much did she
have in the beginning?
Chameli had Rs 2400 in the beginning Solution:
Let us assume that Chameli had Rs. x in the beginning.
She spent 75% of her money.
So, percentage of money left with her = (100 75) % = 25%.
So, 25% of x = 600 or,
25
100
× x = 600 or, x =
600 100
25
= 2400
Thus, Chameli had Rs 2400 in the beginning.
Question 6
If 60% people in a city like cricket only, 30% like football only and the remaining like
Other games, then what percent of the people like other games?
If the total number of people is 50 lakh, find the exact number who like each type of
game.
30 lakh like cricket, 15 lakh like football, 5 lakh like other games
Solution:
According to the question,
60% people like cricket only, 30% people like football only and remaining like
other games.
Therefore, the percentage of people who like other games
= (100 (60 + 30)) % = 10%
Total no. of people = 50 lakh
No. of people who like cricket only = 60% of 50 lakh =
× 50 lakh = 30 lakh
No. of people like football only = 30% of 50 lakh =
30
100
× 50 lakh =15 lakh
No. of people like who like other games = 10% of 50 lakh =
× 50 lakh = 5 lakh.
Exercise 8.2
Question 1
A man got a 10% increase in his salary.
If his new salary is Rs 1, 54, 000, find his original salary.
The original salary is Rs 1, 40, 000 Solution:
10% increase in salary means:
If previous salary is Rs 100 then increased salary = Rs 110
If new salary is Rs. 110, then original salary = Rs 100
The new salary is 1, 54, 000, so the original salary =
100
× Rs
110
1, 54,000 = Rs 1, 40,000
Thus, the 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?
80%
Solution:
On Sunday, the no. of zoo visitors = 845
On Monday, the no. of zoo visitors = 169
Decrease in the no. of people visiting zoo on Monday = 845 169 = 676
Percentage decrease on Monday
=
Decrease 676
× 100 × 100
VisitoronSund 845
%%
ay
= 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.
Rs 34.80
Solution:
C.P. of all 80 articles = Rs 2400
16% profit means if C.P. is Rs. 100, then S.P. is Rs 116
Therefore, if C.P. is Rs 2400, then S.P. =
116
× Rs
100
2400 = Rs 2784
S.P. of 80 articles = Rs 2784.
S.P. of one article = Rs
2784
80
= Rs 34.80. Question 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.
The required S.P. is Rs 18342.50
Solution:
C.P. of the article = Rs 15,500
Charge on its repairing = Rs 450
New C.P. = Rs (15500 + 450) = Rs 15950
As at profit of 15% it is sold, so S.P. = 115% of C.P.
=
115
100
× Rs 15950 = Rs 18342.50
Thus, the required S.P. is Rs 18342.50.
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.
Gain Percent = 2%
Solution:
C.P. of the TV = Rs 8,000
C.P. of the VCR = Rs 8,000
4% loss on the VCR means,
If CP is Rs 100, then S.P. is Rs 96
If C.P. is Rs 8000, then S.P. will be
× Rs 8000 = Rs 7680.
S.P. of the VCR = Rs 7680
8% profit on T.V. means,
If C.P. is Rs 100, then S.P. is Rs 108.
So, if C.P. is Rs 8000, then S.P. will be
× Rs 8000 = Rs 8640. S.P. of the T.V. = Rs 8640
Overall C.P. of the T.V. and the VCR = Rs (8000 + 8000) = Rs 16,000.
Overall S.P. of the TV and the VCR = Rs (7680 + 8640) = Rs 16,320
From above, Overall S.P. > Overall C.P.
Profit = Rs (16320 16000) = Rs 320
Gain present =
320
100 2%
16000

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 1450 and two
shirts marked at Rs 850 each?
Rs 2835
Solution:
10% discount on the marked price (M.P.)
Article of M.P. Rs. 100 has S.P. = Rs 90
So, the pair of jeans having M.P. as Rs 1450 has a S.P.
= Rs
× 1450 = Rs 1305
M.P. of the two shirts= Rs 850 × 2 = Rs 1700
The shirts having MP as Rs 1700 has a S.P. =
90
100
× 1700 = Rs 1530
So, the customer has to pay for Jeans and two shirts
= Rs (1305 + 1530) = Rs 2835. 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.
The overall gain is = Rs 1270
Solution:
S.P. of each buffalo = Rs 20,000
5% gain means a profit of Rs 5 on the C.P. of Rs 100.
If the S.P. is Rs 105, and then C.P. is Rs 100.
If S.P. is Rs 20,000, then C.P. =
100
105
× Rs 20,000 = Rs 19048
Gain = Rs (20000 19048) = Rs 952
10% loss on the other buffalo means:
If S.P. is 90, then C.P. is Rs 100
So, if S.P. is Rs 20,000, then C.P. =
100
105
× Rs 20,000 = Rs 22222
C.P. of the other buffalo = Rs 22222
Loss = Rs (22222 20000) = Rs 2222
Overall Loss = Rs (2222 952) = Rs 1270
Question 8
The price of a T.V. 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.
Rs. 14560
Solution:
Price of the T.V. = Rs 13,000, Sale tax charge = 12%
Sale tax on the T.V. = 12% of Rs 13,000 = Rs
× 13000 = Rs 1560
Amount paid by Vinod for the T.V.= Rs (13000 + 1560) = Rs 14560. Question 9
Arun bought a pair of skates of a sale where the discount given was 20%.
If the amount he pays is Rs. 1,600 find the marked price.
M.P. of skates is Rs 2000
Solution:
20% discount means,
For M.P. of Rs. 100, the S.P. = Rs (100 20) = Rs 80.
So, if S.P. = Rs 80, then M.P. = Rs100
If S.P. = 1600, then M.P. will be = Rs
100
80
× 1600 = Rs 2000.
Thus, M.P. of the skates is Rs 2000.
Question 10
I purchased a hair-dryer for Rs 5,400 including 8% VAT.
Find the price before VAT was added.
Price before VAT = Rs 5000
Solution:
8% VAT included means; Rs 8 is a added to the original price of Rs 100.
If including VAT, the price in Rs 108, the original price = 100
So, if including VAT, the price is Rs 5400, the original price
= Rs
× 5400 = Rs 5000
Price before VAT = Rs. 5000. Exercise 8.3
Question 1
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 half yearly.
(d) Rs 8,000 for 1 year at 9% per annum compounded half-yearly.
(You could use the year by year calculation using S.I. formula to verify).
(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.
(a) Rs 15377.34, Rs 4577.34
(b) Rs 22869, Rs 4869
(c) Rs 70304, Rs 7804
(d) Rs 8736.20, Rs 736.20
(e) Rs 10816, Rs 816
Solution:
(a) Given, P = Rs 10,800, T = 3 years,
R = 12
1
2
% per annum =
25
2
% per annum
We know that,
A = P
T
R
1+
100



= Rs 10800
3
25 1
1 + ×
2 100




= Rs 10800
3
9
8


 = Rs 10800
999
888
= Rs 15377.34
Now, C.I. = A P = Rs (15377.34 10800) = Rs 4577.34(b) Given, P = Rs
18000, T = 2
1
2
years, R = 10% per annum compounded annually.
We know that, A = P
T
R
1+
100



So, the amount at the end of 2 years is given by
A = Rs 18000
2
10
1+
100



= Rs 18000
2
1
1+
10



=Rs18000 ×
11
10
×
11
10
= Rs 180 × 121 = Rs 21780
Rs 21780 would act as a principal for next 1/2 year.
We find S.I. on Rs 21780 for 1/2 year.
S.I. = Rs
1
21780 × × 10
2
100
= Rs 1089
Interest for two years = Rs (21780 18000) = Rs 3780
Interest for the next 1/2 years = Rs 1089
Total compounded interest = Rs (3780 + 1089) = Rs 4869
Now A = P + C.I = Rs (18000 + 4869) = Rs 22869
(b) Given, P = Rs 18000, T = 2
1
2
years, R
= 10% per annum compounded annually.
We know that, A = P
T
R
1+
100



So, the amount at the end of 2 years is given by
A = Rs 18000
2
10
1+
100



= Rs 18000
2
1
1+
10



= Rs18000 ×
11
10
×
11
10 = Rs 180 × 121 = Rs 21780
Rs 21780 would act as a principal for next 1/2 year.
We find S.I. on Rs 21780 for 1/2 year.
S.I. = Rs
1
21780 × × 10
2
100
= Rs 1089
Interest for two years = Rs (21780 18000) = Rs 3780
Interest for the next 1/2 years = Rs 1089
Total compounded interest = Rs (3780 + 1089) = Rs 4869
Now A = P + C.I = Rs (18000 + 4869) = Rs 22869
(c) Given, P = Rs 62500, T = 1
1
2
years = 3 half years,
R = 8% per annum = 4% half-yearly
We know that, A = P
T
R
1+
100



= Rs 2500
3
4
1+
100



= Rs 62500 ×
26 26 26
25 25 25

= Rs 70304
Now, C.I. = A P = Rs (70304 62500) = Rs 7804.
(d) Given, P = Rs 8000, T = 1 year = 2 half years,
R = 9% per annum =
9
2
% per half year
We know that, A = P
T
R
1+
100



= Rs 8000
2
9
1+
2×100



= 8000
209 209
××
200 200
= Rs 8736.20
Now, C.I. = A P = Rs (8736.20 8000) = Rs 736.20. (e) Given, P = Rs 10000, T = 1 year = 2 half years,
R = 8% per annum = 4% per half year
We know that, A = P
T
R
1+
100



= Rs 10000
2
4
1+
100



= Rs
26 26
10000 × ×
25 25
= Rs 10816
Now, C.I. = A P = Rs (10816 10000) = Rs 816.
Question 2
Kamla 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?
Rs 36659.70
Solution:
Given, P = Rs 26400, R = 15% per annum, T = 2 years and 4 months,
We know that,
Amount at the end of 2 years =
T
R
1+
100



=
22
15 3 23 23
26400 1 + 26400 1 + 26400 × ×
100 20
Rs R
20 20
s Rs

= Rs 66 × 23 × 23 = Rs 34914.
Rs 34914 would act as principal for next 4 months i.e.,
1
3
year
So, S.I. =
P × R × T
100
= Rs 34914 ×
15 1
100 3
= Rs 1745.70
Amount paid by Kamla after 2 years = Rs (34914 + 1745.70) = Rs 36659.70
Question 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?
Fabina paid Rs 362.50 more than Radha Solution:
In case of Fabina:
Here, P = Rs 12500, R = 12% per annum, T = 3 years
We know that, S.I.=
P × R × T
100
= Rs 12500 ×
× 3 = Rs 4500.
Here, P = Rs 12500, R = 10% per annum, T = 3 years,
We know that, A =
T
R
P 1 +
100



= Rs 12500
3
10
1+
100



= Rs 12500
11 11 11
× × ×
10 10 10
= Rs 16637.50
Now,C.I. = A P = Rs (16637.50 12500) = Rs 4137.50
It is obvious that Fabina would pay more interest by
Rs (4500 4137.50) = Rs 362.50
Thus, Fabina paid Rs 362.50 more than Radha.
Question 4
I borrowed Rs. 12000 from Jam shed at 6% per annum simple interest for 2 year.
Had I borrowed this sum at 6% per annum compound interest, what extra amount
would I have to pay?
Rs 43.20
Solution:
Given, P = Rs 12000. R = 6% per annum, Time (T) = 2 years
In case of S.I.
We know that, S.I. =
P × R × T
100
= Rs 12000 ×
× 2 = Rs 1440
In case of compound interest,
A =
T
R
P 1 +
100



= Rs12000
2
6
1+
100



= 12000
106 106
××
100 100
= Rs 13483.20
Now,C.I. = A P = Rs (13483.20 12000) = Rs. 1483.20
It is obvious that C.I. > S.I.,
Thus extra amount paid in case of C.I. = Rs. (1483.20 1440) = Rs 43.20 Question 5
Vasudevan invested Rs. 60000 at an interest rate of 12% per annum compounded
half yearly.
What amount would he get
(i) After 6 moths (ii) After 1 year?
(i) Rs 63600 (ii) Rs 67416
Solution:
Given, P = Rs 60000
(i) R = 12% per annum = 6% half yearly and T = 6 months = 1 half year
Amount after 6 months, A
6
=
T
R
P 1 +
100



= Rs 60000
1
6
1+
100



= Rs 60000 ×
106
100
= Rs 63600
(ii) In the second case, T = 1 year = 2 half years Amount after
1 year, A =
T
R
P 1 +
100



= Rs
2
6
60000 1 +
100



= Rs
106 106
60000 × × = Rs 67416
100 100
Question 6
Arif took a loan of Rs. 80,000 from a bank. If the rate of interest is 10% per annum,
find the amount he would pay after 1
1
2
years if the interest is
(i) Compounded annually (ii) Compounded half yearly.
(i) Rs 92400 (ii) Rs 92610
Solution:
Given, P = Rs 80000, R = 10% per annum
= 5% half yearly, and T = 1
1
2
years = 3 half years
Case I:
When amount calculated is compounded annually. A =
T
R
P 1 +
100



So, amount at the end of 1 year is given as
Rs
1
10 11
80000 1 + 80000 ×
100 1
Rs
0



= Rs. 88000
Rs 88000 would be the principal to calculate S.I. for
1
2
year
S.I. =
P × R × T
100
= Rs 88000 ×
×
1
2
= Rs 4400
C.I. for 1
1
2
years on given principal
= Rs (88000 80000) + Rs 4400 = Rs 12400
A = P + C.I. = Rs (80000 + 12400) = Rs 92400
Case II:
A =
T
R
P 1 +
100



= Rs 80000
3
5
1+
100



= Rs 80000
3
21
20



= Rs 80000
21 21 21
20 20 20
= Rs 92610
Difference in the amount in both the cases = Rs (92610 92400) = Rs 210
Question 7
Marina invested Rs. 8000 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.
(i) Rs. 8820 (ii) Rs. 441
Solution:
(i) Given, P = Rs 80000, R = 5% per annum, T = 2 year
Amount after 2 years,
A
2
=
T
R
P 1 +
100



= Rs. 8000
2
5
1+
100



= Rs 8000 ×
21 21
×
20 20
= Rs 8820
Thus, amount credited after 2 years is Rs 8820. (ii) Amount credited after 3 years,
A
3
= 8000
3
5
1+
100



= Rs 8000 ×
21 21 21
××
20 20 20
= Rs 9261
Interest for the 3rd year = A
3
A
2
= Rs (9261 8820) = Rs. 441
Question 8
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?
Rs 11576.25, Rs 1576.25, Yes by Rs 26.25
Solution:
Given, P = Rs 10000, T = 1
1
2
years = 3 half years,
R = 10% per annum = 5% half yearly
When the interest is calculate half yearly, then amount
A
h
=
T
R
P 1+
100



= Rs 10000
3
5
1+
100



= 10000 ×
21 21 21
××
20 20 20
= Rs 11576.25
Now, C.I. = A P = Rs (11576.25 10000) = Rs. 1576.25
When interest is compounded annually,
The amount at the end of 1 year is given by
= Rs. 10000
1
10
1+
100



= Rs 10000 ×
11
10
= Rs 11000
Now, Rs. 11000 would be the principal to calculate the S.I. for
1
2
year,
S.I. =
P × R × T 1000 × 10 × 1
100
1
s
2
R.
00 ×
= Rs 550
C.I. for 1
1
2
years on the given principal
= Rs. (11000 10000) + Rs. 550 = Rs.1550
It is obvious that C.I.
When compounded half yearly is greater than that when compounded yearly.
Extra amount of interest = Rs (1576.25 1550) = Rs 26.25 Question 9
Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at
12
1
2
% per annum, interest being compounded half yearly.
Rs 4913
Solution:
Given P = Rs 4096, T = 18 months = 3 half yearly,
R = 12
1
2
% per annum =
25
4
per half year.
We know that, A =
T
R
P 1 +
100



= Rs 4096
3
25
1+
4 × 100



= Rs
17 17 17
4096 × × ×
16 16 16
= Rs 4913
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?
(i) 48980 (ii) 59535
Solution:
(i) Given, P =? A = 54000, R = 5% per annum, T = 2 years
T
R
A = P 1 +
100



2
5
54000 = P 1 +
100



21 21
54000 = P × ×
20 20
54000 20 20
P 48980
21 21


(ii) For the population in 2005, i.e., 2 years after 2003
Here, P = 54000, T = 2 years, R = 5% per annum We know that, population in 2005,
A = P
T
R
1+
100



= 54000
2
5
1+
100



=
21 21
54000 × ×
20 20
= 9535
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.
The count of bacterial after 2 hours is 531616.
Solution:
Initial count of bacteria, P = 506000, T = 2 hour,
Rate of increasing bacteria, R = 2.5% per hour or 5/2% per hour
We know that, A = P
T
R
1+
100



= 506000
2
5
1+
2 × 100



=
41 41
506000 × ×
40 40
= 531616
Thus, the count of bacteria after 2 hours is 531616.
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.
Rs 38640
Solution:
Given, P = Rs 42000, R = 8% per annum, T = 1 year
We know that, A =
T
R
P 1 –
100



= Rs 42000
1
8
1–
100



= Rs 42000 ×
23
25
= Rs. 38640