ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 7 Percentage and Its applications Check Your Progress
Question 1.
Convert the following percentages into fractions in the simplest form:
(i) 12\(\frac { 1 }{ 2 }\) %
(ii) 66\(\frac { 2 }{ 3 }\) %
(iii) 8\(\frac { 1 }{ 3 }\) %
Solution:
Question 2.
Express each of the following fractions as a percentage:
(i) \(\frac { 5 }{ 8 }\)
(ii) \(\frac { 13 }{ 40 }\)
(iii) \(\frac { 7 }{ 6 }\)
Solution:
Question 3.
Express each of the following percentages as a decimal:
(i) 122%
(ii) 2.2%
(iii) 3\(\frac { 1 }{ 8 }\) %
Solution:
Question 4.
Express 0.0345 as a percentage.
Solution:
0.0345 as percentage = \(\frac { 0.0345 }{ 100 }\) = 3.45 %
Question 5.
Convert each part of the ratio 5 : 6 : 9 to a percentage.
Solution:
5 : 6 : 9
Sum of ratio = 5 + 6 + 9 = 20
Question 6.
(i) What percent of a day is half an hour?
(ii) What percent is \(\frac { 3 }{ 4 }\) metres of 4\(\frac { 1 }{ 2 }\) metres?
Solution:
(i) Half an hour = \(\frac { 1 }{ 2 }\) hour
Hours in a day = 24 hours
Question 7.
The population of a town decreased from 25000 to 24500. Find the percentage decrease.
Solution:
Decrease in population of a town = 25000 – 24500 = 500
Decrease % = \(\frac { 500\times 100 }{ 25000 }\) = 2%
Question 8.
Arun bought a car for ₹ 350000. The next year, the price went upto ₹ 370000. What was the precentage increase in the price?
Solution
C.P. of a car = ₹ 350000
Next year, price of car was = ₹ 320000
Increase = ₹ 370000 – ₹ 350000 = ₹ 20000
Increase % = \(\frac { 20600\times 100 }{ 350000 }\)
= \(\frac { 40 }{ 7 }\) % = 5\(\frac { 5 }{ 7 }\) %
Question 9.
The population of a village has decreased by 6%. If the original population was 3650, find the population after decrease.
Solution:
Decrease in population of a village = 6%
Original population = 3650
Question 10.
43% of the students in a school are girls. If the number of boys is 1482, find:
(i) the total strength of the school
(ii) number of girls in the school.
Solution:
In a school, percentage of girls = 43%
Percentage of boys = 100 – 43 = 57%
Number of boys = 1482
(i) 52% of total students = 1482
Total students = \(\frac { 1482\times 100 }{ 57 }\) = 2600 students
(ii) Number of girls = 2600 – 1482 = 1118 girls
Question 11.
Rohan bought a calculator for ₹ 760 and sold it for ₹ 874. Find his profit and profit percentage.
Solution:
C.P. of a calculator = ₹ 760 and S.P. = ₹ 874
Profit = S.P. – C.P. = ₹ 874 – ₹ 760 = ₹ 114
Question 12.
On selling an article for ₹ 1027, Meena suffered a loss of ₹ 273. Find her loss percentage.
Solution:
S.P. of an article = ₹ 1027
Loss = ₹ 273
C.P. = S.P. + Loss = ₹ 1027 + ₹ 273 = ₹ 1300
Question 13.
By selling a fan for ₹ 710, a trader suffers a loss of ₹ 40. Find the cost price of the fan. At what price this fan should be sold in order to gain 10%?
Solution:
S.P. of a fan = ₹ 710
Loss = ₹ 40
C.P. = S.P. + Loss = ₹ 710 + ₹ 40 = ₹ 750
If Gain = 10%
Question 14.
A shopkeeper sells an article at ₹ 300, thus earning a profit of 20%. Find the cost price of the article.
Solution:
S.P. of an article = ₹ 300
Profit = 20%
Question 15.
A shopkeeper sells an article at ₹ 320, thus suffering a loss of 20%. Find the cost price of the article.
Solution:
S.P. of an article = ₹ 320
Loss = 20%
Question 16.
If ₹ 6000 is borrowed at 6.5% per annum simple interest, find the interest and the amount to be paid at the end of 3 years.
Solution:
Principal (P) = ₹ 1600
Rate (R) = 6.5% p.a.
Time (T) = 3 years
and amount = P + S.I. = ₹ 6000 + ₹ 1170 = ₹ 7170
Question 17.
How long will it take for ₹ 1860 invested at the rate of 9.5% per annum simple interest to amount to ₹ 2449?
Solution:
Principal (P) = ₹ 1860
Amount (A) = ₹ 12449
S.I. = A – P = ₹ 2445 – ₹ 1860 = ₹ 589
Question 18.
At what rate will ₹ 7200 fetch a simple interest of ₹ 3024 in 4 years?
Solution:
Principal (P) = ₹ 7200
S.I. = ₹ 3024
Time = 4 years
Question 19.
What sum of money will yield a simple interest of ₹ 1155in 3 years 6 months at 11% p.a.?
Solution:
S.I. = ₹ 1155
Rate (R) = 11% p.a.
Time (T) = 3 years 6 months
