Comparing Quantities Class 7 Extra Questions Maths Chapter 8
Extra Questions for Class 7 Maths Chapter 8 Comparing Quantities
Comparing Quantities Class 7 Extra Questions Very Short Answer Type
Question 1.
Find the ratio of:
(a) 5 km to 400 m
(b) 2 hours to 160 minutes
Solution:
(a) 5 km = 5 × 1000 = 5000 m
Ratio of 5 km to 400 m
= 5000 m : 400 m
= 25 : 2
Required ratio = 25 : 2
(b) 2 hours = 2 × 60 = 120 minutes
Ratio of 2 hours to 160 minutes
= 120 : 160
= 3 : 4
Required ratio = 3 : 4
Question 2.
State whether the following ratios are equivalent or not?
(a) 2 : 3 and 4 : 5
(b) 1 : 3 and 2 : 6
Solution:
(a) Given ratios = 2 : 3 and 4 : 5
Hence 2 : 3 and 4 : 5 are not equivalent ratios.
(b) Given ratios = 1 : 3 and 2 : 6
LCM of 3 and 6 = 6
Hence, 1 : 3 and 2 : 6 are equivalent ratios.
Question 3.
Express the following ratios in simplest form:
(a) 6\(\frac { 1 }{ 5 }\) : 2\(\frac { 1 }{ 3 }\)
(b) 42 : 56
Solution:
Question 4.
Compare the following ratios:
3 : 4, 5 : 6 and 3 : 8
Solution:
Given: 3 : 4, 5 : 6 and 3 : 8
or \(\frac { 3 }{ 4 }\) , \(\frac { 5 }{ 6 }\) and \(\frac { 3 }{ 8 }\)
LCM of 4, 6 and 8 = 24
Hence, 3 : 8 < 3 : 4 < 5 : 6
Question 5.
State whether the following ratios are proportional or not:
(i) 20 : 45 and 4 : 9
(ii) 9 : 27 and 33 : 11
Solution:
(i) 20 : 45 and 4 : 9
Product of extremes = 20 × 9 = 180
Product of means = 45 × 4 = 180
Here, the product of extremes = Product of means
Hence, the given ratios are in proportion.
(ii) 9 : 27 and 33 : 11
Product of extremes = 9 × 11 = 99
Product of means = 27 × 33 = 891
Here, the product of extremes ≠ Product of means
Hence, the given ratios are not in proportion.
Question 6.
24, 36, x are in continued proportion, find the value of x.
Solution:
Since, 24, 36, x are in continued proportion.
24 : 36 :: 36 : x
⇒ 24 × x = 36 × 36
⇒ x = 54
Hence, the value of x = 54.
Question 7.
Find the mean proportional between 9 and 16.
Solution:
Let x be the mean proportional between 9 and 16.
9 : x :: x : 16
⇒ x × x = 9 × 16
⇒ x
2
= 144
⇒ x = √144 = 12
Hence, the required mean proportional = 12.
Question 8.
Find:
(i) 36% of 400
(ii) 16\(\frac { 2 }{ 3 }\)% of 32
Solution:
Question 9.
Find a number whose 6\(\frac { 1 }{ 4 }\)% is 12.
Solution:
Let the required number be x.
Hence, the required number = 192.
Question 10.
What per cent of 40 kg is 440 g?
Solution:
Let x% of 40 kg = 440 g
Hence, the required Percentage = 1.1%
Comparing Quantities Class 7 Extra Questions Short Answer Type
Question 11.
Convert each of the following into the decimal form:
(а) 25.2%
(b) 0.15%
(c) 25%
Solution:
Question 12.
What per cent of
(a) 64 is 148.48?
(b) 75 is 1225?
Solution:
Question 13.
A machine costs ₹ 7500. Its value decreases by 5% every year due to usage. What will be its price after one year?
Solution:
The cost price of the machine = ₹ 7500
Decrease in price = 5%
Decreased price after one year
= 75 × 95
= ₹ 7125
Hence, the required price = ₹ 7125.
Question 14.
What sum of money lent out at 12 per cent p.a. simple interest would produce ₹ 9000 as interest in 2 years?
Solution:
Here, Interest = ₹ 9000
Rate = 12% p.a.
Time = 2 years
Principal = ?
Hence, the required principal amount = ₹ 37500.
Question 15.
Rashmi obtains 480 marks out of 600. Rajan obtains 560 marks out of 700. Whose performance is better?
Solution:
Rashmi obtains 480 marks out of 600
Marks Percentage = \(\frac { 480 }{ 600 }\) × 100 = 80%
Rajan obtains 560 marks out of 700
Marks Percentage = \(\frac { 560 }{ 700 }\) × 100 = 80%
Since, both of them obtained the same per cent of marks i.e. 80%.
So, their performance cannot be compared.
Question 16.
₹ 9000 becomes ₹ 18000 at simple interest in 8 years. Find the rate per cent per annum.
Solution:
Here, Principal = ₹ 9000
Amount = ₹ 18000
Interest = Amount – Principal = ₹ 18000 – ₹ 9000 = ₹ 9000
Hence, the required rate of interest = 12\(\frac { 1 }{ 2 }\)%.
Question 17.
The cost of an object is increased by 12%. If the current cost is ₹ 896, what was its original cost?
Solution:
Here, rate of increase in cost = 12%
Increased Cost = ₹ 896
Original Cost = ?
Let the Original Cost be ₹ x
Hence, the required cost = ₹ 800.
Comparing Quantities Class 7 Extra Questions Long Answer Type
Question 18.
Radhika borrowed ₹ 12000 from her friends. Out of which ₹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years? (NCERT Exemplar)
Solution:
Total amount borrowed by Radhika = ₹ 12,000
The amount borrowed by her at 18% p.a. = ₹ 4000
Total interest = ₹ 2160 + ₹ 3600 = ₹ 5760
Hence, the total interest = ₹ 5760.
Question 19.
Bhavya earns ₹ 50,000 per month and spends 80% of it. Due to pay revision, her monthly income increases by 20% but due to price rise, she has to spend 20% more. Find her new savings. (NCERT Exemplar)
Solution:
Monthly income of Bhavya = ₹ 50,000
Money spent by her = 80% of ₹ 50,000
= \(\frac { 80 }{ 100 }\) × 50,000 = ₹ 40,000
Due to pay revision, income is increased by 20%
So, the new savings = ₹ 60,000 – ₹ 48,000 = ₹ 12,000
Question 20.
The simple interest on a certain sum at 5% per annum for 3 years and 4 years differ by ₹ 82. Find the sum.
Solution:
Let the required sum be ₹ P.
Simple interest for 3 years
Alternate Method
Simple Interest gained from 3rd to 4th year = ₹ 82
Time (4th year – 3rd year) = 1 year
Required sum = ₹ 1640
Comparing Quantities Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type
Question 21.
Rajan’s monthly income is 20% more than the monthly income of Sarita. What per cent of Sarita’s income is less than Rajan’s monthly income?
Solution:
Let the monthly income of Sarita be ₹ 100.
Rajan’s monthly income
Now, Sarita’s monthly income is less than the monthly income of Raj an by = ₹ 120 – ₹ 100 = ₹ 20
Per cent of less in Rajan’s monthly income
= \(\frac { 20\times 100 }{ 120 }\) = \(\frac { 50 }{ 3 }\)% = 16\(\frac { 2 }{ 3 }\)%
Hence, the required per cent = 16\(\frac { 2 }{ 3 }\)%
Question 22.
If 10 apples are bought for ₹ 11 and sold at the rate of 11 apples for ₹ 10. Find the overall gain or loss per cent in these transactions.
Solution:
CP of 10 apples = ₹ 11
CP of 1 apple = ₹ \(\frac { 11 }{ 10 }\)
SP of 11 apples = ₹ 10
SP of 1 apple = ₹ \(\frac { 10 }{ 11 }\)
Question 23.
If 25 men can do a work in 36 hours, find the number of men required to do the same work in 108 hours.
Solution:
Let the number of men required to be x.
Men : Hours :: Men : Hours
25 : 36 :: x : 108
Product of extremes = 25 × 108
Product of means = 36 × x
Product of means = Product of extremes
36 × x = 25 × 108
⇒ x = 25 × 3 = 75
Hence, the required number of men = 75.
Question 24.
A machine is sold by A to B at a profit of 10% and then B sold it to C at a profit of 20%. If C paid ₹ 1200 for the machine, what amount was paid by A to purchase the machine?
Solution:
Cost price of machine for C = Selling price of the machine for B = ₹ 1200
Hence, the required cost price = ₹ 909\(\frac { 10 }{ 11 }\) or ₹ 909.09 (approx)
