These Class 7 Maths Extra Questions Chapter 7 Comparing Quantities will help students prepare well for the exams.
Class 7 Maths Chapter 7 Extra Questions Comparing Quantities
Class 7 Maths Comparing Quantities Extra Questions – 2 Marks
Question 1.
a) 70%
Solution:
70% = \(\frac { 70 }{ 100 }\) = \(\frac { 7 }{ 10 }\)
b) 36%
Solution:
36% = \(\frac { 36 }{ 100 }\) = \(\frac { 18 }{ 50 }\) = \(\frac { 9 }{ 25 }\)
Question 2.
a) 17%
Solution:
17% = \(\frac { 17 }{ 100 }\)
b) 98%
Solution:
98% = \(\frac { 98 }{ 100 }\) = \(\frac { 49 }{ 50 }\)
Question 3.
a) 75%
Solution:
75% = \(\frac { 75 }{ 100 }\) = \(\frac{3 \times 25}{4 \times 25}\) = \(\frac{3}{4}\)
b) 90%
Solution:
90% = \(\frac { 90 }{ 100 }\) = \(\frac { 9 }{ 10 }\)
Question 4.
Divide ₹ 1200 between Jaya and Atual in the ratio 2 : 3
Solution:
Given ratio = 2 : 3
Let the first part = 2x
second part = 3x
Given 2x + 3x = 1200
5x = 1200
x = \(\frac { 1200 }{ 5 }\) ⇒ x = 240
Share of Jaya = 2 × 240 = ₹480
Share of Atual = 3 × 240 = ₹720
Question 5.
(3x + 1 ) : (x + 4) = 7 : 5, then find x.
Solution:
3x + 1 : x + 4 = 7 : 5
\(\frac{3 x+1}{x+4}\) = \(\frac{7}{5}\)
5(3x + 1) = 7(x + 4)
15x + 5 = 7x + 28
15x – 7x = 28 – 5
8x = 23
x = \(\frac{23}{8}\)
Question 6.
Express each of the following into per cent.
a) 0.735
Solution:
0.735 = \(\frac{735}{1000}\) × 100 = \(\frac{735}{10}\) = 73.5%
b) 3.44
Solution:
3.44 = \(\frac{344}{100}\) × 100 = 344%
Question 7.
Express the following in simplest form
a) 0.5%
Solution:
0.5% = \(\frac{0.5}{100}\) = \(\frac{5}{1000}\) = \(\frac{1}{200}\) = 1 : 200
b) 51%
Solution:
51% = \(\frac{51}{100}\) = 51 : 100
Question 8.
What per cent of 20 kg is 3 kg ?
Solution:
Required per cent = \(\frac{3}{20}\) × 100
= 3 × 5 = 15%
∴ 3kg is 15% of 20 kg
Question 9.
If 6% is 60% of number. What is that number?
Solution:
6 = 60% of x ⇒ 6 = \(\frac{60}{100}\) × x
6 = \(\frac{6}{10}\) × x ⇒ 1 = \(\frac{1}{10}\) × x ⇒ x = 10
∴ The number = 10
Question 10.
Find 20% less than 145.
Solution:
20% of 145 = \(\frac{20}{100}\) × 145 = \(\frac{1}{5}\) × 145 = 29
20% less than 145 = 145 – 29 = 116
Question 11.
If \(\frac{4}{7}\) of 56% of K is 21 , find K.
Solution:
\(\frac{4}{7}\) of 56% of K = 21
\(\frac{4}{7}\) × \(\frac{56}{100}\) × K = 21
\(\frac{8}{25}\) × K = 21
K = 21 × \(\frac{25}{8}\) ⇒ K = \(\frac{525}{8}\)
Question 12.
If x% of 2050 is 650 , then find x.
Solution:
x% of 2050 is 650
\(\frac{x}{100}\) × 2050 = 650
x × 20.5 = 650
x = \(\frac{650}{20.5}\) = \(\frac{6500}{205}\) = 31.7
Question 13.
Find
i) 20% of 400
Solution:
20% of 400 = \(\frac{20}{100}\) × 400 = 20 × 4 = 80%
ii) 16\(\frac{2}{3}\) % of 32.
Solution:
16\(\frac{2}{3}\) % of 32
= \(\frac{50}{100 \times 3}\) × 32 = \(\frac{32}{2 \times 3}\) = \(\frac{8}{2}\) = 4
Question 14.
Find the simple interest on ₹8000 at 3% p.a. in 2 years.
Solution:
P = ₹ 8000; R = 3% ; T = 2years
1 = \(\frac{PTR}{100}\) = \(\frac{8000 \times 2 \times 3}{100}\)
= 80 × 2 × 3 = 80 × 6 = 480
∴ Simple interest = ₹480
Question 15.
Kajol has ₹ 2000 with her, and she spent 75% from the amount she had. Calculate the money left with her.
Solution:
Total amount = ₹ 2000
75% of 2000 = \(\frac{75}{100}\) × 2000 = \(\frac{3}{4}\) × 2000
= 3 × 500 = ₹ 1500
Amount left with Kajol
= ₹ 2000 – ₹1500 = ₹ 500
Comparing Quantities Extra Questions Class 7 – 3 Marks
Question 16.
Find a number whose 6\(\frac{1}{4}\)% is 20 .
Solution:
Let the number be x.
6\(\frac{1}{4}\)% of x = 20
\(\frac{25}{4 \times 100}\) × x = 20
x = \(\frac{20 \times 4 \times 100}{25} \)
x = 20 × 4 × 4 ⇒ x = 320
∴ Required number = 320
Question 17.
40% of a number is 100 find the number.
Solution:
Let the number = x
40% of x = 100
\(\frac{40}{100}\) × x = 100
x = \(\frac{100 \times 100}{40}\)
x = \(\frac{100 \times 10}{4}\) ⇒ x = 25 × 10 ⇒ x = 250
∴ The required number = 250
Question 18.
Mandana scored 480 marks out of 600 marks and Rashmi scored 890 marks out of 1000 marks whose performance is better? Why?
Solution:
Mandana scored 480 out 600% of marks = \(\frac{40}{100}\) × 100 = \(\frac{480}{6}\) = 80%
Rashmi score 890 out of 1000% of marks = \(\frac{890}{1000}\) × 100 = \(\frac{890}{10}\) = 89%
∴ Rashmi score better than Mandana.
Question 19.
In a school there are 456 girls. Calculate the total number of students of 24% of the total students are boys.
Solution:
Total number of students = 456
% of boys = 24%
% of girls = 100% – 24% = 76%
∴ 76% – 456; 100% —?
\(\frac{100 \times 456}{76}\) = 100 × 6 = 600
∴ Total number of students = 600
Question 20.
Express 18 hours as per cent of 3 days.
Solution:
One day = 24 hours
3 days = 3 × 24 = 72 hours
x% of 72 = 18
\(\frac{x}{100}\) × 72 = 18
x = \(\frac{18 \times 100}{72}\) ⇒ x = \(\frac{100}{4}\) = 25
25% of 3 days is equal to 18 hours.
Question 21.
A box contains 60 tomatoes, out of which 15\(\frac{1}{2}\)% are rotten ones. How many are rotten ? (to the nearest value).
Solution:
Total number of tomatoes = 60
15\(\frac{1}{2}\)% of 60 = \(\frac{31}{2 \times 100}\) × 60
= \(\frac{31 \times 30}{100}\) = \(\frac{31 \times 3}{10}\) = \(\frac{93}{10}\) = 9.3
∴ There are 9 rotten tomatoes.
Question 22.
Mrs Navya saves 28% of her income. If she saves ₹840 per month. Find her monthly income.
Solution:
% of savings = 28%
Amount saved per month = ₹ 840
28% ___ 840; 100% ___?
? = \(\frac{100 \times 840}{28}\)
? = 100 × 30 = ₹ 3000
∴ Monthly income of mrs Navya = ₹3000
Question 23.
The selling price of 10 articles is the same as the cost price of 11 articles. Find gain percent.
Solution:
Let the CP = ₹x
SP of 10 articles = CP of 11 articles = ₹ 11x
CP of 10 articles = 10 x
Gain = ₹ 11x – ₹ 10x = ₹ x
gain% = \(\frac{g}{C P}\) × 100 = \(\frac{x}{10 x} \times 100\) = 10%
Question 24.
Gaurav sells his machine for ₹18,000 making a profit of 20%. How much did the machine cost him?
Solution:
Let CP of machine =₹ 100
Profit = 20%
SP = ₹ 100 + ₹ 20 = ₹ 120
If SP is ₹ 120, then CP = ₹ 100
If SP is ₹ 18,000, then
CP = \(\frac{100}{120}\) × 18000 = ₹ 15000
∴ Cost of machine = ₹ 15000
Question 25.
Find the interest on ₹ 1200 at 6% p.a. for 146 days.
Solution:
P = ₹ 1200 ; R = 6%
T = 146 days = \(\frac{146}{365}\) year = \(\frac{2}{5}\) year
I = \(\frac{\text { PTR }}{100}\) = \(\frac{1200 \times 2 \times 6}{5 \times 100}\)
= \(\frac{12 \times 12}{5}\) = ₹\(\frac{144}{5}\) = ₹28.80
Extra Questions of Comparing Quantities Class 7 – 5 Marks
Question 26.
Find the S.I. on ₹ 1000 for 6 months at the rate of 5 paise per rupee per month.
Solution:
P = ₹ 1000; T = 6 months
Interest on ₹ 1 for 1 month = 5 paise
Interest on ₹ 1 for 6 months
= (5 × 6) paise = 30 paise = ₹\(\frac{30}{100}\)
Interest on ₹ 1000 for 6 months
= ₹\(\frac{300}{100} \times 1000\) = ₹ 300
∴ Simple interest = ₹ 300
Question 27.
In how many years will ₹ 750 amount to ₹ 900 at 4% per annum.
Solution:
P = ₹ 750 ; A = ₹ 900 ; R = 4% p.a.
I = A – P = ₹ 900 – ₹ 750 = ₹ 150
I = \(\frac{P T R}{100}\)
150 = \(\frac{750 \times 4 \times T}{100}\)
T = \(\frac{150 \times 100}{750 \times 4}\) = \(\frac{100}{5 \times 4}\) = \(\frac { 20 }{ 4 }\)
T = 5 Years
∴ In 5 years will ₹ 750 amount to ₹ 900 at 4% p.a.
Question 28.
On selling an article for ₹ 329, a person lost 6%. Find the CP of the article.
Solution:
Question 29.
₹ 9000 becomes ₹ 18000 at S.I. in 8 years. Find the rate of interest.
Solution:
P = ₹ 9000; A = ₹ 18,000
I = A – P = ₹ 18,000 – ₹ 9,000 = ₹ 9,000
T = 8 years
I = \(\frac{\mathrm{PTR}}{100}\)
9000 = \(\frac{9000 \times 8 \times \mathrm{R}}{100}\)
I = \(\frac{\mathrm{R} \times 8}{100}\)
R = \(\frac{100}{8}\) = \(\frac{50}{4}\) = \(\frac{25}{4}\) = 12\(\frac{1}{2}\)%
Question 30.
What sum of money lent out at 12% p.a. simple interest would give ₹9000 as interest in 2 years.
Solution:
Let P be the principle
R = 12%; I = ₹ 9000 ; T = 2 years
I = \(\frac{P T R}{100}\)
9000 = \(\frac{P \times 2 \times 12}{100}\)
P = \(\frac{9000 \times 100}{2 \times 12}\) = \(\frac{4500 \times 50}{6}\)
P = \(\frac{1500 \times 50}{2}\) =1500 × 25
P = ₹ 37,500
∴ Sum of money lent out = ₹ 37,500