NCERT Solutions Class 7 Mathematics Chapter 8 Comparing Quantities

Exercise 8.1

Q.1) Find the ratio of :
a) Rs 5 to 50 paise b) 15 kg to 210 g
c) 9 m to 27 cm d) 30 days to 36 hour
Sol.1) To find ratios, both quantities should be in same unit.
(a) Rs. 5 to 50 paise
⇒ 5 × 100 𝑝𝑎𝑖𝑠𝑒 𝑡𝑜 50 𝑝𝑎𝑖𝑠𝑒 [ Rs. 1 = 100 paise]
⇒ 500 paise to 50 paise
Thus, the ratio is = 500/50 = 10/1 = 10 ∶ 1

(b) 15 kg to 210 g
⇒ 15 × 1000 𝑔 𝑡𝑜 210 𝑔 [ 1 kg = 1000 g]
⇒ 15000 g to 210 g
Thus, the ratio is = 15000/210 = 500/7 = 500 ∶ 7

(c) 9 m to 27 cm
⇒ 9 × 100 𝑐𝑚 𝑡𝑜 27 𝑐𝑚 [ 1 m = 100 cm]
⇒ 900 cm to 27 cm
Thus, the ratio is
900/27 = 100/3 = 100: 3

(d) 30 days to 36 hours
⇒ 30 × 24 ℎ𝑜𝑢𝑟𝑠 𝑡𝑜 36 ℎ𝑜𝑢𝑟𝑠 [ 1 day = 24 hours]
⇒ 720 hours to 36 hours
Thus, the ratio is 720/36 = 20/1 = 20: 1

Q.2) In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Sol.2) 6 students need = 3 computers
∴ 1 student needs = 3/6 computers
∴ 24 students need = 3/6 x 24 = 12 computers
Thus, 12 computers will be needed for 24 students.

Q.3) Population of Rajasthan = 570 lakhs and population of U.P. = 1660 lakhs. Area of Rajasthan = 3 lakh 𝑘𝑚2 and area of U.P. = 2 lakh 𝑘𝑚2.
(i) How many people are there per 𝑘𝑚2 in both states? (ii) Which state is less populated?
Sol.3) (i) People present per 𝑘𝑚2 = Population Area
In Rajasthan = 570 lakhs/ 3 lakhs per 𝑘𝑚2 = 190 people 𝑘𝑚2
In U.P. = 1660 lakhs / 2 lakh per 𝑘𝑚2 = 830 people per 𝑘𝑚2
(ii) Rajasthan is less populated

Exercise 8.2

Q.1) Convert the given fractional numbers to per cents.
a) 1/8
b) 5/4
c) 3/40
d) 2/7
So.1)

Q.2) Convert the given decimal fractions to per cents:
(a) 0.65 (b) 2.1 (c) 0.02 (d) 12.35
Sol.2) (a) 0.65
65/100 × 100% = 65%
(b) 2.1
2.1/00 × 100% = 210%
(c) 0.02
2/100 × 100% = 210%
(d) 12.35
12.35/100 × 100% = 1235%

Q.3) Estimate what part of the figures is coloured and hence find the per cent which is coloured.

Sol.3) i) Coloured part = 1/4
∴ percent of coloured part = 1/4
× 100% = 25%
ii) Coloured part = 3/5
∴ percent of coloured part = 3/5
× 100% = 60%
iii) Coloured part = 3/8
∴ percent of coloured part = 3/8 × 100% = 3/2 × 25% = 37.5%

Q.4) Find:
(a) 15% of 250 (b) 1% of 1 hour (c) 20% of Rs.2500 (d) 75% of 1 kg
Sol.4) (a) 15% of 250 
15/100 × 250 = 15 × 2.5 = 37.5
(b) 1% of 1 hour
1% of 60 minutes =1% of (60 × 60)𝑠𝑒𝑐𝑜𝑛𝑑𝑠
= 1/100 × 60 × 60 = 6 × 6 = 36 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
(c) 20% of Rs.2500 
20/100 × 2500 = 20 × 25 = 𝑅𝑠. 500
(d) 75% of 1 kg
75% of 1000g = 75/100 × 1000 = 750𝑔 = 0.750 𝑘𝑔

Q.5) Find the whole quantity if:
(a) 5% of it is 600 (b) 12% of it is Rs.1080 (c) 40% of it is 500 km
(d) 70% of it is 14 minutes (e) 8% of it is 40 litres
Sol.5) Let the whole quantity be 𝑥 in given questions:
(a) 5% of 𝑥 = 600
⇒ 5/100 x 𝑥 = 600
⇒ 𝑥 = 600×100/5 = 12,000

(b) 12% of 𝑥 =Rs.1080
⇒ 12/100 × 𝑥 = 1080
⇒ 𝑥 = 1080×100/12 = 𝑅𝑠. 9,000

(c) 40% of 𝑥 = 500 𝑘𝑚
⇒ 40/100 × 𝑥 = 500
⇒ 𝑥 = 500×100 40 = 1,250 𝑘𝑚

(d) 70% of 𝑥 = 14 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
⇒ 70/100 × 𝑥 = 14
⇒ 𝑥 = 14×100/70 = 20 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

(e) 8% of 𝑥 = 40 𝑙𝑖𝑡𝑟𝑒𝑠
⇒ 8/100 × 𝑥 = 40
⇒ 𝑥 = 40×100/8 = 500 𝑙𝑖𝑡𝑟𝑒𝑠

Q.6) Convert given per cents to decimal fractions and also to fractions in simplest forms:
(a) 25% (b) 150% (c) 20% (d) 5%

Q.7) In a city, 30% are females, 40% are males and remaining are children. What percent are children?
Sol.7) Given: Percentage of females = 30%
Percentage of males = 40%
Total percentage of females and males = 30 + 40 = 70%
Percentage of children = Total percentage – Percentage of males and females
= 100% – 70% = 30%
Hence, 30% are children.

Q.8) Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Sol.8) Total voters = 15,000
Percentage of voted candidates = 60%
Percentage of not voted candidates = 100 – 60 = 40%
Actual candidates, who did not vote = 40% of 15000
= 40/100 × 15000 = 6,000
Hence, 6,000 candidates did not vote.

Q.9) Meeta saves Rs.400 from her salary. If this is 10% of her salary. What is her salary?
Sol.9) Let Meera’s salary be 𝑅𝑠. 𝑥.
Now, 10% of salary = 𝑅𝑠. 400
⇒ 10% 𝑜𝑓 𝑥 = 𝑅𝑠. 400
⇒ 10/100 × 𝑥 = 400
⇒ 𝑥 = 400×100/10 = 4,000
⇒ 𝑥 = 4,000
Hence, Meera’s salary is 𝑅𝑠. 4,000

Q.10) A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Sol.10) Number of matches played by cricket team = 20
Percentage of won matches = 25%
Total matches won by them = 25% of 20
= 25/100 × 20 = 5
Hence, they won 5 matches.

Exercise 8.3

Q.1) Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
a) Gardening shears bought for Rs.250 and sold for Rs.325.
b) A refrigerator bought for Rs.12,000 and sold at Rs.13,500.
c) A cupboard bought for Rs.2,500 and sold at 3,000.
d) A skirt bought for Rs.250 and sold at Rs.150.
Sol.1) (a) Cost price of gardening shears = 𝑅𝑠. 250
Selling price of gardening shears = 𝑅𝑠. 325
Since, 𝑆. 𝑃. > 𝐶. 𝑃., therefore here is profit.
∴ Profit = 𝑆. 𝑃. – 𝐶. 𝑃.
= 𝑅𝑠. 325 – 𝑅𝑠. 250 = 𝑅𝑠. 75
Now Profit% = 𝑃𝑟𝑜𝑓𝑖𝑡/𝐶.𝑃. × 100 
= 75/250 × 100 = 30%
Therefore, Profit = 𝑅𝑠. 75 and Profit% = 30%

(b) Cost price of refrigerator = Rs. 12,000
Selling price of refrigerator = Rs.13,500
Since, S.P. > C.P., therefore here is profit.
∴ Profit = S.P. – C.P. = Rs.13500 – Rs.12000 = Rs.1,500
Now Profit% = 𝑃𝑟𝑜𝑓𝑖𝑡/𝐶.𝑃. × 100
= 1500/12000 × 100 = 12.5%
Therefore, Profit = Rs.1,500 and Profit% = 12.5%

(c) Cost price of cupboard = Rs. 2,500
Selling price of cupboard = Rs. 3,000
Since, S.P. > C.P., therefore here is profit.
∴ Profit = S.P. – C.P. = Rs.3,000 – Rs.2,500 = Rs. 500
Now Profit% = 𝑃𝑟𝑜𝑓𝑖𝑡/𝐶.𝑃. × 100
= 500/2500 × 100 = 20%
Therefore, Profit = Rs. 500 and Profit% = 20%

(d) Cost price of skirt = Rs. 250
Selling price of skirt = Rs. 150
Since, C.P. > S.P., therefore here is loss.
∴ Loss = C.P. – S.P. =Rs.250 – Rs.150 = ₹100
Now Loss% = 𝐿𝑜𝑠𝑠/𝐶.𝑃. × 100
= 100/250 × 100 = 40%
Therefore, Profit = Rs. 100 and Profit% = 40%

Q.2) Convert each part of the ratio to percentage:
(a) 3 : 1 (b) 2 : 3 : 5 (c) 1 : 4 (d) 1 : 2 : 5
Sol.2) (a) 3 : 1 Total part = 3 + 1 = 4
Therefore, Fractional part = 3/4 : 1/4
⇒ Percentage of parts = 3/4 × 100: 1/4 × 100
⇒ Percentage of parts = 75% ∶ 25%

(b) 2 : 3 : 5
Total part = 2 + 3 + 5 = 10
Therefore, Fractional part = 2/10 : 3/10 : 5/10
⇒ Percentage of parts = 2/10 × 100: 3/10 × 100: 5/10 × 100
⇒ Percentage of parts = 20% ∶ 30% ∶ 50%

(c) 1 : 4
Total part = 1 + 4 = 5
Therefore, Fractional part 1/5 : 4/5
⇒ Percentage of parts = 1/5 × 100: 4/5 × 100
⇒ Percentage of parts = 20% ∶ 80%

(d) 1 : 2 : 5
Total part = 1 + 2 + 5 = 8
Therefore, Fractional part 1/ 8 : 2/8 : 5/8
⇒ Percentage of parts = 1/8 × 100: 2/8 × 100: 5/8 × 100
⇒ Percentage of parts = 12.5% ∶ 25% ∶ 62.5%

Q.3) The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Sol.3) The decreased population of a city from 25,000 to 24,500.
Population decreased = 25,000 – 24,500 = 500
Decreased Percentage = 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑑 / 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 x 100
= 500/25000 × 100 = 2%
Hence, the percentage decreased is 2%.

Q.4) Arun bought a car for Rs.3,50,000. The next year, the price went up to Rs.3,70,000. What was the percentage of price increase?
Sol.4) Increased in price of a car from Rs.3,50,000 to Rs.3,70,000.
Amount change = 𝑅𝑠. 3,70,000 – 𝑅𝑠. 3,50,000 = 𝑅𝑠. 20,000.
Therefore, Increased percentage = 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 / 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 x 100
= 20000/350000 × 100 = 5(5/7)%
Hence, the percentage of price increased is 5(5/7)%

Q.5) I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?
Sol.5) The cost price of T.V. = 𝑅𝑠. 10,000
Profit percent = 20%
Now, Profit = 𝑃𝑟𝑜𝑓𝑖𝑡% 𝑜𝑓 𝐶. 𝑃. = 20/100 × 100 = 𝑅𝑠. 2,000
Selling price = 𝐶. 𝑃. + 𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑅𝑠. 10,000 + 𝑅𝑠. 2,000 = 𝑅𝑠. 12,000
Hence, he gets 𝑅𝑠. 12000 on selling his T.V.

Q.6) Juhi sells a washing machine for Rs.13,500. She loses 20% in the bargain. What was the price at which she bought it?
Sol.6) Selling price of washing machine = Rs.13,500
Loss percent = 20%
Let the cost price of washing machine be 𝑅𝑠. 𝑥
Since, Loss = Loss% of C.P.
⇒ 𝐿𝑜𝑠𝑠 = 20% 𝑜𝑓 𝑅𝑠. 𝑥 = 20/100 × 𝑥 = 𝑥/5
Therefore, S.P. = C.P. – Loss
⇒ 13500 = 𝑥 − 𝑥/5
⇒ 13500 = 4𝑥/5
⇒ 𝑥 = 13500×5/4 = 𝑅𝑠. 16,875
Hence, the cost price of washing machine is 𝑅𝑠. 16,875.

Q.7) (i) Chalk contains Calcium, Carbon and Oxygen in the ratio 10:3:12. Find the percentage of Carbon in chalk.
(ii) If in a stick of chalk, Carbon is 3 g, what is the weight of the chalk stick?
Sol.7) (i) Given ratio = 10 : 3 : 12
Total part = 10 + 3 + 12 = 25
Part of Carbon = 3/25
Percentage of Carbon part in chalk = 3/25 × 100 = 12%
(ii) Quantity of Carbon in chalk stick = 3 𝑔
Let the weight of chalk be 𝑥 𝑔.
Then, 12% 𝑜𝑓 𝑥 = 3
⇒ 12/100 × 𝑥 = 3
⇒ 𝑥 = 3×100/12 = 25𝑔
Hence, the weight of chalk stick is 25 𝑔.

Q.8) Amina buys a book for Rs.275 and sells it at a loss of 15%. How much does she sell it for?
Sol.8) The cost of a book = Rs. 275
Loss percent = 15%
Loss = Loss% of C.P. = 15% of Rs.275 = 15/100 × 275 = 𝑅𝑠. 41.25
Therefore, 𝑆. 𝑃. = 𝐶. 𝑃. – 𝐿𝑜𝑠𝑠
= 𝑅𝑠. 275 – 𝑅𝑠. 41.25 = 𝑅𝑠. 233.75
Hence, Amina sells a book for 𝑅𝑠. 233

Q.9) Find the amount to be paid at the end of 3 years in each case:
(a) Principal = Rs.1,200 at 12% p.a. (b) Principal = Rs.7,500 at 5% p.a
Sol.9) (a) Here, Principal (P) = ₹1,200, Rate (R) = 12% p.a., Time (T) = 3 years
Simple interest = 𝑃×𝑅×𝑇 / 100 = (1200×12×3) / 100 = 𝑅𝑠. 432
Now, Amount = Principal + Simple Interest
= 𝑅𝑠. 1200 + 𝑅𝑠. 432
= 𝑅𝑠. 1,632
(b) Here, Principal (P) = ₹7,500, Rate (R) = 5% p.a., Time (T) = 3 years
Simple interest = 𝑃×𝑅×𝑇 / 100 = 7500×5×3 / 100 = 𝑅𝑠. 1,125
Now, Amount = Principal + Simple Interest
= 𝑅𝑠. 7,500 + 𝑅𝑠. 1,125
= 𝑅𝑠. 8,625

Q.10) What rate gives Rs. 280 as interest on a sum of Rs.56,000 in 2 years?
Sol.10) Here, Principal (P) = Rs.56,000, Simple Interest (S.I.) = Rs.280, Time (T) = 2 years
Simple interest = 𝑃×𝑅×𝑇 / 100
⇒ 280 = (56000×𝑅×2) / 100
⇒ 𝑅 = 280×100 / 56000×2 ⇒ 𝑅 = 0.25%
Hence, the rate of interest on sum is 0.25%

Q.11) If Meena gives an interest of Rs.45 for one year at 9% rate p.a. What is the sum she has borrowed?
Sol.11) Simple Interest = Rs.45, Rate (R) = 9% p.a., Time (T) = 1 years
Simple interest = 𝑃×𝑅×𝑇 / 100
⇒ 45 = (𝑃×9×1) / 100
⇒ 𝑃 = 45×100/9×1
⇒ 𝑃 = 𝑅𝑠. 500
Hence, she borrowed 𝑅𝑠. 500.