NCERT solutions for class 7 Maths Chapter 8 Comparing Quantities

Solution for 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 hours


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Answer :

(a) Rs. 5 to 50 paise
Converting the given quantities into same units, we have
Rs 5 = 5 × 100 = 500 paise
\(\therefore \) 500 paise : 50 paise
= 10 : 1
(b) 15 kg to 210 g
Converting the given quantities into same units, we have 15 kg = 15 × 1000 = 15000 g
\(\therefore \)15 kg : 210 g = 15000 g : 210 g
= 1500 : 21 = 500 : 7
(c) 9 m to 27 cm Converting the given quantities into same units, we have
9 m = 9 × 100 = 900 cm
\(\therefore \)9m: 27 cm = 900 cm : 27 cm
= 100 : 3 .
(d) 30 days to 36 hours
Converting the given quantities into same units, we have
30 days = 30 × 24 hours
\(\therefore \)30 days : 36 hours
= 720 hours : 36 hours = 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?


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Answer :

\(\because \)6 students require 3 computers
\(\therefore \)1 student will require = \(\frac{3}{6} \)computers
\(\therefore \)24 students will require = \(\frac{3}{6} \)x 24 computers
= 3 × 4 computers = 12 computers
Hence the number of computers required = 12.

Q.3 Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs.
Area of Rajasthan = 3 lakh \( km^2 \) and area of UP = 2 lakh \( km^2 \).
(i) How many people are there per \( km^2 \) in both these States?
(ii) Which State is less populated?


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Answer :

Population of Rajasthan = 570 lakhs
Population of UP = 1660 lakhs
Area of Rajasthan = 3 lakh\( km^2 \)
Area of UP = 2 lakh \( km^2 \)
(i) Number of people per \( km^2 \)of Rajasthan = \(\frac{570 lakhs}{3 lakh km^2} \)= 190 per \(km^2 \)
Number of people in UP = 1660 lakhs
Area of UP = 2 lakh\( km^2 \)
Number of people per\( km^2 \)of UP
= \(\frac{1660 lakhs}{2 lakh km^2} =830 per km^2 \)
Since 190 per \(km^2\)< 830 per \(km^2 \)
(ii) Rajasthan is less populated state.

Solution for Exercise 8.2

Q.1 Convert the given fractional numbers to percent.
(a) \(\frac{1}{8} \)
(b) \(\frac{5}{4} \)
(c) \(\frac{3}{40} \)
(d) \(\frac{2}{7} \)


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Answer :

(a) \(\frac{1}{8} \)= \(\frac{1×100}{8×100} \)=\(\frac{100}{8} \)%=\(12\frac{1}{2} \)%
(b) \(\frac{5}{4} \)= \(\frac{5×100}{4×100} \)=\(\frac{5×100}{4} \)%=125%
(c) \(\frac{3}{40} \)= \(\frac{3×100}{40×100} \)=\(\frac{3×100}{40} \)%=\(\frac{15}{2}% \)=\(7\frac{1}{2} \) %
(d) \(\frac{2}{7} \)= \(\frac{2×100}{7×100} \)=\(\frac{200}{7} \)%=\(28\frac{4}{7} \)%

Q.2 Convert the given decimal fractions to per cents:
(?) 0.65
(b) 2.1
(c) 0.02
(d) 12.35


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Answer :

(a) 0.65 =\(\frac{0.65×100}{100} \) = 0.65 × 100% = 65% (b) 2.1 = \(\frac{2.1×100}{100} \)= 2.1 ×100% = 210% (c) 0.02 =\(frac{0.02×100}{100} \)= 0.02 × 100% = 2% (d) 12.35 = \(\frac{12.35×100}{100} \) = 12.35 × 100% = 1235%

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


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Answer :

(i) Fraction of coloured part = \(\frac{1}{4} \)
\(\therefore \) Percentage of coloured parts
= \(\frac{1×100}{4×100} \) = \(\frac{100}{4} \)%= 25%
(ii) Fraction of coloured part = \(\frac{3}{5} \)
\(\therefore \)Percentage of coloured parts
= \( \frac{3×100}{5×100} \)= \(\frac{300}{5} \) %= 60%
(iii) Fraction of coloured part = \(\frac{3}{8} \)
\(\therefore \) Percentage of coloured parts
\(\frac{3×100}{8×100} \)=\(\frac{300}{8} \)% = 37.5%

Q.4 Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of Rs. 2500
(d) 75% of 1 kg


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Answer :

(a) 15% of 250 =\(\frac{15}{100} \)×250=\(\frac{75}{2} \)=37.5 (b) 1% of 1 hour = 1% of 60 min =1% of( 60 × 60 ) sec = 1% 3600 sec = \(\frac{1}{100} \)×3600 min=36 sec
(c) 20% of Rs. 2500 = \(\frac{20}{100} \) × 2500 = Rs. 500
(d) 75% of 1 kg = 75% of 1000g = \(\frac{75}{100} \) ×1000 = 750 g

Q.5 Find the whole quantity if
(a) 5% of it is 600
(b) 12 % of it is? 1080
(c) 40% of it is 500 km
(d) 70% of it is 14 minutes
(e) 8% of it is 40 litres


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Answer :

(a) 5% of x = 600 = \(\frac{5}{100} \)×x = 600
Or, x = \(\frac{600×100}{5} \) = 12000
(b) 12% of x = 1080 = \(\frac{12}{100} \) × x = 1080
Or, x = \(\frac{1080×100}{12} \) = 9000
(c) 40% of x =500 = \(\frac{40}{100} \) × x = 500
Or, x = \(\frac{500×100}{40} \) = 1250 km
(d) 70% of x = 14min = \(\frac{70}{100} \)×x = 14min
Or, x = \(\frac{14×100}{70} \) = 20
(e) 8% of x = 40 liters = \(\frac{8}{100} \)×x = 40 liters
Or, x = \(\frac{40×100}{8} \) = 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%


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Answer :

(a) 25% = 25/100 = 0.25
(b) 150% = 150/100 = 1.5
(c) 20% = 20/100 = 0.2
(d) 5% = 5/100 = 0.05

Q.7 In a city, 30% are females, 40% are males and remaining are children. What per cent are children?


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Answer :

Given: 30% are females
40% are males
Total Percentage of females and males
= 30% + 40% = 70%
\(\therefore \)Percentage of children
= (100 – 70)% = 30%

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?


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Answer :

Total number of voters = 15,000
Percentage of the voters who voted = 60%
\(\therefore \)Percentage of the voters who did not vote = (100 – 60)% = 40%
Actual number of voters who did not vote
= 40% of 15,000
=\(\frac{40}{100} \)×15,000=6,000

Q.9 Meena saves ? 400 from her salary. If this is 10% of her salary. What is her salary?


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Answer :

Let Meena’s salary by Rs. x.
\(\therefore \)10% of x = Rs. 400
Or, \(\frac{10}{100} \) × x = Rs. 400
Or,, x = \(\frac{400×100}{10} \)=Rs. 400
Thus, her salary is ? 4000.

Q.10 A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?


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Answer :

Number of matches played by the cricket team = 20
Percentage of the matches won by them = 25%
i.e. \(\frac{25}{100} \)×20=5 matches
Thus, the number of matches won by them = 5

Solution for 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 Rs.3,000.
(d) A skirt bought for Rs. 250 and sold at Rs.150.


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Answer :

(a) Here, CP = Rs.250
SP = Rs. 325
Since SP > CP
\(\therefore \)Profit = SP – CP
= Rs. 325 – Rs. 250 = Rs. 75
\(\therefore \) profit% = \(\frac{profit ×100}{CP} \)
= \(\frac{75}{100} \) ×100 = 30 %
(b) Here, CP = Rs.12000
SP = Rs. 13500
Since SP > CP
\(\therefore \)Profit = SP – CP
= Rs. 13500– Rs. 12000 = Rs. 1500
\(\therefore \) profit% = \(\frac{profit ×100}{CP} \)
= \(\frac{1500}{12000} \) ×100 = \(12\frac{1}{2} \)%
(c) Here, CP = Rs.2500
SP = Rs. 3000
Since SP > CP
\(\therefore \)Profit = SP – CP
= Rs. 3000– Rs. 2500 = Rs. 500
\(\therefore \) profit% = \(\frac{profit ×100}{CP} \)
= \(\frac{500}{2500} \) ×100 = 20%
(d) Here, CP = Rs.250
SP = Rs. 150
Since SP < CP
\(\therefore \)loss = CP– SP
= Rs. 250– Rs. 150 = Rs. 100
\(\therefore \) LOSS% = \(\frac{LOSS×100}{CP} \)
= \(\frac{100}{250} \) ×100 = 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


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Answer :

(a) 3 : 1
Sum of the ratio parts = 3 + 1 = 4
Percent of first part = \(\frac{3}{4} \) × 100 = 75%
Percent of second part = \(\frac{1}{4} \) × 100 = 25 %
(b) 2:3:5
Sum of the ratio parts = 2 + 3 +5 = 10
Percent of first part = \(\frac{2}{10} \) × 100 = 20%
Percent of second part = \(\frac{3}{10} \) × 100 = 30 %
Percent of second part = \(\frac{5}{10} \) × 100 = 50 %
(c) 1 : 4
Sum of the ratio parts = 1 + 4= 5
Percent of first part = \(\frac{1}{5} \) × 100 = 20%
Percent of second part = \(\frac{4}{5} \) × 100 = 80 %
(d) 1:2:5
Sum of the ratio parts = 1+ 2 +5 = 8
Percent of first part = \(\frac{1}{8} \) × 100 = 12.5%
Percent of second part = \(\frac{2}{8} \) × 100 = 25 %
Percent of third part = \(\frac{5}{8} \) × 100 = 62.5%

Q.3 The population of a city decreased from 25,000 to 24,500. Find the Percentage decrease.


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Answer :

Initial population = 25,000
Decreased population = 24,500
Decrease in population
= 25,000 – 24,500 = 500
Percentage of decrease =\(\frac{500×100}{25000} \) = 2%
Hence the Percentage of decrease in population = 2%.

Q.4 Arun bought a car for ? 3,50,000. The next year, the price went upto ? 3,70,000. What was the Percentage of price increase?


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Answer :

Original price of the car = Rs.3,50,000
Price increased next year = Rs.3,70,000
Increase in price
= Rs. 3,70,000 – Rs. 3,50,000 = Rs. 20,000
\(\therefore \)Percentage of the increase in the price
= \(\frac{20000×100}{350000} \) = \(\frac{40}{7} \) % = \(5\frac{5}{7} \) %

Q.5 I buy a TV for Rs.10,000 and sell it at a profit of 20%. How much money do I get for it?


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Answer :

Here, CP = Rs. 10,000
Profit = 20%
SP = \(\frac{CP×(100+P%)}{100} \)
=\(\frac{10,000×(100+20)}{100} \) = \(\frac{10,000×120}{100} \) =Rs. 12,000

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?


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Answer :

SP of the washing machine = ? 13,500 Loss = 20% CP = \(\frac{SP×100}{(100-loss%)} \) =\(\frac{13,500×100}{(100-20)} \)= \(\frac{13,500×100}{(80)} \)= Rs.16875

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?


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Answer :

(i) Sum of the ratio parts = 10 + 3 + 12 = 25
\(\therefore \) Percentage of carbon in chalk
=\(\frac{3}{25} \) ×100%=12%
Hence, the Percentage of carbon in chalk = 12%
(ii) Weight of carbon = 3 g
\(\therefore \)Weight of chalk =\(\frac{3}{3} \) × 25 g = 25 g
Hence, the weight of chalk = 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?


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Answer :

CP of book = Rs. 275
Loss = 15%
SP = \(\frac{CP×(100-L%)}{100} \)
= \(\frac{275×(100-15)}{100} \) = \(\frac{275×(85)}{100} \) = Rs.233.75

Q.9 Find the amount to be paid at the end of 3 years in each case.
(a) Principal = Rs. 1200 at 12% p.a.
(b) Principal = Rs. 7500 at 5% p.a.


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Answer :

(a) Given: Principal = Rs. 1200
Rate of interest = 12% p.a., T = 3 years
\(\therefore \) Interest = \(\frac{P×R×T}{100} \) =\(\frac{1200×12×3}{100} \)= Rs. 432
Amount = Principal + Interest
= Rs 1200 + Rs 432 = ? 1632
(b) Given: Principal = Rs. 7500
Rate of interest = 5% p.a., T = 3 years
\(\therefore \) Interest = \(\frac{P×R×T}{100} \) =\(\frac{7500×5×3}{100} \) =Rs.1125
Amount = Principal + Interest
= Rs. 7500 + 1125 = Rs. 8625

Q.10 What rate gives Rs. 280 as interest on a sum of Rs.56,000 in 2 years?


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Answer :

Given: Principal = Rs. 56,000
Interest = Rs.280
Time = 2 years
Rate = \(\frac{Interest ×100}{P×T} \)= \(\frac{280×100}{56000×2} \) = 1/4% = 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?


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Answer :

Interest = Rs. 45
Time = 1 year
Rate = 9% p.a.
P= \(\frac{I×100}{R×T} \) = \(\frac{45×100}{9×1} \) = Rs.500



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