Check NCERT Solutions for Class 7 Mathematics, Chapter 7 - Congruence of Triangles. This chapter is one of the most important chapters of CBSE Class 7 Mathematics subject. Students of Class 7 are advised to prepare this chapter well, as concepts given in this chapter are required to solve questions of other chapters.
NCERT Solutions for Class 7 Mathematics, Chapter 8 - Comparing Quantities:
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
Solution:
(a) Rs. 5 to 50 paise
As, 1 rupee = 100 paise
So, 5 rupee = 500 paise
∴ Ration of Rs. 5 to 50 paise = Rs. 5/50 Paise = 500 paise /50 paise = 10/1.
(b) 15 kg to 210 g
As, 1 kg = 1000 g
So, 15 kg = 15000 g
∴Ratio of 15 kg to 210 g = 15kg/210 g = 15000g/210g = 500/7.
(c) 9 m to 27 cm
As, 1 m = 100 cm
So, 9 m = 900 cm
∴ Ratio of 9 m to 27 cm = 900cm/27cm = 900/27 = 100/3.
(d) 30 days to 36 hours
As, 1 days = 24 hrs
So, 30 days = 24 × 30 = 720 hrs
∴ Ratio of 30 days to 36 hours = 30days/36hrs = 720 hrs/36 hrs = 20/1.
2. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solutions:
Given, number of computers required for 6 students = 3
So, the number of computers required for 1 student = 3/6 = 1/2
Now, for 24 students, the number of computers required = 24 × (½) = 12
3. Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km2 and area of UP = 2 lakh km2.
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
Solutions:
(i) Given, population of Rajasthan in 3 km2 area = 570 lakh
So the population of Rajasthan in 1 km2 area = 570/3 = 190 lakh
Now, Population of U.P in 2 km2 area = 1660 lakh
So, Population of U.P in 1 km2 area = 1660/2 = 830 lakh
(ii) On comparing the numbers we can easily observe that Rajasthan is less populated.
Exercise 8.2
1. Convert the given fractional numbers to percents.
(a) 1/8
(b) 5/4
(c) 3/40
(d) 2/7
Solution:
(a) 1/8 = (1/8) × 100/100
= 1/8 × 100 % = 12.5%
(b) 5/4 = (5/4) × 100/100
= 500/4 % = 125 %
(c) 3/40 = (3/40) × 100/100
= 300/40 % = 7.5 %
(d) 2/7 = (2/7) × 100/100
= 200/7 % = 28.57 %
2. Convert the given decimal fractions to per cents.
(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Solutions:
(a) 0.65 = 0.65 × 100%
= (65/100) × 100%
= 65%
(b) 2.1 = 2.1 × 100%
= (21/10) × 100%
= 210%
(c) 0.02 = 0.02 × 100%
= (2/100) × 100%
= 2%
(d) 12.35 = 12.35 × 100%
= (1235/100) × 100%
= 1235%
3. Estimate what part of the figures is coloured and hence find the per cent which is coloured.
Solutions:
(i) 1 out of 4 equal parts is shaded so it represents the fraction 1/4.
Now, 1/4 = (1/4) × 100% = 25%.
(ii) 3 out of 5 equal parts are shaded so it represents the fraction 3/5.
Now, 3/5 = (3/5) × 100% = 60%.
(iii) 3 out of 8 equal parts are shaded so it represents the fraction 3/8.
Now, 3/8 = (3/8) × 100% = 300/8 % = 37.5%.
4. Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of Rs. 2500
(d) 75% of 1 kg
Solutions:
(a) 15% of 250 = (15/100) × 250 = 75/2 = 37.5
(b) As, 1 hour = 60 minutes
So,1% of 60 Minutes = (1/100) × 60 Minutes= 3/5 Minutes
(c) 20% of Rs. 2500 = (20/100) × Rs. 2500 = Rs. 500
(d) 75% of 1 Kg = (75/100) × 1 Kg = 0.75 Kg
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.
Solutions:
(a) Let the quantity be x
According to the question, 5% of x = 600
⇒ (5/100) × x = 600
⇒ x = 600 × (100/5) = 12000
(b) Let the quantity be x
According to the question, 12% of x = Rs. 1080
⇒ (12 × 100) × x = Rs. 1080
⇒ x = Rs. 1080 × (100/12) = Rs. 9000
(c) Let the quantity be x
According to the question, 40% of x = 500 Km
⇒ (40/100) × x = 500 Km
⇒ x = 500 × (100/40) = 1250 Km
(d) Let the quantity be x
According to the question, 70% of x = 14 min
⇒ x × (70/100) = 14 min
⇒ x = 14 × (100/70) = 20 min
(e) Let the quantity be x
According to the question, 8% of x = 40 L
⇒ x × (8/100) = 40 L
⇒ x = 40 × (100/8) = 500 L
6. Convert given per cents to decimal fractions and also to fractions in simplest forms:
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Solutions:
(a) 25% = 25/100
Decimal form = 0.25
Fraction form = 1/4
(b) 150%
Decimal form = 150/100 = 1.5
Fraction form = 3/2
(c) 20% = 20/100
Decimal form = 0.2
Fraction form = 1/5
(d) 5% = 5/100
Decimal form = = 0.05
Fraction form = 1/20
7. In a city, 30% are females, 40% are males and remaining are children. What percent are children?
Solutions:
According to the question, 30% are females and 40% are males.
So % of children are = (100 − 30 − 40) % = 30%
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?
Solutions:
Percentage of voters who voted = 60%
Percentage of those who did not vote = 100% − 60% = 40%
Number of people who did not vote = 40% of 15000
= (40/100) × 15000 = 6000
Therefore, 6000 people did not vote.
9. Meeta saves Rs. 400 from her salary. If this is 10% of her salary. What is her salary?
Solutions:
Let Meeta’s salary be Rs x.
Given that,
10% of x = 400
⇒(10/100) × x = 400
⇒ x/10 = 400
⇒ x = 400 × 10 = Rs. 4000
Therefore, Meeta’s salary is Rs 4000.
10. A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solutions:
Given, number of games won = 25% of 20 = (25/100) × 20 = 5
Hence, the team won 5 matches.
Exercise 8.3
1. Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent 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.
Solutions:
(a) According to the statement,
Cost price = Rs 250
Selling price = Rs 325
So, Profit = Selling Price - Cost Price 325 − 250 = Rs 75
Now, Profit % = (Profit/Cost Price) × 100 = 75/250 × 100 = 30%
(b) According to the statement,
Cost price = Rs 12000
Selling price = Rs 13,500
So, Profit = 13500 − 12000 = Rs 1500
Now, Profit % = (Profit/Cost Price) × 100 = (500/2500) × 100 = 20%
(d) According to the statement,
Cost price = Rs 250
Selling price = Rs 150
So, loss = 250 − 150 = Rs 100
Now, loss % = (Loss/CP) × 100 = (100/250) × 100 = 40%
2. Convert each part of the ratio to percentage:
(a) 3:1
(b) 2:3:5
(c) 1:4
(d) 1:2:5
Solutions:
(a) 3: 1
First part = 3/4 = (3/4) × 100% = 75%
Second part = 1/4 = (1/4) × 100% = 25%
(b) 2: 3: 5
First part = 2/10 = (2/10) × 100% = 20%
Second part = 3/10 = (3/10) × 100% = 30%
Third part = 5/10 = (5/10) × 100% = 50%
(c) 1: 4
First part = 1/5 = (1/5) × 100% = 20%
Second part = 4/5 = (4/5) × 100% = 80%
(d) 1: 2: 5
First part = 1/8 = (1/8) × 100% = 12.5%
Second part = 2/8 = (2/8) × 100% = 25%
Third part = 5/8 = (5/8) × 100% = 62.5%
3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solutions:
According to the question,
Initial population = 25000
Final population = 24500
Net Decrease in population = 500
% Decrease = [(Net Decrease in population)/Initial population] × 100= (500/25000) × 100 = 2%
4. Arun bought a car for Rs. 3, 50,000. The next year, the price went upto Rs. 3, 70,000. What was the Percentage of price increase?
Solutions:
According to the question,
Initial price = Rs 350000
Final price = Rs 370000
Increase in price = Rs 20000
% increase = [(Increase in price)/(Initial price)] × 100= (20000/350000) × 100= 5.71 %
5. I buy a T.V. for Rs. 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solutions:
According to the question,
Cost price = Rs 10000
Now, profit = 20% of 10000 = (20/100) × 10000 = Rs 2000
Also, Selling price = Cost price + Profit = 10000 + 2000 = Rs 12,000
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?
Solutions:
According to the question,
Selling price = Rs 13,500 & Loss % = 20%
Suppose the cost price be x.
∴ Loss = 20% of x
Also, Cost price − Loss = Selling price
⇒x – (20/100) × x = 13500
⇒x – (1/5)x = 13500
⇒(4/5)x = 13500
⇒x = 13500 × (5/4) = 16875
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 3g, what is the weight of the chalk stick?
Solutions:
(i) According to the question,
Ratio of calcium, carbon, and oxygen = 10: 3: 12
Therefore, percentage of carbon = [3/(10 + 3 + 12)] × 100 = 12 %
(ii) Suppose the weight of the stick be x (in gram)
12 % of x = 3
⇒(12/100) × x = 3
⇒x = 3 × (100/12) = 25 g
8. Amina buys a book for Rs. 275 and sells it at a loss of 15%. How much does she sell it for?
Solutions:
Given, Cost price = Rs 275 & Loss % = 15%
Therefore, loss = 15% of 275
Now, Selling price = Cost price − Loss
⇒Selling Price = 275 – (15/100) × 275
⇒Selling Price = 275 – (4125/100)
⇒Selling Price =275 − 41.25
Therefore, Selling price = Rs 233.75
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.
Solutions:
(a) Given, Principal = Rs 1200, Rate = 12 % p.a., Time = 3 years
Simple Interest = (Principal × Rate × Time)/100 = (1200 × 12 × 3)/100 = Rs 432
Amount = Principal + Simple Interest = 1200 + 432 = Rs 1632
(b) Given, Principal = Rs 7500, Rate = 5% p.a., Time = 3 years
Simple Interest = (Principal × Rate × Time)/100 = (7500 × 5 × 3)/100 = Rs 1125
Amount = Principal + Simple Interest = 7500 + 1125 = Rs 8625
10. What rate gives Rs. 280 as interest on a sum of Rs. 56,000 in 2 years?
Solutions:
Simple Interest = (Principal × Rate × Time)/100
⇒280 = (56000 × Rate × 2)/100
Rate = 280/(560 × 2) = 1/4 = 0.25
11. If Meena gives an interest of Rs. 45 for one year at 9% rate p.a.. What is the sum she has borrowed?
Solutions:
Simple Interest = (Principal × Rate × Time)/100
⇒ 45 = (Principal × 9 × 1)/100
Principal = (45 × 100)/9 = Rs 500.
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