# NCERT Solutions for Class 7 Maths Exercise 8.3 Chapter 8 Comparing Quantities

NCERT Solutions for Class 7 Maths Exercise 8.3 Chapter 8 Comparing Quantities in simple PDF are given here. This exercise of NCERT Solutions for Class 7 Chapter 8 contains topics related to converting ratios to percent, increase or decrease as per cent, prices related to an item or buying and selling, profit or loss as a percent and charges given on borrowed money or simple interest. We at BYJU’s have prepared these NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities in such a way that while solving the solutions, students will understand properly without getting confused.

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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 ₹ 250 and sold for ₹ 325.

Solution:-

From the question, it is given that

Cost price of gardening shears = ₹ 250

Selling price of gardening shears = ₹ 325

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= ₹ (325 – 250)

= ₹ 75

Profit % = {(Profit/CP) × 100}

= {(75/250) × 100}

= {7500/250}

= 750/25

= 30%

(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.

Solution:-

From the question, it is given that

Cost price of refrigerator = ₹ 12000

Selling price of refrigerator = ₹ 13500

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= ₹ (13500 – 12000)

= ₹ 1500

Profit % = {(Profit/CP) × 100}

= {(1500/12000) × 100}

= {150000/12000}

= 150/12

= 12.5%

(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.

Solution:-

From the question, it is given that

Cost price of cupboard = ₹ 2500

Selling price of cupboard = ₹ 3000

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= ₹ (3000 – 2500)

= ₹ 500

Profit % = {(Profit/CP) × 100}

= {(500/2500) × 100}

= {50000/2500}

= 500/25

= 20%

(d) A skirt bought for ₹ 250 and sold at ₹ 150.

Solution:-

Since (SP) < (CP), so there is a loss

Loss = (CP) – (SP)

= ₹ (250 – 150)

= ₹ 100

Loss % = {(Loss/CP) × 100}

= {(100/250) × 100}

= {10000/250}

= 40%

2. Convert each part of the ratio to percentage:

(a) 3 : 1

Solution:-

We have to find total parts by adding the given ratio = 3 + 1 = 4

1st part = ¾ = (¾) × 100 %

= 3 × 25%

= 75%

2nd part = ¼ = (¼) × 100%

= 1 × 25

= 25%

(b) 2: 3: 5

Solution:-

We have to find total parts by adding the given ratio = 2 + 3 + 5 = 10

1st part = 2/10 = (2/10) × 100 %

= 2 × 10%

= 20%

2nd part = 3/10 = (3/10) × 100%

= 3 × 10

= 30%

3rd part = 5/10 = (5/10) × 100%

= 5 × 10

= 50%

(c) 1:4

Solution:-

We have to find total parts by adding the given ratio = 1 + 4 = 5

1st part = (1/5) = (1/5) × 100 %

= 1 × 20%

= 20%

2nd part = (4/5) = (4/5) × 100%

= 4 × 20

= 80%

(d) 1: 2: 5

Solution:-

We have to find total parts by adding the given ratio = 1 + 2 + 5 = 8

1st part = 1/8 = (1/8) × 100 %

= (100/8) %

= 12.5%

2nd part = 2/8 = (2/8) × 100%

= (200/8)

= 25%

3rd part = 5/8 = (5/8) × 100%

= (500/8)

= 62.5%

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

Solution:-

From the question, it is given that

Initial population of the city = 25000

Final population of the city = 24500

Population decrease = Initial population – Final population

= 25000 – 24500

= 500

Then,

Percentage decrease in population = (population decrease/Initial population) × 100

= (500/25000) × 100

= (50000/25000)

= 50/25

= 2%

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?

Solution:-

From the question, it is given that

Arun bought a car for = ₹ 350000

The price of the car in the next year, went up to = ₹ 370000

Then increase in price of car = ₹ 370000 – ₹ 350000

= ₹ 20000

The percentage of price increase = (₹ 20000/ ₹ 350000) × 100

= (2/35) × 100

= 200/35

= 40/7

=

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?

Solution:-

From the question, it is given that

Cost price of the T.V. = ₹ 10000

Percentage of profit = 20%

Profit = (20/100) × 10000

= ₹ 2000

Then,

Selling price of the T.V. = cost price + profit

= 10000 + 2000

= ₹ 12000

∴ I will get it for ₹ 12000.

6. Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?

Solution:-

From the question, it is given that

Selling price of washing machine = ₹ 13500

Percentage of loss = 20%

Now, we have to find the cost price washing machine

By using the formula, we have:

CP = ₹ {(100/ (100 – loss %)) × SP}

= {(100/ (100 – 20)) × 13500}

= {(100/ 80) × 13500}

= {1350000/80}

= {135000/8}

= ₹ 16875

7. (i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.

Solution:-

From the question it is given that,

The ratio of calcium, carbon and oxygen in chalk = 10: 3: 12

So, total part = 10 + 3 + 12 = 25

In that total part amount of carbon = 3/25

Then,

Percentage of carbon = (3/25) × 100

= 3 × 4

= 12 %

(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?

Solution:-

From the question it is given that,

Weight of carbon in the chalk = 3g

Let us assume the weight of the stick be x

Then,

12% of x = 3

(12/100) × (x) = 3

X = 3 × (100/12)

X = 1 × (100/4)

X = 25g

∴The weight of the stick is 25g.

8. Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?

Solution:-

From the question, it is given that

Cost price of book = ₹ 275

Percentage of loss = 15%

Now, we have to find the selling price book,

By using the formula, we have:

SP = {((100 – loss %) /100) × CP)}

= {((100 – 15) /100) × 275)}

= {(85 /100) × 275}

= 23375/100

= ₹ 233.75

9. Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹ 1,200 at 12% p.a.

Solution:-

Given: – Principal (P) = ₹ 1200, Rate (R) = 12% p.a. and Time (T) = 3years.

If interest is calculated uniformly on the original principal throughout the loan period, it is called Simple interest (SI).

SI = (P × R × T)/100

= (1200 × 12 × 3)/ 100

= (12 × 12 × 3)/ 1

= ₹432

Amount = (principal + SI)

= (1200 + 432)

= ₹ 1632

(b) Principal = ₹ 7,500 at 5% p.a.

Solution:-

Given: – Principal (P) = ₹ 7500, Rate (R) = 5% p.a. and Time (T) = 3years.

If interest is calculated uniformly on the original principal throughout the loan period, it is called Simple interest (SI).

SI = (P × R × T)/100

= (7500 × 5 × 3)/ 100

= (75 × 5 × 3)/ 1

= ₹ 1125

Amount = (principal + SI)

= (7500 + 1125)

= ₹ 8625

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

Solution:-

Given: – P = ₹ 56000, SI = ₹ 280, t = 2 years.

We know that,

R = (100 × SI) / (P × T)

= (100 × 280)/ (56000 × 2)

= (1 × 28) / (56 × 2)

= (1 × 14) / (56 × 1)

= (1 × 1) / (4 × 1)

= (1/ 4)

= 0.25%

11. If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?

Solution:-

From the question it is given that, SI = ₹ 45, R = 9%, T = 1 year, P =?

SI = (P × R × T)/100

45 = (P × 9 × 1)/ 100

P = (45 ×100)/ 9

= 5 × 100

= ₹ 500

Hence, she borrowed ₹ 500.