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# Mathematics – Class 7 – Chapter 8 – Comparing Quantities – Exercise 8.3 – NCERT Exercise Solution

1. Tell what is the profit or loss in the following transactions. Also, find profit percent or loss per cent in each case.

(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.

Solution:

Given,

Cost price (CP) of gardening shears = ₹ 250

Selling price (SP) of gardening shears = ₹ 325

Here, (SP) > (CP)

So, there is a profit

Now,

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:

Given,

Cost price of refrigerator = ₹ 12000

Selling price of refrigerator = ₹ 13500

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

Then,

Profit = (SP) – (CP)

= ₹ (13500 – 12000)

= ₹ 1500

Now,

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:

Given,

Cost price of cupboard = ₹ 2500

Selling price of cupboard = ₹ 3000

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

Profit = (SP) – (CP)

= ₹ (3000 – 2500)

= ₹ 500

Now,

Profit % = {(Profit/CP) × 100}

= {(500/2500) × 100}

= {50000/2500}

= 500/25

= 20%

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

Solution:

CP of skirt = ₹ 250

SP of skirt = ₹ 150

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

Loss = (CP) – (SP)

= ₹ (250 – 150)

= ₹ 100

Now,

Loss % = {(Loss/CP) × 100}

= {(100/250) × 100}

= {10000/250}

= 40%

2. Convert each part of the ratio to percentage:

(a) 3 : 1

Solution:

First, we add the given ratio to find total parts,

3 + 1 = 4

Now,

1st part = 3/4 = (3/4) × 100 %

= 3 × 25%

= 75%

2nd part = 1/4 = (1/4) × 100%

= 1 × 25

= 25%

(b) 2: 3: 5

Solution:

First, we add the given ratio to find total parts

2 + 3 + 5 = 10

Now,

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:

First, we add the given ratio to find total parts

1 + 4 = 5

Now,

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:

First, we add the given ratio to find total parts

1 + 2 + 5 = 8

Now,

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:

Given,

The initial population of the city = 25000

The final population of the city = 24500

So,

Population decrease = Initial population – Final population

= 25000 – 24500

= 500

Now,

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:

Given,

Arun bought a car for = ₹ 350000

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

Then,

Increase in price of car = ₹ 370000 – ₹ 350000

= ₹ 20000

Now,

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

= (2/35) × 100

= 200/35

= 40/7

= 5.714 %

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:

Given,

Cost price (CP) of the T.V. = ₹ 10000

Percentage of profit = 20%

Then,

Profit = (20/100) × 10000

= ₹ 2000

Now,

Selling price (SP) of the T.V. = cost price + profit

= 10000 + 2000

= ₹ 12000

Hence, 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:

Given,

Selling price of washing machine = ₹ 13500

Percentage of loss = 20%

Now,

we have to find the cost price washing machine

We know that,

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:

Given,

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

So, we have to add them to find the total part

10 + 3 + 12 = 25

In the 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:

Given,

Weight of carbon in the chalk = 3g

Now,

Let the weight of the stick be x

According to the question,

12% of x = 3

(12/100) × (x) = 3

X = 3 × (100/12)

X = 1 × (100/4)

X = 25g

Hence, 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:

Given,

Cost price of book = ₹ 275

Percentage of loss = 15%

Now, we have to find the selling price book,

We know that,

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.

Time (T) = 3years.

Then,

SI = (P × R × T)/100

= (1200 × 12 × 3)/ 100

= (12 × 12 × 3)/ 1

= ₹432

Now,

Amount = (principal + SI)

= (1200 + 432)

= ₹ 1632

(b) Principal = ₹ 7,500 at 5% p.a.

Solution:

Given,

Principal (P) = ₹ 7500

Rate (R) = 5% p.a.

Time (T) = 3years.

Then,

SI = (P × R × T)/100

= (7500 × 5 × 3)/ 100

= (75 × 5 × 3)/ 1

= ₹ 1125

Now,

Amount = (principal + SI)

= (7500 + 1125)

= ₹ 8625

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

Solution:

Given,

Principal (P) = ₹ 56000

Simple Interest (SI) = ₹ 280

Time (T) = 2 years.

From the formula, 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:

Given,

Simple Interest (SI) = ₹ 45

Rate (R) = 9%

(Time)T = 1 year

Principal (P) =?

From the formula, we know that,

SI = (P × R × T)/100

45 = (P × 9 × 1)/ 100

P = (45 ×100)/ 9

= 5 × 100

= ₹ 500

Hence, she borrowed ₹ 500.

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