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 BYJUâ€™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 â‚¹ 250 and sold for â‚¹ 325.**

**Solution:-**

From the** **question, it is given that

Cost price of gardening shears = â‚¹ 250

Selling price of gardening shears = â‚¹ 325

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

Profit = (SP) â€“ (CP)

= â‚¹ (325 â€“ 250)

= â‚¹ 75

Profit % = {(Profit/CP) Ã— 100}

= {(75/250) Ã— 100}

= {7500/250}

= 750/25

= 30%

**(b) A refrigerator bought for â‚¹ 12,000 and sold at â‚¹ 13,500.**

**Solution:-**

From the** **question, it is given that

Cost price of refrigerator = â‚¹ 12000

Selling price of refrigerator = â‚¹ 13500

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

Profit = (SP) â€“ (CP)

= â‚¹ (13500 â€“ 12000)

= â‚¹ 1500

Profit % = {(Profit/CP) Ã— 100}

= {(1500/12000) Ã— 100}

= {150000/12000}

= 150/12

= 12.5%

**(c) A cupboard bought for â‚¹ 2,500 and sold at â‚¹ 3,000.**

**Solution:-**

From the** **question, it is given that

Cost price of cupboard = â‚¹ 2500

Selling price of cupboard = â‚¹ 3000

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

Profit = (SP) â€“ (CP)

= â‚¹ (3000 â€“ 2500)

= â‚¹ 500

Profit % = {(Profit/CP) Ã— 100}

= {(500/2500) Ã— 100}

= {50000/2500}

= 500/25

= 20%

**(d) A skirt bought for â‚¹ 250 and sold at â‚¹ 150.**

**Solution:-**

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

Loss = (CP) â€“ (SP)

= â‚¹ (250 â€“ 150)

= â‚¹ 100

Loss % = {(Loss/CP) Ã— 100}

= {(100/250) Ã— 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

1^{st} part = Â¾ = (Â¾) Ã— 100 %

= 3 Ã— 25%

= 75%

2^{nd }part = Â¼ = (Â¼) Ã— 100%

= 1** **Ã— 25

= 25%

**(b) 2: 3: 5 **

**Solution:-**

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

1^{st} part = 2/10 = (2/10) Ã— 100 %

= 2 Ã— 10%

= 20%

2^{nd }part = 3/10 = (3/10) Ã— 100%

= 3** **Ã— 10

= 30%

3^{rd }part = 5/10 = (5/10) Ã— 100%

= 5** **Ã— 10

= 50%

**(c) 1:4 **

**Solution:-**

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

1^{st} part = (1/5) = (1/5) Ã— 100 %

= 1 Ã— 20%

= 20%

2^{nd }part = (4/5) = (4/5) Ã— 100%

= 4** **Ã— 20

= 80%

**(d) 1: 2: 5**

**Solution:-**

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

1^{st} part = 1/8 = (1/8) Ã— 100 %

= (100/8) %

= 12.5%

2^{nd }part = 2/8 = (2/8) Ã— 100%

= (200/8)

= 25%

3^{rd }part = 5/8 = (5/8) Ã— 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 â€“ Final population

= 25000 â€“ 24500

= 500

Then,

Percentage decrease in population = (population decrease/Initial population) Ã— 100

= (500/25000) Ã— 100

= (50000/25000)

= 50/25

= 2%

**4. Arun bought a car for â‚¹ 3,50,000. The next year, the price went upto â‚¹ 3,70,000. What was the Percentage of price increase?**

**Solution:-**

From the question, it is given that

Arun bought a car for = â‚¹ 350000

The price of the car in the next year, went up to = â‚¹ 370000

Then increase in price of car = â‚¹ 370000 â€“ â‚¹ 350000

= â‚¹ 20000

The percentage of price increase = (â‚¹ 20000/ â‚¹ 350000) Ã— 100

= (2/35) Ã— 100

= 200/35

= 40/7

=

**5. I buy a T.V. for â‚¹ 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. = â‚¹ 10000

Percentage of profit = 20%

Profit = (20/100) Ã— 10000

= â‚¹ 2000

Then,

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

= 10000 + 2000

= â‚¹ 12000

âˆ´ I will get it for â‚¹ 12000.

**6. Juhi sells a washing machine for â‚¹ 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 = â‚¹ 13500

Percentage of loss = 20%

Now, we have to find the cost price washing machine

By using the formula, we have:

CP = â‚¹ {(100/ (100 â€“ loss %)) Ã— SP}

= {(100/ (100 â€“ 20)) Ã— 13500}

= {(100/ 80) Ã— 13500}

= {1350000/80}

= {135000/8}

= â‚¹ 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) Ã— 100

= 3 Ã— 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) Ã— (x) = 3

X = 3 Ã— (100/12)

X = 1 Ã— (100/4)

X = 25g

âˆ´The weight of the stick is 25g.

**8. Amina buys a book for â‚¹ 275 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 = â‚¹ 275

Percentage of loss = 15%

Now, we have to find the selling price book,

By using the formula, we have:

SP = {((100 â€“ loss %) /100) Ã— CP)}

= {((100 â€“ 15) /100) Ã— 275)}

= {(85 /100) Ã— 275}

= 23375/100

= â‚¹ 233.75

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

**(a) Principal = â‚¹ 1,200 at 12% p.a. **

**Solution:-**

Given: â€“ Principal (P) = â‚¹ 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 Ã— R Ã— T)/100

= (1200 Ã— 12 Ã— 3)/ 100

= (12 Ã— 12 Ã— 3)/ 1

= â‚¹432

Amount = (principal + SI)

= (1200 + 432)

= â‚¹ 1632

**(b) Principal = â‚¹ 7,500 at 5% p.a. **

**Solution:-**

Given: â€“ Principal (P) = â‚¹ 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 Ã— R Ã— T)/100

= (7500 Ã— 5 Ã— 3)/ 100

= (75 Ã— 5 Ã— 3)/ 1

= â‚¹ 1125

Amount = (principal + SI)

= (7500 + 1125)

= â‚¹ 8625

**10. What rate gives â‚¹ 280 as interest on a sum of â‚¹ 56,000 in 2 years?**

**Solution:-**

Given: â€“ P = â‚¹ 56000, SI = â‚¹ 280, t = 2 years.

We know that,

R = (100 Ã— SI) / (P Ã— T)

= (100 Ã— 280)/ (56000 Ã— 2)

= (1 Ã— 28) / (56 Ã— 2)

= (1 Ã— 14) / (56 Ã— 1)

= (1 Ã— 1) / (4 Ã— 1)

= (1/ 4)

= 0.25%

**11. If Meena gives an interest of â‚¹ 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 = â‚¹ 45, R = 9%, T = 1 year, P =?

SI = (P Ã— R Ã— T)/100

45 = (P Ã— 9 Ã— 1)/ 100

P = (45 Ã—100)/ 9

= 5 Ã— 100

= â‚¹ 500

Hence, she borrowed â‚¹ 500.