NCERT Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

NCERT Solutions Class 7 Mathematics 
Chapter – 8 (Comparing Quantities)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 8 Comparing Quantities Exercise 8.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 8 Comparing Quantities

Exercise – 8.3 

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.
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.

Solution –

(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
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
NCERT Class 7 Maths Solution

= \frac{75\times 100}{250}%
= 3 × 10 %
= 30%

(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
Cost price of refrigerator = ₹ 12000
Selling price of refrigerator = ₹ 13500
Since (SP) > (CP), so there is a profit
Profit = (SP) – (CP)
= ₹ (13500 – 12000)
= ₹ 1500
NCERT Class 7 Maths Solution

= \frac{1500\times 100}{12000}%

= \frac{150}{12}%

= 12.5%

(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
Cost price of cupboard = ₹ 2500
Selling price of cupboard = ₹ 3000
Since (SP) > (CP), so there is a profit
Profit = (SP) – (CP)
= ₹ (3000 – 2500)
= ₹ 500

NCERT Class 7 Maths Solution
= \frac{500\times 100}{2500}%

= 20%

(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Cost price of skirt = ₹ 250
Selling price of skirt = ₹ 150
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (250 – 150)
= ₹ 100
NCERT Class 7 Maths Solution

= \frac{100\times 100}{250}%

= 40%

2. Convert each part of the ratio to percentage:
(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5

Solution –

(a) 3 : 1
Sum of the ratio parts = 3 + 1 = 4
Percentage of First part = \frac{3}{4}\times 100 %
= 3 × 25%
= 75%

Percentage of Second part =  \frac{1}{4}\times 100%
= 1 × 25%
= 25%

(b) 2 : 3 : 5
Sum of the ratio parts = 2 + 3 + 5 = 10
Percentage of First part = \frac{2}{10}\times 100%
= 2 × 10%
= 20%

Percentage of Second part = \frac{3}{10}\times 100 %
= 3 × 10%
= 30%

Percentage of Third part = \frac{5}{10}\times 100 %
= 5 × 10%
= 50%

(c) 1 : 4
Sum of the ratio parts = 1 + 4 = 5

Percentage of First part = \frac{1}{5}\times 100 %
= 1 × 20%
= 20%

Percentage of Second part = \frac{4}{5}\times 100 %
= 4 × 20%
= 80%

(d) 1 : 2 : 5
Sum of the ratio parts = 1 + 2 + 5 = 8
Percentage of First part = \frac{1}{8}\times 100%
= \frac{100}{8} %
= 12.5%

Percentage of Second part = \frac{2}{8}\times 100%
= \frac{1}{4}\times 100
= 25%

3rd part = \frac{5}{8}\times 100%
= \frac{500}{8}%
= 62.5%

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

Solution –
Initial population of the city = 25000
Final population of the city = 24500
Population decrease = Initial population – Final population
= 25000 – 24500
= 500
Then,
NCERT Class 7 Maths Solution

= \frac{500\times 100}{25000}%

= \frac{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 –
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 = \frac{20000\times 100}{350000}%
= \frac{2\times 100}{35} %

= \frac{40}{7}% = 5\frac{5}{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 –
Cost price of the T.V. = ₹ 10000
Percentage of profit = 20%
Profit = \frac{20\times 10000}{100} = ₹ 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 –
Selling price of washing machine = ₹ 13500
Percentage of loss = 20%
Cost Price washing machine = \frac{100\times SP}{100-loss%}
= \frac{100\times 13500}{100-20}

= \frac{1350000}{80}

= ₹ 16875

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

Solution –
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 = \frac{3}{25}
Then,
Percentage of carbon = \frac{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 –
Weight of carbon in the chalk = 3g
Let us assume the weight of the stick be x
Then,
12% of x = 3

\frac{12}{100} × (x) = 3

x = 3 × \frac{100}{12}

x = 1 × \frac{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 –
Cost price of book = ₹ 275
Percentage of loss = 15%
Selling Price of book = \frac{(100-loss%)}{100}\times CP

= \frac{(100-15)}{100}\times 275

= \frac{85}{100}\times 275

= \frac{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.
(b) Principal = ₹ 7,500 at 5% p.a.

Solution –

(a) Principal = ₹ 1,200 at 12% p.a.
Principal (P) = ₹ 1200,
Rate (R) = 12% p.a.
Time (T) = 3 years.
Simple interest (SI) = \frac{P\times R\times T}{100}

= \frac{1200\times 12\times 3}{100}

= 12 × 12 × 3
= ₹432
Amount = (principal + SI)
= (1200 + 432)
= ₹ 1632

(b) Principal = ₹ 7,500 at 5% p.a.
Principal (P) = ₹ 7500,
Rate (R) = 5% p.a.
Time (T) = 3 years.

Simple Interest (SI) = \frac{P\times R\times T}{100}
= \frac{7500\times 5\times 3}{100}

= 75 × 5 × 3
= ₹ 1125
Amount = (principal + SI)
= (7500 + 1125)
= ₹ 8625

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

Solution –
P = ₹ 56000,
SI = ₹ 280,
T = 2 years.

R = \frac{100\times SI}{P\times T}

= \frac{100\times 280}{56000\times 2}

= \frac{1\times 28}{56\times 2}

= \frac{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 –
SI = ₹ 45,
R = 9%,
T = 1 year,
P =?
SI = \frac{P\times R\times T}{100}

45 = \frac{P\times 9\times 1}{100}

P = \frac{45\times 100}{9}

P = 5 × 100
P = ₹ 500
Hence, she borrowed ₹ 500.

 

 

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