# Ncert Solutions For Class 7 Maths Ex 8.3

## Ncert Solutions For Class 7 Maths Chapter 8 Ex 8.3

QUESTION 1:

What is the profit and theloss for the following transaction.Also find what is the profit percentage and the loss percentage in the cases given:

(i) Gardening shears were bought for Rs 250 and were sold for Rs 325

(ii) A refrigerator was bought for Rs 12,000 and sold for Rs 13,500

(iii) A cupboard was bought for Rs 2,500 and sold for Rs 3,000.

(iv) A shirt was bought for Rs 250 and sold for Rs 150

Solution;

(i)the cost price of the gardening shears: Rs 250

The selling price of the gardening shears: Rs 325

Since,  SP> CP  therefore here it is a profit.

Therefore profit = SP – CP = Rs 325 – Rs 250

= Rs 75

Now profit % = profitC.P×100$\frac{profit}{C.P}\times 100$

= 75250×100=$\frac{75}{250}\times 100=$ %

Therefore, Profit = Rs 75 and Profit % = 30%

(ii) Cost prince of refrigerator = Rs 12,000

Selling price of Refrigerator = Rs 13,500

Since, S.P > C.P Therefore, it is a profit.

Therefore, Profit = SP – CP = Rs 13500 – Rs 12000 = Rs 1500

Now profit % = profitC.P×100$\frac{profit}{C.P}\times 100$

= 150012000×100$\frac{1500}{12000}\times 100$ = 12.5%

Therefore, Profit = Rs 1,500 and Profit % = 12.5 %

(iii) Cost price of cupboard = Rs 2500

Selling price of cupboard = Rs 3,000

Since, SP> CP, Therefore it is a profit.

Profit = SP – CP = Rs 3,000 – Rs 2,500 = Rs 500

150012000×100$\frac{1500}{12000}\times 100$

= 5002500×100$\frac{500}{2500}\times 100$ =20%

Therefore, Profit = 500 and profit % = 20%

(iv) Cost price of shirt = Rs 250

Selling price of shirt = Rs 150

Since CP > SP, therefore here it is a loss.

Therefore Loss = CP – SP = Rs 250 – Rs 150

= Rs 100

Now loss % = LossC.P×100$\frac{Loss}{C.P}\times 100$

= 100250×100$\frac{100}{250}\times 100$ = 40%

Therefore , Loss = Rs 100 and Loss% = 40%

QUESTION 2:

Convert the following ratio to its percentages:

(i) 3 : 1

(ii) 2 : 3 : 5

(iii) 1 : 4

(iv) 1 : 2 : 5

Solution;

(i)3 : 1

Total part = 3 + 1 = 4

Therefore, Fractional part = 34:14$\frac{3}{4}:\frac{1}{4}$

=> Percentage of parts = 34×100:14×100$\frac{3}{4}\times 100:\frac{1}{4}\times 100$

=> Percentage of parts = 75 % : 25 %

(ii)2 : 3 : 5

Total part = 2 + 3 + 5 = 10

Therefore, Fractional part = 210:310:510$\frac{2}{10}:\frac{3}{10}:\frac{5}{10}$

=> Percentage of parts = 210×100:310×100:510×100$\frac{2}{10}\times 100:\frac{3}{10}\times 100:\frac{5}{10}\times 100$

=> Percentage of parts = 20 % : 30 % : 50 %

(iii)1 : 4

Total part = 1 + 4 = 5

Therefore, Fractional part = 15:45$\frac{1}{5}:\frac{4}{5}$

=> Percentage of parts = 15×100:45×100$\frac{1}{5}\times 100:\frac{4}{5}\times 100$

=> Percentage of parts = 20% : 80%

(iv)1 : 2 : 5

Total part = 1 + 2 + 5 = 8

Therefore, Fractional part = 18:28:58$\frac{1}{8}:\frac{2}{8}:\frac{5}{8}$

=> Percentage of parts = 18×100:28×100:58×100$\frac{1}{8}\times 100:\frac{2}{8}\times 100:\frac{5}{8}\times 100$

=> Percentage of parts = 12.5 % : 25 % : 62.5 %

QUESTION 3:

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

Solution;

The decrease in population of a city is from 25,000 to 24,500.

Population Decreased = 25000 – 24500 = 500

Decreased percentage = PopulationDecreasedOriginalPopulation×100$\frac{Population Decreased}{Original Population}\times 100$

= 50025000×100$\frac{500}{25000}\times 100$ = 2 %

Hence, the percentage of decrease in the population is 2 %.

QUESTION 4:

Reah and Reuben bought a caravan for Rs. 3,50,000. The next year, the price raised to Rs 3,70,000. What is the percentage of price increased?

Solution;

Increased in price of the caravan from Rs 3, 50,000 to 3, 70,000

Amount change = Rs 3, 70,000 – Rs 3, 50,000 = Rs 20,000

Therefore, Increased percentage = AmountofchangeOriginalAmount×100$\frac{Amount of change}{Original Amount}\times 100$

20000350000×100=557$\frac{20000}{350000}\times 100 = 5\frac{5}{7}$ %

Hence, The percentage of price increased is 557$5\frac{5}{7}$ %

QUESTION 5:

Harismitha bought a T.V for Rs 10,000 and sold it at a profit of 20%. How much money did she get for it?

Solution;

The cost price of T.V = Rs 10000

Profit percent = 20%

Now, Profit = Profit % of CP

= 20100×10000$\frac{20}{100}\times 10000$

= Rs 2000

Selling price = CP + Profit = Rs 10000 + Rs 2000 = Rs 12000

She gets Rs 12000 on selling the T.V

QUESTION 6:

Nancy sold a washing machine for Rs 13500. She Lost 20% in the bargain. What is the price that she bought the washing machine for?

Solution;

Selling price of washing machine= Rs 13,500

Loss percent = 20%

Let the cost price of washing machine be Rs x

Since, Loss = Loss % of CP

=> Loss = 20 % of Rs x = 20100×x=x5$\frac{20}{100}\times x=\frac{x}{5}$

Therefore, SP = CP – Loss

=>13500=xx5=>13500=4x5=>x=13500×54=Rs16,875$\\=>13500=x-\frac{x}{5}\\=>13500=\frac{4x}{5}\\=>x=\frac{13500\times 5}{4}=Rs16,875$

Hence, The cost price of washing machine is Rs 16,875.

QUESTION 7:

(a) Chalk contains Calcium, Carbon and Oxygen in the ratio 10: 3 : 12. Find the percentage of carbon in chalk.

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

Solution;

(a) Given ratio = 10 : 3 : 12

Total part = 10 + 3 + 12 = 25

Part of carbon = 325$\frac{3}{25}$

Percentage of Carbon part in Chalk = 325$\frac{3}{25}$ x 100 = 12 %

(b) Quality of Carbon in chalk stick = 3g

Let the weight of chalk be x g.

Then, 12 % of x = 3

=> 12100$\frac{12}{100}$ x X = 3

=> x = 3×10012$\frac{3\times 100}{12}$ = 25g

Hence, the weight of chalk stick is 25g.

QUESTION 8:

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

Solution;

The cost of a book = Rs 275

Loss percentage = 15 %

Loss = Loss % of CP = 15 % of 275

= 15100×275$\frac{15}{100}\times 275$ = Rs 41.25

Therefore , SP = CP – Loss = 275 – 41.25 = Rs 233.75

Hence, Amulya sold the book for Rs 233.75

QUESTION 9:

Find the amount to be paid at the end of 3 years in the following cases:

(i ) Principal = Rs 1200 at 12 % p. a

(ii) Principal = Rs 7500 at 5 % p. a

Solution;

(i) Here, Principal (p) = 1200

Rate ( r ) = 12 % p a

Time ( T ) = 3 years

Simpleinterest=P×R×T100=1200×12×3100$Simple interest = \frac{P \times R\times T}{100} = \frac{1200\times 12\times 3}{100}$

=432

Now, Amount = Principal + Simple Interest

= Rs 1200 + Rs 432

= Rs 1632

(ii) Here, Principal (p) = 7500

Rate ( r ) =  5 % p a

Time ( T ) = 3 years

Simpleinterest=P×R×T100=7500×5×3100$Simple interest = \frac{P \times R\times T}{100} = \frac{7500\times 5\times 3}{100}$

= 1125

Now, Amount = Principal + Simple Interest

= Rs 7500 + Rs 1125

= Rs 8625

QUESTION 10:

What rate gives Rs 280 as interest on a sum onRs 56,000 in 2 years?

Solution;

Here, Principal (P ) = Rs 56000

Simple Interest (SI ) = Rs 280

Time ( T) = 2 yrs

Simple interest = P×R×T100$\frac{P \times R\times T}{100}$

=> 280 = 56000×R×2100$\frac{56000\times R\times 2}{100}$

=> R = 280×10056000×2$\frac{280\times 100}{56000\times 2}$

=> Hence, the rate of interest on sum is 0.25%

QUESTION 11:

If Deepak gives an interest of Rs 45 for one year at 9% rate per annum. What is the sum he has borrowed from Nancy?

Solution;

Simple Interest = Rs 45

Rate (R ) = 9%

Time ( T) = 1 year

Simple interest = P×R×T100$\frac{P \times R\times T}{100}$

=> 45 = P×9×1100$\frac{P\times 9\times 1}{100}$

=> P = 45×1009$\frac{45\times 100}{9}$

=> P = Rs 500

Hence, she had borrowed Rs 500.