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NCERT Exemplar · Class 7 · Maths

Chapter 7: Comparing Quantities

Practice each question, then open the answer or worked solution.

Questions

70 questions
Q41Written answer

Sonal bought a bed sheet for ₹ 400 and sold it for ₹ 440. Her ________% is ________.

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Solution:-

Sonal bought a bed sheet for ₹ 400 and sold it for ₹ 440. Her Profit% is 10.

From the question, it is given that, cost price of bed = ₹ 400

Selling price of bed = ₹ 440

SP is more than CP so, Profit = SP – CP

= 440 – 400

= ₹ 40

We know that, Profit percentage = (Profit/CP) × 100

= (40/400) × 100

= 10%

Q42Written answer

Nasim bought a pen for ₹ 60 and sold it for ₹ 54. His ________% is ________.

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Solution:-

Nasim bought a pen for ₹ 60 and sold it for ₹ 54. His loss% is 10.

From the question, it is given that, cost price of pen = ₹ 60

Selling price of pen = ₹ 54

CP is more than SP so, Loss = CP – SP

= 60 – 54

= ₹ 6

We know that, loss percentage = (loss/CP) × 100

= (6/60) × 100

= 10%

Q43Written answer

Aahuti purchased a house for ₹ 50,59,700 and spent ₹ 40300 on its repairs. To make a profit of 5%, she should sell the house for ₹ ________.

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Solution:-

Aahuti purchased a house for ₹ 50,59,700 and spent ₹ 40300 on its repairs. To make a profit of 5%, she should sell the house for ₹ 5355000.

From the question it is give that,

Cost price of house purchased by Aahuti = ₹ 50,59,700

Amount spent to repair the house = ₹ 40,300

Then, total cost price of house = ₹ 50,59,700 + 40,300

= ₹ 5100000

Profit % = (profit/CP) × 100

5 = ((SP – CP)/CP) × 100

5 = ((SP – 5100000)/5100000) × 100

(5 × 5100000)/100 = SP – 5100000

SP = ₹ 5355000

Q44Written answer

If 20 lemons are bought for ₹ 10 and sold at 5 for three rupees, then ________ in the transaction is ________%.

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Solution:-

If 20 lemons are bought for ₹ 10 and sold at 5 for three rupees, then profit in the transaction is 20%.

From the question, it is given that, cost price of 20 lemons = ₹ 10

Selling price of 5 lemons = ₹ 3

So, selling price of 1 lemon = ₹ 3/5

Then, selling price of 20 lemons = ₹ (3/5) × 20

= 12

SP is more than CP so, Profit = SP – CP

= 12 – 10

= ₹ 2

We know that, Profit percentage = (Profit/CP) × 100

= (2/10) × 100

= 20%

Q45Written answer

Narain bought 120 oranges at ₹ 4 each. He sold 60 % of the oranges at ₹ 5 each and the remaining at ₹ 3.50 each. His ________ is ________%.

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Solution:-

Narain bought 120 oranges at ₹ 4 each. He sold 60 % of the oranges at ₹ 5 each and the remaining at ₹ 3.50 each. His profit is 10%.

From the question, it is given that, cost price of one orange = ₹ 4

cost price of 120 oranges = ₹ 4 × 120 = ₹ 480

So, selling price of 60% of oranges = ₹ 5 per each

60% of 120 oranges = (60/100) × 120

= 7200/100

= 72 oranges

Selling price of 72 oranges = 72 × 5 = ₹ 360

SP of remaining oranges = (120 – 72) × 3.50

= 48 × 3.50

= ₹ 168

Therefore total selling price of 120 oranges = ₹ 360 + ₹ 168

= ₹ 528

SP is more than CP so, Profit = SP – CP

= 528 – 480

= ₹ 48

We know that, Profit percentage = (Profit/CP) × 100

= (48/480) × 100

= 10%

Q46Written answer

A fruit seller purchased 20 kg of apples at ₹ 50 per kg. Out of these, 5% of the apples were found to be rotten. If he sells the remaining apples at ₹ 60 per kg, then his _________is _________%.

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Solution:-

A fruit seller purchased 20 kg of apples at ₹ 50 per kg. Out of these, 5% of the apples were found to be rotten. If he sells the remaining apples at ₹ 60 per kg, then his profit is 14 %.

From the question it is given that, A fruit seller purchased 20 kg of apples at ₹ 50 per kg

In that 20 kg of apples 5% were rotten, so good apples = 20 – 5% of 20kg

= 20 – (5/100) × 20

= 20 – (100/100)

= 20 – 1

= 19 kg

So, 19 kg apples were sold at ₹ 60 per kg

Then, total Selling price of apples = 19 × 60

= ₹ 1140

Given cost price of 20 kg of apples = 20 × 50

= ₹ 1000

SP is more than CP so, Profit = SP – CP

= 1140 – 1000

= ₹ 140

We know that, Profit percentage = (Profit/CP) × 100

= (140/1000) × 100

= 14%

Q47Written answer

Interest on ₹ 3000 at 10% per annum for a period of 3 years is ________.

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Solution:-

Interest on ₹ 3000 at 10% per annum for a period of 3 years is ₹ 900.

From the question it is given that,

Principal= ₹ 3000

Time = 3 years

Rate = 10%

Then, we know the formula of Simple interest I = (P × R × T)/100

I= (3000 × (10) × 3)/100

I = ₹ 900

Q48Written answer

Amount obtained by depositing ₹ 20,000 at 8 % per annum for six months is ________.

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Solution:-

Amount obtained by depositing ₹ 20,000 at 8 % per annum for six months is ₹ 20800

From the question it is given that,

Principal= ₹ 20000

Time = 6 months = 6/12 = ½

Rate = 8%

Then, we know the formula of Simple interest I = (P × R × T)/100

I= (20000 × 8 × ½ ) /100

I = ₹ 800

Amount received = Principal + Interest

= 20000 + 800

= ₹ 20800

Q49Written answer

Interest on ₹ 12500 at 18% per annum for a period of 2 years and 4 months is ________.

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Solution:-

Interest on ₹ 12500 at 18% per annum for a period of 2 years and 4 months is ₹ 5250.

From the question it is given that,

Principal= ₹ 12500

Time = 2 years 4 months = (2 + (4/12))

= (2 + (1/3))

= 7/3 year

Rate = 18%

Then, we know the formula of Simple interest I = (P × R × T)/100

I= (12500 × 18 × (7/3)) /100

I = ₹ 5250

Q50Written answer

25 ml is _________ percent of 5 litres.

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Solution:-

25 ml is 0.5 percent of 5 litres

Let us assume 25 ml be Q % of 5 liters.

So, 25 ml = Q % of 5l

25 = (Q/100) × 5 × 1000

Q = (25 × 100)/(5 × 1000)

Q = 0.5%