Class 7 - Maths - Comparing Quantities

**Exercise 8.1**

**Question 1:**

Find the ratio of:

(a) ₹5 to 50 paise (b) 15 kg to 210 g (c) 9 m to 27 cm (d) 30 days to 36 hours

Answer:

To find ratios, both quantities should be in same unit.

(a) ₹5 to 50 paise

Since ₹1 = 100 paise

So, ₹5 = 5 * 100 = 500 paise

Thus, the ratio = 500/50 = 10/1 = 10 : 1

(b) 15 kg to 210 g

Since 1 kg = 1000 g

So, 15 kg = 15 * 1000 = 15000 g

Thus, the ratio = 15000/210 = 500/7 = 500 : 7

(c) 9 m to 27 m

Since 1 m = 100 cm

So, 9 m = 9 * 100 = 900 cm

Thus, the ratio = 900/27 = 100/3 = 100 : 3

(d) 30 days to 36 hours

Since 1 day = 24 hours

So, 30 days = 30 * 24 = 720 hours

Thus, the ratio = 720/36 = 20/1 = 20 : 1

**Question 2:**

In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?

Answer:

Since 6 students need = 3 computers

So, 1 student needs = 3/6 = 1/2 computers

Now, 24 students need = 24 * 1/2 = 24/2 = 12 computers

Thus, 12 computers will be needed for 24 students.

**Question 3:**

Population of Rajasthan = 570 lakhs and population of U.P. = 1660 lakhs. Area of Rajasthan = 3 lakh km^{2} and area of U.P. = 2 lakh km^{2}.

(i) How many people are there per km^{2} in both states?

(ii) Which state is less populated?

Answer:

(i) People present per km^{2} = Total Population/Area

In Rajasthan = 570 lakhs/3 lakhs per km^{2} = 190 people km^{2}

In U.P. = 1660 lakhs/2 lakh per km^{2} = 830 people per km^{2}

(ii) It can be observed that population per km^{2} area is lesser for Rajasthan. Therefore,

Rajasthan is less populated.

**Exercise 8.2**

**Question 1:**

Convert the given fractional numbers to percent:

(a) 1/8 (b) 5/4 (c) 3/40 (d) 2/7

Answer:

(a) 1/8 = (1/8) * 100 % = 100/8 % = 12.5 %

(b) 5/4 = (5/4) * 100 % = 500/4 % = 125 %

(c) 3/40 = (3/40) * 100 % = 300/40 % = 15/2 % = 7.5 %

(d) 2/7 = (2/7) * 100 % = 200/7 % = 28.57 %

**Question 2:**

Convert the given decimal fractions to per cents:

(a) 0.65 (b) 2.1 (c) 0.02 (d) 12.35

Answer:

(a) 0.65 = (65/100) * 100% = 65%

(b) 2.1 = (21/100) * 100% = 210%

(c) 0.02 = (2/100) * 100 % = 2%

(b) 12.35 = (1235/100) * 100% = 1235%

**Question 3:**

Estimate what part of the figures is coloured and hence find the percent which is coloured.

Answer:

(i) Coloured part = 1/4

So, Percent of coloured part = (1/4) *100% = 25%

(ii) Coloured part = 3/5

So, Percent of coloured part = (3/5) * 100% = 300/5% = 60%

(iii) Coloured part = 3/8

So, Percent of coloured part = (3/8) * 100% = 300/8 % = 37.5%

**Question 4:**

Find:

(a) 15% of 250 (b) 1% of 1 hour (c) 20% of ₹2500 (d) 75% of 1 kg

Answer:

(a) 15% of 250 = (15/100) * 250 = 15 * 2.5 = 37.5

(b) 1% of 1 hour = 1% of 60 minutes

= 1% of (60 * 60) seconds

= (1/100) * 60 * 60 seconds

= 3600/100 seconds = 36 seconds

(c) 20% of ₹2500 = (20/100) * 2500 = 2500/5 = ₹500

(d) 75% of 1 kg = 75% of 1000 g

= (75/100) * 1000

= (3/4) * 1000

= 3000/4

= 750 g

= 750/1000 kg = 0.750 kg [Since 1000 g = 1 kg]

**Question 5:**

Find the whole quantity if:

(a) 5% of it is 600 (b) 12% of it is ₹1080 (c) 40% of it is 500 km

(d) 70% of it is 14 minutes (e) 8% of it is 40 litres

Answer:

Let the whole quantity be x in given questions.

(a) 5% of x = 600

=> (5/100) * x = 600

=> 5x/100 = 600

=> 5x = 600 * 100

=> x = (600 * 100)/5

=> x = 600 * 20

=> x = 12000

(b) 12% of x = ₹1080

=> (12/100) * x = 1080

=> 12x/100 = 1080

=> 12x = 1080 * 100

=> x = (1080 * 100)/12

=> x = 90 * 100

=> x = ₹9000

(c) 40% of = 500 km

=> (40/100) * x = 500

=> 40x/100 = 500

=> 40x = 500 * 100

=> x = (500 * 100)/40

=> x = 50000/40

=> x = 1250 km

(d) 70% of x = 14 minutes

=> (70/100) * x = 14

=> 70x/100 = 14

=> 70x = 14 * 100

=> x = (14 * 100)/70

=> x = 1400/70

=> x = 20 minutes

(e) 8% of x = 40 litres

=> (8/100) * x = 40

=> 8x/100 = 40

=> 8x = 40 * 100

=> x = (40 * 100)/8

=> x = 5 * 100

=> x = 500 litres

**Question 6:**

Convert given per cents to decimal fractions and also to fractions in simplest forms:

(a) 25% (b) 150% (c) 20% (d) 5%

Answer:

**Question 7:**

In a city, 30% are females, 40% are males and remaining are children. What percent are children?

Answer:

Given: Percentage of females = 30%

Percentage of males = 40%

Total percentage of females and males = 30 + 40 = 70%

Percentage of children = Total percentage – Percentage of males and females

= 100% – 70%

= 30%

Hence, 30% are children.

**Question 8:**

Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

Answer:

Total voters = 15,000

Percentage of voted candidates = 60%

Percentage of not voted candidates = 100 – 60 = 40%

Actual candidates, who did not vote = 40% of 15000

= (40/100) * 15000

= 40 * 150

= 6,000

Hence, 6,000 candidates did not vote.

**Question 9:**

Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?

Answer:

Let Meera’s salary be ₹x.

Now, 10% of salary = ₹ 400

=> 10% of x = ₹ 400

=> (10/100) * x = 400

=> x/10 = 400

=> x = 400 * 10

=> x = 4,000

Hence, Meera’s salary is ₹ 4,000.

**Question 10:**

A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

Answer :

Number of matches played by cricket team = 20

Percentage of won matches = 25%

Total matches won by them = 25% of 20

= (25/100) * 20

= (1/4) * 20

= 20/4

= 5

Hence, they won 5 matches.

**Exercise 8.3**

**Question 1:**

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

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

(b) A refrigerator bought ₹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.

Answer:

(a) Cost price of gardening shears = ₹ 250

Selling price of gardening shears = ₹ 325

Since, S.P. > C.P., therefore here is profit.

So, Profit = S.P. – C.P. = ₹325 – ₹250 = ₹ 75

Now Profit% = (Profit/CP) * 100

= (75/250) * 100

= (3/10) * 100

= (3 * 100)/10

= 300/10

= 30%

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

(b) Cost price of refrigerator = ₹ 12,000

Selling price of refrigerator = ₹13,500

Since, S.P. > C.P., therefore here is profit.

So, Profit = S.P. – C.P. = ₹13500 – ₹12000 = ₹1,500

Now Profit% = (Profit/CP) * 100

= (1500/12000) * 100

= (15/120) * 100

= (1/8) * 100

= 100/8

= 12.5%

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

(c) Cost price of cupboard = ₹ 2,500

Selling price of cupboard = ₹ 3,000

Since, S.P. > C.P., therefore here is profit.

So, Profit = S.P. – C.P. = ₹3,000 – ₹2,500 = ₹ 500

Now Profit% = (Profit/CP) * 100

= (500/2500) * 100

= (1/5) * 100

= 100/5

= 20%

Therefore, Profit = ₹ 500 and Profit% = 20%

(d) Cost price of skirt = ₹ 250

Selling price of skirt = ₹ 150

Since, C.P. > S.P., therefore here is loss.

So, Loss = C.P. – S.P. =₹250 – ₹150 = ₹100

Now Loss% = (Loss/CP) * 100

= (100/250) * 100

= (10/25) * 100

= (2/5) * 100

= 2 * 10

= 40%

Therefore, Profit = ₹ 100 and Profit% = 40%

**Question 2:**

Convert each part of the ratio to percentage:

(a) 3 : 1 (b) 2 : 3 : 5 (c) 1 : 4 (d) 1 : 2 : 5

Answer:

(a) 3 : 1

Total part = 3 + 1 = 4

1^{st} part = (3/4) = (3/4) * 100 = 75%

2^{nd} part = (1/4) = (1/4) * 100 = 25%

So, percentage of parts = 75% : 25%

(b) 2 : 3 : 5

Total part = 2 + 3 + 5 = 10

1^{st} part = (2/10) = (2/10) * 100 = 20%

2^{nd} part = (3/10) = (3/10) * 100 = 30%

3^{rd} part = (5/10) = (5/10) * 100 = 50%

So, percentage of parts = 20% : 30% : 50%

(c) 1 : 4

Total part = 1 + 4 = 5

1^{st} part = (1/5) = (1/5) * 100 = 20%

2^{nd} part = (4/5) = (4/5) * 100 = 80%

So, percentage of parts = 20% : 80%

(d) 1 : 2 : 5

Total part = 1 + 2 + 5 = 8

1^{st} part = (1/8) = (1/8) * 100 = 12.5%

2^{nd} part = (2/8) = (2/8) * 100 = 25%

3^{rd} part = (5/8) = (5/8) * 100 = 62.5%

So, percentage of parts = 12.5% : 25% : 62.5%

**Question 3:**

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

Answer:

The decreased population of a city from 25,000 to 24,500.

Population decreased = 25,000 – 24,500 = 500

Decreased Percentage = (Population decreased/Original population) * 100

= (500/25000) * 100

= (5/250) * 100

= 500/250

= 2%

Hence, the percentage decreased is 2%.

**Question 4:**

Arun bought a car for ₹3,50,000. The next year, the price went up to ₹3,70,000. What was the percentage of price increase?

Answer:

Increased in price of a car from ₹ 3,50,000 to ₹ 3,70,000.

Amount change = ₹ 3,70,000 – ₹ 3,50,000 = ₹ 20,000.

Therefore, Increased percentage = (Amount of change/Original amount) * 100

= (20000/350000) * 100

= (2/35) * 100

= 200/35

= 40/7

= 5 %

Hence, the percentage of price increased is 5 %.

**Question 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?

Answer:

The cost price of T.V. = ₹ 10,000

Profit percent = 20%

Now, Profit = Profit% of C.P.

= (20/100) * 10000

= 20 * 100

= ₹ 2,000

Selling price = C.P. + Profit = ₹10,000 + ₹2,000 = ₹ 12,000

Hence, he gets ₹12,000 on selling his T.V.

**Question 6:**

Juhi sells a washing machine for ₹13,500. She loses 20% in the bargain. What was the price at which she bought it?

Answer:

Selling price of washing machine = ₹13,500

Loss percent = 20%

Let the cost price of washing machine be ₹x.

Since, Loss = Loss% of C.P.

=> Loss = 20% of ₹x

=> Loss = (20/100) * ₹x = x/5

Therefore, S.P. = C.P. – Loss

=> 13500 = x – x/5

=> 13500 = 4x/5

=> 4x = 13500 * 5

=> x = (13500 * 5)/4

=> x = 3375 * 5

=> x = ₹16,875

Hence, the cost price of washing machine is ₹16,875.

**Question 7:**

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

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

Answer:

(i) Given ratio = 10 : 3 : 12

Total part = 10 + 3 + 12 = 25

Part of Carbon = 3/25

Percentage of Carbon part in chalk = (3/25) * 100 = 3 * 4 = 12%

(ii) Quantity of Carbon in chalk stick = 3 g

Let the weight of chalk be x g.

Then, 12% of x = 3

(12/100) * x = 3

=> 12x/100 = 3

=> 12x = 3 * 100

=> 12x = 300

=> x = 300/12

=> x = 25 g

Hence, the weight of chalk stick is 25 g.

**Question 8:**

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

Answer:

The cost of a book = ₹275

Loss percent = 15%

Loss = Loss% of C.P. = 15% of ₹275

= (15/100) * 275

= (15 * 275)/100

= 4125/100

= ₹ 41.25

Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75

Hence, Amina sells a book for ₹233.75.

**Question 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.

Answer:

(a) Here, Principal (P) = ₹1,200, Rate (R) = 12% p.a., Time (T) = 3 years

Simple Interest = (P * R * T)/100

= (1200 * 12 * 3)/100

= 12 * 12 * 3

= ₹ 432

Now, Amount = Principal + Simple Interest

= ₹1200 + ₹432 = ₹1,632

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

Simple Interest = (P * R * T)/100

= (7500 * 5 * 3)/100

= 75 * 5 * 3

= ₹1,125

Now, Amount = Principal + Simple Interest

= ₹7,500 + ₹1,125 = ₹ 8,625

**Question 10:**

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

Answer:

Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years

Simple Interest = (P * R * T)/100

=> 280 = (56000 * R * 2)/100

=> R = (280 * 100)/(56000 * 2)

=> R = 28/(56 * 2)

=> R = 1/(2 * 2)

=> R = 1/4

=> R = 0.25%

Hence, the rate of interest on sum is 0.25%.

**Question 11:**

If Meena gives an interest of ₹45 for one year at 9% rate p.a. What is the sum she has borrowed?

Answer:

Simple Interest = ₹45, Rate (R) = 9% p.a., Time (T) = 1 years

Simple Interest = (P * R * T)/100

=> 45 = (P * 9 * 1)/100

=> 45 = 9P/100

=> 9P = 45 * 100

=> P = (45 * 100)/9

=> P = 5 * 100 => P = 500

Hence, she borrowed ₹ 500.

.