7. Comparing Quantities

Solutions to Ex. 7.1

Q1. Convert the given fractional numbers to per cents.

Ans:

(a) \( \frac{1}{8} \)

= \( \frac{1}{8} \) x \( \frac{100}{100} \) = \( \frac{100}{8} \) %

= \( 12\frac{1}{2} \) % or 12.5% 

(b) \( \frac{5}{4} \)

= \( \frac{5}{4} \) x \( \frac{100}{100} \) = \( \frac{500}{4} \) %

= 125% 

(c) \( \frac{3}{40} \)

= \( \frac{3}{40} \) x \( \frac{100}{100} \) = \( \frac{300}{40} \) %

= \( 7\frac{1}{2} \) % or 7.5% 

(d) \( \frac{2}{7} \)

= \( \frac{2}{7} \) x \( \frac{100}{100} \) = \( \frac{200}{7} \) %

= \( 28\frac{4}{7} \)% 

Q2. Convert the given decimal fractions to per cents.

Ans:

(a) 0.65 = (0.65 x 100)% = 65%

(b) 2.1 = (2.1 x 100)% = 210%

(c) 0.02 = (0.02 x 100)% = 2%

(d) 12.35 = (12.35 x 100)% = 1235%

Q3. Estimate what part of the figures is coloured and hence find the per cent which is coloured.

Ans:

(i) \( \frac{1}{4} \) is the coloured part.

= (\( \frac{1}{4} \) x 100)% = 25%

(ii) \( \frac{3}{5} \) is the coloured part.

= (\( \frac{3}{5} \) x 100)% = 60%

(iii) \( \frac{3}{8} \) is the coloured part.

= (\( \frac{3}{8} \) x 100)% = 37.5%

Q4. Find:

Ans:

(a) 15% of 250

= \( \frac{15}{100} \) x 250

= 37.5

(b) 1% of 1 hour

As 1 hour = 60 minutes

= \( \frac{1}{100} \) x 60

= \( \frac{3}{5} \) minutes or 36 seconds.

(c) 20% of ₹2500

= \( \frac{20}{100} \) x 2500

= ₹500

(d) 75% of 1 kg

= \( \frac{75}{100} \) x 1

= 0.75kg or 750g

Q5. Find the whole quantity if:

Ans:

(a) 5% of it is 600

Let the whole quantity be a.

\( \frac{5}{100} \) x a = 600

a = 600 x \( \frac{100}{5} \) = 12000

The whole quantity is 12,000.

(b) 12% of it is ₹1080

Let the whole quantity be a.

\( \frac{12}{100} \) x a = 1080

a = 1080 x \( \frac{100}{12} \) = ₹9,000

The whole quantity is ₹9,000.

(c) 40% of it is 500 km

Let the whole quantity be a.

\( \frac{40}{100} \) x a = 500

a = 500 x \( \frac{100}{40} \) = 1250km

The whole quantity is 1,250km.

(d) 70% of it is 14 minutes

Let the whole quantity be a.

\( \frac{70}{100} \) x a = 14

a = 14 x \( \frac{100}{70} \) = 20 minutes

The whole quantity is 20 minutes.

(e) 8% of it is 40 litres

Let the whole quantity be a.

\( \frac{8}{100} \) x a = 40

a = 40 x \( \frac{100}{8} \) = 500 litres

The whole quantity is 500 litres.

Q6. Convert given per cents to decimal fractions and also to fractions in simplest forms:

Ans:

(a) 25%

Decimal form = \( \frac{25}{100} \) = 0.25

Fraction form = \( \frac{25}{100} \) = \( \frac{1}{4} \)

(b) 150% 

Decimal form = \( \frac{150}{100} \) = 1.5

Fraction form = \( \frac{15}{100} \) = \( \frac{3}{2} \)

(c) 20% 

Decimal form = \( \frac{20}{100} \) = 0.2

Fraction form = \( \frac{20}{100} \) = \( \frac{1}{5} \)

(d) 5%

Decimal form = \( \frac{5}{100} \) = 0.05

Fraction form = \( \frac{5}{100} \) = \( \frac{1}{20} \)

Q7. In a city, 30% are females, 40% are males and remaining are children. What per cent are children?

Ans:

Given: Percentage of females = 30%

   Percentage of males = 40%

Total Percentage of males & females = 30% + 40% = 70%

Percentage of children = 100 - 70 = 30%

The per cent of children is 30%.

Q8. 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?

Ans:

Given: Total voters = 15,000

  Percentage voted = 60%

Percentage of voters who did not vote

= (100 - 60)% = 40%

Actual voters = 40% of 15000

    = \( \frac{40}{100} \) x 15000

   = 6,000 voters

Number of voters who didn't vote is 6,000 or 40%.

Q9. Meeta saves ₹4000 from her salary. If this is 10% of her salary. What is her salary?

Ans:

Given: Savings = ₹4000

Let the salary be a.

10% of a = 4000

\( \frac{10}{100} \) x a = 4000

a = 4000 x \( \frac{100}{10} \) = ₹40,000

Meeta's salary is ₹40,000.

Q10. A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

Ans:

Given: Total matches played = 20

   Matches won = 25%

The number of matches won

= 25% of 20

= \( \frac{25}{100} \) x 20

= 5

The cricket team won 5 matches.

Solutions to Ex. 7.2

Q1.Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

Ans:

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

Given: CP = ₹250

   SP = ₹325

Profit = SP - CP = 325 - 250

Profit = ₹75

%Profit = \( \frac{Profit}{CP} \) x 100

    = \( \frac{75}{250} \) x 100 = 30%

Profit is ₹75 & Percent profit is 30%.

(b) A refrigerator bought for ₹12,000 and sold at ₹13,500.

Given: CP = ₹12,000

   SP = ₹13,500

Profit = SP - CP = 12,000 - 13,500

Profit = ₹1500

%Profit = \( \frac{Profit}{CP} \) x 100

    = \( \frac{1500}{12000} \) x 100

    = \( 12\frac{1}{2} \)% or 12.5%

Profit is ₹75 & Percent profit is 12.5%.

(c) A cupboard bought for ₹2,500 and sold at ₹3,000.

Given: CP = ₹2500

   SP = ₹3000

Profit = SP - CP = 3000 - 2500

Profit = ₹500

%Profit = \( \frac{Profit}{CP} \) x 100

    = \( \frac{500}{2500} \) x 100 = 20%

Profit is ₹500 & Percent profit is 20%.

(d) A skirt bought for ₹250 and sold at ₹150.

Given: CP = ₹250

   SP = ₹150

As, CP > SP, it is a loss.

Loss = CP - SP =  250 = 150

Loss = ₹100

%Loss = \( \frac{Loss}{CP} \) x 100

    = \( \frac{100}{250} \) x 100 = 40%

Loss is ₹100 & Percent loss is 40%.

Q2. Convert each part of the ratio to percentage:

Ans:

(a)  3 : 1 

Sum of the ratio = 3 +1 = 4

Percentage of first part is

= \( \frac{3}{4} \) x 100 = 75%

Percentage of second part is

= \( \frac{1}{4} \) x 100 = 25%

(b) 2 : 3 : 5 

Sum of the ratio = 2 + 3 + 5 = 10

Percentage of first part is

= \( \frac{2}{10} \) x 100 = 20%

Percentage of second part is

= \( \frac{3}{10} \) x 100 = 30%

Percentage of third part is

= \( \frac{5}{10} \) x 100 = 50%

(c) 1:4 

Sum of the ratio = 1 + 4 = 5

Percentage of first part is

= \( \frac{1}{5} \) x 100 = 20%

Percentage of second part is

= \( \frac{4}{5} \) x 100 = 80%

(d) 1 : 2 : 5

Sum of the ratio = 1 + 2 +5 = 8

Percentage of first part is

= \( \frac{1}{8} \) x 100 = 12.5%

Percentage of second part is

= \( \frac{2}{8} \) x 100 = 25%

Percentage of third part is

= \( \frac{5}{8} \) x 100 = 62.5%

Q3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Ans:

Given: Initial population = 25,000

   New population = 24,500

Decrease in population = 25000 - 24000 = 500

Percentage decrease = \( \frac{500}{25000} \) x 100 = 2%

Q4. 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?

Ans:

Given: Initial price = ₹3,50,000

   New price = ₹3,70,000

Increase in price = 3,70,000 - 3,50,000

= 20,000

Percentage increase = \( \frac{20,000}{3,50,000} \) x 100

     = \( 5\frac{5}{7} \)%

Q5. I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?

Ans:

Given: CP = ₹10,000

   Profit = 20%

Profit = 20% of ₹10,000

= \( \frac{20}{100} \) x 10,000 = ₹2000

Money got = CP + Profit = ₹10,000 + ₹2000

= ₹12000

Q6. Juhi sells a washing machine for ₹13,500. She loses 20% in the bargain. What was the price at which she bought it?

Ans:

Given: SP = ₹13,500

   Loss = 20%

If CP is ₹100,

20% loss = \( \frac{20}{100} \) x 100 = ₹20

SP = 100 - 20 = ₹80

Thus, when SP is ₹80, CP is ₹100

So, when SP is ₹13,500, 

CP = \( \frac{13,500}{80} \) x 100 = ₹16875

Juhi bought the washing machine for ₹16875.

Q7. (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 3g, what is the weight of the chalk stick?

Ans:

(i) Given: Ratio = 10:3:12

     Sum of the ratio = 10 + 3 +12 = 25

     Percentage of carbon = \( \frac{3}{25} \) x 100 = 12%

(ii) Given: Carbon = 3g

      12% of chalk weight = 3

      \( \frac{12}{100} \) x Chalk weight = 3

      Chalk weight = 3 x \( \frac{100}{12} \) = 25g

Q8. Amina buys a book for ₹275 and sells it at a loss of 15%. How much does she sell it for?

Ans:

Given: CP = ₹275

   Loss = 15%

If CP is ₹100,

20% loss = \( \frac{15}{100} \) x 100 = ₹15

SP = 100 - 15 = ₹85

Thus, when CP is ₹100, SP is ₹85

So, when CP is ₹275,

SP = \( \frac{275 \times 85 }{100} \)= ₹233.75

Amina sold the book for ₹233.75.

Q9. Find the amount to be paid at the end of 3 years in each case:

Ans:

(a) Given: Principal (P) = ₹1,200

        Rate of Interest (R) = 12% p.a.

Simple Interest = \( \frac{P \times R \times T }{100} \)

      = \( \frac{1200 \times 12 \times 3 }{100} \) = ₹432

Amount to be paid = 1,200 + 432 = ₹1,632

(b) Given: Principal (P) = ₹7,500

        Rate of Interest (R) = 5% p.a.

Simple Interest = \( \frac{P \times R \times T }{100} \)

      = \( \frac{7500 \times 5 \times 3 }{100} \) = ₹1,125

Amount to be paid = 7500 + 1125 = ₹8,625

Q10. What rate gives ₹280 as interest on a sum of ₹56,000 in 2 years?

Ans:

Given: Principal (P) = ₹56,000

   Simple Interest (SI) = ₹280

   Time period (T) = 2 years

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

280 = \( \frac{56000 \times R \times 2 }{100} \)

R = \( \frac{280 \times 100 }{56000 \times 2 } \) = 0.25%

Rate of Interest is 0.25%.

Q11. If Meena gives an interest of ₹45 for one year at 9% rate p.a. What is the sum she has borrowed?

Ans:

Given: Simple Interest (SI) = ₹45

   Rate of Interest (R) = 9%

   Time period (T) = 1 year

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

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

P = \( \frac{45 \times 100  }{9} \) = ₹500

The sum borrowed by Meena is ₹500.