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Chapter 8: Comparing Quantities

EXERCISE- 8.1:

QUESTION 1:

Calculate the ratio:

            (i) Rs10 to 50 paise                      (ii) 10kg to 500g

           (iii) 60m to 20cm                          (iv) 20days to 50 hours

Solution;

(i) Rs.10 to 50 paise

=> 10 x 100 = 1000 paise            (Since , Rs.1 = 100 paise)

Thus, the ratio =  1000/50 = 20 : 1

(ii) 10kg to 500g

10 x 1000= 10000g (Since 1kg is 1000g)

Thus, the ratio = 10000/500 = 20 : 1

(iii) 60m to 20cm

60 x 100 = 6000cm (Since 1m = 100cm)

Thus, the ratio =6000/20 = 300 : 1

(iv) 20days to 50 hours

20 x 24 = 480 hours

Thus, the ratio = 480/50 = 48 : 5

 

QUESTION 2:

A computer lab has 2 computers for every 4 students. For 20 students how many computers will be required?

Solution;

Given,

4 students = 2 computers

1 student requires = 2/4 computers.

Therefore, 20 students require = (2/4) x 20 = 10 computers.

 

QUESTION 3:

Mizoram has a population of 10 lakh and Sikkim has a population of 6 lakh. Area of Mizoram is  21,000 km2 and the area of Sikkim is 7000 km2.

            (i) Find the number of people per km2 in both the states.

            (ii) Which state has a lesser population?

Solution;

(i) People per km2 = Population/ Area

In Mizoram = 1000000/21000 = 47.61 ≃48 people/km2

In Sikkim      = 600000/7000         = 85.71 ≃ 86 people/km2

(ii) Sikkim has a lesser population.

 

EXERCISE- 8.2

QUESTION 1:

Convert the given fractional numbers to percent:

(a) 1/3                       (b)6/5                        (c)4/25                      (d)3/7

Solution;

(a) 1/3 = (1/3) x 100% = 100/3 = 33.33%

(b) 6/5 = (6/5) x 100%  = 6 x 20 = 120%

(c)  4/25 = (4/25) x 100% = 16%

(d) 2/7 = (2/7) x 100% = 28.57%

 

QUESTION 2:

Express the given decimal numbers in percentages.

(a) 0.31         (b)0.03                      (c)19.62        (d)0.007

Solutions;

(a) 0.31 = (31/100) x 100% = 31%

(b) 0.03 = (3/100) x 100%   = 3%

(c) 19.62 = (1962/100) x 100% = 1962%

(d) 0.007 = (7/1000) x 100% = 0.7%

 

 QUESTION 3:

Find what part of the following figures is colored; also find the percentage of the colored part.

Comparing

Solutions;

(i) Colored part = 1/4

Therefore, percentage of the colored portion = (1/4) x 100%  = 25%

(ii) Colored portion = 3/5

Therefore, percentage of the colored portion = (3/5) x 100 =60%

(iii) Colored portion = (3/8) x 100% = 37.5%

 

QUESTION 4:

Calculate :

(a) 15% of 450                    (b) 2% of 2 hour

 (c) 10% of ₹2600              (d) 50% of 5 kg

Solutions;

(a) 15% of 450 = (15/100) x 450 = 67.5

(b) 2% of 2 hours= 2% of (2×3600) secs         (Since 1 hour = 60 x 60 secs)

= (2/100) x 7200 =144secs.

(c) 10% of Rs.2600 = (10/100) x 2600

=Rs.260

(d) 50% of 5kg         = 50% of (5000g)              (Since 1kg = 1000g)

= (50/100) x 5000

= 2500 g or 2.5kg

 

 QUESTION 5:

Calculate the total quantity if:

 (a) 5% of it is 100             

(b) 10% of it is ₹1060

(c) 20% of it is 800 km

(d) 90% of it is 18 minutes

(e) 9% of it is 60 liters

Solutions;

let the total quantity be Y in all above cases,

(a)5% of Y  = 100

=> (5/100) x Y =100

=> Y=(100×100)/5 =2000

(b)10% of Y = 1060

=> (10/100) x Y = 1060

=> Y = (1060 x 100)/10 = Rs.10600

(c)20% of Y = 800

=> (20/100) x Y = 800

=> Y = (800 x 100)/20 = 4000km

(d) 90% of Y = 18min

=> (90/100) x Y = 18

=> Y=(18 x 100)/90 =20mins.

(e) 9% of Y = 60

=> (9/100) x Y = 60

=> Y = (60 x 100)/9 =666.66liters.

 

QUESTION 6:

Convert the following percentages to fractions , simplest fractions and decimal numbers:

(a)20%                      (b)25%                      (c)5%             (d)120%

Solution;

Percentage Fractions Simplest Fraction Decimal Numbers.
20% 20/100 1/5 0.2
25% 25/100 1/4 0.25
5% 5/100 1/20 0.05
120% 120/100 6/5 1.2

 

QUESTION 7:

12000 voters constituted a constituency, out of which 80% voted. What percentage and number of voters did not vote?

Solution;

Given,

Total no. of voters = 12000

Percentage who voted = 80%

Thus, percentage who did not vote = 100 -80 = 20%

Therefore, the number of people who did not vote = 20% of 12000

= (20/100) x 12000 = 2400

 

QUESTION 8:

Sanjay saves Rs.5000 every month, if this is 20% of his salary. What is his monthly salary?

Solution;

Let Sanjay’s salary be Y

Now, 20% of Y = 5000

(20/100) x Y = 5000

Y = Rs.25000

Therefore,Sanjay’s salary is Rs.25000

 

QUESTION 9:

A localdota team played 10 matches in a season. If they won 40% of their matches, in how many matches do they win?

Solution;

Given,

Number of matches played by the dota2 team =10

Percentage of matches won = 40%

Therefore, total matches they were victorious in = 40% of 10

= (40/100) x 10

=4.

 

EXERCISE- 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 % = \(\frac{profit}{C.P}\times 100\)

= \(\frac{75}{250}\times 100=\) %

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 % = \(\frac{profit}{C.P}\times 100\)

= \(\frac{1500}{12000}\times 100\) = 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

\(\frac{1500}{12000}\times 100\)

= \(\frac{500}{2500}\times 100\) =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 % = \(\frac{Loss}{C.P}\times 100\)

= \(\frac{100}{250}\times 100\) = 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 = \(\frac{3}{4}:\frac{1}{4}\)

=> Percentage of parts = \(\frac{3}{4}\times 100:\frac{1}{4}\times 100\)

=> Percentage of parts = 75 % : 25 %

(ii)2 : 3 : 5

Total part = 2 + 3 + 5 = 10

Therefore, Fractional part = \(\frac{2}{10}:\frac{3}{10}:\frac{5}{10}\)

=> Percentage of parts = \(\frac{2}{10}\times 100:\frac{3}{10}\times 100:\frac{5}{10}\times 100\)

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

(iii)1 : 4

Total part = 1 + 4 = 5

Therefore, Fractional part = \(\frac{1}{5}:\frac{4}{5}\)

=> Percentage of parts = \(\frac{1}{5}\times 100:\frac{4}{5}\times 100\)

=> Percentage of parts = 20% : 80%

(iv)1 : 2 : 5

Total part = 1 + 2 + 5 = 8

Therefore, Fractional part = \(\frac{1}{8}:\frac{2}{8}:\frac{5}{8}\)

=> Percentage of parts = \(\frac{1}{8}\times 100:\frac{2}{8}\times 100:\frac{5}{8}\times 100\)

=> 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 = \(\frac{Population Decreased}{Original Population}\times 100\)

= \(\frac{500}{25000}\times 100\) = 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 = \(\frac{Amount of change}{Original Amount}\times 100\)

\(\frac{20000}{350000}\times 100 = 5\frac{5}{7}\) %

Hence, The percentage of price increased is \(5\frac{5}{7}\) %

 

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

= \(\frac{20}{100}\times 10000\)

= 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 = \(\frac{20}{100}\times x=\frac{x}{5}\)

Therefore, SP = CP – Loss

\(\\=>13500=x-\frac{x}{5}\\=>13500=\frac{4x}{5}\\=>x=\frac{13500\times 5}{4}=Rs16,875\)

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 = \(\frac{3}{25}\)

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

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

Let the weight of chalk be x g.

Then, 12 % of x = 3

=> \(\frac{12}{100}\) x X = 3

=> x = \(\frac{3\times 100}{12}\) = 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

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

\(Simple interest = \frac{P \times R\times T}{100} = \frac{1200\times 12\times 3}{100}\)

=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

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

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

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

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

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

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

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

=> P = Rs 500

Hence, she had borrowed Rs 500.



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2 Comments

  1. Samarth barsainya Samarth barsainya
    April 26, 2018    

    Book mai 1.3 mai 2 question verify the following ka 2 question

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