Chapter 12 - Percentage (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 12.9:

Question 1:

Find:
 (i) 22% of 120
 (ii) 25% of Rs 1000
 (iii) 25% of 10 kg
 (iv) 16.5% of 5000 metre
 (v) 135% of 80 cm
 (vi) 2.5% of 10000 ml

ANSWER:

(i) 22% of 120 = 22/100 × 120

= 26.4

(ii) 25% of 1000 = 25/100 × 1000

= 250

(iii) 25% of 10 kg = 25/100 × 10 kg

= 25/10kg

= 5/2kg

= 2.5 kg

(iv) 16.5% of 5000 m

= 16.5/100 × 5000 m

= 16.5 × 50 m

= 825 m

(v) 135% of 80 cm

= 135/100 × 80 cm

= (135 × 8)/10cm

= 108 cm

= 1.08 m

(vi) 2.5% of 10000 ml 

=  2.5/100 × 10000 ml

= 250 ml

PAGE NO 12.9:

Question 2:

Find the number a, if
 (i) 8.4% of a is 42
 (ii) 0.5% of a  is 3
 (iii) 12% of a is 50
 (iv) 100% of a is 100

ANSWER:

(i) 8.4 % of a is 42

.i.e., 8.4/100 × a = 42

⇒ a = (42 × 100)/8.4

∴ a = 500

(ii) 0.5% of a is 3

.i.e., 0.5/100 × a = 3

⇒ a = (3 × 100)/0.5

∴ a = 600

(iii) 12% of a is 50

.i.e., 12 × 1/100 × a = 50

⇒ a = 50 × 200

∴ a = 10000

(iv) 100% of a is 100

.i.e., 100/100 × a = 100

⇒ 1 × a = 100

∴ a = 100

PAGE NO 12.9:

Question 3:

x is 5% of y, y is 24% of z. If x  = 480, find the values of y and z.

ANSWER:

Given: x = 5/100y 

⇒ y = 100/5x

∴ y = 20x

Also, y = 24/100z

∴ z = 100/24y

Putting x = 480 in the above equations, we get:

y = 20(480) = 9600

z = 100/24 × 9600 = 40000

PAGE NO 12.9:

Question 4:

A coolie deposits Rs 150 per month in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income.

ANSWER:

Let his monthly income be Rs x .

Then, 15% of x  =  150

⇒15/100x = 150

⇒x = (150 × 100)/15 = 1000

∴ His monthly income is Rs 1000.

PAGE NO 12.9:

Question 5:

Asha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.

ANSWER:

Let x be the total number of marks of the examination.

Then, 86.875% of x = 695

⇒ 86.875/100x = 695

⇒ x = (695 × 100)/86.875 = 800

∴ The total number of marks of the examination is 800.

PAGE NO 12.9:

Question 6:

Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened.

ANSWER:

Let the school opened for x days in the given year.

We have:90% of x  = 216

⇒ 90/100x = 216

⇒x = (216 × 100)/90 = 240

∴ The school opened for 240 days in the given year.

PAGE NO 12.9:

Question 7:

A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.

ANSWER:

Let the number of orange trees be x.It is given that there are 2,000 trees.

 While 12% of them are mango trees, 18% are lemon trees and the rest are orange trees.

Now, (12% of 2000)  + (18% of 2000) + x = 2000

⇒ 12/100 × 2000 + 18/100 × 2000 + x = 2000

⇒ 240 + 360 + x = 2000

⇒ 600 + x = 2000

⇒ x = 2000−600

⇒ x = 1400

∴ There are 1,400 orange trees.

PAGE NO 12.10:

Question 8:

Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.

ANSWER:

In a balanced diet of 2600 calories, 12% is protein.

∴ Amount of protein in food intake  =  12% of 2600

= 12/100 × 2600

= 312

Similarly, a balanced diet contains 25% fats.

∴ Amount of fats in food intake  =  25% of 2600

= 25/100 × 2600

= 650

Similarly, a balanced diet contains 63% carbohydrates.

∴ Amount of carbohydrates in food intake  =  63% of 2600

= 63/100 × 2600

= 1638 

PAGE NO 12.10:

Question 9:

A cricketer scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in
 (i) Sixes
 (ii) fours
 (iii) twos
 (iv) singles

ANSWER:

(i) The cricketer hits 3 sixes.i.e., 3 × 6 = 18

∴ 18/62 × 100

= 29.03%

(ii) The cricketer hits 8 fours.i.e., 8 × 4 = 32

∴ 32/62 × 100

= 51.61%

(iii) The cricketer hits 2 twos.i.e., 2 × 2 = 4

∴ 4/62 × 100

= 6.45%

(iv) The cricketer hits 8 singles.i.e., 8 × 1 = 8

∴ 8/62 × 100

= 12.90%

PAGE NO 12.10:

Question 10:

A cricketer hit 120 runs in 150 balls during a test match. 20% of the runs came in 6's, 30% in 4's, 25% in 2's and the rest in 1's. How many runs did he score in
 (i) 6's
 (ii) 4's
 (iii) 2's
 (iv) singles
 What % of his shots were scoring ones?

ANSWER:

(i) Let the cricketer score w runs in 6's

∴ 20% of 120  = w

⇒ 20/100 × 120 = w

⇒ w = 24

(ii) Let the cricketer score x runs in 4's.

∴ 30% of 120  = x

⇒ 30/100 × 120 = x

⇒ x = 36

(iii) Let the cricketer score y runs in 2's.

∴ 25% of 120  = y

⇒ 25/100 × 120 = y

⇒ y = 30

(iv) Let the cricketer score z runs in 1's.

Since, the cricketer scores the rest of the runs in 1's, we have:24 + 36 + 30 + z = 120

⇒ z + 90 = 120

⇒ z = 30 

AGE NO 12.10:

Question 11:

Radha earns 22% of her investment. If she earns Rs 187, then how much did she invest?

ANSWER:

Let the investment be Rs x.i.e., 22% of x  =  187

⇒22/100x = 187

⇒ x = 18700/22 = 850

∴ Radha invests Rs 850.

PAGE NO 12.10:

Question 12:

Rohit deposits 12% of his income in a bank. He deposited Rs 1440 in the bank during 1997. What was his total income for the year 1997?

ANSWER:

Let the total income for the year 1997 be Rs x.

i.e., 12% of x  = 1440

⇒ 12/100x = 1440

⇒ x = 144000/12

⇒ x = 12000

∴ Rohit's total income was Rs 12,000.

PAGE NO 12.10:

Question 13:

Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur?

ANSWER:

Let the amount of gunpowder that contains 9 kg nitre be x kg.

i.e., 75/100 ×  x = 9

⇒ 100/75 = x9

⇒ x = 900/75

⇒ x = 12

∴ The amount of gunpowder containing 9 kg nitre is 12 kg.

Let the amount of gunpowder containing 2.3 kg sulphur be y kg. 

i.e., 10/100 × y = 2.3

⇒ 100/10 = y/2.3

⇒ y = 230/10

⇒ y = 23

∴ The amount of gunpowder containing 2.3 kg sulphur is 23 kg.

PAGE NO 12.10:

Question 14:

An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy?

ANSWER:

Composition of the alloy  =  15 parts of tin  +  105 parts of copper = 120 parts

∴ Percentage of tin =  15/120 × 100 = 12.5%

Also, percentage of copper =  105/120 × 100 = 87.5%

PAGE NO 12.10:

Question 15:

An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.

ANSWER:

Percentage of copper in the alloy = 32

Percentage of nickel in the alloy = 40

Percentage of zinc in the alloy = 100−32−40 = 28

∴ Amount of zinc in 1 kg of the alloy  =  (0.28 × 1) kg = 280 gm 

PAGE NO 12.10:

Question 16:

A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride?

ANSWER:

Let the length of total ride be x km.Then, 10% of x  =  122

⇒ 10/100x = 122

⇒ x = 1220

∴ The length of total ride is 1,220 km.

PAGE NO 12.10:

Question 17:

A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female?

ANSWER:

Total number of female students  =  300−142 = 158

Total number of female teachers  =  30−12 = 18

Total number of females =  158 + 18 = 176

Total population of the school =  300 + 30 = 330

Percentage of females  =  176/330 × 100 = 160/3% 

or 53.33%

PAGE NO 12.10:

Question 18:

Aman's income is 20% less than that of Anil. How much percent is Anil's income more than Aman's income?

ANSWER:

Let Anil's income be x.

Then, Aman's income  =  x−20x/100 = 8x/10

Difference in the incomes of Anil and Aman  =  x−8x/10

= 2x/10

∴ Percentage of the difference in the incomes of Anil and Aman to that of Aman's income  =  (2x/10)/(8x/10) × 100

= 1/4 × 100

= 25%

PAGE NO 12.10:

Question 19:

The value of a machine depreciates every year by 5%. If the present value of the machine be Rs 100000, what will be its value after 2 years?

ANSWER:

It is given that the value of the machine depreciates by 5% every year.

Present value of the machine  =  Rs 100000

∴ For the first year, 5% of 100000  =  5/100 × 100000 = Rs 5000

∴ Value of the machine after one year  =  100000−5000 = Rs 95000

Value of the machine in the second year  =  Rs 95000

∴ 5% of 95000 = 5/100 × 95000 = Rs 4750

∴ Value of the machine after 2 years  = 95000−4750 = Rs 90250

∴ After two years, the value of the machine will be Rs 90,250.

PAGE NO 12.10:

Question 20:

The population of a town increases by 10% annually. If the present population is 60000, what will be its  population after 2 years?

ANSWER:

Present population  =  60000

It increases by 10% annually

∴ Increase in the population in the first year  =  10% of 60000  = 10/100 × 60000

= 6000

∴ Population after 1 year  =  60000 + 6000

= 66000

Increase in the population in the second year, 10% of 66000  = 10/100 × 66000

= 6600

Thus, population after 2 years  =  66000 + 6600

= 72600

PAGE NO 12.10:

Question 21:

The population of a town increases by 10% annually. If the present population is 22000, find its population a year ago.

ANSWER:

Let the population of the town one year ago  =  x

Now, it is given that population of the town increases by 10% annually.

Present population  =  x  +  10% of x

= x  +  10/100 × x

= x  +  x/10

= 11x/10

But present population of the town  =  22000

So, 11x/10  =  22000

⇒ 11x  =  220000

⇒ x  =  220000/11

⇒ x  =  20000

So, population of the town 1 year ago  =  20000

PAGE NO 12.10:

Question 22:

Ankit was given an increment of 10% on his salary. His new salary is Rs 3575. What was his salary before increment?

ANSWER:

Let the initial salary be Rs x.

We know that:Salary before increment  +  increment given on salary  =  new salary

∴ x  +  10% of x  =  3575

⇒ x + 10/100x = 3575

⇒ 100x + 10x = 357500

⇒ 110x = 357500

⇒ x = 357500/110

⇒ x = 3250

∴ Salary before increment  =  Rs 3,250

PAGE NO 12.10:

Question 23:

In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase?

ANSWER:

We have to reduce the consumption such that the expenditure does not increase. 

For this, we use the following formula:

(r/(r + 100)) × 100, where r  =  percentage rise in the price of the commodity 

∴ Percentage reduction in the consumption  =  (10/(10 + 100)) × 100 = (10/110) × 100    (∵ r  =  10%) = 9(1/11)

PAGE NO 12.10:

Question 24:

Mohan's income is Rs 15500 per month. He saves 11% of his income. If his income increases by 10%, then he reduces his saving by 1%, how much does he save now?

ANSWER:

Mohan's saving  =  11% of 15500

= 11/100 × 15500

= Rs. 1705

Mohan's saving  =  11% of 15500

= 11/100 × 15500

= Rs. 1705

It is given that Mohan's income increases by 10%.

 ∴ Increase in income  = 10/100 × 15500 = Rs 1550

Increased income  = Rs (15500 + 1550) = Rs 17050

Now, percentage of saving = (11−1) = 10%

∴ Saving  =  10/100 × 17050 = Rs 1705

Thus, the amounts of his present and earlier savings are the same.

PAGE NO 12.10:

Question 25:

Shikha's income is 60% more than that of Shalu. What percent is Shalu's income less than Shikha's?

ANSWER:

Let Shalu's income be Rs x.

∴ Shikha's income = Rs (x + (60x/100)) = Rs 160x/100 = Rs 16x/10

Difference in the incomes of Shikha and Shalu  = 16x/10−x = (16x−10x)/10 = Rs 6x/10

Percentage of the difference in the incomes of Shikha and Shalu to that of Shikha's income      = 37.5%

∴ The income of Shalu is less than that of Shikha by 37.5%.

PAGE NO 12.10:

Question 26:

Rs 3500 is to be shared among three people so that the first person gets 50% of the second, who in turn gets 50% of the third. How much will each of them get?

ANSWER:

Let x, y and z be the amounts received by the first, second and the third person, respectively.

We have:x = 50% of y 

= 50/100.y = 1/2.y

∴ y = 2x

Again, y = 50% of z

= 1/2.z

∴ z = 2y = 2(2x) = 4x

Also, x + y + z = 3500

Substituting the values of z and y, we get:

x + 2x + 4x = 3500

⇒ 7x = 3500

⇒ x = 500

∴ y = 2x = 2(500)

= 1000

∴ z = 4x = 4(500)

= 2000

Thus, the amounts received by the first, second and the third person are Rs 500, Rs 1000 and Rs 2000, respectively. 

PAGE NO 12.10:

Question 27:

After a 20% hike, the cost of Chinese Vase is Rs 2000. What was the original price of the object?

ANSWER:

Let the original price of the object be x.

According to the question, we have:

20%of x  + x  =  2000

⇒ 20x/100 + x = 2000

⇒ 120/100.x = 2000

⇒ x = 200000/120

⇒ x = 1666.67

Thus, the original price of the object was Rs 1,666.67.