Class 8 RD Sharma Solutions - Chapter 12 Percentage - Exercise 12.2 | Set 1

Chapter 12 of RD Sharma’s Class 8 Mathematics textbook focuses on the concept of percentages a fundamental topic in mathematics that finds wide application in daily life. Exercise 12.2 | Set 1 aims to help students develop a strong understanding of how to calculate percentages convert fractions and decimals into percentages and apply percentage calculations in the various practical scenarios.

What is Percentage?

A percentage is a way of expressing a number as a fraction of 100. The term "percent" comes from the Latin phrase "per centum" which means "by the hundred". Percentages are used to compare quantities, express ratios, and describe changes in values. For example: if a student scores 75% on a test it means they answered 75 out of every 100 questions correctly. Percentages are used in various real-life situations such as calculating discounts, interest rates, and statistics.

Question 1. Find:
(i) 22% of 120
(ii) 25% of Rs 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 meter
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml

Solution:

(i) 22% of 120

22% of 120 can be termed as (22/100) X 120 = 26.4

(ii) 25% of Rs 1000

25% of Rs 1000 can be termed as (25/100) X 1000 = Rs 250

(iii) 25% of 10 kg

25% of 10 kg can be termed as (25/100) X 10 = 2.5 kg

(iv) 16.5% of 5000 meter

16.5% of 5000 meter can be termed as (16.5/100) X 5000 = 825 meter

(v) 135% of 80 cm

135% of 80 cm can be termed as (135/100) X 80 = 108 cm

(vi) 2.5% of 10000 ml

2.5% of 10000 ml can be termed as (2.5/100) X 10000 = 250 ml

Question 2. Find the number ‘a’, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) 1/2 % of a is 50
(iv) 100% of a is 100

Solution:

(i) Given, 8.4% of a is 42

So, (8.4/100) × a = 42

a = 4200/8.4 = 500

Hence, a = 500

(ii) Given, 0.5% of a is 3

So, (0.5/100) × a = 3

a = 300/0.5 = 600

Hence, a = 600

(iii) Given, 1/2 % of a is 50

So, (0.5/100) × a = 50

a = 5000/0.5 = 10000

Hence, a = 10000

(iv) Given, 100% of a is 100

So, (100/100) × a = 100

a = 100

Hence, a = 100

Question 3. x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.

Solution:

Given, x is 5% of y

y is 24% of z

x = 480

Since, x = 5% of y

We can write, 480 = 5% of y

So, 480 = (5/100) × y

y = 9600

Also, y = 24% of z

So, 9600 = (24/100) × z

z = 40000

Hence, y is 9600 and z is 40000

Question 4. A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income.

Solution:

Let a be the monthly income of coolie

Given, a coolie deposits Rs. 150 per month in his post and this is equal to 15% of his monthly income

So, 15 % of a = Rs 150

(15/100) × a = 150

a = Rs 1000

Hence, the monthly income of coolie is Rs 1000 

Question 5. Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination.

Solution:

Let a be the total number of marks in the Exam 

Given, marks scored by Asha is 695 and this is 86.875% marks

So, 86.875% of a = 695

(86.875/100) × a = 695

a = 800 marks

Hence, the examination is of 800 marks

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?

Solution:

Given, Number of days Deepti went to school is 216 and this is 90 % of her attendance

Let a be the total number of days, the school was opened.

So, 90 % of a = 216

(90/100) × a = 216

a = 240 days

Hence, the school was opened for 240 days

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.

Solution:

Given, the total number of trees = 2000

12% of these are mango trees, 18% lemon and the rest are orange trees

So, 12 % of 2000 are mango trees

12 % of 2000 = Number of Mango trees 

Number of Mango trees = (12/100) × 2000 = 240 trees

Number of Lemon trees = 18 % of 2000 = (18/100) × 2000 = 360 trees

And, the number of Orange trees = Total number of trees — (Number of Mango trees+ Number of Lemon trees)

= 2000 - (240 + 360) = 1400

Hence, there are 1400 orange trees in the garden

Question 8. Balanced diet should contain 12% of protein, 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.

Solution:

Given, total amount of calories needed = 2600

So, the amount of Protein needed = 12 % of 2600

= (12/100) × 2600 = 312 calories

The amount of Fats needed = 25 % of 2600

= (25/100) × 2600 = 650 calories

The amount of Carbohydrate needed = 63 % of 2600

= (63/100) × 2600 = 1638 calories 

Hence, the amount of calories needed in Protein, Fats and Carbohydrate is 312, 650 and 1638 calories respectively

Question 9. A cricketer scored a total of 62 runs in 96 balls. He hits 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

Solution:

Given, a cricketer scored a total of 62 runs in 96 balls

i) The total runs scored in form of Sixes = 3 × 6 = 18

So, percentage of runs scored in form of Sixes = (18/62) × 100 = 29.03%

ii) The total runs scored in form of Fours = 8 × 4 = 32

So, percentage of runs scored in form of Fours = (32/62) × 100 = 51.61%

iii) The total runs scored in form of Twos = 2 × 2 = 4

So, percentage of runs scored in form of Twos = (4/62) × 100 = 6.45%

iv) The total runs scored in form of Singles = 1 × 8 = 8

So, percentage of runs scored in form of Singles = (8/62) × 100 = 12.9%

Question 10. A cricketer hits 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?

Solution:

Given, a cricketer hits 120 runs in 150 balls

i) Number of runs scored in form of 6's = 20 % of 120

= (20/100) × 120 = 24 runs

ii) Number of runs scored in form of 4's = 30 % of 120

= (30/100) × 120 = 36 runs

iii) Number of runs scored in form of 2's = 25 % of 120

= (25/100) × 120 = 30 runs

iv) Number of runs scored in form of singles = 120 - (24 + 36 + 30) = 30 runs

Now, the percentage of runs scored in singles = (30/100) × 120 = 25%

Hence, 25 % of runs scored in form of singles

Question 11. Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest?

Solution:

Given, the amount of money Radha earned through investment = Rs 187

Let a be the amount of money she invested

So, we can say that 22 % of a = 187

(22/100) × a = 187

a = Rs 850

Hence, Radha invested Rs 850  

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

Solution:

Given, Rohit deposits 12% of his income in a bank. 

Let a be the total income of Rohit in year 1997

So, we can say 12 % of a = 1440

(12/100) × a = 1440

a = Rs 12000

Hence, the total income of Rohit in year 1997 is Rs 12000

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?

Solution:

Given, Gunpowder contains 75% nitre and 10% sulphur

Let a be the total amount of gunpowder

i) So, we can say 75 % of a = 9 kg Nitre

(75/100) × a = 9

a = 12 kg

Hence, 12 kg gunpowder carries 9 kg nitre

ii) So, we can say 10 % of a = 2.3 kg Sulphur

(10/100) × a = 2.3

a = 23 kg

Hence, 23 kg gunpowder carries 2.3 kg sulphur

Chapter 12 Percentage - Exercise 12.2 | Set 2

Conclusion

Understanding percentages is crucial for solving a variety of the mathematical problems and for the making informed decisions in everyday life. Exercise 12.2 | Set 1 in Chapter 12 helps students practice and master the calculation and application of the percentages laying the groundwork for the more advanced mathematical concepts in the future.