One of the interesting chapters in RD Sharma is Percentage. In Exercise 12.2 of Chapter 12, we shall discuss problems based on “Finding a percentage of a number”. The solutions given here are explained in simple and easily understandable language by the subject experts at BYJU’S, keeping in mind the latest CBSE guidelines. Students who are preparing for the annual exam and who wish to present their answers without any mistakes can refer to RD Sharma Solutions, which is formulated by our subject experts. Students can easily download the PDF from the links given below.
RD Sharma Solutions for Class 8 Maths Exercise 12.2 Chapter 12 Percentage
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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
Solution:
(i) 22% of 120
Here, 22% of 120 can be expressed as (22 × 120) / 100 = 2640/100 = 26.40
(ii) 25% of Rs 1000
Here, 25% of Rs 1000 can be expressed as (25 × 1000) / 100 = 25000/100 = Rs 250
(iii) 25% of 10 kg
Here, 25% of 10 Kg can be expressed as (25 × 10) / 100 = 250/100 = 2.5 Kg
(iv) 16.5% of 5000 metre
Here, 16.5% of 5000 metre can be expressed as (16.5 × 5000) / 100 = 16.5 × 50 = 825 m
(v) 135% of 80 cm
Here, 135% of 80 cm can be expressed as (135 × 80) / 100 = (135 × 4) / 5 = 108 cm
(vi) 2.5% of 10000 ml
Here, 2.5% of 10000 ml can be expressed as (2.5 × 10000) / 100 = 25000/100 = 250 ml
2. Find the number a, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) ½% of a is 50
(iv) 100% of a is 100
Solution:
(i) 8.4% of a is 42
Here, 8.4% of a is 42 can be expressed as (8.4 × a) / 100 = 42
Which implies, a = (42 × 100) / 8.4 —— (1)
Here, 8.4 can been written as 84/10
Substituting the above in (1)
a = (42 × 100 × 10) / 84
a = 42000/84
a = 500
(ii) 0.5% of a is 3
Here, 0.5 of a is 3 can be expressed as (0.5 × a) / 100 = 3
Which implies, a = (3 × 100) / 0.5 —— (1)
Here, 0.5 can been written as 5/10
Substituting the above in (1)
a = (3 × 100 × 10) / 5
a = 3000/5
a = 600
(iii) ½% of a is 50
Here, 0.5 of a is 50 can be expressed as (0.5 × a) / 100 = 50
Which implies, a = (50 × 100) / 0.5 —— (1)
Here, 0.5 can been written as 5/10
Substituting the above in (1)
a = (50 × 100 × 10) / 5
a = 50000/5
a = 10000
(iv) 100% of a is 100
Here, 100% of a is 100 can be expressed as (100 × a) / 100 = 100
Which implies, a = (100 × 100) / 100
a = 10000/100
a = 100
3. x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.
Solution:
Given value x = 480
And x is 5% of y
Here, x is 5% of y can be expressed as (5 × y) / 100 = x
Which implies, x = y × (5/100)
Substituting x = 480 in the above equation we get,
480 = y × (5/100)
Which implies, y = (480 × 100) / 5
y = 48000/5
y = 9600
It is also given that, y is 24% of z
Which implies, y = z × (24/100)
Substituting y = 9600 in the above equation we get,
9600 = z × (24/100)
Which implies, z = (9600 × 100) / 24
z = 96000/24
z = 40000
∴ y = 9600 and z = 40000
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.
Solution:
Let the monthly income of a coolie be ‘x’
Here, coolie deposits Rs 150 per month which is 15% of his monthly income
From the above we can derive that,
x × (15/100) =150
Which implies, x = (150 × 100) / 15
x = 15000/15
x = 1000
∴ Monthly income is Rs 1000
5. Asha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.
Solution:
Given, Marks scored by Asha is 695
And Percentage of marks Asha got is 86.875%
Now,
Let the total marks be ‘x’
From the above we can derive that,
x × (86.875/100) =695
Here, 86.875 can be expressed as 86875/1000
Which implies, x = (695 × 100 × 1000) / 86875
x = 6950000/86875 = 800
Total no: of Marks, x = 800
∴ Total number of marks is 800marks
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 = 216 days
Deepti Attendance percentage is = 90%
So, let the number of days when school remained opened be x days
Hence,
(x × 90)/100 = 216
By using cross multiplication we get,
x = (216×100)/90
= 240 days
∴ Number of days the school remained opened for 240 days
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 details are,
Total number of trees = 2000
Number of mango trees = 12% of 2000
= (12/100) × 2000
= 240 trees
Number of lemon trees = 18% of 2000
= (18/100) × 2000
= 360 trees
Number of orange trees = 2000 – (Number of mango trees+ Number of lemon trees)
= 2000 – (240+360)
= 2000 – 600
= 1400 trees
∴ Number of orange trees are 1400 trees
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.
Solution:
The given details are,
Amount of calorie daily needed = 2600 calorie
Amount of protein needed = 12% of 2600
= (12/100) × 2600
= 312 calorie
Amount of fats needed = 25% of 2600
= (25/100) × 2600
= 650 calorie
Amount of carbohydrate needed = 63% of 2600
= (63/100) × 2600
= 1638 calorie
∴ Amount of calories required in protein is 312 calories, fat is 650 calories and carbohydrates is 1638 calories
9. A cricketer scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 two’s and 8 singles. What percentage of the total runs came in
(i) Sixes
(ii) 4’s
(iii) 2’s
(iv) Singles
Solution:
The given details are,
Total runs scored by cricketer = 62 runs
(i) Runs scored in 3 sixes = 3×6
= 18
Percentage of runs scored in sixes = (18/62) × 100
= 29.03%
(ii) Runs scored in 8 fours = 8×4
= 32
Percentage of runs scored in fours = (32/62) × 100
= 51.61%
(iii) Runs scored in 2 two’s = 2×2
= 4
Percentage of runs scored in two’s = (4/62) × 100
= 6.45%
(iv) Runs scored in singles = 8 × 1
= 8
Percentage of runs scored in singles = (8/62) × 100
= 12.9%
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?
Solution:
The given details are,
Total number of runs scored by cricketer = 120
(i) 20% of Runs scored in 6’s = (20/100) × 120 = 24 runs
(ii) 30% of Runs scored in 4’s = (30/100) × 120 = 36 runs
(iii) 25% of Runs scored in 2’s = (25/100) × 120 = 30 runs
(iv) Runs scored in singles = 120 – (24+36+30)
= 120 – 90 = 30 runs
Percentage of shots scoring ones = (Runs came in singles/Total runs scored) × 100
= (30/120) × 100
= 25%
11. Radha earns 22% of her investment. If she earns Rs 187, then how much did she invest?
Solution:
Given, percentage Radha earns = 22% of investment
So, let us consider total investment be Rs x
By calculating, (x/100) × 22 = 187
By cross multiplying we get,
(x/100) = 187/22
x = (187×100)/22
= 850
∴ Radha’s total investment is Rs 850
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?
Solution:
The given details are,
Percentage Rohit deposited in bank = 12% of total income
Rohit deposited money during the year 1997 = Rs 1440
So, let us consider the total income of Rohit as Rs x
By calculating,
(x/100) × 12 = 1440
By cross multiplying
x = (1440×100)/22
= 12000
∴ Rohit’s total income for the year 1997 is Rs 12000
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 details are,
Percentage of nitre in gunpowder = 75%
Amount of nitre in gunpowder = 9 kg
Let us consider the amount of gunpowder be ‘x’ kg
So, by calculating (x/100) × 75 = 9
By cross multiplying
x/100 = 9/75
x = (9×100)/75
= 12kg
Percentage of sulphur in gunpowder = 10%
Amount of sulphur in gunpowder = 2.3 kg
Let us consider amount of gunpowder be ‘x’ kg
So, by calculating (x/100) × 10 = 2.3
By cross multiplying
x/100 = 2.3/10
x = (2.3×100)/10
= 23kg
∴ The amount of gunpowder in nitre is 12kg
The amount of gunpowder in sulphur is 23kg
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?
Solution:
Given details are,
Amount of tin in an alloy = 15 parts
Amount of copper in an alloy = 105 parts
So, total weight of alloy = 15 + 105 = 120 parts
Now, by calculating
Percentage of copper in alloy = (105/120) × 100
= 525/6
= 87.50%
∴ Percentage of copper in an alloy is 87.50%
15. An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.
Solution:
Given details are,
Alloy contains, 32% of copper
40% of nickel
Remaining zinc
Mass of alloy = 1kg = 1000 grams
Mass of copper in alloy = (1000/100) × 32
= 320 grams
Mass of nickel in alloy = (1000/100) × 40
= 400 grams
So, mass of zinc in alloy = 1000 – (320 + 400)
= 1000 – 720
= 280 grams
∴ Mass of zinc in 1kg of alloy is 280 grams
16. A motorist travelled 122 kilometers before his first stop. If he had 10%of his journey to complete at this point, how long was the total ride?
Solution:
Given details are,
Motorist total distance travelled before first stop = 122 km
Journey completed at first stop = 10 %
Let us consider total ride to be travelled be ‘x’ km
So, by calculating
(x/100) × 10 = 122
By cross multiplying we get,
x/100 = 122/10
x = (122 × 100)/10
= 1220 km
∴ Motorist total ride is 1220 km
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?
Solution:
The given details are,
In a school, number of students are = 300
Number of boys = 142
Number of girls = 300 – 142 = 158
In a school, number of teachers are = 30
Number of male teachers are = 12
Number of female teachers are = 30 – 12 = 18
Total number of students and teachers is = 300+30 = 330
Total numbers of female in the school is = 158+18 = 176
Percentage of female in the school = (176/330) × 100
= 160/3%
∴ Total of 160/3% are female in the school.
18. Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income?
Solution:
Given Aman’s income is 20% less than Anil’s income
Let us consider Aman’s and Anil’s income as Rs x
Aman’s income = x – x × (20/100)
= x – x × (1/5)
= x – x/5
= (5x-x)/5
= 4x/5
Let us find the difference between Anil’s and Aman’s income = x – 4x/5
= (5x-4x)/5
= x/5
When, Anil’s income is more than Aman’s income the percentage is = (x/5)/(4x/5) × 100
= 25%
∴ 25% of Anil’s income is more than Aman’s income.
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?
Solution:
Given details are,
Present value of machine is = Rs 100000
Every year the depreciation in price is = 5%
So, value after two years = 100000 × (100-5)/100 × (100-5)/100
= 100000 × 95/100 × 95/100
= 90250
∴ Value of machine after two years is Rs 90250
20. The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years?
Solution:
Given details are,
Present population of town is = 60000
Annually population increases by = 10%
So, Population after 2 years = present population × [(100 + increased %)/100] years
= 60000 × (100+10)/100 × (100+10)/100
= 60000 × 110/100 × 110/100
= 60000 × 11/10 × 11/10
= 72600
∴ After 2 years population will be 72600
21. The population of a town increases by 10% annually. If the present population is 22000, find its population a year ago.
Solution:
Given details are,
Present population of town is = 22000
Let the population of town be 100 a year ago.
Then,
Annual population increase is = 10% of 100 = 10
The present population = 100 + 10 = 110
If present population is 110, population year ago = 100
If present population is 1, population year ago = 100/110
If present population is 22000, population year ago = 100/110 × 22000
= 10/11 × 22000
= 20000
∴ 1 year ago population was 20000
22. Ankit was given an increment of 10% on his salary. His new salary is Rs 3575. What was his salary before increment?
Solution:
Let the salary of Ankit before increment be = Rs x
New salary of Ankit = Rs 3575
Increase in salary is = 10% of 100 = 10
Present salary = 100 + 10 = 110
So, Salary of Ankit before increment is x × 110/100 = 3575
By calculating for x we get,
x × 110 = 3575 × 100
x = (3575 × 100)/110
= 3250
∴ Salary of Ankit before increment is Rs 3250
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?
Solution:
Given details are,
Increase in petrol price by = 10%
Reduction in consumption while having same expenditure =
(increase%)/(100+increase%) × 100
= 10/(100+10) × 100
= 1000/110
= 100/11
=
∴ at the cost of same expenditure one can reduce
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?
Solution:
Mohan monthly income is = Rs 15500
Mohan savings is = 11% of 15500
= 15500 × 11/100
= Rs 1705
Monthly income increases by = 10%
New monthly income is = 15500 + 10/100 × 15500
= 15500 + 1550
= Rs 17050
When savings reduced by 1% will result in = 11 – 1 = 10% of 17050
New savings = (10/100) × 17050
= Rs 1705
∴ Savings is Rs 1705, which remains the same even after increment.
25. Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s?
Solution:
Let us consider Shikha’s and Shalu’s income be Rs x
So, Shikha’s income is 60% more of Shalu’s = x + x × 60/100
= x + 3x/5
= (5x+3x)/5
= 8x/5
Difference between Shikha’s and Shalu’s income will be = 8x/5 – x
= (8x-5x)/5
= 3x/5
When Shalu’s income is less than Shikha’s income (in %) = (3x/5)/(8x/5) × 100
= 3x/8x × 100
= 300/8
= 37.5%
∴ By 37.5%, Shalu’s income is less than Shikha’s income.
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?
Solution:
We know that the total money to be shared is = Rs 3500
Let us consider third person get = Rs x
So, second person gets (50% of third) = 50% of x
= 50/100 × x
= Rs x/2
Now, first person gets (50% of second) = 50% of x/2
= 50/100 × x/2
= Rs x/4
We know that,
x/4 + x/2 + x = 3500
by taking 4 as LCM
(x+2x+4x)/4 = 3500
By cross multiplying
x+2x+4x = 3500 × 4
7x = 14000
x = 14000/7
= 2000
∴ Each of the person gets,
First person (x/4) gets = x/4 = 2000/4 = Rs 500
Second person (x/2) gets = x/2 = 2000/2 = Rs 1000
Third person (x) gets = x = Rs 2000
27. After a 20% hike, the cost of Chinese Vase is Rs 2000. What was the original price of the object?
Solution:
Let cost price of Chinese Vase before hike be = Rs x
The hike is = 20% of 100 = 20/100
The cost price of Chinese Vase after hike is = Rs 2000
So, let’s calculate for x,
x + x×20/100 = 2000
x + x/5 = 2000
(5x+x)/5 = 2000
6x = 2000×5
x = 10000/6
= 1666.6667
∴ Original price of Chinese Vase is = Rs. 1666.67
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