Solution RD Sharma Maths Class 8 Percentage Exercise 12.2
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 1
Percentage of
a) 22% of 120
b) 25% of 1000
c) 25% of 10 kg
d) 16.5 % of 5000 metre
e) 135 % of 80 cm
f) 2.5 % of 10000 ml
Solution:
a) 22% of 120 = \( {22 \over 100} \times 120 = 26.4 \)
b) 25% of 1000 = \( {25 \over 100} \times 1000 = 250 \)
c) 25% of 10 kg = \( {25 \over 100} \times 10 = 2.5 \)
d) 16.5 % of 5000 metre = \( {16.5 \over 100} \times 5000 = 825 meter \)
e) 135 % of 80 cm = \( {135 \over 100} \times 80 = 108 cm \)
f) 2.5 % of 10000 ml = \( {2.5 \over 100} \times 10000 = 250 cm \)
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 2
a) 8.4% of a is 42
b) 0.5% of a is 3
c) (1/2) % of a is 50
d) 100% of a is 100
Solution:
a) 8.4% of a is 42 = \( {8.4 \over 100} \times x = 42 \)
\( \implies x = {42 \times 100 \over 8.4} = 500 \)
b) 0.5% of a is 3 = \( {0.5 \over 100} \times x = 3 \)
\( \implies x = {3 \times 100 \over 0.5} = 600 \)
c) (1/2) % of a is 50 = \( {{1 \over 2} \over 100} \times x = 50 \)
\( \implies x = {50 \times 100 \over {1 \over 2}} = 10000 \)
d) 100% of a is 100 = \( {100 \over 100} \times x = 100 \)
\( \implies x = {100 \times 100 \over 100} = 100 \)
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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
\( \implies x= {5 \over 100} \times y \)
\( \implies y = {100 \over 5} \times x \)
\( \implies y = 20 \times x \)
\( \implies y = 20 \times 480 \)
\( \implies y = 9600 \)
and
y is 24% of z
\( \implies y= {24 \over 100} z \)
\( \implies z= {100 \over 24} y \)
\( \implies z = {100 \over 24} \times 9600 \)
\( \implies z = 40000 \)
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 4
A coolie deposits Rs.150 per month in his post office savings bank account. If this is 15% of this monthly income, find his monthly income.
Solution:
Let his monthly income be x
Given in question,
15 % of x = 150
\( \implies {15 \over 100} \times x = 150 \)
\( \implies x = 150 \times {100 \over 15} \)
\( \implies x = 1000 \)
Monthly income of coolie is Rs.1000
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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.
Solution:
Let x be the total marks in the examination
According to the question,
86.875% of x = 695
\( \implies {86.875 \over 100} \times x = 695 \)
\( \implies x = 695 \times {100 \over 86.875} \)
\( \implies x = 800 \)
The total marks in the examination is 800.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 6
Deepti went to school for 216 days in a full year. If her attendance is 90%, find her number of days on which the school was opened?
Solution:
Let the school was opened for x days
According to the question,
90 % of x = 216
\( \implies {90 \over 100} \times x = 216 \)
\( \implies x = 216 \times {100 \over 90} \)
\( \implies x = 240 \)
The school was opened for 240 days in the given year.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 7
A garden has 2000 trees. 12% of these are mango trees, 18% are lemon and the rest are orange trees. Find the number of orange trees?
Solution:
Let the number of orange trees be x
There are total 2000 trees.
Given in question,
Number of mango trees = 12% of total trees = 12% of 2000
\( = {12 \over 100} \times 2000 = 240 \)
Given in question,
Number of mango trees = 18% of total trees = 18% of 2000
\( = {18 \over 100} \times 2000 = 360 \)
Given, Rest all are orange trees.
Number of orange trees = (2000 – 240 - 360)
= 1400
Number of orange trees are 1400.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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?
Solution:
In a balanced diet of 2600 calories
Given, 12 % protein.
Amount of protein intake = 12% of 2600 = \( {12 \over 100} \times 2600 = 312 \) calories
Given, 25 % fats.
Amount of fats intake = 25% of 2600 = \( {25 \over 100} \times 2600 = 650 \) calorie
Rest intake is carbohydrates.
Amount of carbohydrates intake = 2600 - (315 + 650) = 1638 calories
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 9
A cricketer diet scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 twos and 8 singles. What is the percentage of total runs came in:
a) Sixes
b) Fours
c) Twos
d) Singles
Solution:
a) Sixes
Cricketer hits 3 sixes.
Run scored by sixes = \( 3 \times 6 = 18 \)
Run percentage scored in sixes \(= {18 \over 62} \times 100 = 29.03 \) %
b) Fours
Cricketer hits 8 fours.
Run scored by sixes = \( 8 \times 4 = 32 \)
Run percentage scored in sixes \(= {32 \over 62} \times 100 = 51.61 \) %
c) Twos
Cricketer hits 2 twos.
Run scored by sixes = \( 2 \times 2 = 4 \)
Run percentage scored in sixes \(= {4 \over 62} \times 100 = 6.45 \) %
d) Singles
Cricketer hits 8 singles.
Run scored by sixes = \( 8 \times 1 = 8 \)
Run percentage scored in sixes \(= {8 \over 62} \times 100 = 12.90 \) %
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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 rest in 1’s. How many rubs did he score in?
Solution:
Let us assume the cricketer scored x runs in 6’s.
Given, 20% of 120 = x
\( \implies {20 \over 100} \times 120 = x \)
\(\implies x = 24 \)
Let us assume the cricketer scored y runs in 4’s.
30% of 120 = y
\( \implies {30 \over 100} \times 120 = y \)
\(\implies y = 36 \)
Let us assume the cricketer scored z runs in 2’s.
25% of 120 = z
\( \implies {25 \over 100} \times 120 = z \)
\(\implies z = 30 \)
Let us assume the cricketer scored t runs in 1’s.
24 + 36 + 30 + t = 120
\( \implies t = 30 \)
The cricketer scored 30 runs by taking singles.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 11
Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest?
Solution:
Let the Radha's investment be Rs. x
According to the question
22% of x = 187
\( \implies {22 \over 100 } \times x = 187 \)
\( \implies x = 187 \times {100 \over 22} = 850 \)
Radha invested Rs. 850
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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?
Solution:
Let Rohit's total income of the year 1997 be Rs .x
According to the question,
12% of x = 1440
\( \implies {12 \over 100 } \times x = 1440 \)
\( \implies x = 1440 \times {100 \over 12} = 12000 \)
Rohit’s total income during 1997 is Rs.12, 000.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 13
Gun powder contains 75% nitre and 10% sulphur. Find the amount of the gun powder which carries 9 kg of nitre. What amount of gun powder would contain 2.3 kg of sulphur?
Solution:
Let the amount of gun powder that contains 9 kg nitre be x kg (75%)
\( = { 75 \over 100 } \times x = 9 \)
\( \implies x = 9 \times {100 \over 75} = 12 \)
Let the amount of gun powder that contains 2.3 kg sulphur be y kg(10%)
\( = { 10 \over 100 } \times x = 2.3 \)
\( \implies x = 2.3 \times {100 \over 10} = 23 \)
The amount of gun powder containing 2.3 kg sulphur is 23 kg.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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?
Solution:
Composition of the alloy = 15 parts of tin + 105 parts of copper = 120 parts of both.
i.e. we can say 15 part out of 120 is tin and 105 pars is copper out of 120.
Now we will find both metals out of 100, because percentage means out of 100.
Therefore, the percentage of tin = \( { 15 \over 120 } \times 100 = 12.5 \) %
Let the amount of gun powder that contains 9 kg nitre be x kg
Percentage of copper =\( { 105 \over 120 } \times 100 = 87.5 \) %
The percentage of copper is 87.5%
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 15
An alloy contains 32% of copper, 40% of nickel and rest zinc. Find the mass of the zinc in 1 kg of alloy?
Solution:
Percentage of copper in the alloy = 32% = 32 out of 100
Percentage of nickel in the alloy = 40% = 40 out of 100
So, percentage of zinc in the alloy = 100 - (32 + 40) = 28% = 28 out of 100.
Amount of zinc in 1 kg of alloy \( = {28 \over 100} \times 1 = 0.280 kg = 280 gm \)
The mass of zinc in 1 kg of the alloy is 280 gm.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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?
Solution:
Let the length of the total ride be x km.
i.e. 10 % of x is 122 km.
10% of x = 122
\( \implies {10 \over 100} \times 122 = 1220 \) km.
The total length of the total ride is 1220 km.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 17
A certain school has 30 students, 142 of whom are boys. It has 30 teachers, 12 of which are men. What percent of total number of students and teachers in the school is female?
Solution:
Total number of female students = 300 - 142 = 158
Number of female teachers = 30 - 12 = 18
To tal number of females = 158 + 18 = 176
Total population of the school = 300 + 30 = 330
Percentage of teacher in the school is female \(= {176 \over 330} \times 100 = 53.33 \) %
The percentage of total number of students and teachers in the school is female is 53.33%
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 18
Aman’s income is 20% less than that of Anil. How much present is Anil’s income more than Aman’s?
Solution:
Let Anil’s income be 100
Then, Aman’s income = 20% less than Anil = 100 - 20% of 100 \(= 100 − {20 \over 100}\times 100 = 100 - 20 = 80 \)
Difference in the incomes of Anil and Aman to that of Aman’s income = 100- 80 = 20
Anil's income is more than Aman's in percentage = \( {20 \over 80} \times 100 = 25 \) %
Anil’s income is 25% more than that of Aman’s.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 19
The value of the machine depreciates every year by 5%. If the present value of the machine is Rs.100000, what will be it’s value after 2 years?
Solution:
It is given that the value of the machine depreciates by 5% every year. The present value of the machine = Rs.100000
Therefore, 5% of 100000 = Rs.5000
Value of the machine after 1st year = Rs (100000 - 5000)
= Rs. 95000
5 % of 95000 = Rs.4750
Value of the machine in the 2nd year =Rs (95000 - 4750)
= Rs 90250
After two years, the value of the machine will be Rs.90250
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 20
The population of the town increased by 10% annually. If the present population is 60000, what will be its population after 2 years?
Solution:
Present population = 60000
It increases 10% annually.
10 % of 60000 \(= {10 \over 100 } \times 60000 = 6000 \)
Increase in the population in the first year = 60000 + 6000 = 66000
66000 is the increase in the population in the second year = 10% of 66000 = 6600
Thus population after 2 years = 66000 + 6600 = 72600
The population of the town after 2 years is 72600.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 21
The population of the town is increased by 10% annually. If the present population is 22000, find the population a year ago.
Solution:
Let the population of the town one year ago be x
Now, it is given that population of the town increases by 10%
Present population = x + 10% of x \(= x + {10 \over 100} \times x = x + {x \over 10} = {11x \over 10 }
But present population of the town = 22000
\( \implies {11x \over 10 } = 22000 \)
\( \implies x = 22000 \times 10 \over 11 = 20000 \)
The population of the town a year ago is 20000.
Solution RD\times Sharma Maths Class 8 Percentage Exercise 12.2 Question 22
Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. what was his salary before investment?
Solution:
Let the initial salary be Rs. x
We know that salary
Before increment + increment given on salary = new salary
\( \implies \) x +10% of x = 3575
\( \implies x + {10 \over 100 } \times x = 3575 \)
\( \implies {110x \over 100} = 3575 \)
\( \implies x = 3250 \)
Salary before increment is Rs.3250
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 23
In new budget, the price of the petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase?
Solution:
We have to reduce the consumption such that the expenditure does not increase.
For this, we use the following formula:
\( {r \over r +100} \times 100 \)
Where r is the percentage rise in the price of the commodity.
given, r =10%
Therefore, percentage reduction in the consumption = \( {10 \over 10 +100} \times 100 \)
\( = {100 \over 11} = 9{1 \over 11} \)
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 24
Mohan’s income is Rs.15500 per month. He saves 11% of his income. If his income is increased by 10%, then he reduces his saving by 1%, how much does he save now?
Solution:
Saving of Mohan = 11 % of 15500
\( \implies { 11 \over 100} \times 15500 = Rs \ 1705 \)
Mohan's income is increases by 10%
Therefore, increase in income = \( {10 \over 100} \times 15500 = Rs \ 1550 \)
Increased income = 15500+ 1550 = 17050
Now, percentage of saving (according to question) = (11 - 1) % = 10 %
So, current saving \(= {10 \over 100} \times 17050 = Rs \ 1705 \)
So, the amounts of his present and earlier saving are the same.
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 25
Shikha’s income is 60% more than that of shalu. What percent is shalu’s income less than shikha’s?
Solution:
Let shalu’s income be Rs. x
Shikha’s income = Rs x + 60% of x
\(= x + {60x \over 100} = { 160x \over 100 } \)
Difference in Shikha and Shalu incomes
\( {16 \over 10}x - {16 \over 100}x = Rs \ {6 \over 10}x \)
Percentage of the difference in the incomes of shikha and shalu to that of shikha’s income
\( \frac{6x \over 10}{16x \over 10}= 37.5 \) %
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 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 other of them get?
Solution:
first person gets 50 % of second
and second 50 % of third.
Let x be the amounts received by the third person.
so, second person received = 50 % of x = \({50 \over 100}x = {x \over 2} \)
first person received = 50% of \({x \over 2 } = {50 \over 100} \times { x \over 2} = {x \over 4} \)
Total amount \(= x + {x \over 2} + {x \over 4} = {7x \over 4} \)
\({7x \over 4}= 3500 \)
\( x = 3500 \times {4 \over 7} = 2000 \)
So, first person received Rs = \({x \over 4} = {2000 \over 4} = Rs \ 500 \)
Second person received Rs = \({x \over 2} = {2000 \over 2} = Rs \ 1000 \)
Third person received Rs = x = Rs 2000
Solution RD Sharma Maths Class 8 Percentage Exercise 12.2 Question 27
After a 20% hike, the cost of Chinese vase is Rs 2000. What was the original price of the object?
Solution:
Let the original price of the object = Rs. x
According to the question, 20% of x + x = 2000
\( { 20 \over 100} \times x + x = 2000 \)
\( { x \over 5} + x = 2000 \)
\( { 6x \over 5} = 2000 \)
\( x = {2000 \times 5 \over 6} = 1666.67 \)