RD Sharma Solutions Class 8 Percentage Exercise 12.2

RD Sharma Class 8 Solutions Chapter 12 Ex 12.2 PDF Free Download

RD Sharma Solutions Class 8 Chapter 12 Exercise 12.2

Exercise 12.2

Q1: Find:

i) 22% of 120

ii) 25% of 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) Given: To find 22% of 120

= \(\frac{22}{100}\times 120\) = 26.4

Therefore, 22% of 120 is 26.4

ii) Given: To find 25% of 1000

= \(\frac{25}{100}\times 1000\) = 250

Therefore, 25% of 1000 is 250

iii) Given: To find 25% of 10 kg

= \(\frac{25}{100}\times 10\) = 2.5 kg

Therefore, 25% of 10 kg is 2.5 kg

iv) Given: To find 16.5 % of 5000 metre

= \(\frac{16.5}{100}\times 5000\) = 825 metre

Therefore, 16.5 % of 5000 metre is 825 metre

v) Given: To find 135 % of 80 cm

= \(\frac{135}{100}\times 80\) = 108 cm.

Therefore, 135 % of 80 cm is 108 cm

vi) Given: To find 2.5 % of 10000 ml

= \(\frac{2.5}{100}\times 10000\) = 250 ml

Therefore, 2.5 % of 10000 ml is 250 ml

Q2: Find the number a , if :

i) 8.4% of a is 42

ii) 0.5% of a is 3

iii) \(\frac{1}{2}\) % of a is 50

iv) 100% of a is 100

Solution:

i) Given: 8.4% of a is 42

= \(\frac{8.4}{100}\times a=42\) To find a:

= \(a=\frac{42\times 100}{8.4}\)

a= 500

Therefore, 8.4% of 500 is 42

ii)  Given: 0.5% of a is 3

= \(\frac{0.5}{100}\times a=3\)

To find a:

= \(a=\frac{3\times 100}{0.5}\)

a= 600

Therefore, 0.5% of 600 is 3

iii)  Given: \(\frac{1}{2}\) % of a is 50

= \(\frac{1}{200}\times a=50\)

To find a:

= \(a=\frac{50\times 200}{1}\)

a= 10000

Therefore, \(\frac{1}{2}\) % of 10000 is 50

iv)  Given: 100% of a is 100

= \(\frac{100}{100}\times a=100\)

To find a:

= \(a=\frac{100\times 100}{100}\)

a= 100

Therefore, 100% of 100 is 100

Q3: X is 5% of y, y is 24% of z. if x = 480, find the values of y and z.

Solution:

Given that,

x is 5% of y and y is 24% of z

From,

x is 5% of y

x= \(\frac{5}{100}\) y

Therefore, y= \(\frac{100}{5}\) x

= y = 20 x

Subsbtitute the value of x =480

= y = 20(480)

So, the value of y = 9600

Also given that,

y is 24% of z

y= \(\frac{24}{100}\) z

Therefore, z= \(\frac{100}{24}\) y

Now, subsbtitute the value of y = 9600

= z = \(\frac{100}{24}\times 9600\)

z = 40000

Therefore, the value of z = 40000

Q4: 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 the monthly income coolie be “a”

ie, the equation becomes

= 15 % of a = 150

= \(\frac{15}{100}\times x=150\)

\(a=\frac{150\times 100}{15}\)

a= 1000

Therefore, the monthly income of a coolie is Rs.1000

Q5: 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 the total number of marks in the examination be “a”

From the given data, the equation can be written as

86.875% of a = 695

= \(\frac{86.875}{100}\times a=695\)

= \(a=\frac{695\times 100}{86.875}\)

= 800

Therefore, the total number of marks in the examination is 800.

Q6: 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 “a” be the number of days in which the school was opened.

From the given information, the equation can be written as

90 % of a = 216

\(\frac{90}{100}\times a=216\)

= \(a=\frac{216\times 100}{90}\)

a= 240 days

Therefore, the number of days in which the school was opened for a year is 240 days .

Q7: 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:

Given: Total number of trees = 2000 trees.

Number of mango trees = 12% of 2000

Number of lemon trees = 18% of 2000

Let the number of orange trees be “a”

To find the number of mango trees:

Number of mango trees = 12% of 2000

= \(\frac{12}{100}\times 2000\)

= \(a=\frac{12\times 20}{1}\)

No. of mango trees= 240

To find the number of lemon trees:

Number of lemon trees = 18% of 2000

= \(\frac{18}{100}\times 2000\)

= \(a=\frac{18\times 20}{1}\)

No. of lemon trees= 360

So, the number of orange trees =Total number of trees – Number of Mango trees – Number of lemon trees.

= (2000-240-360)

=1400

Therefore, the number of orange trees are 1400.

Q8: A 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:

Total calories of balanced diet = 2600 calories

Out of these 12 % protein, 25 %   fats and 63% of carbohydrates

To find the amount of calories in each.

For, 12 % protein,

Amount of protein intake = 12% of 2600

= \(\frac{12}{100}\times 2600\)

Therefore, the amount of protein = 312 calories

For, 25 % fats,

Amount of fats intake = 25% of 2600

= \(\frac{25}{100}\times 2600\)

Therefore, the amount of fat = 650 calorie

Fro, 63% of carbohydrates,

Amount of carbohydrates intake =Total calories – (Amount of protein + Amount of fat)

=2600-(315+650)

Therefore, the amount of carbohydrate= 1638 calories

Q9: A cricketer 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:

1) Sixes

2) Fours

3) Twos

4) Singles

Solution:

Given that, Total runs = 62 runs

Total number of balls = 96 balls

A cricketer hits 3 sixes, 8 fours, 2 twos and 8 singles

Therefore, to find the percentage, proceed the following

1) Sixes

A cricketer hits 3 sixes = 3(6) = 18

Therefore, percentage of sixes= \(\frac{18}{62}\times 100\)

= 29.03%

2) Fours

A cricketer hits 8 four = 8(4) = 32

Therefore, percentage of fours= \(\frac{32}{62}\times 100\)

= 51.61%

3) Twos

A cricketer hits 2 twos = 2(2) = 4

Therefore, percentage of twos= \(\frac{4}{62}\times 100\)

= 6.45%

4) Singles

A cricketer hits 8 singles = 8(1) = 8

Therefore, percentage of singles= \(\frac{8}{62}\times 100\) = 12.90

Q10: 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 runs did he score in?

(i) 6’s.

(ii) 4’s

(iii) 2’s

(iv) 1’s

Solution:

Assume that the cricketer scored “a” runs in 6’s.

20% of 120 = a

a = 24

Therefore, the cricketer scored 24 runs by hitting 6’s

Assume that the cricketer scored “b” runs in 4’s.

30% of 120 = b

b = 36

Therefore, the cricketer scored 36 runs by hitting  4’s

Assume that the cricketer scored “c” runs in 2’s

25% of 120 = c

c = 30

Therefore, the cricketer scored 30 runs by hitting 2’s

Assume that the cricketer scored “d” runs in 1’s.

24+36+30+d = 120

d = 30

Therefore, the cricketer scored 30 runs by hitting 1’s

Q11: Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest?

Solution:

Let the investment of Radha be Rs. a

From the given information, we can write it as

22% of a = 187

= \(\frac{22}{100}\times a=187\)

a = \(\frac{187\times 100}{22}\)

a= 850

Therefore, the investment of Radha is Rs. 850

Q12: 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 the total income of Rohit for a year, 1997 be Rs .a

From the given information, we can write it as

12% of x =1440

= \(\frac{12}{100}\times a=1440\)

a = \(\frac{1440\times 100}{12}\)

a = 12000

Therefor, the total income of Rohit during 1997 is Rs.12, 000.

Q13: Gun powder contains 75% nitre and 10% sulphur. Find the amount of gun powder which carries 9 kg of nitre. What amount of gun powder would contain 2.3 kg of sulphur?

Solution:

Assume that the amount of gun powder that contains 9 kg nitre be “a” kg

From the given information, we can write it as

\(\frac{75}{100}\times a=9\)

a = \(\frac{9\times 100}{75}\)

a= 12 kg

Therefore, the amount of gun powder that carries 9 kg nitre is 12 kg

Also, assume that the amount of gun powder that contains 2.3 kg sulphur be “b”  kg

= \(\frac{25}{100}\times b=2.3\)

b = \(\frac{2.3\times 100}{2.3}\)

b = 23 kg

Therefore, the amount of gun powder that carries 2.3 kg sulphur is 23 kg.

Q14: 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 that, the total composition of the alloy = 15 parts if tin + 105 parts of copper

Assume that the percentage of tin be “a”

percentage of tin = \(\frac{15}{120}\times 100\)

a = 12.5%

Therefore, the percentage of tin 12.5%

Assume that the percentage of coper be “b”

Percentage of copper = \(\frac{105}{120}\times 100\)

b= 87.5%

Therefore, the percentage of copper is 87.5%

Q15: An alloy contains 32% of copper, 40% of nickel and rest zinc. Find the mass of the zinc in 1 kg of alloy?

Solution

Given that,

Percentage of copper in the alloy = 32%

Percentage of nickel in the alloy = 40%

Therefore, the percentage of zinc in an alloy = 100 – ( Percentage of copper in the alloy+ Percentage of nickel in the alloy)

So, Percentage of zinc in the alloy = 100-(32+40) = 28%

Total Amount of zinc in 1 kg of alloy = 0.28(1) = 280 gm

Therefore, the mass of zinc in 1 kg of the alloy is 280 gm.

Q16: 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 “l” km

From the given information, we can write it as

10% of l = 122

l = \(\frac{122\times 100}{10}\)

l = 1220 km

Hence, the total length of the total ride is 1220 km.

Q17: A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What per cent of the total number of students and teachers in the school is female?

Solution:

Given that

Total number of students in a school = 300

Total number of male students in a school = 142

Therefore,

The total number of female students in a school = 300-142 = 158

Total number of teachers = 30

Total number of male teachers = 12

Therefore, the total number of female teachers = 30-12 = 18

So, the total number of females in a school= total number of female students in a school  +  total number of female teachers

The total number of females in a school  = 158+18 = 176

Also, the total population of the school =Total number of students + Total number of teachers

= 300+30 = 330

So, the Percentage of teacher and student in the school is female  = \(\frac{176}{330}\times 100\)

= 53.33%

Hence, the percentage of the total number of students and teachers in the school is female is 53.33%

Q18: Aman’s income is 20% less than that of anil. How much present is anil’s income more than a man’s?

Solution:

Let Anil’s income be “a”

From the given statement, we can write it as,

Therefore, Aman’s income = \((a-20)100= 8×10\)

Hence. the difference in the incomes of Anil and Aman to that of Aman’s income is given as

= \(\frac{2\times 10}{8\times 10}\times 100\)

= 25%

Hence, the Anil’s income is 25% more than that of Aman’s.

Q19: The value of the machine depreciates every year by 5%. If the present value of the machine is Rs.100000, what will be its value after 2 years?

Solution:

Given that,

The value of the machine depreciates by every year = 5%

The present value of the machine is Rs.100000

So, 5% of 100000 = Rs.5000

Value of the machine after 1st year = Rs (100000-5000)

Therefore, the value of the machine after 1st year = Rs.4750= Rs. 95000

5 % of 95000

Therefore, the value of the machine after 1st year = Rs.4750

Value of the machine in the 2nd year =Rs (95000-4750)

= Rs 90250

Therefore, the value of the machine after 2nd year = Rs.4750

Q20: The population of the town increased by 10% annually. If the present population is 60000, what will be its population after 2 years?

Solution:

Given that, the present population of a town = 60000

Every year the population is increased by 10%

So,  10% of 60000 = 6000

Increase in the population in the first year = 60000+6000 = 66000

So, 66000 is the increase in the population in the 2nd  year = 10% of 66000 = 6600

the population of the after 2 years = 66000+6600 = 72600

Therefore, the population of the town after 2 years is 72600.

Q21: 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 before 1 year be “a”

It is given that the population of the town increases by 10%

So, the present population of the town = a+10% of a

= a+ \(\frac{10}{100}a\)

= \(\frac{110}{100}a\)

Given that, the present population of the town = 22000

From the given information, we can write it as,

\(\frac{10}{100}x\) = 22000

a = 20000

Hence, the population of the town before a year is 20000.

Q22: 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 of Ankit be Rs. a

From the given data, we can say that

New salary of Ankit = Before increment + increment given on salary

= a+ 10% of a = 3575

= \(\frac{110}{100}a\) = 3575

a = 3250

Therefore, the salary of Ankit before increment is Rs.3250

Q23: In a new budget, the price of petrol rose by 10%. By how much per cent must one reduce the consumption so that the expenditure does not increase?

Solution:

Assume that the consumption of petrol be 100 litres and the price is Rs.100, then

So the new price of petrol is Rs 110, because of the price of petrol increases by 10%.

Therefore, now Rs 110 can fetch 100 litres of petrol.

So, Rs 100 can fetch = (100/110)x 100

= 10000/110

=1000/11 litres of petrol.

Therefore, the reduction in the consumption of petrol  = (100- (1000/11))%

= 100/11 %

Therefore, the percentage reduction in consumption =9(1/11)%

Q24: 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:

From the given statement, we can write it as

Savings of Mohan  = 11% of 15500

= \(\frac{11}{100}\times 15500\)

= Rs. 1705

Therefore, the savings of Mohan is Rs. 1705

Also, given that Mohan’s income increases by 10%

So, the increase in an income of Mohan = \(\frac{10}{100}\times 15500\)

= Rs 1550

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

Saving = \(\frac{10}{100}\times 17050\)

Thus, the Mohan present and earlier savings are the same as before.

Q25: Shikha’s income is 60% more than that of shalu. What per cent is Shalu’s income less than Shikha’s?

Solution

Let Shika’s income be a

Let Shalu’s income be b

From the given statement, we can wite the expression as

a = b + (60/100)b

100a = 160b

therefore, b = 100a/160

b = 5a/8

b = a- (3a/8)

b = a – 60a/160

Therefore, Shalu’s income = Shika’s salary – 37.5% of Shika’s salary

Q26: 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

Let a, b, and c be the amounts received by the first, second, third person respectively.

Given that

a= 50% of b

= \(\frac{50}{100}y\) = 1/2b

b = 2a

Again, b = 50% of c

b = 1/2 c

c= 2b

Therefore, c= 2b=4a

= a+b+c =3500

Substituting the value of z and y we get,

= a+2b+4c = 3500

= 7a = 3500

= a= 500

= b=2a = 2(500) = 1000

=c= 4a = 4(500) = 2000

Therefore, the amount received by three persons are 500, 1000 and 2000 respectively.

Q27: After a 20% hike, the cost of Chinese vase is Rs2000. What was the original price of the object?

Solution

Let the object original price be Rs. a

From the given information, we can write the expression as

We have 20% of a+a = 2000

= \(\frac{20}{100}a\) + a = 2000

= \(\frac{120}{100}a\) = 2000

= a = 1666.67

Therefore, the original price of the object is Rs.1666.67.

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