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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Life Mathematics Ex 1.1

Question 1.

Fill in the blanks:

Question (i)

If 30% of x is 150, then x is ………….

Answer:

500

Hint:

Given 30% of x is 150

Question (ii)

2 minutes is ………… % to an hour.

Answer:

3\(\frac{1}{3}\)%

Hint:

Let 2 min be x% of an hour

and 1 hr = 60 min

x% = \(\frac{2}{60}\) x 100 = \(\frac{200}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)

x = 3\(\frac{1}{3}\)

Question (iii)

If x % of x = 25, then x = …………

Answer:

50

Hint:

Given that x % of x is 25

∴ \(\frac{x}{100}\) = 25

x^{2} = 25 x 100 = 2500

∴ x = \(\sqrt{2500}\) = 50

Question (iv)

In a school of 1400 students, there are 420 girls. The percentage of boys in the school is ………….

Answer:

70

Hint:

Given total number of students in school = 1400

Number of girls in school = 420

Number of boys in school = 1400 – 420 = 980

% of boys = 70%

Question (v)

0.5252 is ………….. %.

Answer:

52.52%

Hint:

Given a number, and to express as a percentage, we need to multiply by 100

∴ to express 0.5252 as percentage, we should multiply by 100

∴ 0.5252 x 100 = 52.52%

Question 2.

Rewrite each underlined portion using percentage language.

Question (i)

One half of the cake is distributed to the children.

Answer:

Answer:

50% of the cake is distributed to the children

Hint:

One half is nothing but \(\frac{1}{2}\) as percentage, we need to multiply by 100

\(\frac{1}{2}\) x 100 = 50%

Question (ii)

Aparna scored 7.5 points out of 10 in a competition.

Answer:

Aparna scored 75% in a competition

Hint:

7.5 points out of 10 is \(\frac{7.5}{10}\) = 0.75

For percentage, we need to multiply by 100

∴ We get 0.75 x 100 = 75%

Question (iii)

The statue was made of pure silver.

Answer:

96% students participated in sports

Hint:

Pure silver means there are no other metals

so, 100 out of 100 parts is made of silver = \(\frac{100}{100}\)

∴ to express as percentage, \(\frac{100}{100}\) x 100% = 100%

Question (iv)

48 out of 50 students participated in sports

Answer:

96% students participated in sports

Hint:

48 out of 50 students in fraction form is \(\frac{48}{50}\).

As a percentage, we need to multiply by 100

Question (v)

Only 2 persons out of 3 will be selected in the interview

Answer:

Only 66\(\frac{2}{3}\)% persons will be selected in the interview.

Hint:

2 out of 3 in fraction form is \(\frac{2}{3}\) to express as percentage, we need to multiply by 100

\(\frac{2}{3}\) x 100 = \(\frac{200}{3}\) = 66\(\frac{2}{3}\)%

Question 3.

48 is 32% of what number?

Solution:

Let the number required to be found be ‘x’

Given that 32% of x is 48

Question 4.

A bank pays ₹ 240 as interest for 2 years for a sum of ? 3000 deposited as savings. Find the rate of interest given by the bank.

Solution:

The formula for simple interest is given by

Interest (I) = \(\frac{PxRxT}{100}\) 100

Where P is principal amount; R is rate of interest in percentage; t is time period Substituting given values in formula, we get

P = 3000, t = 2 yrs, I = Rs 240

Question 5.

A Welfare Association has a sports club where 30% of the members play cricket, 28% play volleyball, 22% play badminton and the rest play indoor games. If 30 member play indoor games.

- How many members are there in the sports club?
- How many play cricket, volleyball and badminton?

Solution:

Given data

Members who play Cricket = 30% …..(1)

Members who play Vollyball = 28% …..(2)

Members who play Batminton = 22% ……(3)

Members who play outdoor games = (1) + (2) + (3) = 30

Members who play indoor games = 100 – 80 = 20%

Also given 30 members play indoor games

Let total no of members be ‘x’

∴ \(\frac{30}{x}\) = 20

∴ x = \(\frac{30×100}{20}\) = 15

1. Total number of members is 150

2. Members who play Cricket = 30% of total = \(\frac{30}{100}\) x 150

Question 6.

What is 25% of 30% of 400?

Solution:

Required to find 25% of 30% of 400

Question 7.

If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of the number.

Solution:

Given that 75% of number less 60% of number is 82.5. Let the number be ‘x’

∴ \(\frac{75}{100}\) × x = 82.5

∴ 0.75 – 0.6.60x = 82.5

∴ 0.15x = 82.5

∴ x = \(\frac{82.5}{0.15}\) = \(\frac{8250}{15}\) = 550

Required to find 20% of number ie 20% of x.

Question 8.

If a car is sold for ₹ 2,00,000 from its original price of? 3,00,000 find the percentage for decrease in the value of the car.

Solution:

Original price of car = ₹ 3,00,000

actual selling price of car = ₹ 2,00,000

Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000

Question 9.

A number when increased by 18% gives 236. Find the number.

Solution:

Let the number be x. Given that when it is increased by 18%, we get 236.

x + \(\frac{18}{100}\) = 236

\(\frac{100x+18x}{100}\) = 236

\(\frac{118}{100}\)x = 236

The number = x = \(\frac{236×100}{118}\) = 200

Question 10.

A number when decreased by 20% gives 80. Find the number.

Solution:

Let the number be x. Given that when it is increased by 20% we get 80.

Question 11.

A number is increased by 25% and then decreased by 20%. Find the change in that number.

Solution:

Method 1:

Let the number be x. First it is increased by 25%

Method 2:

[to understand, let us assume that number is 100]

So, first when we increase by 25%, we get

We get back 100 ⇒ No change

Question 12.

If the numerator of a fraction is increased by 25% and the denominator is increased by 10%, it becomes \(\frac{2}{5}\). Find the original fraction.

Solution:

Let the fraction be \(\frac{N}{D}\) where N is numerator & D is denominator

It is given that numerator is include by 25%

Question 13.

A fruit vendor bought some mangoes of which 10% were rotten. He sold 33 \(\frac{1}{3}\)% of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.

Solution:

Let the number of mangoes bought by fruit seller initially be x.

Given that 10% of mangoes were rotten

∴ Number of rotten mangoes = \(\frac{10}{100}\) × x

Number of good mangoes = x – no. of rotten mangoes

= x – \(\frac{10}{100}\)x = \(\frac{100x-10x}{100}\) = \(\frac{90}{100}\)x …(1)

Number of mangoes sold = 33\(\frac{1}{3}\)% of good mangoes = \(\frac{100}{3}\)%

∴ Mangoes sold = \(\frac{100}{3}\) x \(\frac{90}{100}\)x × \(\frac{1}{100}\) = \(\frac{30}{100}\)x

Number of mangoes remaining = No. of good mangoes – No. of mangoes sold

From (1) and (2)

Question 14.

A student gets 31% marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.

Solution:

Let the maximum marks in the exam be ‘x’. Pass percentage is given as 35%

∴ Pass mark = \(\frac{35}{100}\) × x = \(\frac{35}{100}\)x

Student gets 31 % marks = \(\frac{31}{100}\) × x = \(\frac{31}{100}\)x

But student fails by 12 marks → meaning his mark is 12 less than pass mark.

∴ \(\frac{31}{100}\)x = \(\frac{35}{100}\)x – 12

∴ \(\frac{35}{100}\)x – \(\frac{31}{100}\)x = 12

∴ \(\frac{35x-31x}{100}\) = 12 ⇒ \(\frac{4x}{100}\) = 12

∴ \(\frac{12×100}{4}\) = 300

Question 15.

The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.

Solution:

Let number of boys be ‘6’ and number of girls be ‘g’

Ratio of boys and girls is given as 5:3

b:g = 5:3 ⇒ \(\frac{b}{g}\) = \(\frac{5}{3}\) …..(A)

Failure in boys = 16% = \(\frac{16}{100}\) x b = \(\frac{16b}{100}\)

Failure in girls = 8% = \(\frac{8}{100}\) x g = \(\frac{8g}{100}\)

Pass in boys = 100 – 16% = 84% \(\frac{84}{100}\)b …..(1)

Pass in girls = 100 – 8% = 92% = \(\frac{92}{100}\)g

From A, we have \(\frac{b}{g}\) = \(\frac{5}{3}\), = adding 1 on both sides, we get

\(\frac{b}{g}\) + 1 = \(\frac{5}{3}\) + 1

\(\frac{b+g}{g}\) = \(\frac{5+3}{3}\) = \(\frac{8}{3}\)

∴ g = \(\frac{3}{8}\)(b + g)

Similarly b = \(\frac{5}{8}\)(b + g)

Total pass = Pass in girls + Pass in boys

= (1) + (2) = \(\frac{84}{100}\)b + \(\frac{92}{100}\)g

Objective Type Questions

Question 16.

12% of 250 litres is the same as ………. of 150 liters

(a) 10%

(b) 15%

(c) 20%

(d) 30%

Answer:

(c) 20%

Hint:

12% of 250 = \(\frac{12}{100}\) x 250 = 30 lit

Percentage : \(\frac{30}{150}\) x 100 = 20%

Question 17.

If three candidates A, B and C is 3 in a school election got 153, 245 and 102 votes respectively, the percentage of votes for the winner is ……..

(a) 48%

(b) 49%

(c) 50%

(d) 45%

Answer:

(b) 49%

Hint:

Candidate 1: 153

Candidate 2: 245 – winner [as maximum votes]

Candidate 3: 102

Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500

Question 18.

15% of 25% of 10000 = ………..

(a) 375

(b) 400

(c) 425

(d) 475

Answer:

(a) 375

Hint:

15% of 25% of 10000 is

First let us do 25% of 10,000, which is

Next 15% of the above is \(\frac{15}{100}\) x 2500 = 375

Question 19.

When 60 is subtracted from 60% of a number to give 60, the number is –

(a) 60

(b) 100

(c) 150

(d) 200

Answer:

(d) 200

Hint:

Let the number be ‘x’.

60% of the number is \(\frac{60}{100}\) × x = \(\frac{60x}{100}\)

Given that when 60 is subtracted from 60%, we get 60

i.e \(\frac{60}{100}\)x – 60 = 60

∴ \(\frac{60}{100}\)x = 60 + 60 = 120

∴ x = \(\frac{120×100}{60}\) = 200

Question 20.

If 48% of 48 = 64% of x, then x =

(a) 64

(b) 56

(c) 42

(d) 36

Answer:

(d) 36

Hint:

Given that 48% of 48 48 = 64% of x