Fractions — Exercise 4(G)

A man earns ₹ 18,720 per month and spends $\dfrac{5}{6}$ of his income. Find his monthly

(i) expenditure

(ii) saving.

Answer

(i) Given, a man earns = ₹ 18,720 per month

$\text{Expenditure }= \dfrac{5}{6} \text{ of his income}\\[1em] = \dfrac{5}{6} \times 18720\\[1em] = \dfrac{5 \times 18720}{6}\\[1em] = 5 \times 3120\\[1em] = 15,600.$

Hence, expenditure = ₹ 15,600.

(ii) Saving = Income - Expenditure

= 18,720 - 15,600

= ₹ 3,120.

Hence, saving = ₹ 3,120.

A water tank can hold $56\dfrac{1}{4}$ litres of water. How much water is contained in the tank when it is $\dfrac{8}{15}$ full?

Answer

Given,

Total capacity of tank = $56\dfrac{1}{4}$ = $\dfrac{225}{4}$ litres

Fraction full = $\dfrac{8}{15}$

$\text{Water in the tank }= \dfrac{8}{15} \times \dfrac{225}{4}\\[1em] = \dfrac{8 \times 225}{15 \times 4}\\[1em] = \dfrac{1800}{60}\\[1em] = 30.$

Hence, water in the tank = 30 litres.

After reading $\dfrac{5}{8}$ of a book, 168 pages are left. How many pages are there in all in the book?

Answer

Let the total number of pages be x.

Given, number of pages read = $\dfrac{5}{8}$ x

Pages are left = 168

Number of pages left

$\Rightarrow 1x - \dfrac{5}{8}x = 168\\[1em] \Rightarrow \dfrac{8x}{8} - \dfrac{5x}{8} = 168\\[1em] \Rightarrow \dfrac{8x - 5x}{8} = 168\\[1em] \Rightarrow \dfrac{3x}{8} = 168\\[1em] \Rightarrow x = \dfrac{168 \times 8}{3}\\[1em] \Rightarrow x = \dfrac{1344}{3}\\[1em] \Rightarrow x = 448$

Hence, total number of pages = 448.

$\dfrac{2}{3}$ of the students in a class are boys and the rest are 17 girls.

(i) How many boys are there in the class?

(ii) What is the total strength of the class?

Answer

(i) Given,

Number of boys in a class = $\dfrac{2}{3}$ of the students

Number of girls in a class = 17

Let the number of the students in the class be x.

Number of girls = Total students - number of boys

$\Rightarrow x - \dfrac{2}{3}x = 17\\[1em] \Rightarrow \dfrac{3}{3}x - \dfrac{2}{3}x = 17\\[1em] \Rightarrow \dfrac{3 - 2}{3}x = 17\\[1em] \Rightarrow \dfrac{1}{3}x = 17\\[1em] \Rightarrow x = 17\times 3\\[1em] \Rightarrow x = 51$

Number of boys = $\dfrac{2}{3} \times 51 = \dfrac{2 \times 51}{3} = \dfrac{102}{3} = 34$

Hence, number of boys in the class = 34.

(ii) Hence, number of students in the class = 51.

A man earns ₹ 25,440 per month. He spends $\dfrac{1}{4}$ of it on house rent, $\dfrac{3}{8}$ on food and clothes, $\dfrac{1}{10}$ on insurance and $\dfrac{1}{5}$ on other items. The rest he saves. How much does he save each month?

Answer

Given,

Total salary = ₹ 25440

House rent = ​$\dfrac{1}{4}$ part of the salary

Food & clothes = ​$\dfrac{3}{8}$ part of the salary

Insurance = $\dfrac{1}{10}$ part of the salary

Other items = $\dfrac{1}{5}$​ part of the salary

$\text{Total fraction spent } = \dfrac{1}{4} + \dfrac{3}{8} + \dfrac{1}{10} + \dfrac{1}{5}\\[1em] = \dfrac{1 \times 10}{4 \times 10} + \dfrac{3 \times 5}{8 \times 5} + \dfrac{1 \times 4}{10 \times 4} + \dfrac{1 \times 8}{5 \times 8}\\[1em] = \dfrac{10}{40} + \dfrac{15}{40} + \dfrac{4}{40} + \dfrac{8}{40}\\[1em] = \dfrac{10 + 15 + 4 + 8}{40}\\[1em] = \dfrac{37}{40}\\[1em]$

Saving fraction = 1 - Total spent fraction

$\Rightarrow \text{Saving fraction } = 1 - \dfrac{37}{40} \\[1em] = \dfrac{40}{40} - \dfrac{37}{40} \\[1em] = \dfrac{40 - 37}{40} \\[1em] = \dfrac{3}{40} \\[1em]$

Saving amount = Saving fraction × Salary

= $\dfrac{3}{40} \times 25,440 = \dfrac{3 \times 25,440}{40} = \dfrac{76,320}{40} = 1,908$.

Hence, saving per month = ₹ 1,908.

An objective test was given to a group of 168 students. It was found that $\dfrac{5}{6}$ of the students gave all correct answers. How many students made 1 or more mistakes?

Answer

Given,

Total students = 168

Part of students who gave all correct answers = $\dfrac{5}{6}$

Part of students who made 1 or more mistakes = $1 - \dfrac{5}{6} = \dfrac{6 - 5}{6} = \dfrac{1}{6}$.

Students who made 1 or more mistakes = $\dfrac{1}{6}$ × Total number of students

= $\dfrac{1}{6} \times 168$

= 28.

Hence, no. of students who made 1 or more mistakes = 28.

In an orchard, $\dfrac{1}{3}$ of the trees are guava trees, $\dfrac{1}{8}$ are banana trees and the rest are mango trees. If there are 117 mango trees in the orchard, how many trees in all are there?

Answer

Given,

Part of guava trees = $\dfrac{1}{3}$

Part of banana trees = $\dfrac{1}{8}$

Part of mango trees = 1 - (part of guava trees + part of banana trees)

$= 1 - \Big(\dfrac{1}{3} + \dfrac{1}{8}\Big) \\[1em] = 1 - \Big(\dfrac{8 + 3}{24}\Big) \\[1em] = 1 - \dfrac{11}{24} \\[1em] = \dfrac{24 - 11}{24} \\[1em] = \dfrac{13}{24}.$

No. of mango trees = Part of mango trees × Total number of trees

$117 = \dfrac{13}{24} \times \text{Total number of trees} \\[1em] \text{Total number of trees} = 117 \times \dfrac{24}{13} \\[1em] \text{Total number of trees} = 9 \times 24 = 216.$

Hence, total number of trees = 216.

In a school, $\dfrac{1}{25}$ of the students were absent on a certain day. If 720 students were present on that day, what is the total number of students in the school?

Answer

Given,

Total students present on a certain day = 720

Part of students absent on a certain day = $\dfrac{1}{25}$

Part of students present on a certain day = $1 - \dfrac{1}{25}$

$= \dfrac{25 - 1}{25} \\[1em] = \dfrac{24}{25}.$

Number of students present on a certain day = Part of students present on a certain day × Total number of students

$\Rightarrow 720 = \dfrac{24}{25} \times \text{ Total number of students} \\[1em] \Rightarrow \text{ Total number of students} = 720 \times \dfrac{25}{24} \\[1em] \Rightarrow \text{ Total number of students} = 30 \times 25 = 750.$

Hence, total number of students in the school = 750.

The product of three numbers is $7\dfrac{1}{2}$. If two of them are $1\dfrac{1}{7}$ and $3\dfrac{3}{4}$, find the third number.

Answer

Given, product of three numbers = $7\dfrac{1}{2}$

Two numbers = $1\dfrac{1}{7}$ and $3\dfrac{3}{4}$

Let the other number be x.

$\Rightarrow 1\dfrac{1}{7} \times 3\dfrac{3}{4} \times x = 7\dfrac{1}{2}\\[1em] \Rightarrow \dfrac{8}{7} \times \dfrac{15}{4} \times x = \dfrac{15}{2}\\[1em] \Rightarrow \dfrac{8 \times 15}{7 \times 4} \times x = \dfrac{15}{2}\\[1em] \Rightarrow \dfrac{120}{28} \times x = \dfrac{15}{2}\\[1em] \Rightarrow x = \dfrac{15 \times 28}{2 \times 120}\\[1em] \Rightarrow x = \dfrac{420}{240}\\[1em] \Rightarrow x = \dfrac{7}{4}\\[1em] \Rightarrow x = 1\dfrac{3}{4}.$

Hence, the third number = $1\dfrac{3}{4}$.