Fractions — Exercise 6(E)

From a rope $10\dfrac{1}{2}$ m long, $4\dfrac{5}{8}$ m is cut off. Find the length of the remaining rope.

Answer

Total length of the rope = $10\dfrac{1}{2}$ m.

Length cut off = $4\dfrac{5}{8}$ m.

Length of remaining rope = Total length − Length cut off

$= 10\dfrac{1}{2} - 4\dfrac{5}{8} \\[1em] = \dfrac{21}{2} - \dfrac{37}{8} \\[1em] \text{L.C.M. of 2 and 8 = 8}\\[1em] = \dfrac{84}{8} - \dfrac{37}{8} \\[1em] = \dfrac{47}{8} \\[1em] = 5\dfrac{7}{8} \text{ m}.$

Hence, the length of the remaining rope = $5\dfrac{7}{8}$ m.

A piece of cloth is 5 m long. After washing, it shrinks by $\dfrac{1}{25}$ of its length. What is the length of the cloth after washing?

Answer

Length of the cloth = 5 m.

Shrinkage = $\dfrac{1}{25}$ of its length

$= \dfrac{1}{25} \times 5 \\[1em] = \dfrac{5}{25} \\[1em] = \dfrac{1}{5} \text{ m}.$

Length of the cloth after washing = 5 − $\dfrac{1}{5}$

$= \dfrac{25}{5} - \dfrac{1}{5} \\[1em] = \dfrac{24}{5} \\[1em] = 4\dfrac{4}{5} \text{ m}.$

Hence, the length of the cloth after washing = $4\dfrac{4}{5}$ m.

I bought wheat worth ₹ $12\dfrac{1}{2}$, rice worth ₹ $25\dfrac{3}{4}$ and vegetables worth ₹ $10\dfrac{1}{4}$. I gave a hundred-rupee note to the shopkeeper; how much money did he return to me?

Answer

Price of wheat = ₹ $12\dfrac{1}{2}$

Price of rice = ₹ $25\dfrac{3}{4}$

Price of vegetables = ₹ $10\dfrac{1}{4}$

Money given to Shopkeeper = ₹ 100

Total money spent = $12\dfrac{1}{2} + 25\dfrac{3}{4} + 10\dfrac{1}{4}$

$= \dfrac{25}{2} + \dfrac{103}{4} + \dfrac{41}{4} \\[1em] = \dfrac{50}{4} + \dfrac{103}{4} + \dfrac{41}{4} \\[1em] = \dfrac{194}{4} \\[1em] = \dfrac{97}{2} \\[1em] = 48\dfrac{1}{2}.$

So, total money spent = ₹ $48\dfrac{1}{2}$.

Money returned by the shopkeeper = 100 − $48\dfrac{1}{2}$

$= \dfrac{200}{2} - \dfrac{97}{2} \\[1em] = \dfrac{103}{2} \\[1em] = 51\dfrac{1}{2}.$

Hence, the shopkeeper returned ₹ $51\dfrac{1}{2}$.

Out of 500 oranges in a box, $\dfrac{3}{25}$ are rotten and $\dfrac{1}{5}$ are kept for some guests. How many oranges are left in the box?

Answer

Total oranges = 500.

Number of rotten oranges = $\dfrac{3}{25} \text{ of } 500 = \dfrac{3}{25} \times 500$ = 60.

Number of oranges kept for guests = $\dfrac{1}{5} \text{ of } 500 = \dfrac{1}{5} \times 500$ = 100.

Oranges used = 60 + 100 = 160.

Oranges left = Total oranges − Oranges used

Oranges left = 500 − 160 = 340.

Hence, the number of oranges left in the box = 340.

An ornament piece is made of gold and copper. Its total weight is 96 g. If $\dfrac{1}{12}$ of the ornament is copper, find the weight of gold in it.

Answer

Total weight of the ornament = 96 g.

Weight of copper = $\dfrac{1}{12} \text{ of } 96 = \dfrac{1}{12} \times 96$ = 8 g.

Weight of gold = Total weight − Weight of copper

= 96 − 8

= 88 g.

Hence, the weight of gold = 88 g.

A girl did half of some work on Monday and one-third of it on Tuesday. How much will she have to do on Wednesday in order to complete the work?

Answer

Let the total work = 1.

Work done on Monday = $\dfrac{1}{2}$.

Work done on Tuesday = $\dfrac{1}{3}$.

Total work done in two days = $\dfrac{1}{2} + \dfrac{1}{3}$

$= \dfrac{3}{6} + \dfrac{2}{6} \\[1em] = \dfrac{5}{6}.$

Work left for Wednesday = Total work - Work done in two days

Work left for Wednesday = 1 − $\dfrac{5}{6}$

$= \dfrac{6}{6} - \dfrac{5}{6} \\[1em] = \dfrac{1}{6}.$

Hence, she will have to do $\dfrac{1}{6}$ of the work on Wednesday.

A man spends $\dfrac{3}{8}$ of his money and still has ₹ 720 left with him. How much money did he have at first?

Answer

Part of money spent = $\dfrac{3}{8}$.

Part of money left = 1 − $\dfrac{3}{8} = \dfrac{8}{8} - \dfrac{3}{8} = \dfrac{5}{8}$.

Given, $\dfrac{5}{8}$ of the money = ₹ 720.

$\Rightarrow \text{Total money} = 720 \div \dfrac{5}{8} \\[1em] = 720 \times \dfrac{8}{5} \\[1em] = 144 \times 8 \\[1em] = 1152.$

Hence, the man had ₹ 1,152 at first.

In a school, $\dfrac{4}{5}$ of the students are boys, and the number of girls is 100. Find the number of boys.

Answer

Part of students who are boys = $\dfrac{4}{5}$.

Part of students who are girls = 1 − $\dfrac{4}{5} = \dfrac{1}{5}$.

Given, $\dfrac{1}{5}$ of the students = 100 girls.

$\Rightarrow \text{Total students} = 100 \div \dfrac{1}{5} \\[1em] = 100 \times 5 \\[1em] = 500.$

Number of boys = $\dfrac{4}{5} \text{ of } 500 = \dfrac{4}{5} \times 500$ = 400.

Hence, the number of boys = 400.

After finishing $\dfrac{3}{4}$ of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?

Answer

Given, $\dfrac{3}{4}$ of the journey = 12 km.

$\Rightarrow \text{Total journey} = 12 \div \dfrac{3}{4} \\[1em] = 12 \times \dfrac{4}{3} \\[1em] = 16 \text{ km}.$

Distance left = Total journey − Distance covered

= 16 − 12

= 4 km.

Hence, the distance still left to be covered = 4 km.

When Ajit travelled 15 km, he found that one-fourth of his journey was still left. What was the full length of the journey?

Answer

Part of journey left = $\dfrac{1}{4}$.

Part of journey travelled = 1 − $\dfrac{1}{4} = \dfrac{3}{4}$.

Given, $\dfrac{3}{4}$ of the journey = 15 km.

$\Rightarrow \text{Total journey} = 15 \div \dfrac{3}{4} \\[1em] = 15 \times \dfrac{4}{3} \\[1em] = 20 \text{ km}.$

Hence, the full length of the journey = 20 km.

In a particular month, a man earns ₹ 7,200. Out of this income, he spends $\dfrac{3}{10}$ on food, $\dfrac{1}{4}$ on house rent, $\dfrac{1}{10}$ on insurance and $\dfrac{2}{25}$ on holidays. How much did he save in that month?

Answer

Total income = ₹ 7,200.

Fraction of income spent = $\dfrac{3}{10} + \dfrac{1}{4} + \dfrac{1}{10} + \dfrac{2}{25}$

LCM of 10, 4, 10 and 25 = 100.

$= \dfrac{30}{100} + \dfrac{25}{100} + \dfrac{10}{100} + \dfrac{8}{100} \\[1em] = \dfrac{73}{100}.$

Fraction of income saved = 1 − $\dfrac{73}{100} = \dfrac{27}{100}$.

Money saved = $\dfrac{27}{100}$ of 7,200

$= \dfrac{27}{100} \times 7,200 \\[1em] = 27 \times 72 \\[1em] = 1,944.$

Hence, the man saved ₹ 1,944 in that month.