Question

8 men and 20 boys, working 10 hours a day, can do a piece of work in 60 days. If a man's work is equal
to that of 2 boys, then how many boys will be required to help 19 men to do twice the work in 50 days
working 9 hours a day?
​

Answer 1

Given:

8 men and 20 boys, working 10 hours a day, can do a piece of work in 60 days.

A man's work is equal to that of 2 boys,

To find:

How many boys will be required to help 19 men to do twice the work in 50 days working 9 hours a day?

Solution:

Let "x" represents the required no. of boys to help 19 men.

A man's work is equal to that of 2 boys

i.e., M = 2B

So,

8M + 20 B = (8 × 2B) + 20 B = 36B

and

19M + xB = (19 × 2B) + xB = 38B + xB

Also, the 19 men + x boys do twice the work than 8 men and 20 boys

i.e., $\frac{W_1}{W_2} = \frac{1}{2}$

The formula of Man-hours-days is as follows:

$\boxed{\bold{\frac{M_1 \times H_1 \times D_1}{W_1} = \frac{M_2 \times H_2 \times D_2}{W_2} }}$

Now, using the above formula, we get

$\frac{(8M + 20B) \times 10 \times 60}{1} = \frac{(19M+xB) \times 9 \times 50}{2} }}$

$\implies \frac{36B \times 10 \times 60}{1} = \frac{(38B+xB) \times 9 \times 50}{2} }}$

$\implies \frac{36 \times 10 \times 60}{1} = \frac{(38+x) \times 9 \times 50}{2} }}$

$\implies 12 \times 2 \times 2 \times 2 = (38+x)$

$\implies 96= 38+x$

$\implies x = 96 - 38$

$\implies \bold{x = 58}$

Thus, 58 boys will be required to help 19 men to do twice the work in 50 days working 9 hours a day.

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