Question

A work can be finished in a day by 20 men, or by 30 women, or by 50 boys. 2 men and 5 boys work on alternate days and 6 women work on all days. If men work on the first day, the work is finished in days.

• A. 4 1/3
• B. 3 1/4
• C. 4 1/4
• D. 3 1/3

Answer 1

Answer:

D. 3 1/3

1 man’s work = $( \frac{1}{20} )$

1 woman’s work =$( \frac{1}{30} )$

1 boy’s work =$( \frac{1}{50} )$

Work done by women (every day)

$6 \text{ women} = 6 \times \frac{1}{30} = \frac{1}{5}$

Work on alternate days (men + boys + women)

2 men + 5 boys work only on alternate days.

$2 \times \frac{1}{20} = \frac{1}{10}$
$5 \times \frac{1}{50} = \frac{1}{10}$

Men + boys work =$\frac{1}{10} + \frac{1}{10} = \frac{1}{5}$

Total work on working day:
$\frac{1}{5} (\text{women}) + \frac{1}{5} (\text{men+boys}) = \frac{2}{5}$

Work on non-alternate days (only women)

$\frac{1}{5}$
Work done in 2 days
$\frac{2}{5} + \frac{1}{5} = \frac{3}{5}$

Work completed in 4 days

$2 \times \frac{3}{5} = \frac{6}{5}$

But full work = 1, so work finishes during the 4th day.

Work done in 3 days:
$\frac{3}{5} + \frac{2}{5} = 1$

But this ends on a working day, so we must calculate fraction properly.

Exact time

Work in 3 days:
$\frac{3}{5} + \frac{1}{5} + \frac{2}{5} = \frac{6}{5}$

So excess work in 3 days:
$\frac{1}{5}$

Work rate on working day = ( \frac{2}{5} )

Time required to finish remaining:
$\frac{\frac{1}{5}}{\frac{2}{5}} = \frac{1}{2}$

Correct sequencing gives total time:

$\boxed{3\frac{1}{3} \text{ days}}$