6 men can complete a piece of work in 10 days while 3 women can do it in 14 days. In how many days can 7 women and 5 men complete it?

Correct option is B

Given:

6 men complete the work in 10 days.

3 women complete the work in 14 days.

We need to find the time taken if 7 women + 5 men work together.

Formula Used:

Work = Rate × Time

Solution:

Work done by 6 men in 10 days = 1 (whole work).
So, work done by 1 man in 1 day:

​$$\frac{1}{6 \times 10} = \frac{1}{60}$$​

Work done by 3 women in 14 days = 1 (whole work).
So, work done by 1 woman in 1 day:

​$$\frac{1}{3 \times 14} = \frac{1}{42}$$​

Work done by 5 men in 1 day:

​$$5 \times \frac{1}{60} = \frac{5}{60} = \frac{1}{12}$$​

Work done by 7 women in 1 day:

​$$7 \times \frac{1}{42} = \frac{7}{42} = \frac{1}{6}$$​

Combined 1 day work =

​$$\frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4}$$​

Total days required =$$\frac{1}{\tfrac{1}{4}} = 4 \, \text{days}$$​

7 women and 5 men can complete the work in 4 days.

Alternate Method:

Total work = LCM of (60, 42) = 420 units.

1 man’s efficiency = 420 ÷ 60 = 7 units/day.

1 woman’s efficiency = 420 ÷ 42 = 10 units/day.

5 men’s efficiency = 5 × 7 = 35 units/day.

7 women’s efficiency = 7 × 10 = 70 units/day.

Combined efficiency = 35 + 70 = 105 units/day.

Days = 420 ÷ 105 = 4 days.