Question

A can do a piece of work in 32 days and B in 48 days. They work together for 8 days and then A goes away.In how much time (in days) will B finish the 60% of the remaining work?

• A. 18 2/5
• B. 19 3/4
• C. 16 4/5
• D. 17 3/7

Answer 1

Answer:

C. 16 4/5

A’s 1 day work $(= \frac{1}{32})$
B’s 1 day work $(= \frac{1}{48})$

Work done together in 1 day:

$\frac{1}{32} + \frac{1}{48} = \frac{3+2}{96} = \frac{5}{96}$

Work done in 8 days together:

$8 \times \frac{5}{96} = \frac{40}{96} = \frac{5}{12}$

Remaining work:

$1 - \frac{5}{12} = \frac{7}{12}$

60% of the remaining work:

$60$

Time taken by B alone:

$B’s 1 day work (= \frac{1}{48})$

$\text{Time} = \frac{\frac{7}{20}}{\frac{1}{48}} = \frac{7}{20} \times 48 = \frac{336}{20} = 16.8$

Answer: 16.8 days