Question

A and B together can paint a room in 18 days. They work together for 8 days and then B goes away. A finishes the rest of work in 15 days. How long would each take to finish the work separately?

Answer 1

A will alone take 27 days to paint a room and B alone will take 54 days to finish the same work.  

Step-by-step explanation:

A and B together can paint a room in 18 days

So, work done by (A+B) in 1 day = $\frac{1}{18}$

∵ They have worked together for only 8 days

∴ Work done by (A+B) in 8 days = $\frac{8}{18} = \frac{4}{9}$

∴ The remaining amount of work = 1 - $\frac{4}{9} = \frac{5}{9}$ ← done by A alone in 15 days

So,

If $\frac{5}{9}$th work is done by A alone in 15 days

Then, 1 work will be done by A in = 15 × $\frac{9}{5}$ = 3 × 9 = 27 days

∴ Work done A alone in 1 day = $\frac{1}{27}$

Now,

The work done By B alone to finish the same work will be given by,

= [work done by (A+B) in 1 day] - [work done by A alone in 1 day]

= $\frac{1}{18} - \frac{1}{27}$

= $\frac{3 - 2}{54}$

= $\frac{1}{54}$

Thus, A would take 27 days and B would take 54 days to finish the work separately.

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Answer 2

A alone can finish the work in 27 days

B alone can finish the work in 54 days

Let the total work = x

A and B can together paint in 18 days .

A can finish 10 days work of A and B together in 15 days .

Therefore , 1 day work of A and B = 1.5 days work of A

A can finish the total work alone in = 1.5 × 18 = 27 days

One day work of A and B together = one day work of A + one day work of B

One days work = total work / time taken to complete the work

=> x/18 = x/27 + x/t [t - time taken by B alone to complete the work]

=> 1/18- 1/27 = 1/t

=> t = 54 days

Time taken by B alone = 54 days