Question

A and B together complete a task in 18 hours. After 6 hours A leaves. B takes 36 hours to finish rest of the task. How many hours would A have taken to do the task if he worked alone ?
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Answer 1

27 hour....................

Answer 2

A alone can complete the work in 27 hours

Solution:

A and B together complete a task in 18 hours

Therefore,

A + B can do in one hour is $\frac{1}{18}$

After 6 hours A leaves

Therefore, for 6 hours, A + B work is:

$6 \times \frac{1}{18} = \frac{1}{3}$

$Remaining\ work = 1 - \frac{1}{3} = \frac{2}{3}$

2/3 of work has to be done by B

B takes 36 hours to finish rest of the task

So complete work will be done by B in:

$36 \times \frac{3}{2} = 54\ hours$

Therefore,

$\frac{1}{A} + \frac{1}{54} = \frac{1}{18}\\\\\frac{1}{A} = \frac{1}{18} - \frac{1}{54}\\\\\frac{1}{A} = \frac{3}{54} - \frac{1}{54}\\\\\frac{1}{A} = \frac{2}{54}\\\\\frac{1}{A} = \frac{1}{27}$

Thus A alone can complete the work in 27hours

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