Question

A can do a piece of work in 25 days and B can do it in 20 days. They work at it together for 5 days and then A goes away. In how many days will B finish the remaining work?
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Answer 1

Answer:

Let B takes x more days . Now Follow the question

$5 \times ( \frac{1}{20} + \frac{1}{25} ) + x \times \frac{1}{25} = 1 \\ 25 + 20 + 4x = 100 \\ 4x = 55 \\ x = \frac{55}{4} = 13.75$

So x will complete the work in 14th day (exactly 13.75 days)

Answer 2

A can do a piece of work in 25 days.

Work done by A in one day = $\sf{\frac{1}{25}}$

B do a piece of work in 20 days.

Work done by B in one day = $\sf{\frac{1}{20}}$

Now,

Work done by both A and B in one day = $\sf{\frac{1}{25}}$ + $\sf{\frac{1}{20}}$

=> $\sf{\frac{4\:+\:5}{100}}$

=> $\sf{\frac{9}{100}}$

Both A and B work it together for 5 days.

So,

=> $\sf{\frac{9}{100}\:\times\:5}$

=> $\sf{\frac{9}{20}}$

Remaining work = $\sf{1\:-\:\frac{9}{20}}$

=> $\sf{\frac{11}{20}}$

B alone do a piece of work in 20 days.

So,

Time taken by B to complete the remaining work = $\sf{20\:\times\:\frac{11}{20}}$

=> $\sf{11\:days}$