A can do a piece of work in 24 days. If B is 60% more efficient than A, then find the number of days required by B to do the twice as large as the earlier work.

Question

Sunil Raikwar · Answer

Let the whole work be one unit.

So work done by A in one day=1/24

Work done by B in one day=(1/24)+ 60% of (1/24)=(1/24)+(1/40)=8/120=1/15

So B can complete the whole work in 15 days.

Thanks & Regards

Sunil Raikwar

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A can do a piece of work in 24 days. If B is 60% more efficient than A, then find the number of days required by B to do the twice as large as the earlier work.

A can do a piece of work in 24 days. If B is 60% more efficient than A, then find the number of days required by B to do the twice as large as the earlier work.

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A can do a piece of work in 24 days. If B is 60% more efficient than A, then find the number of days required by B to do the twice as large as the earlier work.

A can do a piece of work in 24 days. If B is 60% more efficient than A, then find the number of days required by B to do the twice as large as the earlier work.

1 Answers

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Other Related Questions on 10 grade maths

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