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

Chapter/Topic: Comparing Quantities

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answer is D.

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Detailed Solution

We have to find the number of days B will take to finish a work that is twice as big as the earlier one.
If a person can do a work in x days, then, we can say that in 1 day that person can do 1/x.
A's 1 day's work =

\frac{1}{24}

We will now find the number of days required by B to do the same work as A. We know that B is 60% more efficient than A. So, the amount of work done by B in one day can be calculated as follows,
B's 1 day's work =

\frac{1}{24}

+(60% of

\frac{1}{24})

\frac{1}{24}

\frac{160}{100}

\frac{1}{15}

B can do same work in 15 days
Therefore, the number of days required by B to complete the same quantity of work as A is 15 days.
Now, we are asked to find the number of days B will take to do a work which is twice as large as the earlier one. So, we will multiply the days taken by B to finish the earlier work by 2.
We know that the number of days required by B to finish the original work is 15 days. So, to finish a work which is 2 times the previous work
B can do double work in2×15 = 30 days
Hence, Option (4) is the correct answer.

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