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 isChapter/Topic:  Comparing Quantities

# 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 isChapter/Topic:  Comparing Quantities

1. A
24
2. B
36
3. C
15
4. D
30

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### 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 $\frac{1}{24}\right)$​)​= $\frac{1}{24}$​ ​×$\frac{160}{100}$​​=$\frac{1}{15}$
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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