A can do 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?
Answers
Answered by
2
Answer:
28.8 days
Step-by-step explanation:
Given, Time taken by A to do a work
= 24 days
B is 60% more efficient than B
Therefore, Time taken by A to do twice the work = 24 * 2 = 48 days
Hence, Time taken by B to do twice as large as earlier work = 60% of 48
= 60 * 48/100 = 144/5 = 28.8 days or
28 days 19 hours 12 minutes
Hence B will take 28.8 days to do the required work
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Answered by
1
Answer:
Step-by-step explanation:
Given, Time taken by A to do a work
= 24 days
B is 60% more efficient than B
Therefore, Time taken by A to do twice the work = 24 * 2 = 48 days
Hence, Time taken by B to do twice as large as earlier work = 60% of 48
= 60 * 48/100 = 144/5 = 28.8 days or
28 days 19 hours 12 minutes
Hence B will take 28.8 days to do the required work
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