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 to do the work twice as large the earlier work
Answers
Answered by
52
Answer:
The number of days required by B is 72/5 or 14.4 days.
Step-by-step explanation:
60%of 24= 3×24/5=72/5=14.4 Days
Answered by
117
Step-by-step explanation:
W.K.T ,A can do work in 24 days
B is 60% efficient than A
therefore
24/100 × 60
=24 / 5 × 3
=72 / 5
=14.4
given,
2 times larger wark than earlier
14.4 × 2
=28.8 approximately 30
days required by B to do the twice as large as earlier work is 30
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