Question

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​

Answer 1

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

Answer 2

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