Question

A can do a piece of work in 80 days he works for 10 days and then b alone fininshes the remaining work in 42 days the two together could complete the work in

Answer 1

A’s Efficiency = 1/80

A works for 10 days therefor completed work by A = 10* (1/80) =10/80 = 1/8

Remaining work = 1-(1/8) = 7/8

B completed the remaining job in 42 days

Therefor B alone can complete the job in =

(7/8) /1/b = 42 , b is the efficiency of B

(7/8 )*b = 42

b= 42*8/7 =48

Efficiency of B = 1/48

Therefor A & B can complete the work in = (1/80) + (1/48)

= 1/30 They can complete the work in 30 days

Answer 2

A's 1 days' work = 1/80

A's 10 days' work = 10/80 = 1/8

Remaining work =1 − 1/8 = 7/8

7/8 of the work is completed by B in 42 days

∴ The whole work is completed by B in

$\frac{42 \times 8}{7} = 48 \: \: days$

∴ B's 1 days' work = 1/48

(A + B)'s 1 days' work = 1/80 + 1/48 = 3+5/240=8/240=1/30

∴ A and B can complete the whole work together in 30 days