Correct Answer - Option 4 : 90
Given:
A and B can do a piece of work in 36 days,
B and C can do the same work in 60 days,
A and C can do the same work in 45 days.
Concept used:
Efficiency = 1/(Total time)
Calculation:
(A + B)’s 1 day work = 1/36 ----(i)
(B + C)’s 1 day work = 1/60 ----(ii)
(C + A)’s 1 day work = 1/45 ----(iii)
From equations (i), (ii) and (iii),
2(A + B + C)’s 1 day work = 1/36 + 1/60 + 1/45
⇒ (5 + 3 + 4)/180 = 1/15
⇒ (A + B + C)’s 1 day work = 1/(15 × 2) = 1/30 ----(iv)
From equations (iii) and (iv) we get
B’s 1 day work = 1/30 – 1/45 = (3 – 2)/90 = 1/90
Time taken by B alone to complete the work = 90
∴ B alone can complete the same work in 90 days
Alternate method:
Let us find the LCM of 36, 60 and 45
LCM of 36, 60 and 45 = 180
Let the total work be 180 units.
A + B can do = 180/36 = 5 units/day ----(i)
B + C can do = 180/60 = 3 units/day ----(ii)
C + A can do = 180/45 = 4 units/day ----(iii)
Adding the above equations, we get
2(A + B + C) = 5 + 3 + 4
⇒ A + B + C = 12/2 = 6 ----(iv)
Subtracting equation (iii) from (iv)
B = 6 - 4 = 2units/day
⇒ Time taken by B = 180/2 = 90days
∴ B alone can complete the same work in 90 days
