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A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 12 days and C alone in 24 days, then B alone could do it in:
24 days
48 days
36 days
72 days
48 days
Given: Time for A+B = 12 days, time for C = 24 days. A completes work in same time as B+C.
Time for A+B = 12 days, time for C = 24 days. A completes work in same time as B+C.
Efficiency=TimeTotal┬аWorkтАЛ
Use the LCM of 12 and 24 which is 24 units. Determine individual efficiencies, solve the linear system by equating total work rate of A with B+C.
Students often assume A, B, and C have equal efficiencies or confuse the A = B+C relation with A+B+C = 1.
Define efficiencies
Let total work be the LCM of 12 and 24, which is 24 units. Thus, efficiency of A+B is 24/12=2 units/day and efficiency of C is 24/24=1 unit/day.
EAтАЛ+EBтАЛ=2,ECтАЛ=1
Use the A = B+C condition
We are given that A takes the same time as B+C, meaning their efficiencies are equal: EAтАЛ=EBтАЛ+ECтАЛ.
EAтАЛтИТEBтАЛ=ECтАЛ=1
Solve for EBтАЛ
We have a system of two equations: EAтАЛ+EBтАЛ=2 and EAтАЛтИТEBтАЛ=1. Subtracting the second from the first gives 2EBтАЛ=1, so EBтАЛ=0.5 units/day.
2EBтАЛ=1тЯ╣EBтАЛ=0.5
Calculate time for B
Time taken by B = Total work / Efficiency of B, which is 24/0.5=48.
TimeBтАЛ=0.524тАЛ=48
B is correct because the calculation of B's efficiency (0.5 units/day) leads to a total time of 48 days.
This logic is identical to mixture and allegation problems where the sum of two rates equals a third rate; always look for the sum of variables.