A man has three sons. First and second son can complete a work in 24 d
| A man has three sons. First and second son can complete a work in 24 days and 36 days respectively. The man himself can complete the same work in 36/11 days. If the man does twice the work in the same time his three sons complete the whole work together, then in how many days his third son will complete the work alone?
A. 10 days
B. 12 days
C. 15 days
D. 8 days
Right Answer is: B
SOLUTION
Let the total work = LCM(24,36) = 72 units
Efficiency of first son = 3 units
Efficiency of second son = 2 units
Efficiency of man himself = 72/(36/11) = 22 units
According to the question,
Efficiency of man = Twice the sum of efficiencies of his sons
22 = 2(Efficiency of first son + Efficiency of second son + Efficiency of third son)
11 = 2 + 3 + Efficiency of third son
Efficiency of third son = 11-5 = 6 units
Time taken by third son to complete the work = 72/6 = 12 days
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A man has three sons. First and second son can complete a work in 24 d