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1. Let eff. be A, B and C respectively.
If B takes three times as long as A and C together then B=(A+C)/3 as efficiency is inversely proportional to time.

and C twice as long as A and B together--> C=(A+B)/2

Total work done = (A+B+C)*10 = A*t1 = B*t2 =C*t3
From the options, i can see that value of C is different in each option.
(A+B+C)*10 = C*t3
(2C+C)*10 = C*t3
3C*10= C*t3
Therefore, t3=30. Only option B corresponds to it. hence the answer.
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I took a simpler approach: All three men completed the job in 10 days. Simply use the options to get to the answer:
Individually the finished in 24, 40, 30 days respectively (Option B) so together they'd have completed the work in 1/24+1/40+1/30 days taking LCM; (5+3+4)/12=12/120=10 days Therefore B
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