Question
A alone can finish the work in 10days and B alone can
finish the same work in 15 days. All A, B and C together take 5 days to finish the work. Find the time in which the same work can be completed in a sequence such that A starts first then B and then C.Solution
Total amount of work = LCM(10,15,5) = 30 Efficiency of A: 30/10 = 3units/day Efficiency of B: 30/15 = 2units/day Efficiency of A+B+C = 5 units/day => Efficiency of C=6-3-2=1 unit/day Work done by A+B+C together in 3 days = (3+2+1)×1 = 6 unit Total number of days required to finish the work: 30/6 => 5×3 = 15 days
Statements: N > U < I ≤ R < D < A; K ≥ O = G > X = E > A
Conclusions:
I. R < E
II. K > D
III. U ≥ X
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