Question 58
A, B and C can individually complete a piece of work in 24 days, 15 days and 12 days,respectively. B and C started the work and worked for 3 days and left. The number of days required by A alone to complete the remaining work, is:
Solution
Let the total work be 120 units.
$$(\because$$ LCM of 24, 15 and 12 is 120.)
Efficiency of A = work/time = 120/24 = 5 units/day
Efficiency of B = 120/15 = 8 units/day
Efficiency of C = 120/12 = 10 units/day
Work done by B and C in 3 days = (8 + 10) $$\times 3 = 18 \times$$3 = 54 units
Remaining work = 120 - 54 = 66 units
Time taken by A to complete the remaining work = 66/5 = 13 $$\frac{1}{5}$$
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