Question : A alone can do a piece of work in 10 days. B alone can do the same work in 12 days. They work on alternate days starting with A. In how many days will they complete $\frac{3}{4}$th of the total work?
Option 1: $\frac{47}{3}$ days
Option 2: $\frac{50}{3}$ days
Option 3: $\frac{53}{6}$ days
Option 4: $\frac{49}{6}$ days
Correct Answer: $\frac{49}{6}$ days
Solution :
Given: A alone can do a piece of work in 10 days. B alone can do the same work in 12 days. They work on alternate days starting with A.
Total work = Time × Efficiency
The total work done overall = LCM(10, 12) = 60
A's efficiency $=\frac{60}{10}=6$
B's efficiency $=\frac{60}{12}=5$
Total work of two days = 5 + 6 = 11
The $\frac{3}{4}$ of the total work = $60\times \frac{3}{4}$ = 45
The time taken to complete 44 work = $\frac{44}{11}$ = 4 pair = 4 × 2 = 8 days
And A will work once more the following day to finish the remaining unit work = $\frac{1}{6}$
The number of days will they complete $\frac{3}{4}$ of the total work $=8+\frac{1}{6}= \frac{48+1}{6}=\frac{49}{6}$ days
Hence, the correct answer is $\frac{49}{6}$ days.
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