Question : A, B, and C can do a job in 6 days, 12 days and 15 days respectively. After $\frac{1}{8}$th of the work is completed, C leaves the job. The rest of the work is done by A and B together. The time taken to finish the work is:
Option 1: $5\frac{5}{6}$ days
Option 2: $5\frac{1}{4}$ days
Option 3: $3\frac{1}{2}$ days
Option 4: $3\frac{3}{4}$ days
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Correct Answer: $3\frac{1}{2}$ days
Solution :
Given: A, B, and C can do a job in 6 days, 12 days, and 15 days respectively.
Total work = Efficiency × Time
Total work is the LCM of (6, 12, 15) = 60 units
Efficiency of A $=\frac{60}{6}=10$
Efficiency of B $=\frac{60}{12}=5$
Efficiency of C $=\frac{60}{15}=4$
C has done $\frac{1}{8}$th of the work, which is $\frac{1×60}{8} = \frac{15}{2}$ units
Work left is given as $60-\frac{15}{2} = \frac{105}{2}$ units
Left work is done by both A and B together.
$\therefore$ Time taken by A and B $=\frac{\text{Work}}{\text{Efficiency}} = \frac{105}{2×(10+5)}=\frac{105}{30} = 3\frac{1}{2}$ days
Hence, the correct answer is $3\frac{1}{2}$ days.
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