Question : A can do a certain task in 1.5 times the number of days in which B and C together can do it. If A and B together can do the task in 30 days, and C alone can do it in 120 days, how many days will B take to do this task alone?
Option 1: 75
Option 2: 45
Option 3: 60
Option 4: 50
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Correct Answer: 60
Solution :
Work done by (A + B) in 1 day $=\frac{1}{30}$
Work done by C in 1 day $=\frac{1}{120}$
Work done by (A + B + C) in 1 day $=\frac{1}{30} + \frac{1}{120} = \frac{4 + 1}{120} = \frac{5}{120} = \frac{1}{24}$
The ratio of the time taken by A to the time taken by B and C together = 1.5 : 1 = 3 : 2
Let A complete the task in 3$k$ days and B and C together, in 2$k$ days.
Work done by (A + B + C) in 1 day $=\frac{1}{2k}+\frac{1}{3k} =\frac{1}{24}$
$⇒\frac{5}{6k} =\frac{1}{24}$
$\therefore k=20$
Work done by A in 1 day = $\frac{1}{60}$
Work done by B in 1 day = Work done by (A + B) in 1 day – Work done by A in 1 day $=\frac{1}{30} - \frac{1}{60}=\frac{1}{60}$
$\therefore$ B completed the task in 60 days.
Hence, the correct answer is 60 days.
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