Question : A is three times as good a worker as B and therefore, able to finish a piece of work in 60 days less than B. The time (in days) in which they can do it working together is:
Option 1: 22 days
Option 2: $22\frac{1}{2}$ days
Option 3: 23 days
Option 4: $23\frac{1}{4}$ days
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Correct Answer: $22\frac{1}{2}$ days
Solution :
Given: A : B = 3 : 1
A takes = $x$ days and
B takes = $x$ + 60 days
If the difference of time is 2 days, B takes 3 days.
If the difference in time is 60 days,
B takes = $\frac{3}{2}$ × 60 = 90 days
So, A takes = 30 days
Total work = LCM of 90 and 30 is 90.
Efficiency of A = $\frac{90}{30}$ = 3 units
Efficiency of B = $\frac{90}{90}$ = 1 unit
We know that,
Work = efficiency × time
⇒ Time $=\frac{\text{Total work}}{\text{Efficiency (A+B)}}=\frac{90}{4}=22\frac{1}{2}$ days
Hence, the correct answer is $22\frac{1}{2}$ days.
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