Question : S does half as much work as T in $\frac{1}{8}$th of the time taken by T. If they together complete a work in 60 days, then how many days shall S alone take to complete that work?
Option 1: 100 days
Option 2: 80 days
Option 3: 90 days
Option 4: 75 days
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Correct Answer: 75 days
Solution :
Let the time T takes to complete the work = $x$ days
⇒ the time T takes to complete half of the work = $\frac{x}{2}$ days
⇒ So, S will take to complete half of the work = $\frac{1}{8x}$
So, $\frac{\text{Efficiecncy of T}}{\text{Efficiency of S}}=\frac{\text{Time taken by S}}{\text{Time taken by T}}$
= $\frac{\frac{1}{8x}}{\frac{x}2}$
= $\frac{1}{4}$
Together they complete the work in 60 days,
⇒ Total work = 60 × (1 + 4) = 300 units
S will take to finish the whole work = $\frac{300}4$ = 75 days
Hence, the correct answer is 75 days.
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