Question : A and B can do a piece of work in 10 days and 15 days, respectively. They worked together, but 5 days before the completion of the work, A left. In how many days was the work completed?
Option 1: 4
Option 2: 8
Option 3: 6
Option 4: 9
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Correct Answer: 9
Solution :
Given: Time taken by A = 10 days
Time taken by B = 15 days
A left the work 5 days before the completion of the work.
Let the total work be 1 unit.
⇒ A's 1 day work = $\frac{1}{10}$
And B's 1 day work = $\frac{1}{15}$
Both (A + B)'s 1 day's work = $\frac{1}{10}+\frac{1}{15}$ = $\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$
Let $t$ be the number of days A and B worked together before A left.
Now, A left 5 days before the completion of the work
⇒ For $(t-5)$ days, A and B worked together.
Work done in $(t-5)$ days = $\frac{1}{6}\times (t-5)$
Work done by B in 5 days = $5\times\frac{1}{15}$ = $\frac{1}{3}$
Now, Work done together in $(t-5)$ days + work done by B alone in 5 days = 1
⇒ $\frac{1}{6}\times (t-5)+\frac{1}{3}=1$
⇒ $\frac{t-5+2}{6}=1$
⇒ $t-3= 6$
$\therefore t=9$ days
Hence, the correct answer is 9.
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