Question : To do a certain work, B would take time thrice as long as A and C together and C twice as long as A and B together. The three men together complete the work in 10 days. The time taken by A to complete the work separately is:
Option 1: 22 days
Option 2: 24 days
Option 3: 30 days
Option 4: 20 days
Correct Answer: 24 days
Solution :
If B does the work in 3$x$ days, (A + C) will do the same work in $x$ days.
If C does that work in 2$y$ days. (A + B) will do it in $y$ days.
According to the question,
$\frac{1}{x}+\frac{1}{3x}=\frac{1}{10}$
⇒ $\frac{4}{3x}=\frac{1}{10}$
⇒ $3x=40$
⇒ $x=\frac{40}{3}$
So, B takes $3×\frac{40}{3}$ = 40 days
Again, $\frac{1}{y}+\frac{1}{2y}=\frac{1}{10}$
⇒ $\frac{3}{2y}=\frac{1}{10}$
⇒ $y=15$
So, C takes $2×15$ = 30 days
(A + B + C)'s 1 day work = $\frac{1}{10}$
⇒ $\frac{1}{A}+\frac{1}{40}+\frac{1}{30}=\frac{1}{10}$
⇒ $\frac{1}{A}=\frac{1}{10}-\frac{1}{40}-\frac{1}{30}$
⇒ $\frac{1}{A}=\frac{12-3-4}{120} = \frac{1}{24}$
$\therefore$ A alone will complete work in 24 days.
Hence, the correct answer is 24 days.
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