A takes twice as long as B and C together take to do a piece of work and B takes thrice as long as A and C together. However, the 3 men together can complete the work in 60 days. How long would C alone take to do the work?
Solution
Correct Answer: Option B
Let’s assume that A, B and C can individually finish the work alone in a, b and c days respectively.
∴ Part of work finished by A in one day = 1/a
∴ Part of work finished by A in one day = 1/b
∴ Part of work finished by A in one day = 1/c
Now, as per the given information –
A takes twice as long as B and C together ⇒ A does half the work of B and C together in equal time.
\(\Rightarrow \frac{1}{a} = \frac{1}{2} \times \left( {\frac{1}{b} + \frac{1}{c}} \right)\)
\(\Rightarrow \frac{2}{a} = \frac{1}{b} + \frac{1}{c}\) -----(i)
Similarly, B takes thrice as long as A and C together ⇒ B does 1/3 of (A & C) in equal time
\(\Rightarrow \frac{1}{b} = \frac{1}{3} \times \left( {\frac{1}{a} + \frac{1}{c}} \right)\)
\(\Rightarrow \frac{3}{b} = \frac{1}{a} + \frac{1}{c}\) -----(ii)
Also, A, B, and C together can complete the work in 60 days.
\(\therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{60}}\) ------(iii)
Putting the value of (i) into (iii), we get,
\(\begin{array}{l} \Rightarrow \frac{1}{a}\; + \frac{2}{a}\; = \frac{1}{{60}}\\ \Rightarrow \frac{3}{A} = \frac{1}{{60}}\\ \Rightarrow \frac{1}{A} = \frac{1}{{180}} \end{array}\)
∴ A takes 180 days to complete the entire work alone.
\(\begin{array}{l} \frac{2}{a} - \frac{1}{b} = \frac{3}{b} - \frac{1}{a}\\ \Rightarrow \frac{3}{a} = \frac{4}{b} \end{array}\)
⇒ b = (4 × 180)/3 = 240
∴ B takes 240 days to complete the entire work alone.
Now, putting value of a and b in equation (iii) we get,
\(\begin{array}{l} \Rightarrow \;\frac{1}{{180}}\; + \;\frac{1}{{240}}\; + \;\frac{1}{c}\; = \;\frac{1}{{60}}\\ \Rightarrow \;\frac{1}{C}\; = \;\frac{1}{{60}}\;--\;\left( {\frac{1}{{180}}\; + \;\frac{1}{{240}}} \right)\\ \Rightarrow \;\frac{1}{C}\; = \;\frac{1}{{144}} \end{array}\)
Hence, C takes 144 days to complete the entire work alone.
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