Question

John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

Answer 1

Let the individual times taken by John, Jack and Jim to complete the works be \(a, b\) and \(c\) respectively
Given, \(a = 2b\)....(1)
\(\frac{bc}{b+c} = \frac{1}{3}(a)\)....(2)

\(a-\bigg(\frac{abc}{ab+bc+ac}\bigg)=3\)...(3)
From \((1)\) and \((2)\), we've \(c\) = \(2b\)
\(∴\)  \(2b - \bigg(\frac{2b×b×2b}{(2b)(b)+b(2b)+(2b)(2b)}\bigg) = 3\)
\(⇒ b = 2\)
\(∴ c = 4\)

Answer 2

John = \(x\) units/day
Jack = \(2x\) units/day
Jim = \(x\) units/day (1/3rd of John does = \(3x\) units, out of which jack does \(2x\))
Total = \(4x\) units/day
John’s time taken = \(x(n+3) = 4x × n\)
Then n = 1 day for \(4x\) units

For \(x\) units, Jim will take 4 days