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?
[This Question was asked as TITA]

• A. 3
• B. 4
• C. 6
• D. 9

Answer 1

The Correct Option is B

Let the individual times taken by John, Jack and Jim to complete the works be a, b and c respectively.
Given, 
\(a = 2b \)       …....(1)
\(\frac {bc}{b+c}= \frac 13(a)\)      ......(2)

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

So, the correct option is (B): \(4\)

Answer 2

Let the time taken by Jack to complete the work be x. 
⇒ Time taken by John is 2x
Let the time taken by Jim be y to complete the work.
As per the question, ⇒ 2x/3 = (x × y)/(x + y) ⇒ 2/3 = y/(x + y)
⇒ 2x + 2y = 3y ⇒ 2x = y
John takes three days more than, taken by all of them working together One day work,
⇒ 1/x + 1/2x + 1/2x ⇒ 4/2x
⇒ 2/x Time taken by all = x/2
Now according to the question,
⇒ 2x – x/2 = 3 
⇒ 4x – x = 6
⇒ 3x = 6 
⇒ x = 6/3 
⇒ x = 2
Time taken by Jim is y, 
⇒ y = 2x
⇒ y = 2 × 2
⇒ y = 4

∴ Time taken by Jim to complete the work is 4 days.