Question

Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is:

• A. \(240\)
• B. \(260\)
• C. \(300\)
• D. \(340\)

Hint:

In work and time problems with different efficiencies, it is often easiest to: Assume one person's efficiency as a variable, Express others' efficiencies in terms of that variable using the given ratios, Compute total work done, then divide by an individual’s efficiency for “alone” time.

Answer 1

The Correct Option is D

Approach: Assign total work as a convenient LCM-style number so the rates are whole, then solve for the job size from the two phases directly.

Step 1: Take efficiencies in the ratio Ankita : Bipin : Chandan $=4:2:1$. Let the whole job be $W$ units (unknown for now), with Chandan's rate $=c$ units/day, Bipin $=2c$, Ankita $=4c$.

Step 2: Work in phase 1 (all three, 20 days) $=20(4c+2c+c)=140c$. Work in phase 2 (Ankita + Chandan, 40 days) $=40(4c+c)=200c$.

Step 3: The whole job is done: $140c+200c=W$, so $W=340c$.

Step 4: Chandan alone works at $c$ units/day, so time $=\dfrac{W}{c}=\dfrac{340c}{c}=340$ days. The unknown unit $c$ cancels, which is exactly why we never needed its actual value.

Answer: 340 days.