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:

Convert the efficiency comparisons into a clean ratio first: since Ankita is twice as efficient as Bipin, and Bipin is twice as efficient as Chandan, the ratio Ankita : Bipin : Chandan is 4 : 2 : 1. Then split the 60 days into two separate phases, before and after Bipin leaves, since each phase has a different combined work rate.

Answer 1

The Correct Option is D

Approach: Fix Chandan's one-day work as the base unit, write everyone's rate as a multiple of it, then add up the total work in the two phases (all three, then two).

Step 1: Let Chandan do \(1\) unit/day. Bipin is twice as efficient, so \(2\) units/day; Ankita is twice Bipin, so \(4\) units/day. Together they do \(4+2+1=7\) units/day.

Step 2: For the first \(20\) days all three work: \[ 20\times 7 = 140 \text{ units}. \] Bipin then leaves.

Step 3: The job finishes in \(60\) days, so Ankita and Chandan work the remaining \(40\) days at \(4+1=5\) units/day: \[ 40\times 5 = 200 \text{ units}. \]

Step 4: Total work \(=140+200=340\) units \(=\) one full job. Chandan alone does \(1\) unit/day, so he needs \[ \frac{340}{1}=340 \text{ days}. \]

Answer: 340 days (option 4).

Answer 2

Step 1: Let Chandan’s efficiency be \(x\) units of work per day. \[ E_C = x. \] Bipin is twice as efficient as Chandan: \[ E_B = 2x. \] Ankita is twice as efficient as Bipin: \[ E_A = 2 \cdot (2x) = 4x. \] 
Step 2: Work done by each person. Total time taken for the job to finish is 60 days. Bipin: works for 20 days (then leaves): \[ W_B = E_B \times 20 = 2x \times 20 = 40x. \] \underline{Ankita:} works for all 60 days: \[ W_A = E_A \times 60 = 4x \times 60 = 240x. \] Chandan: also works for all 60 days: \[ W_C = E_C \times 60 = x \times 60 = 60x. \] 
Step 3: Total work. \[ W_{\text{total}} = W_A + W_B + W_C = 240x + 40x + 60x = 340x. \] 
Step 4: Time taken by Chandan alone. If Chandan works alone at efficiency \(x\), then \[ \text{Time} = \frac{W_{\text{total}}}{E_C} = \frac{340x}{x} = 340 \text{ days}. \] Therefore, Chandan would need \(340\) days to complete the job alone.

Answer 3

Concept:

• Assume the whole job equals $1$, and let $D$ be the number of days Chandan alone needs to finish it; then the rate of Chandan is $1/D$ per day.
• Since Ankita is twice as efficient as Bipin, and Bipin is twice as efficient as Chandan, the rate ratio Ankita : Bipin : Chandan is $4:2:1$, so Ankita works at $4/D$ per day and Bipin at $2/D$ per day.
• Build one equation for the total work done across the two phases (before and after Bipin leaves), and solve directly for $D$ instead of introducing an extra placeholder variable.

Step 1: Set up rates in terms of D.
Chandan works at $1/D$ per day, Bipin at $2/D$ per day, Ankita at $4/D$ per day.

Step 2: Work done in the first phase.
For the first $20$ days all three work together, at a combined rate of $\dfrac{4}{D}+\dfrac{2}{D}+\dfrac{1}{D}=\dfrac{7}{D}$ per day.
Work done in this phase $= 20\times\dfrac{7}{D}=\dfrac{140}{D}$.

Step 3: Work done in the second phase.
Bipin leaves after day $20$, so for the remaining $40$ days only Ankita and Chandan work, at a combined rate of $\dfrac{4}{D}+\dfrac{1}{D}=\dfrac{5}{D}$ per day.
Work done in this phase $= 40\times\dfrac{5}{D}=\dfrac{200}{D}$.

Step 4: Add the two phases and set equal to the whole job.
$\dfrac{140}{D}+\dfrac{200}{D}=1 \Rightarrow \dfrac{340}{D}=1 \Rightarrow D=340$.

Final Answer: Chandan alone needs $340$ days.