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:
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:
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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.
Updated On: Jul 24, 2026
- \(240\)
- \(260\)
- \(300\)
- \(340\)
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The Correct Option is D
Approach Solution - 1
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).
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).
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Approach Solution -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.
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