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:
Show 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.
- \(240\)
- \(260\)
- \(300\)
- \(340\)
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to understand the efficiency of Ankita, Bipin, and Chandan and how they contribute to completing the job alone or together.
- Ankita is twice as efficient as Bipin, which means if Bipin completes \( x \) work in a day, Ankita completes \( 2x \) work.
- Bipin is twice as efficient as Chandan, implying if Chandan completes \( y \) work in a day, then Bipin completes \( 2y \) work.
- From this, we establish that Ankita completes \( 4y \) work in a day.
The combined efficiency of all three together is:
\((4y) + (2y) + (y) = 7y\)
Bipin works for the first 20 days, so his contribution with all three working together initially is:
- In 20 days, the work done by all three together is \( 20 \times 7y = 140y \).
The total time taken for the whole job to be completed is 60 days.
The remaining 40 days have only Ankita and Chandan working since Bipin leaves after 20 days. Therefore, the combined efficiency of Ankita and Chandan is:
\(4y + y = 5y\)
In the next 40 days, they complete:
\(40 \times 5y = 200y\)
The total work done by Ankita, Bipin, and Chandan is:
\(140y + 200y = 340y\)
This total work (340y) is equivalent to 1 job. Therefore, Chandan alone would need 340 days to complete the whole job, as he completes \( y \) work in one day.
Thus, the number of days Chandan would need to complete the job alone is 340 days.
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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