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A can do a piece of work in 15 days. B is 25% more efficient than A and C is 40% more efficient than B. A and C work together for 3 days and then C leaves. A and B together will compete the remaining work in:
1. \(2\frac{1}{2}\) days
2. \(3\frac{1}{2}\) days
3. 4 days
4. 3 days

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Correct Answer - Option 4 : 3 days

Given :-

A can do a piece of work in 15 days

B is 25% more efficient than A and

C is 40% more efficient than B

Concept :-

Ratio of efficiency is reciprocal of time.

Number of days required to complete a certain work = (Work to be completed/Work efficiency)

Calculation :-

Let efficiency of A be 100 

⇒ Efficiency of B = 100 + (100 × 25%)

⇒ Efficiency of B = 100 + 25 = 125 

⇒ Efficiency of C = 125 + (125 × 40%)

⇒ Efficiency of C = 125 + 50 = 175

⇒ Ratio of efficiency of A ∶ B ∶ C = 100 ∶ 125 ∶ 175

⇒ Ratio of efficiency of A ∶ B ∶ C = 4 ∶ 5 ∶ 7

A work for 15 days with efficiency 4 

⇒ Total work done by A = 15 × 4 = 60 unit 

Now, 

(A + C) together work for 3 days.

⇒ Total efficiency of A and C together = 4 + 7 = 11 unit/day

⇒ Total work done by A and C together = 11 × 3 = 33 unit 

⇒ Remaining work = Total work - Work done by A and C together 

⇒ Remaining work = 60 - 33 

⇒ Remaining work = 27 unit 

⇒ Total efficiency of A and B together = 4 + 5 = 9 unit/day 

⇒ Required number of days = (27/9) = 3 days

⇒ Time required to complete remaining work = 3 days.

∴ Time required to complete remaining work is 3 days.

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