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The time taken by A and B to finish a work is 12 days. The time taken by B and C to finish double the previous work is 28 days. The ratio of the efficiencies of A and C is 3 : 1. What is the time (in days) taken by C alone to finish the work?

The time taken by A and B to finish a work is 12 days. The time taken by B and C to finish double the previous work is 28 days. The ratio of the efficiencies of A and C is 3 : 1. What is the time (in days) taken by C alone to finish the work? Correct Answer 168

Given:

The time taken by A and B to finish a work is 12 days. The time taken by B and C to finish double the previous work is 28 days. The ratio of the efficiencies of working of A and C is 3 : 1.

Concept used:

The ratio of time taken to finish the work is inversely proportional to the ratio of their efficiencies.

Calculation:

The time taken by A and B to finish a work is 12 days.

⇒ 1/a + 1/b = 1/12     ….(i)

The time taken by B and C to finish double the previous work is 28 days

So, the time taken by B and C to finish the work is 14 days.

⇒ 1/b + 1/c = 1/14     ….(ii)

Comparing 1/b from both the equations, we get,

⇒ 1/b = 1/14 – 1/c = 1/12 – 1/a

⇒ 1/a – 1/c = 1/84

Also, the ratio of the efficiencies of working of A and C is 3 : 1.

The ratio of time taken will be 1 : 3

Let the time taken by A and C be ‘t’ and ‘3t’ respectively.

⇒ 1/t – 1/(3t) = 1/84

⇒ 2/(3t) = 1/84

⇒ t = 56 days

∴ C takes (56 × 3) days = 168 days

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