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A can do a piece of work in 10 days while B can do it in 30 days. A works on the first day, B works on the second day and this sequence continues till the completion of the work. In total how many days does it take to complete the work?
1. 14\(\frac{2}{3}\)
2. 15
3. 14\(\frac{1}{3}\)
4. 14

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Correct Answer - Option 1 : 14\(\frac{2}{3}\)

Given:

Time taken by A to do the job = 10 days 

Time taken by B to do the job = 30 days 

Formula used:

Total Work = Time Taken × Efficiency 

⇒ Efficiency = Total Work/Time Taken 

Calculation:

Let the total work be the LCM of 10 and 30 units 

⇒ Total Work = 30 

One day efficiency of A = 30/10 = 3 units 

One day efficiency of B = 30/30 = 1 unit  

Work done by A and B in 2 days working alternatively = (3 + 1) = 4 

⇒ 2 × 7 days = 4 × 7 

⇒ 14 days = 28 units 

Now, A will do the remaining (30 - 28) 2 units in = 2/3 days 

∴ The total time taken by A and B = (14 + 2/3)

⇒ 14\(\frac{2}{3}\) days 

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