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A completes 36% work in 12 days.

B is twice as fast as A.

C is twice as fast as B.

How many approx days are required for finishing the remaining 64% work if all of them work together.?

a)2 b)4 c)6 d)8

Answer is a but i am getting option b...

i have shown my solution below..

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up vote 3 down vote accepted

work done by A in 1 day is 36/1200

work done by B in 1 day is 36/600

work done by C in 1 day is 36/300

work done by A,B,C in 1 day is 36/1200 + 36/600 + 36/300

Thus A,B,C need 1/(36/1200 + 36/600 + 36/300) days to finish 100% work

i.e. A,B,C need 4.76 days to finish 100% work

Thus A,B,C need 64x4.76/100 days to finish 64% work

i.e. A,B,C need 3.04 days to finish 64% work

Thus answer should be b as 3.04 which is closer to 4 than 2..

But the answer was given as a... ? vch 1 do u feel is correct

pls Help...

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I agree with your calculation. The fact that none of the answers are even close makes me suspect there is a problem with the question. I would not accept either a or b. – Ross Millikan Feb 7 '13 at 15:39
If the $36$% is changed to $30$% and everything else stays the same, the answer is b. – Peter Phipps Feb 7 '13 at 16:13

I agree with you. Let A's speed be a units of work per day. Then B's speed = 2a, and C's speed = 4a. So, working together, A,B,C can complete (1 + 2 + 4)a = 7a units of work per day.

A has completed 36% of the work in 12 days. I.e. 36% of the work = 12a. So the original amount of work is 12a / .36, and (1 - .36) = .64 of the original amount remains. So 0.64/0.36 * 12a remains

The three of them working together at the rate of 7a per day can complete this in $$\frac{0.64 * 12a \text{ work units}}{0.36 * 7a \text{ work units/day}} = 3.05 \text{ days}$$

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