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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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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 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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