Thursday, 7 November 2013

algebra precalculus - A completes 36% work in 12 days.



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..


Answer



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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