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Study the information given below and answer the following questions:
There are n men and m women working at a construction job. The n men are ordered 1 to n such that each man ranked k (where k>1) takes as much time to do the job as the time taken by all men ranked lower than k working together would take to complete the job. The m women are ordered 1 to m such that each woman ranked k (where k>2) takes as much time to do the job as the time taken by the previous two women (k-1 and k-2) working together would take.
Question 3
If the first two men and first two women each working individually would take the same time to do the job and the seventh man takes 50 days to complete the work how much time would be taken the first man and fifth woman to do the job?
is it a cat level ques ..I mean on the difficulty index of 100 . where this will ques rank ?
Hi Vismit. I can't rank the question on an index but I can vouch for the fact that this is a CAT level question albeit a little tricky one.
Let's see how to solve this question.
Let the men be M1, M2, M3,..., Mn and the women be W1, W2, W3, ..., Wn.
Let the time taken by a man Mx be T(Mx) and a woman Wx be T(Wx).
Let the time taken by the first two men and first two women each working individually be 'x'.
T(M1)=T(M2)=T(W1)=T(W2)=x
T(M3)=T(M1)+T(M2)=2x
T(M4)=T(M1)+T(M2)+T(M3)=4x
.
.
T(M7)=32x
It is given that M7 takes 50 days => 32x=50 => x=25/16
T(W3)=T(W2)+T(W1)=2x
T(W4)=T(W3)+T(W2)=3x
T(W5)=T(W4)+T(W3)=5x
The times taken by first man and fifth woman individually are x and 5x. But when working together the time taken is the reciprocal of the sum of reciprocals.
T=5x/6=125/96 days
