virtualanimosity wrote:
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
The fastest and easiest way to solve this question was already proposed by IanStewart.
I am trying another algebraic approach.
Denote by \(W\) the rate of a woman, by \(M\) that of a men, and by \(T\) the time it takes 9 women to complete the work.
We have the following equations:
\(6(3W + 2M) = 9WT = 3M(T-5)\), or, after reducing by 3, \(2(3W + 2M) = 3WT = M(T - 5).\)
We are looking for the ratio \(M/W\) which we can denote by \(n.\) Substituting in the above equations \(M = nW,\) we can write:
\(2(3W + 2nW) = 3WT = nW(T - 5).\)
Divide through by \(W,\) so \(6 + 4n = 3T = nT - 5n.\) Solving for \(T\) (equality between the last two expressions) we obtain \(T=\frac{5n}{n-3}.\)
Taking the equality of the first two expressions, we get \(6+4n=\frac{3\cdot{5}n}{n-3}.\)
From the possible answer choices we can deduce that \(n\) must be a positive integer.
We need \(\frac{15n}{n-3}\) to be a positive integer. We can see that \(n\) cannot be odd and it must be greater than 3.
We have to choose between B and D.
Only \(n = 6\) works.
Answer D.
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