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It takes 6 days for 3 women and 2 men working together to [#permalink]

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19 Aug 2009, 13:33

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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

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? 1] 3 times 2] 4 times 3] 5 times 4] 6 times 5] 7 times

Clear D:

x: days 9 women doing the job --> 1 woman works W=1/9x per day x-5: days 3 men doing the job --> 1 man works M=1/[3*(x-5)] per day

The answer to the question is: M/W

And, how much is x?

As per first data:

2*M+3*W=1/6 You can solve x, and obtain 2 values, 10 and 1 (1 is impossible because that would imply that the 3 men take -4 days), so x=10. So M/W=6 times
_________________

Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?

Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?

Below is another solution which is a little bit faster.

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

Let one woman complete the job in \(w\) days and one man in \(m\) days. So the rate of 1 woman is \(\frac{1}{w}\) job/day and the rate of 1 man is \(\frac{1}{m}\) job/day.

It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\).

3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

Solving: \(m=15\) and \(w=90\). \(\frac{w}{m}=6\).

Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?

Below is another solution which is a little bit faster.

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

Let one woman complete the job in \(w\) days and one man in \(m\) days. So the rate of 1 woman is \(\frac{1}{w}\) job/day and the rate of 1 man is \(\frac{1}{m}\) job/day.

It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\).

3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

Solving: \(m=15\) and \(w=90\). \(\frac{w}{m}=6\).

Answer: D.

Hm, i got stuck cuz I got something a little different: YOURS: 3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

MINE: 3 men would do the same work 5 days sooner than 9 women --> \(\frac{3}{m}=\frac{9}{w}+5\)

In the above equation you also have for 2 men: \(\frac{2}{m}\) - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...

Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?

Below is another solution which is a little bit faster.

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

Let one woman complete the job in \(w\) days and one man in \(m\) days. So the rate of 1 woman is \(\frac{1}{w}\) job/day and the rate of 1 man is \(\frac{1}{m}\) job/day.

It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\).

3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

Solving: \(m=15\) and \(w=90\). \(\frac{w}{m}=6\).

Answer: D.

Hm, i got stuck cuz I got something a little different: YOURS: 3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

MINE: 3 men would do the same work 5 days sooner than 9 women --> \(\frac{3}{m}=\frac{9}{w}+5\)

In the above equation you also have for 2 men: \(\frac{2}{m}\) - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...

Let one woman complete the job in \(w\) days and one man in \(m\) days.

First equation: It takes 6 days for 3 women and 2 men working together to complete a work: As the rate of 1 woman is \(\frac{1}{w}\) job/day, then the rate of 3 women will be \(\frac{3}{w}\) job/day. As the rate of 1 man is \(\frac{1}{m}\) job/day, then the rate of 2 men will be \(\frac{2}{m}\) job/day. Combined rate of 3 women and 2 men in one day will be: \(\frac{3}{w}+\frac{2}{m}\) job/day.

As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\).

Second equation: 3 men would do the same work 5 days sooner than 9 women: As 1 man needs \(m\) days to do the job 3 men will need \(\frac{m}{3}\) days to do the job. As 1 woman needs \(w\) days to do the job 9 women will need \(\frac{w}{9}\) days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence \(\frac{m}{3}\) is 5 less than \(\frac{w}{9}\) --> \(\frac{m}{3}+5=\frac{w}{9}\).

Bunel-I got a quadratic equation while solving these two eqn. Is there a simple way of solving them?

I also got quadratic equation (\(m^2-3m-180=0\)) and it wasn't too hard to solve (discriminant would be perfect square \(d=3^3+4*180=729=27^2\)) --> \(m=-12\) or \(m=15\).
_________________

First set : 3/w+2/m=1/6 Second set: 3/m = 1/x and 9/w=1/(x+5) ==> 1/m = 1/3x and 1/w = 1/9(x+5) Enter in first equation and solve for x 1/(3x+15) + 2/3x = 1/6 simplify => x^2-x-20=0 solve for x : x=(1 +- (1+80)^,5)/2= (1+- 9)/2. X can only be positive ==> x= 5 enter in 2nd set ==> 3/m=1/5 and 9/w=1/10 => 3/2m=9/w => m/w=1/6

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? 1] 3 times 2] 4 times 3] 5 times 4] 6 times 5] 7 times

Let women do \(w\) units of work per day and men do \(m\) units of work per day, then question asks, what is m/w

Total units of work to be done = \(6*(3*w+2*m)\)

Time taken by 9 women to do this work on their own = \(6*(3*w+2*m)\)/\((9*w)\) = \(2 + 4/3*m/w\)

Time taken by 3 men to do this work on their own = \(6*(3*w+2*m)/(3*m)\) = \(6/(m/w) + 4\)

Let m/w be x

then we know

\(2 + (4/3)*x -6/x -4 = 5\)

or

\((4/3)*x - 6/x= 7\)

Now substituting for x from choices will quickly give us x = 6 so D

I like this as it reduces quickly to the required form of m/w and it obviates the need for a quadratic equation and also lest w and m remain in numerator most of the time

Bunuel I saw both the solutions. However, please tell me where I go wrong -

Lets assume M work / day and W work / day are the rates of men and women respectively.

First stimulus -It takes 6 days for 3 women and 2 men working together to complete a work Second stimulus - 3 men would do the same work 5 days sooner than 9 women

1/(3W) + 1/(2M) = 6 1/(9W) - 1/(3M) = 5

I am getting negative value of M Please correct.

Bunuel wrote:

nonameee wrote:

Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?

Below is another solution which is a little bit faster.

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

Let one woman complete the job in \(w\) days and one man in \(m\) days. So the rate of 1 woman is \(\frac{1}{w}\) job/day and the rate of 1 man is \(\frac{1}{m}\) job/day.

It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\).

3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

Solving: \(m=15\) and \(w=90\). \(\frac{w}{m}=6\).

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? 1] 3 times 2] 4 times 3] 5 times 4] 6 times 5] 7 times

I received a PM asking me to respond. First, this is definitely not a realistic GMAT question. For one thing, it's horribly written (the phrase 'to complete a work' is not English, the question should read '*By* how many times...', the question needs to make clear that each man works at the same rate, as does each woman, the word 'sooner' is non-idiomatic, the word 'output' is misused, etc). For another, it's terribly contrived, and altogether tedious if you take any normal approach; I don't see any direct way to solve that will allow you to avoid a quadratic equation. Real GMAT questions are never designed in such a way, so you can confidently move on to better material and ignore this question (incidentally, where is it from?).

While it isn't especially fast either, you can work backwards from the answers here relatively easily. This might at least be less confusing for some than a direct (algebraic) approach. Say we get 1 unit of work per woman per day. If you test, say, answer C, we'd then get 5 units of work per man per day. The job would then require 6(3 + 2*5) = 78 units of work. Notice that, to find how long it will take 9 women to do the job, we'll need to get an integer when we divide 78 by 9, so C cannot be right. If you move next to D, we have 6 units of work per man per day, and the job requires 6(3 + 2*6) = 90 units of work. Thus 9 women do the job in 10 days, and 3 men would do the job in 90/(6*3) = 5 days, so D is correct.
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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

Let one woman complete the job in \(w\) days and one man in \(m\) days. So the rate of 1 woman is \(\frac{1}{w}\) job/day and the rate of 1 man is \(\frac{1}{m}\) job/day.

It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\).

3 men would do the same work 5 days sooner than 9 women --> \(\frac{m}{3}+5=\frac{w}{9}\).

Solving: \(m=15\) and \(w=90\). \(\frac{w}{m}=6\).

Answer: D.

Bunuel,

When the question says "3 men would do the same work 5 days sooner than 9 women".

Do the colored words refer to the word work mentioned in the previous sentence; Total Work to complete the job?

Bunel-I got a quadratic equation while solving these two eqn. Is there a simple way of solving them?

I also got quadratic equation (\(m^2-3m-180=0\)) and it wasn't too hard to solve (discriminant would be perfect square \(d=3^3+4*180=729=27^2\)) --> \(m=-12\) or \(m=15\).

Hello Bunuel,

I honestly did not understand the simplification...i normally go wrong this part of the problem. Kindly can u suggest more simplified method to do during a time-constraint?

Regards
_________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Bunel-I got a quadratic equation while solving these two eqn. Is there a simple way of solving them?

I also got quadratic equation (\(m^2-3m-180=0\)) and it wasn't too hard to solve (discriminant would be perfect square \(d=3^3+4*180=729=27^2\)) --> \(m=-12\) or \(m=15\).

Just a smal typo: in the discriminant, it should be \(3^2\) and not \(3^3.\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: It takes 6 days for 3 women and 2 men working together to [#permalink]

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15 Sep 2012, 11:53

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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PhD in Applied Mathematics Love GMAT Quant questions and running.