Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
It takes 6 days for 3 women and 2 men working together to [#permalink]
12 Mar 2010, 22:33
2
This post received KUDOS
5
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
42% (03:24) correct
58% (02:47) wrong based on 90 sessions
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
Re: Interesting Work Problem [#permalink]
13 Mar 2010, 02:17
6
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) days and one man in \(m\) days.
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 \(w=90\). \(\frac{w}{m}=6\).
Re: Interesting Work Problem [#permalink]
13 Mar 2010, 02:22
2
This post received KUDOS
Answer is D. This is my explanation:
Let 9W take x days. Question gives us that 3M take x-5 days.
Therefore 1W does 1/9x of work in 1 day. ------> A 1M does 1/3(x-5) of work in 1 day. ------> B
We need to find what the above values in bold are to measure output. Hence we need to find x.
Using A and B 2M do 2/3(x-5) work in 1 day. 3W do 3/9x work in 1 day.
Therefore (2M + 3W) do 2/3(x-5) + 3/9x work in 1 day.
The above amount is 1/6th of work (Question says that 2M + 3W finish work in 6 days). 2/3(x-5) + 3/9x = 1/6 --------> Solve this to get x = 1 or x = 10. Take x=10 because x=1 cannot be substituted in B.
Substitute x=10 in A and B we get 1W does 1/90 work in 1 day => 1W does the work in 90 days. 1M does 1/15 work in 1 day => 1M does the work in 15 days.
1M does work 6 times (90/15) faster than 1W which implies 1M's output is 6 times that of 1W.
Re: Interesting Work Problem [#permalink]
13 Mar 2010, 02:56
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) hours and one man in \(m\) hours.
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.
But I think you have mixed days with hours. (w & m hours) _________________
Re: Interesting Work Problem [#permalink]
13 Mar 2010, 05:33
Hussain15 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
Please provide the reasoning with your answers.
Lets Try it working another smart way....
Let there be a total of 180 pages to be written (let it be the work that man and women and doing together and isolatedly. I have taken 180 for simplicity of calculations based on multiple of 6, 2, 3 and 9)
Now we have
3w+2m = 6 days
or 3w or 2m do 30 pages per day....(180pages/6days = 30 pages)
Now lets work with choices knowing that W or M can only be integers. (obviously Men or women can not be fraction numbers)
Only option 4 that gives 6 times can be solved as under
3w+2m= 30 pages
option 4 gives w=6m
so 3(6M) + 2m = 30
or m = 2
rest all options give fractional numbers for Men and women...
Hope this helps !!! correct me if i am wrong. Answer D... _________________
Re: Interesting Work Problem [#permalink]
03 Aug 2013, 11:30
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) hours and one man in \(m\) hours.
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.
I mess up in the reciprocals. I mean I did almost the exact same way as you except that I didn't switch the reciprocals. When I formulated the second condition, I did, 3/m = 9/w - 5. Can you please explain, when do we do what? Why are they both different? I kept the same fraction format for both conditions and didn't get the correct answer.
Re: Interesting Work Problem [#permalink]
27 Sep 2013, 04:42
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) hours and one man in \(m\) hours.
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, Shouldn't we be writting
m/3 * (t-5) = w/9 * t where t is the time that women do the job.
Re: Interesting Work Problem [#permalink]
27 Sep 2013, 07:49
Expert's post
Skag55 wrote:
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) hours and one man in \(m\) hours.
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, Shouldn't we be writting
m/3 * (t-5) = w/9 * t where t is the time that women do the job.
m/3 is time, t-5 is also time...
One man completes the job in \(m\) days --> 3 men in m/3 days. One woman completes the job in \(w\) days --> 9 women in w/9 days.
We are told that m/3 is 5 less than w/9 --> \(\frac{m}{3}+5=\frac{w}{9}\).
Re: It takes 6 days for 3 women and 2 men working together to [#permalink]
27 Sep 2013, 22:36
Hussain15 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
1. Let 1 woman do the job in w days and 1 man do the job in m days 2. In 1 day, 1 woman does 1/w of the job and 1 man does 1/m of the job 3. 3 women do 3/w of the job and 2 men do 2/m of the job in 1 day. 4. It is given that 3 women and 2 men do the job in 6 days or in 1 day they do 1/6 of the job. 5. Equating (3) and (4) we have, 3/w + 2/m = 1/6 6. Also 3 men to do the work takes m/3 days. 9 women to do the work takes w/9 days 7. It is given that 3 men do the job 5 days sooner than 9 women. 8. So, m/3 = w/9 - 5 9. Solving (5) and (8) we have m=15 and w=90
Re: Interesting Work Problem [#permalink]
19 Nov 2013, 05:33
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) days and one man in \(m\) days.
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.
Hi Bunuel
I have the two equations well set and ready to go but I'm having some trouble with the algebra Any tips to solve faster? What approach did you use? Combination?
Re: Interesting Work Problem [#permalink]
19 Nov 2013, 05:38
Expert's post
jlgdr wrote:
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) days and one man in \(m\) days.
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.
Hi Bunuel
I have the two equations well set and ready to go but I'm having some trouble with the algebra Any tips to solve faster? What approach did you use? Combination?
Re: Interesting Work Problem [#permalink]
19 Nov 2013, 09:24
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) days and one man in \(m\) days.
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.
I don't get these workrate problems at all, I've gotten all 50 or 60 that I've tried incorrect...can you explain why here: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\) you had the w & m in the denominator, and then here \(\frac{m}{3}+5=\frac{w}{9}\) it was in the numerator?? Every time I see that stuff it makes NO sense to me, it looks like people just arbitrarily decide to put the letters in the numerator or the denominator. In this case, you have the first equation being for work rate, with the letters being in the denominator, and then the second equation for the work rate the letters appear in the numerator. I just don't get it.
Also how did you solve? Did you just look for numbers that made sense? Seems like there'd be a ton of arithmetic to do in such a sort time span to solve otherwise.
Re: Interesting Work Problem [#permalink]
19 Nov 2013, 14:17
Expert's post
AccipiterQ wrote:
Bunuel wrote:
Hussain15 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
Please provide the reasoning with your answers.
Let one woman complete the job in \(w\) days and one man in \(m\) days.
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.
I don't get these workrate problems at all, I've gotten all 50 or 60 that I've tried incorrect...can you explain why here: \(\frac{3}{w}+\frac{2}{m}=\frac{1}{6}\) you had the w & m in the denominator, and then here \(\frac{m}{3}+5=\frac{w}{9}\) it was in the numerator?? Every time I see that stuff it makes NO sense to me, it looks like people just arbitrarily decide to put the letters in the numerator or the denominator. In this case, you have the first equation being for work rate, with the letters being in the denominator, and then the second equation for the work rate the letters appear in the numerator. I just don't get it.
Also how did you solve? Did you just look for numbers that made sense? Seems like there'd be a ton of arithmetic to do in such a sort time span to solve otherwise.
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...