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by silverflamein » Tue Nov 08, 2011 6:34 pm
Can 3 men and 7 women finish a certain piece of work in 15 days???

1) 2men and 4 women can finish the work in 10 days

2) each of the women can do twice the work that each of the men can do in a given time.
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Source: — Data Sufficiency |
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by user123321 » Tue Nov 08, 2011 7:04 pm
1) if 2M & 4W can finish the work in 10 days,
then 3M & 7W can finish the work in even less than 10 days. hence sufficient.
2) since we dont know how much work an individual man can do in a day, it is insufficient.

[spoiler]is it A?[/spoiler]

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by pemdas » Tue Nov 08, 2011 7:20 pm
st(1) (2m+4w)/10 OR (m+2w)/5 is one day's work. 15 days work will be 3m+6w Sufficient to conclude that 3 men and 7 women can finish up the required work with one less woman, answer Yes.

st(2) w=2m this is Not Sufficient, as we are not suggested which time (day, week ...)

a
silverflamein wrote:Can 3 men and 7 women finish a certain piece of work in 15 days???

1) 2men and 4 women can finish the work in 10 days

2) each of the women can do twice the work that each of the men can do in a given time.
Last edited by pemdas on Fri Nov 11, 2011 5:01 am, edited 1 time in total.
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by batmannavneet » Wed Nov 09, 2011 9:56 am
I think the answer is A.
I don't quite understand the calculation of statement 2 below.
pemdas wrote:the question tests average speed: Can it be that 1/3m (throughput of one men for one job is 1/3 of whole unit) and 1/7w (throughput of one women for one job is 1/7 of whole unit) *multiplied* by time, 15 days be enough to complete one job?

st(1) 10*(1/2m+1/4w)=1 means 10 hours is enough to complete one job with the given throughputs, and if we match 10(2+1)/4mw=1, 30=4mw we see that men and women can be only integer and not fractions
out of possible values for m*w (15=2mw). The question asked was: 15(1/3m+1/7w)=1? and can answer No since 15(7+3)=21mw and 50=7mw is not equivalent to 15=2mw, 50/15 =! 7/2 <> 3 1/3 =! 3 1/2 Sufficient

st(2) w=2m although we are not given day-people function, we can find whether 15(1/m + 1/2m)=1 to test 50=7mw; 15*3=2m and m=45/2, w=45. The question asked was 50=7mw and f we plug in m and w we get 50=! 7*(45/2)*45. We can answer No, Sufficient
d
silverflamein wrote:Can 3 men and 7 women finish a certain piece of work in 15 days???

1) 2men and 4 women can finish the work in 10 days

2) each of the women can do twice the work that each of the men can do in a given time.
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by bpdulog » Wed Nov 09, 2011 10:21 am
I think the answer is C.

With A, you still have 2 unknowns. I think it be okay if the same amount of men and women were added, but in this case it is not.
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by rijul007 » Wed Nov 09, 2011 10:23 am
Statement 1
Let the rate at wich men do their work be m and that for women be w.

2m +4w = 10

Multiplying both sides of the equation by 3/2

3m + 6w = 15

Hence addition of one more woman would only reduce the number of days..

Sufficient

Statement 2
w = 2m

Total rate at which they do their work => 3m + 7w => 17m


We dont know the value of m or about the amount of work

Hence, Insufficient

Option A
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by saketk » Wed Nov 09, 2011 11:08 am
bpdulog wrote:I think the answer is C.

With A, you still have 2 unknowns. I think it be okay if the same amount of men and women were added, but in this case it is not.
You don't need to solve this question at all. you can answer this by using common sense.

It is logical to say that when less people can do the job in 10 days then more people will Definitely take lesser time to do the same job.

Answer should be A
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by vaibhavgupta » Thu Nov 10, 2011 5:57 pm
silverflamein wrote:Can 3 men and 7 women finish a certain piece of work in 15 days???

1) 2men and 4 women can finish the work in 10 days

2) each of the women can do twice the work that each of the men can do in a given time.
I went for C

could somebody pls tell OA?
:?
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by pemdas » Fri Nov 11, 2011 5:02 am
just revised/reread q.
closer to @rijul007 in solution and it's A
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