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by EMAN » Sat Nov 14, 2009 3:26 pm
Mary and Nancy can each perform a certain task in m and n hours, respectively. Is m<n?

(1) Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is greater than m.

(2) Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is less than n.

This one baffled me a little bit. Please explain an easy way to solev this problem.
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Source: — Data Sufficiency |
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by life is a test » Sun Nov 15, 2009 1:48 am
hmm interesting question...IMO D?

1. 2(1/m + 1/n)> 1/m ->1/m + 1/n > 2/m --> if m =1 then n<1, if m = 2 then n>2...hence suff

2. 2(1/m + 1/n)< 1/n ->1/m + 1/n < 2/n -> if n =1 then m>1, if n=2, m >2 ...hence suff.

whats the OA?
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by palvarez » Mon Nov 16, 2009 11:58 am
m and n hrs each. Combined: mn/(m+n)

Is m < n

1. 2mn/(m+n) > m
2. 2mn/(m+n) < n


1. mn > m^2 or n > m
2. mn < n^2 or m < n

D
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by enniguy » Tue Nov 17, 2009 4:28 pm
life is a test wrote:hmm interesting question...IMO D?

1. 2(1/m + 1/n)> 1/m ->1/m + 1/n > 2/m --> if m =1 then n<1, if m = 2 then n>2...hence suff

2. 2(1/m + 1/n)< 1/n ->1/m + 1/n < 2/n -> if n =1 then m>1, if n=2, m >2 ...hence suff.

whats the OA?
How are these true?
The RHS should be 1/2m and 1/2n isn't it? 2 can't be on numerator when you move it the other side.

IMO, answer is C. This is how.
1/m < 2{ (m+n)/mn } < 1/n. ->
1/m < 1/n.
m>n.
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by Birgit Anne » Mon Nov 30, 2009 12:04 pm
I agree with Palvarez: Each statement alone is sufficient. Great explanation, thank you!
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by picaj » Sat Dec 19, 2009 6:46 pm
May I ask... where did you get this problem?
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by lunarpower » Sat Dec 19, 2009 10:26 pm
EMAN wrote:Mary and Nancy can each perform a certain task in m and n hours, respectively. Is m<n?

(1) Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is greater than m.

(2) Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is less than n.

This one baffled me a little bit. Please explain an easy way to solev this problem.
conceptual solution is fastest.

background fact:
if some person/machine takes a certain amount of time, then TWO IDENTICAL such people/machines will take HALF the amount of time.

this fact is actually all you need to solve this problem:

(1)
divide by 2 to give -->
mary & nancy together took MORE than half the time for mary by herself.
(mary & mary together, if there were 2 marys, would take EXACTLY half the time)
therefore
mary & nancy together are SLOWER than 2 marys
therefore
nancy is slower than mary
SUFFICIENT

(2)
divide by 2 to give -->
mary & nancy together took LESS than half the time for nancy by herself.
(nancy & nancy together, if there were 2 nancys, would take EXACTLY half the time)
therefore
mary & nancy together are FASTER than 2 nancys
therefore
mary is faster than nancy
SUFFICIENT

ans (d)
Ron has been teaching various standardized tests for 20 years.

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