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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life is a test
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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?
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?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?
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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Birgit Anne
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lunarpower
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conceptual solution is fastest.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.
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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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
