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Source: — Data Sufficiency |
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by cubicle_bound_misfit » Thu Sep 11, 2008 9:22 am
Simplified,
the q asks for

is ab/(a+b) < ab ? where a, b >0

as (a+b)>0

ineqality becomes is ab < ab(a+b) ? where a,b >0

from 1.
a=2b
q is , IS 2b.b < 2b.b.3b ?

or 1< 3b
if B is a positive fraction >= 1/4 it is not true.

hence 1 not suff.

from stmt 2.

if (a+b) > 1
ab(a+b) > ab for any value of a &b hence Sufficient.

ANSWER : B

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by amitdgr » Thu Sep 11, 2008 9:50 am
cubicle_bound_misfit wrote:
if (a+b) > 1
ab(a+b) > ab for any value of a &b hence Sufficient.

Thanks for reply ... But can you explain this step ... how did you deduce that if (a+b) > 1 , ab(a+b) > ab for all a,b ?

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by amitdgr » Thu Sep 11, 2008 9:57 am
The solution just struck me

from the question we get to know that a and b are positive

so ab/(a+b) < ab can be written as ab(a+b) > ab and since ab is positive, we can divide both sides by ab ... that gives us the simplified question

is (a+b) > 1 ?


statement 1 says a = 2b ... this does not help me in telling whether a+b>1

incidentally statement 2 gives us the info directly that (a+b) > 1

so 2 is sufficient
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by cubicle_bound_misfit » Thu Sep 11, 2008 10:05 am
Yep thats the logic
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by smclean23 » Thu Sep 11, 2008 4:18 pm
How did you simply the question to ab/a+b?
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Re: Exponents

by amitdgr » Thu Sep 11, 2008 8:14 pm
amitdgr wrote:If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1 * b^-1)^-1 ?

1) a = 2b

2) a + b > 1
we know x^-1 = 1/x

(a^-1 + b^-1)^-1 can be written as (1/a + 1/b)^-1
further simplified , [(a+b)/ab]^-1 => [ ab/(a+b) ]

hope that helps
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