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If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1

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If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1 [#permalink]

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New post 11 Sep 2008, 09:09
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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


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E

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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 09:28
so this simplifies to ab/(b+a) < ab?

since b and a are positive we can say

ab<ab(b+a)? which basically means its asking if a and b are fraction...and less than 1


1)

a=2b lets just pick number suppose b=3/7 then a=6/7

so 18/49 < (18/49)(9/7)?

yes..

suppose b=1 a=2 2<2(3) Yes

but suppose b=1/4 then a=1/2

1/8<1/8(3/4) NO...insuff

2)
a+b >1

Sufficient


IgnitedMind 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


OA will follow, please discuss

Kudos [?]: 322 [0], given: 2

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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 10:20
Expression in the main question can be simplified to

is ab/(a+b) < ab......and since both a and b are positive, this further simplifies to
is (a+b) > 1........statement 2 clearly answers that.

Hence, B.

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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 10:37
IgnitedMind 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


A
B
C
D
E

OA will follow, please discuss

scthakur wrote:
ab/(a+b) < ab

question is :
ab/(a+b) < ab or ab(1/(a+b) -1 )<0 ????


(1) says a=2b , a=2b does not guarantee a+b>1 hence above expr can be +vbe or -ve INSUFFI

(2)a+b>1 does not say a and b have same signs or opposite ,in both cases different values for the expr.

(1) and (2) => SUFFI both are of same signs and again a+b>1 proves the relation given above

IMO C
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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 11:02
The best way to solve would be, simplify the question as much as possible

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 ...

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

Answer is B
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Last edited by amitdgr on 11 Sep 2008, 21:17, edited 1 time in total.

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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 12:30
i think its B !

because we need either ab<0 (impossible as a N b both are +ve) OR a+b>1

first stamnt doesnt help 2b+b may or may not be >1

2 stmnt proves a+b>1

hence B

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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 12:49
IgnitedMind 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


A
B
C
D
E

OA will follow, please discuss


ab/(a+b) < ab
Question is ab(a+b)>ab?

1) a=2b.. a+b can be <1 or a+b>1
not sufficient
2) a+b>1
also ab is positive.
ab(a+b)>ab is always true.
sufficient.


B
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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 17:58
(a^-1 + b^-1)^-1 less than (a^-1 * b^-1)^-1 ?

simplifyling the expression ab/(a+b) < ab ? or is 1 < ( a+ b) , which is stated in option 2 directly

so IMO B

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Re: DS EXPONENTS [#permalink]

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New post 11 Sep 2008, 18:34
oops i didnt read the questio properly again!!!
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Re: DS EXPONENTS   [#permalink] 11 Sep 2008, 18:34
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