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a and b ^-1

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by dikku07 » Wed Sep 23, 2009 8:07 am
If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1b^-1)^-1?

(1) a = 2b

(2) a + b > 1



A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient
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Source: — Data Sufficiency |
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Re: a and b ^-1

by kaulnikhil » Wed Sep 23, 2009 8:57 am
dikku07 wrote:If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1b^-1)^-1?

(1) a = 2b

(2) a + b > 1



A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient
B
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by mp2437 » Wed Sep 23, 2009 9:01 am
Answer is B.

(a^-1 + b^-1)^-1 can be expressed as ab/(a+b), and you are asked if this term is less than (a^-1b^-1)^-1, which can be expressed as ab.

So, is ab/(a+b) < ab?

Statement 1: a = 2b. Substitute directly. Is 2b^2/(3b) < 2b^2? Well, we are only told that b is positive, so it could be less than or greater than 1. Plug in different values for each case, and you could see that if b < 1, then the first term is in fact greater than the second term, but less than the second term when you plug in b > 1. So Statement 1 is not sufficient.

Statement 2: You are given a+b > 1, therefore, anything greater than 1 in a denominator will lower the value of the whole term. Therefore, the first term is less than the second term, which is sufficient.

You are left with choice B.
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by vaibhav.iit2002 » Wed Sep 30, 2009 10:01 am
mp2437 wrote:Answer is B.

(a^-1 + b^-1)^-1 can be expressed as ab/(a+b), and you are asked if this term is less than (a^-1b^-1)^-1, which can be expressed as ab.

So, is ab/(a+b) < ab?
Statement 2: You are given a+b > 1, therefore, anything greater than 1 in a denominator will lower the value of the whole term. Therefore, the first term is less than the second term, which is sufficient.

You are left with choice B.
what if ab<0? inequality will get reversed.

in fact if (a+b)>1,
ab/(a+b) < ab => ab < ab(a+b) => ab(a+b-1)>0
now if a+b>1 the ques. is is ab>0 which we can't say anything for sure as it depends on sign of a and b.

1 and 2: if a=2b,
we need to prove if
ab(a+b-1)>0 OR 2b^2(a+b-1) > 0

this is true as a+b>1

hence C
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by nithi_mystics » Wed Sep 30, 2009 2:05 pm
Answer is B

(a^-1 + b^-1)^-1 = ab/(a+b)
and (a^-1b^-1)^-1 = ab

Since a+b is positive, ab/(a+b) is greater than ab. This is true even if a*b is negative.

For eg: say a = -4 and b= 6,
then ab/(a+b) = -12 and ab = -24. -12 > -24. Hence B is the answer.

Thanks.
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by vaibhav.iit2002 » Wed Sep 30, 2009 7:07 pm
nithi_mystics wrote:Answer is B

(a^-1 + b^-1)^-1 = ab/(a+b)
and (a^-1b^-1)^-1 = ab

Since a+b is positive, ab/(a+b) is greater than ab. This is true even if a*b is negative.

For eg: say a = -4 and b= 6,
then ab/(a+b) = -12 and ab = -24. -12 > -24. Hence B is the answer.

Thanks.
As u said
1. if a=-4, b=6 => ab/(a+b) > ab [correct]
2. now if a=4, b=6,
ab/(a+b)=24/10=2.4
and ab=24
hence ab/(a+b) < ab

therefore we get different results on the basis of sign of a and b.
isn't?
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by nithi_mystics » Thu Oct 01, 2009 6:42 am
You are right!

I guess I did not pay attention when I tried to solve it :|
Thanks
Nithi
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by mp2437 » Thu Oct 01, 2009 8:52 am
guys, it says a and b are positive. What's the deal here?
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by vaibhav.iit2002 » Thu Oct 01, 2009 8:57 am
mp2437 wrote:guys, it says a and b are positive. What's the deal here?
Bullshit !! I was really sleepy and I guess so was nithi_mystics
:P
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