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15 Jan 2015, 06:59
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:32) correct
36%
(02:34)
wrong
based on 762
sessions
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If ab ≠ 0 and a + b ≠ 0, is 1/(a + b) < 1/a + 1/b ?
(1) |a| + |b| = a + b
(2) a > b
(1) |a| + |b| = a + b
(2) a > b
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
17 Jan 2015, 06:36
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Statement (1): From the information given, we can conclude that both, a and b are not negative. The two absolute values must be positive (the question stem tells us they are not 0), so a + b must also be positive. As the absolute values have the same sign, a and be must also have the same sign, otherwise their sum would not be equal to the sum of their absolute values.
Picking a few number pairs, we quickly realize that that information is sufficient. This also makes sense theoretically. Going from \(\frac{1}{a}\) to \(\frac{1}{(a+b)}\) will always decrease the term for positive a and b, while going from \(\frac{1}{a}\) to \(\frac{1}{a}+\frac{1}{b}\) will always increase the term.
So statement 1 is sufficient.
Statement (2): Here we can easily show that this is not sufficient by picking numbers. Choose e.g. 2 and 2, so we get \(\frac{1}{4}<1\), which is obviously true. But picking 2 and -1 will give \(1<\frac{1}{2}-1\), which is obviously wrong.
Therefore, answer A is correct.
Picking a few number pairs, we quickly realize that that information is sufficient. This also makes sense theoretically. Going from \(\frac{1}{a}\) to \(\frac{1}{(a+b)}\) will always decrease the term for positive a and b, while going from \(\frac{1}{a}\) to \(\frac{1}{a}+\frac{1}{b}\) will always increase the term.
So statement 1 is sufficient.
Statement (2): Here we can easily show that this is not sufficient by picking numbers. Choose e.g. 2 and 2, so we get \(\frac{1}{4}<1\), which is obviously true. But picking 2 and -1 will give \(1<\frac{1}{2}-1\), which is obviously wrong.
Therefore, answer A is correct.
25 Jan 2015, 12:38
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Lets Consider stmt1:
From stmt1 we can say that both a and b are positive. Then only |a| + |b| = a + b .
now form the question we have to prove that
1/(a+b) < 1/a + 1/b
or we can rephrase that we have to show that
[ab-(a+b)^2]/ab(a+b) < 0
In this case denominator cannot be less than zero as since a and b are positive, ab has to be positive and so do a + b
so the numerator has to be less than zero
so ab-(a+b)^2 < 0
which comes down to a^2 + b ^2 > -ab
which will always be true.
We are able to derive a conclusion using statement 1 .
If we consider 2. Straight away we can say that it's not sufficient.
Hence Option A is the answer.
From stmt1 we can say that both a and b are positive. Then only |a| + |b| = a + b .
now form the question we have to prove that
1/(a+b) < 1/a + 1/b
or we can rephrase that we have to show that
[ab-(a+b)^2]/ab(a+b) < 0
In this case denominator cannot be less than zero as since a and b are positive, ab has to be positive and so do a + b
so the numerator has to be less than zero
so ab-(a+b)^2 < 0
which comes down to a^2 + b ^2 > -ab
which will always be true.
We are able to derive a conclusion using statement 1 .
If we consider 2. Straight away we can say that it's not sufficient.
Hence Option A is the answer.
