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If \(a\) and \(b\) are integers and \(a \lt b \lt 0\), what is the value of \(a-b\)?
(1) \(a^2=b^2+7\)
(2) \(ab=12\)
(1) \(a^2=b^2+7\)
(2) \(ab=12\)
16 Sep 2014, 00:44
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Official Solution:
If \(a\) and \(b\) are integers and \(a \lt b \lt 0\), what is the value of \(a-b\)?
(1) \(a^2=b^2+7\).
Rearrange and apply \(a^2 - b^2 = (a - b)(a + b)\) to get \((a - b)(a + b) = 7\). Since \(a\) and \(b\) are integers, both \(a + b\) and \(a - b\) are integers as well. Thus, we have the product of two integers equal to 7. There are only two combinations of such factors possible: (1, 7) and (-1, -7). Given that both \(a\) and \(b\) are negative, the first case is not possible, so \(a - b\) must be either -1 or -7. However, it cannot be -7 because in this case, \(a + b\) must be -1, and no two negative integers yield a sum of -1. Therefore, \(a - b = -1\). Sufficient.
(2) \(ab=12\).
If \(a = -12\) and \(b = -1\), then \(a - b = -11\), but if \(a = -4\) and \(b = -3\), then \(a - b = -1\). Insufficient.
Answer: A
If \(a\) and \(b\) are integers and \(a \lt b \lt 0\), what is the value of \(a-b\)?
(1) \(a^2=b^2+7\).
Rearrange and apply \(a^2 - b^2 = (a - b)(a + b)\) to get \((a - b)(a + b) = 7\). Since \(a\) and \(b\) are integers, both \(a + b\) and \(a - b\) are integers as well. Thus, we have the product of two integers equal to 7. There are only two combinations of such factors possible: (1, 7) and (-1, -7). Given that both \(a\) and \(b\) are negative, the first case is not possible, so \(a - b\) must be either -1 or -7. However, it cannot be -7 because in this case, \(a + b\) must be -1, and no two negative integers yield a sum of -1. Therefore, \(a - b = -1\). Sufficient.
(2) \(ab=12\).
If \(a = -12\) and \(b = -1\), then \(a - b = -11\), but if \(a = -4\) and \(b = -3\), then \(a - b = -1\). Insufficient.
Answer: A
10 Feb 2015, 21:42
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The only issue I have with this problem is that the GMAT does not contradict itself in data sufficiency for the constituent values. However, you get values for a and b from (1) that do not exist in (2).
