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b < a
[#permalink]
02 Sep 2024, 22:37
02 Sep 2024, 22:37
Expert Reply
00:00
Question Stats:
0% (00:00) correct
100% (01:01) wrong based on 4 sessions
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\(b < a\)
Quantity A |
Quantity B |
\(n^2+(n-a)^2\) |
\(n^2+(n-b)^2\) |
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b < a
[#permalink]
03 Sep 2024, 00:44
03 Sep 2024, 00:44
Two cases here:
We can ignore the \(n^2\) on both sides for obvious reasons. So its basically comparing \((n-a)^2\) vs \((n-b)^2\).
1) Let n>0 (say 5) and a=2, b=1.
Clearly, \((n-a)^2 < (n-b)^2\) (9<16).
2) Let n<0 (say -5) and a=2, b=1.
Clearly, Clearly, \((n-a)^2 > (n-b)^2\) (49<36).
Hence, D is correct.
We can ignore the \(n^2\) on both sides for obvious reasons. So its basically comparing \((n-a)^2\) vs \((n-b)^2\).
1) Let n>0 (say 5) and a=2, b=1.
Clearly, \((n-a)^2 < (n-b)^2\) (9<16).
2) Let n<0 (say -5) and a=2, b=1.
Clearly, Clearly, \((n-a)^2 > (n-b)^2\) (49<36).
Hence, D is correct.
gmatclubot
b < a [#permalink]
03 Sep 2024, 00:44
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