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Updated on: 20 Aug 2019, 10:30
Originally posted by BrentGMATPrepNow on 20 Oct 2016, 09:20.
Last edited by BrentGMATPrepNow on 20 Aug 2019, 10:30, edited 2 times in total.
Last edited by BrentGMATPrepNow on 20 Aug 2019, 10:30, edited 2 times in total.
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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:
75%
(hard)
Question Stats:
53% (02:00) correct
47%
(02:14)
wrong
based on 266
sessions
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If \(n\) is an integer, \(xy ≠ 0\), and \(x^n = y^n\), then what is the value of n?
(1) \(\frac{x}{y} = 2\)
(2) \(n < y < x\)
(1) \(\frac{x}{y} = 2\)
(2) \(n < y < x\)
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20 Oct 2016, 10:06
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x^n = y^n --> We can have 3 different cases:
Case 1) \(x = y\) and n is any value
Case 2) \(|x| \neq {|y|}\) and n = 0
Case 3) \(x = -y\) and n is even
St1: x/y = 2 --> \(|x| \neq {|y|}\) --> n = 0
Sufficient
St2: n < y < x --> Can have 2 possibilities
\(x \neq {y}\) --> n = 0
or
\(y = -x\) --> n is even and negative
Not Sufficient
Answer: A
Case 1) \(x = y\) and n is any value
Case 2) \(|x| \neq {|y|}\) and n = 0
Case 3) \(x = -y\) and n is even
St1: x/y = 2 --> \(|x| \neq {|y|}\) --> n = 0
Sufficient
St2: n < y < x --> Can have 2 possibilities
\(x \neq {y}\) --> n = 0
or
\(y = -x\) --> n is even and negative
Not Sufficient
Answer: A
General Discussion
20 Oct 2016, 23:14
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From question we know.
xy ≠ 0 - which means none of them can be equal to zero.
and x^n = y^n meaning either n can be zero or even.
1. Statement 1 : x/y = 2, which means (2)^n = 1, this is only possible for one value of n =0. Sufficient
2. Statement 2 : n < y < x, this means that y would be negative (y = -x) and n would be smaller than y, as we are given
that x^n = y^n, so n would be even negative integer and we can have multiple solutions. So not sufficient.
Option A.
xy ≠ 0 - which means none of them can be equal to zero.
and x^n = y^n meaning either n can be zero or even.
1. Statement 1 : x/y = 2, which means (2)^n = 1, this is only possible for one value of n =0. Sufficient
2. Statement 2 : n < y < x, this means that y would be negative (y = -x) and n would be smaller than y, as we are given
that x^n = y^n, so n would be even negative integer and we can have multiple solutions. So not sufficient.
Option A.
