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Re: If x and y are integers, is ax > ay?
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06 Feb 2017, 10:39
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Top Contributor
1
Bunuel wrote:
If x and y are integers, is ax > ay?
(1) a³x > a³y (2) –ax < –ay
Great question!
Target question:Is ax > ay?
Given: x and y are integers
Statement 1: a³x > a³y First recognize that this statement is quite similar to the target question. Also recognize that, statement 1 tells us that a ≠ 0. This is very useful, because we can now be certain that a² is POSITIVE If a² is POSITIVE, we can safely take the inequality a³x > a³y and divide both sides by a² to get: ax > ay Perfect! This answers the target question. Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: –ax < –ay Let's multiply both sides of the inequality to get: ax > ay[Aside: since we multiplied both sides of the inequality by a NEGATIVE value, we reversed the direction of the inequality sign] Perfect! Once again, we have answered the target question. Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Re: If x and y are integers, is ax > ay?
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31 Oct 2017, 10:42
Raj94* wrote:
Sir, from statement 1 how can we conclude that a^2 is positive
The square of a number (more generally an even power of a number) is always non-negative, so 0 or positive. \(a^3*x > a^3*y\) also implies that \(a \neq 0\), becasue if it were, then we'd have \(a^3*x =0 = a^3*y\). Thus, in this a^2 is not only non-negative but positive.
_________________
Re: If x and y are integers, is ax > ay?
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02 Nov 2017, 10:43
1
Bunuel wrote:
If x and y are integers, is \(ax > ay\)?
(1) \(a^3*x > a^3*y\)
(2) \(–ax < –ay\)
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. We can modify the original condition and question as follows.
\(ax > ay <=> a^3x > a^3y\) if \(a≠0\). \(ax > ay <=> -ax < -ay\).
Condition 1) Since \(a^3x > a^3y\), \(a\) is not zero. By dividing both sides by \(a^2\), we have \(ax > ay\) since \(a^2 > 0\). This is sufficient.
Conditin 2) By multiplying \(-ax < -ay\) by \(-1\), we have \(ax > ay\). This is sufficient.
Re: If x and y are integers, is ax > ay?
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29 Oct 2018, 00:25
1
GMATPrepNow wrote:
Bunuel wrote:
If x and y are integers, is ax > ay?
(1) a³x > a³y (2) –ax < –ay
Great question!
Target question:Is ax > ay?
Given: x and y are integers
Statement 1: a³x > a³y First recognize that this statement is quite similar to the target question. Also recognize that, statement 1 tells us that a ≠ 0. This is very useful, because we can now be certain that a² is POSITIVE If a² is POSITIVE, we can safely take the inequality a³x > a³y and divide both sides by a² to get: ax > ay Perfect! This answers the target question. Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: –ax < –ay Let's multiply both sides of the inequality to get: ax > ay[Aside: since we multiplied both sides of the inequality by a NEGATIVE value, we reversed the direction of the inequality sign] Perfect! Once again, we have answered the target question. Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer: D
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What if a is negative? Will statement 2 still hold?
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Re: If x and y are integers, is ax > ay?
[#permalink]
29 Oct 2018, 00:25