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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seems D is the answer

(1) a^3(x-y)>0
as a and a^3 ,both has same signs thus a(x-y) >0
suff

(2) –ax < –ay
-a(x-y)<0
thus if a is positive than (x-y) also positive so suff
or, if a is negative then (x-y) negative too thus a(x-y)>0
suff

Ans D

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.
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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.

Happy Studying !!!
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What if a is negative? Will statement 2 still hold?

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