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x and y are integers such that x y<0, x>y , and y^2>8 .
[#permalink]
07 Oct 2026, 02:56
07 Oct 2026, 02:56
Expert Reply
00:00
Question Stats:
60% (02:30) correct
40% (03:28) wrong based on 5 sessions
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x and y are integers such that \(x y<0, x>y \), and \(y^2>8\) .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\quad 3 x-3 y\) |
\(\quad x^2-2 x y+y^2\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
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Re: x and y are integers such that x y<0, x>y , and y^2>8 .
[#permalink]
15 Oct 2026, 10:58
15 Oct 2026, 10:58
1
Expert Reply
Solution
Simplify the quantities
Quantity A: \(3x-3y=3(x-y)\)
Quantity B: \(x^2-2xy+y^2=(x-y)^2\)
Determine the value of \(x-y\)
Since \(xy<0, x\) and y have opposite signs.
Given \(x>y,\) it follows that \(x>0\) and \(y<0\).
Also, \(y^2>8\), and \(y\) is an integer, so \(y\le -3\).
Since \(x\ge 1,\)\(\)
\(x-y\ge 1-(-3)=4\)
Compare
Let \(d=x-y,\) where \(d\ge4\).
We compare \(3d\) and\( d^2:\)
\(d^2>3d \quad \text{because } d>3\)
Therefore,
[
\((x-y)^2>3(x-y)\)
Hence, Quantity B is greater.
Simplify the quantities
Quantity A: \(3x-3y=3(x-y)\)
Quantity B: \(x^2-2xy+y^2=(x-y)^2\)
Determine the value of \(x-y\)
Since \(xy<0, x\) and y have opposite signs.
Given \(x>y,\) it follows that \(x>0\) and \(y<0\).
Also, \(y^2>8\), and \(y\) is an integer, so \(y\le -3\).
Since \(x\ge 1,\)\(\)
\(x-y\ge 1-(-3)=4\)
Compare
Let \(d=x-y,\) where \(d\ge4\).
We compare \(3d\) and\( d^2:\)
\(d^2>3d \quad \text{because } d>3\)
Therefore,
[
\((x-y)^2>3(x-y)\)
Hence, Quantity B is greater.
gmatclubot
Re: x and y are integers such that x y<0, x>y , and y^2>8 . [#permalink]
15 Oct 2026, 10:58
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