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x<y, x^2+y^2
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Updated on: 26 Aug 2021, 23:09
Updated on: 26 Aug 2021, 23:09
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Question Stats:
66% (01:32) correct
33% (00:47) wrong based on 24 sessions
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x<y
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x^2+y^2\) |
2xy |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: x<y, x^2+y^2
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25 Aug 2021, 05:53
25 Aug 2021, 05:53
2
cbrelax wrote:
x<y
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x^2+y^2\) |
\(2xy\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
There's actually a perfect square hiding in this question.
Let's start by using matching operations (my favorite Quantitative Comparison strategy)
Given:
Quantity A: \(x^2+y^2\)
Quantity B: \(2xy\)
Subtract \(2xy\) from both quantities to get:
Quantity A: \(x^2-2xy+y^2\)
Quantity B: 0
Factor Quantity A to get:
Quantity A: \((x-y)^2\)
Quantity B: 0
Since \(x<y\), we know that \(x - y\) does not equal 0, which means \((x-y)^2\) must be POSITIVE.
So we have:
Quantity A: Some positive number
Quantity B: 0
Answer: A
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