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12 Jun 2023, 01:30
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Difficulty:
55%
(hard)
Question Stats:
53% (01:33) correct
47%
(01:47)
wrong
based on 51
sessions
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Not Attempted Yet
Is \(x>y?\)
(1) \(x + 2\sqrt{xy}+ y = 0\)
(2) \(x^2-y^2 = 0\)
(1) \(x + 2\sqrt{xy}+ y = 0\)
(2) \(x^2-y^2 = 0\)
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
Updated on: 13 Jun 2023, 02:01
Originally posted by 8.toastcrunch on 12 Jun 2023, 02:05.
Last edited by 8.toastcrunch on 13 Jun 2023, 02:01, edited 1 time in total.
Last edited by 8.toastcrunch on 13 Jun 2023, 02:01, edited 1 time in total.
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Statement (1): x + 2√(xy) + y = 0
We can see that it follows (a+b)^2 where a = √x and b = √y. The resulting equation is:
x + 2√(xy) + y = (√x + √y) ^2 = 0
Since √x and √y are always positive, the only solution is when x = y = 0.
Statement (2): x^2 - y^2 = 0
This equation can be factored as (x - y)(x + y) = 0. From this factorization, we can see that either (x - y) = 0 or (x + y) = 0.
If (x - y) = 0, then x = y. This means that x and y are equal.
If (x + y) = 0, then x = -y. This means that x and y are opposite in sign.
Based on the information from statement (2), we can conclude that x and y can be either equal or opposite in sign, but we cannot determine whether x is greater than y.
Therefore , answer is [A].
We can see that it follows (a+b)^2 where a = √x and b = √y. The resulting equation is:
x + 2√(xy) + y = (√x + √y) ^2 = 0
Since √x and √y are always positive, the only solution is when x = y = 0.
Statement (2): x^2 - y^2 = 0
This equation can be factored as (x - y)(x + y) = 0. From this factorization, we can see that either (x - y) = 0 or (x + y) = 0.
If (x - y) = 0, then x = y. This means that x and y are equal.
If (x + y) = 0, then x = -y. This means that x and y are opposite in sign.
Based on the information from statement (2), we can conclude that x and y can be either equal or opposite in sign, but we cannot determine whether x is greater than y.
Therefore , answer is [A].
12 Jun 2023, 06:21
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Statement (1): x + 2√(xy) + y = 0
We can see that it follows (a+b)^2 where a = √x and b = √y. The resulting equation is:
x + 2√(xy) + y = (√x + √y) ^2 = 0
Since √x and √y are always positive, the only solution is when x = y = 0.
From that we can conclude x is never greater than y.
Statement (2): x^2 - y^2 = 0
follow abhishekvashist reasoning.
Therefore , answer is A.
We can see that it follows (a+b)^2 where a = √x and b = √y. The resulting equation is:
x + 2√(xy) + y = (√x + √y) ^2 = 0
Since √x and √y are always positive, the only solution is when x = y = 0.
From that we can conclude x is never greater than y.
Statement (2): x^2 - y^2 = 0
follow abhishekvashist reasoning.
Therefore , answer is A.
