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If X and Y are positive, is X^2>Y

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If X and Y are positive, is X^2>Y [#permalink]

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New post 27 Jul 2017, 14:54
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

38% (02:02) correct 62% (01:01) wrong based on 55 sessions

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If X and Y are positive, is \(X^2>Y\) ?

(1) \(X>Y^2\)
(2) \(X>Y\)
[Reveal] Spoiler: OA
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Re: If X and Y are positive, is X^2>Y [#permalink]

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New post 27 Jul 2017, 16:40
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ammuseeru wrote:
Can you help me understand following question.

If X and Y are positive, is \(X^2>Y\) ?

1. \(X>Y^2\)
2. \(X>Y\)

Regards,
Ammu


Please see the attached table. You can see different values of X satisfying the conditions given in Statement 1 & Statement 2, yet have different results for \(X^2>Y\).

Hence the answer is E
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If X and Y are positive, is X^2>Y [#permalink]

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New post 15 Mar 2018, 10:59
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]If X and Y are positive, is \(X^2>Y\)?

(1) \(X>Y^2\)

Case 1: If X = 1/2 and Y = 1/3 - \(X>Y^2\) but \(X^2 < Y\)
Case 2: If X = 5 and Y = 2 - \(X>Y^2\) and \(X^2>Y\) (Insufficient)

(2) \(X>Y\)

Case 1: If X = 1/2 and Y = 1/3 - \(X>Y\) but \(X^2 < Y\)
Case 2: If X = 5 and Y = 2 - \(X>Y\) and \(X^2>Y\) (Insufficient)

On combining the information from both the statement,
it is possible that \(X^2>Y\) and \(X^2<Y\) (Insufficient - Option E)
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Re: If X and Y are positive, is X^2>Y [#permalink]

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New post 18 Mar 2018, 23:04
ammuseeru wrote:
If X and Y are positive, is \(X^2>Y\) ?

(1) \(X>Y^2\)
(2) \(X>Y\)


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

If X = 10, Y = 1, then the answer is yes.
If X = 2/3, Y = 1/2, then the answer is no, since X^2 = 4/9 < Y = 1/2.

Therefere, E is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If X and Y are positive, is X^2>Y   [#permalink] 18 Mar 2018, 23:04
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