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

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Senior Manager
Joined: 17 Mar 2014
Posts: 384
If X and Y are positive, is X^2>Y  [#permalink]

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27 Jul 2017, 14:54
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Difficulty:

85% (hard)

Question Stats:

45% (01:52) correct 55% (01:39) wrong based on 94 sessions

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

(1) $$X>Y^2$$
(2) $$X>Y$$
Senior Manager
Joined: 24 Apr 2016
Posts: 333
Re: If X and Y are positive, is X^2>Y  [#permalink]

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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 $$X^2>Y$$.

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

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15 Mar 2018, 10:59
1
]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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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.

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]

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13 Sep 2018, 13:01
ammuseeru wrote:
If X and Y are positive, is $$X^2>Y$$ ?

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

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

if x = 1 and y = 1/2 then $$X^2>Y$$
if x = 1/3 and y = 1/2 the $$X^2<Y$$

insufficient

(2) $$X>Y$$

if x =5 and y =2 then $$X^2>Y$$
if x =1/2 and y = 1/3 then $$X^2<Y$$

insufficient

using both
still insufficient

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

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14 Sep 2018, 03:47
ammuseeru wrote:
If X and Y are positive, is $$X^2>Y$$ ?

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

1) x> y^2 , it works for some values and for other doesn't. x could be just a significantly larger number than y or if they are both in the (0;1) range then x can actually be smaller y. Hence, we cannot conclude if x^2 > y

2) x > y, same as above. We need to consider in which ranges x and y fall into. Remember that if they are factors then squaring x makes it become smaller.

COMBINING con 1 and 2 we have

x > y^2
x > y

=> If x falls into the range (1;inf) then it is obvious that squaring x will make it larger than original x hence will be larger than y and y^2
If x falls into the range (0;1) then we are not sure. Let's pick x=1/3 and y =1/4.
1/3 > 1/4 -> (1/3)^2 = 1/9 < 1/4 -> x^2 > y Wrong

So we have x^2 > y when x is in (1;inf) and x^2 possibly < y when x is in (0;1)
Re: If X and Y are positive, is X^2>Y &nbs [#permalink] 14 Sep 2018, 03:47
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