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# Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4

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Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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06 Sep 2019, 06:45
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Difficulty:

35% (medium)

Question Stats:

68% (01:27) correct 32% (01:34) wrong based on 41 sessions

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Is $$x > y$$ ?

(1) $$x^5 > y^5$$

(2) $$x^4 < y^4$$

Source: Nova GMAT

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Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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06 Sep 2019, 09:38
Statement(1): let x, -1/3 and y, -1/2.x>y and x^5>y^5
Sufficient(2):
Let x, 1 and y,1/2.x>y and x⁴>y⁴.
Not sufficient
Option A

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Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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31 Mar 2020, 09:33
If x=3 and y = 2 the outcome is no
if x=-1 and y=-3 the outcome is yes.

How is the OA A?

SajjadAhmad wrote:
Is $$x > y$$ ?

(1) $$x^5 > y^5$$

(2) $$x^4 < y^4$$

Source: Nova GMAT
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Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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31 Mar 2020, 13:54
TheMongoose wrote:
If x=3 and y = 2 the outcome is no
if x=-1 and y=-3 the outcome is yes.

How is the OA A?

SajjadAhmad wrote:
Is $$x > y$$ ?

(1) $$x^5 > y^5$$

(2) $$x^4 < y^4$$

Source: Nova GMAT

Hi,
when x =3, y = 2. x^5 > y^5 as 3^5 > 2^5.
Again for the second case also x^5 > y^5 as -1 > -243
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Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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31 Mar 2020, 15:48
SajjadAhmad wrote:
Is $$x > y$$ ?

(1) $$x^5 > y^5$$

(2) $$x^4 < y^4$$

Source: Nova GMAT

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since condition 1) tells the exponent is an odd number 5, we have
$$x^5 > y^5$$
$$⇔ \sqrt[5]{x^5}> \sqrt[5]{y^5}$$
$$⇔ x > y$$

Thus condition 1) alone is sufficient.

Condition 2)
Since condition 2) tells the exponent is an even number 4, we have

$$x^4 > y^4$$
$$⇔ \sqrt[4]{x^4}> \sqrt[4]{y^4}$$
$$⇔ |x| > |y|$$

If $$x = 2, y = 1$$, then we have x>y and the answer is 'yes'.
If $$x = -2, y = 1$$, then we have x<y and the answer is 'no'.

Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.
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Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4   [#permalink] 31 Mar 2020, 15:48

# Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4

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