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GMAT Club team member V
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Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 66% (01:28) correct 34% (01:34) wrong based on 38 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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Joined: 28 Jun 2019
Posts: 469
Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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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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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?

Is $$x > y$$ ?

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

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

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

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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?

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
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9016
GMAT 1: 760 Q51 V42
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Re: Is x > y ? (1) x^5 > y^5 (2) x^4 < y^4  [#permalink]

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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
$$x^5 > y^5$$
$$⇔ \sqrt{x^5}> \sqrt{y^5}$$
$$⇔ x > y$$

Thus condition 1) alone is sufficient.

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

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

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.

_________________ 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  