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Is x^7 < 6^y ? (1) x^3 = –125 (2) y^2 = 36

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Is x^7 < 6^y ? (1) x^3 = –125 (2) y^2 = 36  [#permalink]

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New post 22 Nov 2018, 03:12
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (01:08) correct 42% (01:24) wrong based on 187 sessions

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Re: Is x^7 < 6^y ? (1) x^3 = –125 (2) y^2 = 36  [#permalink]

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New post 22 Nov 2018, 12:21
2
I think its a
x=-5 its enough that x^7 = -ve number
if we take 6^1 0r 6^-1 the answer is still positive
so A sufficient
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Re: Is x^7 < 6^y ? (1) x^3 = –125 (2) y^2 = 36  [#permalink]

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New post 22 Nov 2018, 15:09
2
Top Contributor
Bunuel wrote:
Is x^7 < 6^y ?

(1) x³ = –125
(2) y² = 36


KEY CONCEPTS:

#1. ODD exponents preserve the sign of the base.
So, (NEGATIVE)^(ODD integer) = NEGATIVE
and (POSITIVE)^(ODD integer) = POSITIVE

#2. An EVEN exponent always yields a positive result (unless the base = 0)
So, (NEGATIVE)^(EVEN integer) = POSITIVE
and (POSITIVE)^(EVEN integer) = POSITIVE

Target question: Is x^7 < 6^y ?

Statement 1: x³ = –125
This tells us that x = -5
By Rule #1, we know that, since -5 is negative, x^7 = (-5)^7 = some NEGATIVE number
By Rule #2, 6^y = some POSITIVE number
So, it must be the case that x^7 < 6^y
The answer to the target question is YES, x^7 IS less than 6^y
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: y² = 36
This tells us that EITHER y = 6 OR y = -6
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = -1 and y = 6. Here, x^7 = -1 and 6^y = some POSITIVE number. In this case, the answer to the target question is YES, x^7 IS less than 6^y
Case b: x = 1,000,000 and y = 6. Here, x^7 = a VERY VERY big number and 6^y = a KIND OF big number. In this case, the answer to the target question is NO, x^7 is NOT less than 6^y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Re: Is x^7 < 6^y ? (1) x^3 = –125 (2) y^2 = 36  [#permalink]

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New post 24 Dec 2018, 01:46
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Re: Is x^7 < 6^y ? (1) x^3 = –125 (2) y^2 = 36   [#permalink] 24 Dec 2018, 01:46
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