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Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8

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aDAMS
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Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   Updated on: 04 Apr 2023, 10:20
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Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8
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Official Answer
B

Originally posted by aDAMS on 11 Apr 2013, 18:16.
Last edited by Bunuel on 04 Apr 2023, 10:20, edited 2 times in total.
RENAMED THE TOPIC.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   11 Apr 2013, 23:09
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aDAMS wrote:
Q. Is x > 0?

1. x^6>x^7
2. x^7>x^8


Any number with an even exponent is \(>0\), but I wanna give you an alternative solution:
1. \(x^6>x^7\) so \(0>x^7-x^6, 0>x^6(x-1)\) at this point we can divide by x^6 because we know that does not equal 0(otherwise x^6=x^7 and not >) \(0>x-1\) and finally \(1>x\). Is x positive? it could be(0,5) or not (-1)

2.\(x^7>x^8\) becomes \(0>x^6(x^2-x)\) divide \(0>x^2-x\) so \(x(x-1)<0\) and \(0<x<1\). Is x positive? yes
B
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   11 Apr 2013, 18:49
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Stmt 1 is Not Sufficient

x^6 > x^7 means x < 1, so x > 0 when 0< x < 1 and when x < 0 so x < 0

Stmt 2 is Sufficient

x^7 > x ^8 means 0<x<1 so x > 0

A variable with smaller power is more than a variable with larger power iff the variable is in the range 0 < x < 1

Answer is B

//kudos please, if the above explanation is good.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   12 Apr 2013, 02:38
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Is x > 0?

(1) x^6 > x^7. This implies that \(x\neq{0}\), thus \(x^6>0\). Divide both part by x^6: \(1>x\) (\(x\neq{0}\)). Not sufficient.

(2) x^7 > x^8. The same here. Divide both part by x^6: \(x>x^2\) --> \(0<x<1\). Sufficient.

Answer: B.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

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Hope it helps.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   12 Apr 2013, 04:35
Felt the following would be a simpler explanation

The two option would go as follows

1. x^6 > x^7
This option is only possible either when 1>x>0 or x<0
Like if x is <1 but >0 then the values will keep on decreasing with every increasing exponent
and if x<0 then the even powers will be positive and odd ones would be -ve therefore irrespective of the value of |x| the even exponents will always be greater

2. x^7 > x^8
This is only possible when 1>x>0, as the situation would be as in case 1 but when X<0 the even exponents will always be greater

Only 2 is sufficient but 1 is not B
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   12 Apr 2013, 06:59
Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   12 Apr 2013, 07:04
summer101 wrote:
Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??



x*(x-1)<0 implies

either x<0 and x-1 > 0
or x>0 and x-1<0

first is impossible and hence from second we have 0<x<1
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   12 Apr 2013, 07:56
summer101 wrote:
Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??


Till here you are fine x> x^2 so \(x^2-x<0\) we have to solve this, and to solve let me use an old trick.
Lets solve \(x^2-x=0,x(x-1)=0\) so x=1 or x=0. Now because the sign of x^2 is + and the operator is < we take the INTERNAL values: \(0<x<1\).
Remember: to solve inequalities like this (x^2) treat them like equations (replace <,> with = ) then, once you have the results take a look at the sign of x^2 an the operator.
(<,-) or (>,+) take ESTERNAL values. (if they are the "same")
(>,-) or (<,+) take INTERNAL values(like this case).

Let me know if it's clear now
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   10 Jun 2016, 12:53
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AbdurRakib wrote:
Is x>0 ?

1. \(x^6\)>\(x^7\)

2. \(x^7\) >\(x^8\)



1. \(x^6\)>\(x^7\), either 0 < x< 1 or x < -1 (the result of even power of a negative number is positive). Not Sufficient.

2. \(x^7\) >\(x^8\), 0 < x< 1. x cannot be negative, If x were negative, \(x^7\) will be negative and \(x^8\) will be positive and this statement won't hold true. Sufficient.

Answer is B.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   14 Jun 2016, 01:06
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aDAMS wrote:
Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8



\(x^6 > x^7\)

\(x^6 - x^7 > 0\)

\(x^6\)*(1 - x) > 0

\(x^6\) is always a positive value.

so (1-x)> 0
or x < 1

x can be (1/2) a positive value or x can be -2 , a negative value, so not sufficient.


Statement (2)

\(x^7 > x^8\)

\(x^7 - x^8 > 0\)

\(x^7\)*(1 - x) > 0

\(x^6.x.(1-x)\) > 0

Now \(x^6\) is always a positive value.

so x(1-x)> 0

or x(x-1) < 0

0< x <1

Thus x is always positive.

B is sufficient.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   17 Jun 2019, 18:31
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Simply assume values. Let's say x = 0.5 or -0.5

From statement 1, X^6>X^7
Case 1, if X is 0.5, X^6 > X^7 is true
Case 2, if X is -0.5, X^>X^7 is true
We cannot prove is X is >0. Insufficient.

From Statement 2, X^7>X^8
Case 1, if X is 0.5, X^7>X^8 is true
Case 2, if X is -0.5, X^7>X^8 is false
Therefore, sufficient. X>0

Answer is B.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   25 Oct 2019, 02:23
Official Explanation:
Statement (1) tells us that xˆ6>xˆ7
: this is true of any fraction between 0 and 1 and any negative integer; INSUFFICIENT.
Statement (2) tells us that xˆ7>xˆ8
; this is only true of fractions between 0 and 1. Since this guarantees that the lower limit for xis above 0, it definitively answers the question "yes"; SUFFICIENT.
If you’re wondering why statement (2) worked when statement (1) didn’t, note that in the second statement the first exponent is odd and the second is even. The even one must be 0 or positive, so if the odd one is greater, it must be greater than 0.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   08 Aug 2020, 02:05
In statement 2, why are we diving by x^6.

If we take x^7 common. 1>x is what we get which is similar to A.

Can someone help me explain?

Thanks
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   21 Nov 2020, 10:46
aDAMS wrote:
Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8


(1) \(x^6 - x^7 > 0\)
\(x^6 (1 - x) > 0\)
x^6 is always positive, therefore \(1 > x\). \(x < 0\) or \(0 < x < 1\). INSUFFICIENT.

(2) \(x^7 - x^8 > 0\)
\(x^7 (1 - x) > 0\)
\(x^6*x (1 - x) > 0\)
\(x > 0\) and \(1 > x\) OR \(x < 0\) and \(1 < x\); the second range is not possible.

Therefore \(1 > x > 0\). SUFFICIENT.

Answer is B.
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink] New post   11 Apr 2024, 11:33
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Re: Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8 [#permalink]
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