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Is x > 0?

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Is x > 0? [#permalink] New post 11 Apr 2013, 17:16
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

62% (01:46) correct 38% (01:14) wrong based on 53 sessions
Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Apr 2013, 01:24, edited 1 time in total.
RENAMED THE TOPIC.
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Re: DS question on Exponents [#permalink] New post 11 Apr 2013, 17:49
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

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Re: DS question on Exponents [#permalink] New post 11 Apr 2013, 22:09
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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? [#permalink] New post 12 Apr 2013, 01:38
Expert's post
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? [#permalink] New post 12 Apr 2013, 03:35
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? [#permalink] New post 12 Apr 2013, 05:59
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? [#permalink] New post 12 Apr 2013, 06:04
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? [#permalink] New post 12 Apr 2013, 06:56
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? [#permalink] New post 07 May 2014, 23:26
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Re: Is x > 0?   [#permalink] 07 May 2014, 23:26
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