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If x^2 < x, then x must be

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Bunuel
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If x^2 < x, then x must be [#permalink] New post   22 May 2020, 05:55
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A
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C
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E

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Question Stats:

88% (00:35) correct 13% (00:44) wrong based on 328 sessions
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If x^2 < x, then x must be

A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1


PS21231

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Where does p lie if p² < p?

A. Between -1 and 0
B. Between -1 and 1
C. Between 0 and 1
D. It is always less than 0
E. It is always greater than 1
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C
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Ravixxx
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Re: If x^2 < x, then x must be [#permalink] New post   22 May 2020, 06:07
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Bunuel wrote:
If x^2 < x, then x must be

A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1


PS21231


\(x^{2}<x\)

\(x^{2}- x<0\)

\(x(x-1)<0\)

\(0<x<1\)

Answer C
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yashikaaggarwal
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Re: If x^2 < x, then x must be [#permalink] New post   22 May 2020, 06:11
Let's put values of x one by one

A) less than zero = -1/2
=> (-1/2)^2 & -1/2
=> 1/4>-1/2 (incorrect)

B) equal to zero
=> 0^2 & 0
=> 0=0 (incorrect)

C) between 0 & 1 = 1/2
=> (1/2)^2 & (1/2)
=> 1/4<1/2 ((correct))

D) equal to 1
=> 1^2 & 1
=> 1=1 (incorrect)

E) more than one = 2
=> (2)^2 & 2
=> 4>2 (incorrect)

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BrentGMATPrepNow
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Re: If x^2 < x, then x must be [#permalink] New post   24 May 2020, 06:16
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Bunuel wrote:
If x^2 < x, then x must be

A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1
PS21231



Given: x² < x
Subtract x from both sides to get: x² - x < 0
Factor: x(x - 1) < 0

We can already see that, when x = 0, we get: 0(0 - 1) < 0, which results in 0 < 0, which doesn't work. ELIMINATE B
Likewise, when x = 1, we get: 1(1 - 1) < 0, which results in 0 < 0, which doesn't work. ELIMINATE D

At this point we can apply a little bit of number sense.

If x is between 0 and 1, then x is POSITIVE, and (x - 1) is NEGATIVE, which means x(x - 1) < 0 becomes POSITIVE(NEGATIVE) < 0, which works!!

Answer: C

Cheers,
Brent
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Re: If x^2 < x, then x must be [#permalink] New post   24 May 2020, 06:29
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Bunuel wrote:
If x^2 < x, then x must be

A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1


PS21231


x^2 < x

i.e. x^2 - x < 0
i.e. x (x-1) < 0

- Product of two things x and (x-1) is negative i,e, one of them must be negative and one must be positive
- Since x-1 is smaller than x therefore x-1 must be negative and x must be positive

x > 0 and x-1 < 0
x > 0 and x < 1

i.e. 0 < x < 1



Answer: Option C
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Re: If x^2 < x, then x must be [#permalink] New post   31 May 2020, 04:05
Expert Reply
Bunuel wrote:
If x^2 < x, then x must be

A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1


PS21231


In order for x^2 to be less than x, then x must be a number between 0 and 1. For example, if x = 1/2, then x^2 = (½)^2 = 1/4. Since 1/4 < 1/2, we see that x^2 < x for this example, and the equation holds true for any positive fraction (i.e., a value between 0 and 1).

Answer: C
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Re: If x^2 < x, then x must be [#permalink] New post   05 May 2022, 04:52
Ravixxx wrote:
Bunuel wrote:
If x^2 < x, then x must be

A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1


PS21231


\(x^{2}<x\)

\(x^{2}- x<0\)

\(x(x-1)<0\)

\(0<x<1\)

Answer C



If it was x^2>x
Then what would have been the answer?

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Re: If x^2 < x, then x must be [#permalink] New post   03 Sep 2023, 21:20
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Re: If x^2 < x, then x must be [#permalink]
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