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-1<x<0

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-1<x<0 [#permalink] New post   01 Oct 2022, 09:57
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\(-1<x<0\)


Quantity A
Quantity B
\(-x\)
\(-x^3\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


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B
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kk117
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Re: -1<x<0 [#permalink] New post   03 Oct 2022, 10:08
Could someone please explain this question?

I got A as I divided both sides by (-x)

Hence,

A: 1
B: x^2

Now since x is less than 0 but greater than (-1), hence x^2 is less than (1)

Hence, A

But since the answer is B, what is wrong with my approach?
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Re: -1<x<0 [#permalink] New post   12 Oct 2022, 00:14
1
-1 < x <0 is basically saying
x= negative non integer or fraction
Let x= -1/2

Quantity A
-1/2 or -0.5

Quantity B
(-1/2)(-1/2)(-1/2) = -1/8 or -0.125

Quantity A is more negative than Quantity B
Hence B is greater

Posted from my mobile device
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eywilson
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Re: -1<x<0 [#permalink] New post   01 Mar 2023, 12:48
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GREfocus wrote:
-1 < x <0 is basically saying
x= negative non integer or fraction
Let x= -1/2

Quantity A
-1/2 or -0.5

Quantity B
(-1/2)(-1/2)(-1/2) = -1/8 or -0.125

Quantity A is more negative than Quantity B
Hence B is greater

Posted from my mobile device


GREfocus that's not correct though, because it's a double negative situation. You have to sub the variable into the x inside brackets. Since we're dealing with a negative fraction/non-integer, quantity A becomes -(-1/2) or 1/2 and B is -(-1/2)^3 = -(-1/8) = 1/8. Hence A should be bigger
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sree94
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Re: -1<x<0 [#permalink] New post   06 Mar 2023, 14:58
2
here x is negative and the question gives -x, -x^3 which means double negative turns to be positive, so quantity A is bigger
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Re: -1<x<0 [#permalink]
06 Mar 2023, 14:58
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