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|x^3|< 64

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Carcass
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|x^3|< 64 [#permalink] New post   08 Aug 2018, 16:12
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\(|x^3|< 64\)

Quantity A
Quantity B
\(-x\)
\(-|x|\)


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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D
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Re: |x^3|< 64 [#permalink] New post   09 Aug 2018, 12:55
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Carcass wrote:
\(|x^3|< 64\)

Quantity A
Quantity B
-x
-|x|




Let's test some possible values of x

CASE 1
Since |0³| < 64, x COULD equal 0
We get:
Quantity A: -0
Quantity B: -|0|

Evaluate to get:
Quantity A: 0
Quantity B: 0
In this case, the two quantities are EQUAL


CASE 2
Since |(-1)³| < 64, x COULD equal -1
We get:
Quantity A: -(-1)
Quantity B: -|-1|

Evaluate to get:
Quantity A: 1
Quantity B: -1
In this case, Quantity A IS GREATER

Answer: D

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Brent
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Re: |x^3|< 64 [#permalink] New post   09 Aug 2018, 15:47
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Carcass wrote:
\(|x^3|< 64\)

Quantity A
Quantity B
-x
-|x|


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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30 sec approach :

It's impossible to determine the sign of a variable that keeps inside the absolute value sign. X could be positive or negative. U will get know definite value of x.

The best answer is D.
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Re: |x^3|< 64 [#permalink] New post   10 Aug 2018, 03:38
I selected option B because -|x| = -x.

How is it the D? I still do not understand. Could anyone further explain both question and answer?

Thank you.
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Re: |x^3|< 64 [#permalink] New post   10 Aug 2018, 05:07
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ajbozdar wrote:
I selected option B because -|x| = -x.

How is it the D? I still do not understand. Could anyone further explain both question and answer?

Thank you.


It is not always the case that -|x| = -x

If x is a POSITIVE number, then -|x| = -x
For example, if x = 2, we get: -|2| = -2
Simplify to get: -2 = -2

However, if x is a NEGATIVE number, then it is NOT the case that -|x| = -x
For example, if x = -3, we get: -|-3| = -(-3)
Simplify to get: -3 = 3, which is NOT TRUE

Does that help?

Cheers,
Brent
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Re: |x^3|< 64 [#permalink] New post   10 Aug 2018, 06:25
GreenlightTestPrep wrote:

However, if x is a NEGATIVE number, then it is NOT the case that -|x| = -x
For example, if x = -3, we get: -|-3| = -(-3)
Simplify to get: -3 = 3, which is NOT TRUE

Does that help?



I have learnt that any negative number under mod becomes positive.

If so, -|-3| = -(3) i.e. = -3

Did I learn the wrong concept of absolute value?
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Re: |x^3|< 64 [#permalink] New post   10 Aug 2018, 08:13
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ajbozdar wrote:
GreenlightTestPrep wrote:

However, if x is a NEGATIVE number, then it is NOT the case that -|x| = -x
For example, if x = -3, we get: -|-3| = -(-3)
Simplify to get: -3 = 3, which is NOT TRUE

Does that help?



I have learnt that any negative number under mod becomes positive.

If so, -|-3| = -(3) i.e. = -3

Did I learn the wrong concept of absolute value?



You learned the correct concept.
In my solution, you'l see that I also state that -|-3| = -(3) = -3

Cheers,
Brent
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Re: |x^3|< 64 [#permalink] New post   21 Oct 2018, 00:39
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Yes.

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Re: |x^3|< 64 [#permalink] New post   23 Oct 2018, 08:43
is it possible to have negative sign in absolut value
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Re: |x^3|< 64 [#permalink] New post   29 Aug 2021, 20:24
I attempted this sum in a different way:

|x ^ 3| < 64 , hence x needs to be less than 4 . As x can be positive or negative, it can have the values : +1,-1 ; +2 , -2 ;
+3 , -3 .

Lets consider x = 3

QA : -3
QB : -3

a= b

Lets consider x = -3

QA : 3
QB : -3

a>b

hence Ans is D
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HarishKumar
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|x^3|< 64 [#permalink] New post   25 Jun 2023, 20:07
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\(|x^3| < 64 \) \(\text { }\) implies the value of \(x\) could be positive or negative. So, we could write two equations

\(x^3 < 64 \) ......(1)

\((-x)^3 < 64\)......(2)

Now, Quantity B, \(-|x|\) will always be negative.

So, one of the two inequalities above, (1) or (2), has negative value for \(x\) and is equivalent to \(-|x|\), So both quantities are equal.

The other inequality will give \(x\) a positive value and hence it will be greater than Quantity B.

Therefore, the answer is D.
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