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Bunuel
If 5 ≥ |x| ≥ 0, which of the following must be true?

I. x ≥ 0
II. x > –5
III. 25 ≥ x^2 ≥ –25

A. None
B. II only
C. III only
D. I and III only
E. II and III only

I. x could be 4 or -4....hence not must be true
II. x could be -5 or 5 > -5 ........hence not must be true
III. x^2 = 24 then x<5 (either x= -4.9 or x=4.9 say for instance..)..........TRue

Ans C
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Can someone please explain me how C is the correct answer. I chose A.

III. 25 ≥ x^2 ≥ –25

How can we apply square root to -25 and simplify it to 5 ≥ x ≥ –5
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Can someone please explain me how C is the correct answer. I chose A.

III. 25 ≥ x^2 ≥ –25

How can we apply square root to -25 and simplify it to 5 ≥ x ≥ –5

5 ≥ |x| ≥ 0 means that 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5. For any x from these ranges, 25 ≥ x^2 ≥ –25 will be true.
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Bunuel
If 5 ≥ |x| ≥ 0, which of the following must be true?

I. x ≥ 0
II. x > –5
III. 25 ≥ x^2 ≥ –25

A. None
B. II only
C. III only
D. I and III only
E. II and III only

Hi

Option "C" simply can't be correct choice here.

Squire of a real number can never be -ve. It can only be possible only if "x" is an imaginary number. However if "x" is an imaginary number then the condition mentioned in the question itself will not hold true.

Hence option "A-None" should be the correct answer.
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Bunuel
If 5 ≥ |x| ≥ 0, which of the following must be true?

I. x ≥ 0
II. x > –5
III. 25 ≥ x^2 ≥ –25

A. None
B. II only
C. III only
D. I and III only
E. II and III only

Hi

Option "C" simply can't be correct choice here.

Squire of a real number can never be -ve. It can only be possible only if "x" is an imaginary number. However if "x" is an imaginary number then the condition mentioned in the question itself will not hold true.

Hence option "A-None" should be the correct answer.

You did not understand the question.

I'll try to explain again:

5 ≥ |x| ≥ 0 means that 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5. For example, x can be, among infinitely many other values, 0.1, 0.7, 1, 1.7, 4, 5 (because 5 ≥ x ≥ 0) as well as x can be -0.008, -0.4, -3.4, -4, -4.6, -5 (because 0 ≥ x ≥ -5). For ANY possible x, so for ANY x from 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5, it will be true to say that 25 ≥ x^2 ≥ –25.
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Bunuel
If 5 ≥ |x| ≥ 0, which of the following must be true?

I. x ≥ 0
II. x > –5
III. 25 ≥ x^2 ≥ –25

A. None
B. II only
C. III only
D. I and III only
E. II and III only

Hi

Option "C" simply can't be correct choice here.

Squire of a real number can never be -ve. It can only be possible only if "x" is an imaginary number. However if "x" is an imaginary number then the condition mentioned in the question itself will not hold true.

Hence option "A-None" should be the correct answer.

You did not understand the question.

I'll try to explain again:

5 ≥ |x| ≥ 0 means that 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5. For example, x can be, among infinitely many other values, 0.1, 0.7, 1, 1.7, 4, 5 (because 5 ≥ x ≥ 0) as well as x can be -0.008, -0.4, -3.4, -4, -4.6, -5 (because 0 ≥ x ≥ -5). For ANY possible x, so for ANY x from 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5, it will be true to say that 25 ≥ x^2 ≥ –25.

Thanks Bunuel for putting such a simplified explanation.However, I am still not convinced.

25 ≥ x^2 ≥ –25 means x^2 can be -24, -23, -22,-21 etc as well. is there any number "x" for which x^2 can be -ve value (-25,-24 etc) and satisfy 5 ≥ |x| ≥ 0 as well.
Had the answer option been 25 ≥ x^2, I would have selected this as correct answer.

Please let me know if I am missing an important concept here.
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ravi11

Thanks Bunuel for putting such a simplified explanation. I am still not convinced.

25 ≥ x^2 ≥ –25 means x^2 can be -24, -23, -22,-21 etc as well. is there any number "x" for which x^2 can be -ve value (-25,-24 etc) and satisfy 5 ≥ |x| ≥ 0 as well.
Had the answer option been 25 ≥ x^2, I would have selected this as correct answer.

Please let me know if I am missing an important concept here.

You are missing the point. The question asks: if 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5, then which of the options must be true. So, if we choose any possible x from the given ranges (5 ≥ x ≥ 0 or 0 ≥ x ≥ -5) and substitute into the options, which option will be always true for any of the possible x's. Any possible x will satisfy 25 ≥ x^2 ≥ –25.
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ravi11

Thanks Bunuel for putting such a simplified explanation. I am still not convinced.

25 ≥ x^2 ≥ –25 means x^2 can be -24, -23, -22,-21 etc as well. is there any number "x" for which x^2 can be -ve value (-25,-24 etc) and satisfy 5 ≥ |x| ≥ 0 as well.
Had the answer option been 25 ≥ x^2, I would have selected this as correct answer.

Please let me know if I am missing an important concept here.

You are missing the point. The question asks: if 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5, then which of the options must be true. So, if we choose any possible x from the given ranges (5 ≥ x ≥ 0 or 0 ≥ x ≥ -5) and substitute into the options, which option will be always true for any of the possible x's. Any possible x will satisfy 25 ≥ x^2 ≥ –25.

Thanks Bunuel.

I got the point. So basically all the options that have the question condition as subset of range will be valid.
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Hi All,

This question can be solved by TESTing VALUES. Notice the specific inequalities that we're given to work with - based on the information in the prompt, we know that X can be any value from -5 to +5 INCLUSIVE. We're asked which of the following MUST be true.

I. x ≥ 0
II. x > -5

For Roman Numerals 1 and 2, you could consider X = -5. With that value, neither of those two Roman Numerals is true.
Eliminate Answers B, D and E.

III. 25 ≥ x^2 ≥ -25

Roman Numeral 3 asks us to think about SQUARED terms. With the given range of values that we have to work with, the range of the squared terms would be 0 through +25, inclusive. Regardless of the exact value that you choose for X, X^2 will fall into the range provided by Roman Numeral 3 every time, so Roman Numeral 3 IS true.

Final Answer:

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Can someone please explain me how C is the correct answer. I chose A.

III. 25 ≥ x^2 ≥ –25

How can we apply square root to -25 and simplify it to 5 ≥ x ≥ –5

5 ≥ |x| ≥ 0 means that 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5. For any x from these ranges, 25 ≥ x^2 ≥ –25 will be true.


I have a doubt here. Square root of -25 is an imaginary number, 5i. However that lies outside the range of values for x ( Which is [-5,5] )
Doesn't that make Statement-III wrong as well?
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Hi sssjav,

Based on the information we're given at the beginning of the prompt, we know that X can be any value from -5 to +5 INCLUSIVE. That 'restriction' is what we have to work with when trying to determine which of the three Roman Numerals is ALWAYS TRUE.

III. 25 ≥ x^2 ≥ -25

Roman Numeral 3 asks us to think about SQUARED terms. With the given range of values that we have to work with, the range of the squared terms would be 0 through +25, inclusive. Regardless of the exact value that you choose for X, X^2 will ALWAYS fall into the range provided by Roman Numeral 3 every time, so Roman Numeral 3 IS true.

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Sanjeetgujrall
Can someone please explain me how C is the correct answer. I chose A.

III. 25 ≥ x^2 ≥ –25

How can we apply square root to -25 and simplify it to 5 ≥ x ≥ –5

5 ≥ |x| ≥ 0 means that 5 ≥ x ≥ 0 or 0 ≥ x ≥ -5. For any x from these ranges, 25 ≥ x^2 ≥ –25 will be true.


I have a doubt here. Square root of -25 is an imaginary number, 5i. However that lies outside the range of values for x ( Which is [-5,5] )
Doesn't that make Statement-III wrong as well?

Numbers on the GMAT are real by default (GMAT deals with only real numbers), so no need to consider complex roots.
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Hi sssjav,

Based on the information we're given at the beginning of the prompt, we know that X can be any value from -5 to +5 INCLUSIVE. That 'restriction' is what we have to work with when trying to determine which of the three Roman Numerals is ALWAYS TRUE.

III. 25 ≥ x^2 ≥ -25

Roman Numeral 3 asks us to think about SQUARED terms. With the given range of values that we have to work with, the range of the squared terms would be 0 through +25, inclusive. Regardless of the exact value that you choose for X, X^2 will ALWAYS fall into the range provided by Roman Numeral 3 every time, so Roman Numeral 3 IS true.

GMAT assassins aren't born, they're made,
Rich



I agree to the points you've made, but my doubt remains unresolved.
What you're saying is that for the concerned values of x ( [-5,5] ), Statement-III will always be true - Even though the solution set of statement-III alone might include some values other than those with which we are concerned (All As are Bs, but all Bs are not As).

so what you mean to say is that in a Venn-Diagram-Language, the shape/circle representing [-5,5] will lie enclosed within a bigger shape/circle of statement-III. (You can choose to ignore this statement if it it sounds confusing but you understood my point)

However, by the above logic, even statement-I and statement-II will be true for the concerned values of x, even though the solution sets of the statements might include values other than the ones that belong to [-5,5] and/or might not include some of the values from the set [-5,5].

Now the questions asks us which of the given statements are "true" - according to me, there could only be two possible answers : if we go by the above logic, then all the three statements are true, and If we go by the logic that which of the statements truly represent the all the values of x, then none of the statements will be true. (as all of them represent some values which are either more or less than the concerned set)

However, in another case, if it is asked, that which of the given statements will include ALL the concerned values of x, then statement-III is the best option available. I understand that this is what is meant to have been asked from the question. The thing that I need assistance with is understanding the language of the question - and narrow down on the correct meaning of the question. Where exactly did I interpret the question wrongly? Or what is it that I'm missing?



EDIT : Just read bunuel's reply that gmat does not deal with imaginary number values, that clears up my doubt. Thanks to both of you for your resplies!
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Hi sssjav,

From what you describe, the 'issue' seems to be about how you are interpreting the prompt. I'm going to rephrase the prompt a bit - but NOT change the meaning or the question that is asked.

Consider EVERY number from -5 to +5, inclusive (including negatives, 0, fractions, etc.). Those are the ONLY numbers to consider. Now, consider how each of those possible numbers would answer the three questions here:

I. Is EVERY number in that group greater than or equal to 0?

Of course not; none of the negative numbers are greater than 0.

II. Is EVERY number in that group GREATER than -5?

No. There is one value in the group that is NOT (re: -5).

III. When you square EACH number in that group, will you end up with a value that is between -25 and +25, inclusive?

YES. EVERY value in that group, when squared, will result in a number that is between 0 and +25, inclusive (thus, ALL of those results are between -25 and +25, inclusive).

Thus, the only statement that MUST be true is Roman Numeral 3.

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Bunuel
If 5 ≥ |x| ≥ 0, which of the following must be true?

I. x ≥ 0
II. x > –5
III. 25 ≥ x^2 ≥ –25

A. None
B. II only
C. III only
D. I and III only
E. II and III only

Hi

Option "C" simply can't be correct choice here.

Squire of a real number can never be -ve. It can only be possible only if "x" is an imaginary number. However if "x" is an imaginary number then the condition mentioned in the question itself will not hold true.

Hence option "A-None" should be the correct answer.

Since all three portions of 5 ≥ |x| ≥ 0 are nonnegative, we can square the inequality:
\(5^2\) ≥ \(|x|^2\) ≥ \(0^2\)
25 ≥ \(x^2\) ≥ 0

If \(x^2\) is any value in the range above, it will satisfy the inequality in Statement III.
Thus, Statement III must be true.
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The condition 5 ≥ |x| ≥ 0 means “the distance of x from 0 is at most 5,” so x must lie on the number line between -5 and 5, including the end points.

Test each statement
  • Statement I: x ≥ 0
    Counterexample: x = −3 lies in −5 ≤ x ≤ 5 but is not ≥ 0, so I is not always true.
  • Statement II: x > −5
    Counterexample: x = −5 satisfies |x| ≤ 5 but does not satisfy x > −5, so II is not always true.


  • Statement III: 25 ≥ x2 ≥ −25
    From −5 ≤ x ≤ 5, squaring gives 0 ≤ x2 ≤ 25. Since x2 is never negative, if it's written as 25 ≥ x2 ≥ −25, just replace it with 0 ≤ x2 ≤ 25 because x2 ≥ 0 automatically. So III is always true.
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