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Math Expert V
Joined: 02 Sep 2009
Posts: 58427
The entire range of values of x is marked by the dark region on the nu  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 66% (01:47) correct 34% (01:42) wrong based on 260 sessions

HideShow timer Statistics The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment: T6774.png [ 2.39 KiB | Viewed 2869 times ]

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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Given: $$x \geq {-3}$$ and $$x \leq {2}$$
$$(x + 3) \geq {0}$$ and $$(x - 2) \leq {0}$$
$$(x + 3) \geq {0}$$ and $$(2 - x) \geq {0}$$

GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4018
Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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Bunuel wrote: The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

One option is to TEST some values.

According to the diagram, 0 IS a solution to the inequality. Let's test each answer choice to see whether 0 is a solution to the given inequality. If it is NOT a solution, we'll eliminate that answer choice.
A. (3+0)(2−0) = 6 BUT it is NOT the case that 6 ≤ 0 ELIMINATE A
B. (3−0)(2+0) = 6 AND it IS the case that 6 ≥ 0 KEEP B
C. (0−3)(0+2) = -6 AND it IS the case that -6 ≤ 0 KEEP C
D. (0+3)(2−0) = 6 AND it IS the case that 6 ≥ 0 KEEP D
E. (3−0)(2−0) = 6 AND it IS the case that ≥ 0 KEEP E

Test another value. According to the diagram, 3 is NOT a solution to the inequality. Let's test each remaining answer choice to see whether 3 is a solution to the given inequality. If it IS a solution, we'll eliminate it.
B. (3−3)(2+3) = 0 and it IS the case that 0 ≥ 0. So ELIMINATE B
C. (3−3)(3+2) = 0 and it IS the case that 0 ≤ 0. So ELIMINATE C
D. (3+3)(2−3) = -6 and it is NOT the case that -6 ≥ 0 KEEP D
E. (3−3)(2−3) = 0 and it IS the case that 0 ≥ 0. So ELIMINATE E

By the process of elimination, the correct answer is D

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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Bunuel wrote: The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

Looking at the given number line, we see that the dark region on the number line represents all the values of x between -3 and 2, inclusive. In other words, it is -3 ≤ x ≤ 2. Thus:

-3 ≤ x AND x ≤ 2.

Manipulating each inequality we have:

-3 ≤ x

x ≥ -3

x + 3 ≥ 0

x ≤ 2

0 ≤ 2 - x

2 - x ≥ 0

x + 3 ≥ 0 AND 2 - x ≥ 0

The only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0.

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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ScottTargetTestPrep wrote:
Bunuel wrote: The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

Looking at the given number line, we see that the dark region on the number line represents all the values of x between -3 and 2, inclusive. In other words, it is -3 ≤ x ≤ 2. Thus:

-3 ≤ x AND x ≤ 2.

Manipulating each inequality we have:

-3 ≤ x

x ≥ -3

x + 3 ≥ 0

x ≤ 2

0 ≤ 2 - x

2 - x ≥ 0

x + 3 ≥ 0 AND 2 - x ≥ 0

The only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0.

Dear ScottTargetTestPrep,

In the last step, did you simply then multiply the two inequalities? Or how did you arrive at the conclusion that the only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0?
Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8137
Location: United States (CA)
Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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ghnlrug wrote:
ScottTargetTestPrep wrote:
Bunuel wrote: The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

Looking at the given number line, we see that the dark region on the number line represents all the values of x between -3 and 2, inclusive. In other words, it is -3 ≤ x ≤ 2. Thus:

-3 ≤ x AND x ≤ 2.

Manipulating each inequality we have:

-3 ≤ x

x ≥ -3

x + 3 ≥ 0

x ≤ 2

0 ≤ 2 - x

2 - x ≥ 0

x + 3 ≥ 0 AND 2 - x ≥ 0

The only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0.

Dear ScottTargetTestPrep,

In the last step, did you simply then multiply the two inequalities? Or how did you arrive at the conclusion that the only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0?

Yes, I multiplied the inequalities.
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: The entire range of values of x is marked by the dark region on the nu   [#permalink] 04 Feb 2019, 07:54
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