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# M20-33

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Math Expert
Joined: 02 Sep 2009
Posts: 51185

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16 Sep 2014, 00:09
1
7
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:32) correct 42% (01:33) wrong based on 168 sessions

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Is $$|x + 1| \lt 2$$ ?

(1) $$(x - 1)^2 \lt 1$$

(2) $$x^2 - 2 \lt 0$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51185

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16 Sep 2014, 00:09
Official Solution:

"Is $$|x+1| \lt 2$$?" means "is $$-2 \lt x+1 \lt 2$$?" or "is $$-3 \lt x \lt 1$$?"

(1) $$(x-1)^2 \lt 1$$. This statement tells that $$-\sqrt{1} \lt x-1 \lt \sqrt{1}$$ or $$0 \lt x \lt 2$$. So, we can have an YES as well as a NO answer (for example consider $$x=1$$ and $$x=1.1$$). Not sufficient.

(2) $$x^2-2 \lt 0$$. This statement tells that $$-\sqrt{2} \lt x \lt \sqrt{2}$$. Since $$\sqrt{2} \approx 1.4$$, then we still can have an YES as well as a NO answer (consider the same example $$x=1$$ and $$x=1.1$$). Not sufficient.

(1)+(2) From (1) and (2) we have that $$0 \lt x \lt \sqrt{2}$$ and we still can have an YES as well as a NO answer (naturally the example from above works here as well). Not sufficient.

_________________
Intern
Joined: 24 Jun 2015
Posts: 46

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08 Jul 2015, 13:05
Bunuel wrote:
Official Solution:

"Is $$|x+1| \lt 2$$?" means "is $$-2 \lt x+1 \lt 2$$?" or "is $$-3 \lt x \lt 1$$?"

(1) $$(x-1)^2 \lt 1$$. This statement tells that $$-\sqrt{1} \lt x-1 \lt \sqrt{1}$$ or $$0 \lt x \lt 2$$. So, we can have an YES as well as a NO answer (for example consider $$x=1$$ and $$x=1.1$$). Not sufficient.

(2) $$x^2-2 \lt 0$$. This statement tells that $$-\sqrt{2} \lt x \lt \sqrt{2}$$. Since $$\sqrt{2} \approx 1.4$$, then we still can have an YES as well as a NO answer (consider the same example $$x=1$$ and $$x=1.1$$). Not sufficient.

(1)+(2) From (1) and (2) we have that $$0 \lt x \lt \sqrt{2}$$ and we still can have an YES as well as a NO answer (naturally the example from above works here as well). Not sufficient.

Hi Bunuel,

One question... From statement 1 and 2, did you apply the x2−2<0 is equal to ¡x! < 2 (Absolute vale of x) because yo knew both sides are positive???

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
Math Expert
Joined: 02 Sep 2009
Posts: 51185

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09 Jul 2015, 01:03
1
luisnavarro wrote:
Bunuel wrote:
Official Solution:

"Is $$|x+1| \lt 2$$?" means "is $$-2 \lt x+1 \lt 2$$?" or "is $$-3 \lt x \lt 1$$?"

(1) $$(x-1)^2 \lt 1$$. This statement tells that $$-\sqrt{1} \lt x-1 \lt \sqrt{1}$$ or $$0 \lt x \lt 2$$. So, we can have an YES as well as a NO answer (for example consider $$x=1$$ and $$x=1.1$$). Not sufficient.

(2) $$x^2-2 \lt 0$$. This statement tells that $$-\sqrt{2} \lt x \lt \sqrt{2}$$. Since $$\sqrt{2} \approx 1.4$$, then we still can have an YES as well as a NO answer (consider the same example $$x=1$$ and $$x=1.1$$). Not sufficient.

(1)+(2) From (1) and (2) we have that $$0 \lt x \lt \sqrt{2}$$ and we still can have an YES as well as a NO answer (naturally the example from above works here as well). Not sufficient.

Hi Bunuel,

One question... From statement 1 and 2, did you apply the x2−2<0 is equal to ¡x! < 2 (Absolute vale of x) because yo knew both sides are positive???

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

You can take even roots (such as the square root) from an inequality IF both sides are non-negative (check here: inequalities-tips-and-hints-175001.html). Both sides of $$(x-1)^2 \lt 1$$ and $$x^2 \lt 2$$ are non-negative, hence after taking the square root we'll get:

$$|x-1| \lt 1$$ and $$x \lt \sqrt{2}$$ (recall that $$\sqrt{x^2}=|x|$$).
_________________
Intern
Joined: 24 Jun 2015
Posts: 46

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09 Jul 2015, 04:58
Bunuel wrote:
luisnavarro wrote:
Bunuel wrote:
Official Solution:

"Is $$|x+1| \lt 2$$?" means "is $$-2 \lt x+1 \lt 2$$?" or "is $$-3 \lt x \lt 1$$?"

(1) $$(x-1)^2 \lt 1$$. This statement tells that $$-\sqrt{1} \lt x-1 \lt \sqrt{1}$$ or $$0 \lt x \lt 2$$. So, we can have an YES as well as a NO answer (for example consider $$x=1$$ and $$x=1.1$$). Not sufficient.

(2) $$x^2-2 \lt 0$$. This statement tells that $$-\sqrt{2} \lt x \lt \sqrt{2}$$. Since $$\sqrt{2} \approx 1.4$$, then we still can have an YES as well as a NO answer (consider the same example $$x=1$$ and $$x=1.1$$). Not sufficient.

(1)+(2) From (1) and (2) we have that $$0 \lt x \lt \sqrt{2}$$ and we still can have an YES as well as a NO answer (naturally the example from above works here as well). Not sufficient.

Hi Bunuel,

One question... From statement 1 and 2, did you apply the x2−2<0 is equal to ¡x! < 2 (Absolute vale of x) because yo knew both sides are positive???

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

You can take even roots (such as the square root) from an inequality IF both sides are non-negative (check here: inequalities-tips-and-hints-175001.html). Both sides of $$(x-1)^2 \lt 1$$ and $$x^2 \lt 2$$ are non-negative, hence after taking the square root we'll get:

$$|x-1| \lt 1$$ and $$x \lt \sqrt{2}$$ (recall that $$\sqrt{x^2}=|x|$$).

Thanks a lot, I appreciate your help.

Regards.

Luis Navarro
Looking for 700
Senior Manager
Joined: 08 Jun 2015
Posts: 436
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

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28 Oct 2017, 00:10
The prompt asks us if x lies between -3 and 1. Check for each option. The answer is option E , i.e none of the two.
_________________

" The few , the fearless "

Intern
Joined: 10 Dec 2015
Posts: 12
GMAT 1: 590 Q36 V35

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06 Nov 2017, 09:28

(x−1)2<1
x2-2x+1<1
x2-2x<0
x(x-2)<0
x<0
x<2
Math Expert
Joined: 02 Sep 2009
Posts: 51185

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06 Nov 2017, 09:38
nelliegu wrote:

(x−1)2<1
x2-2x+1<1
x2-2x<0
x(x-2)<0
x<0
x<2

x(x - 2) < 0 --> x and x - 2 have the opposite signs.

x > 0 and x - 2 < 0 --> x > 0 and x < 2 --> 0 < x < 2.

x < 0 and x - 2 > 0 --> x < 0 and x > 2 --> no solution.

So, x(x - 2) < 0 holds true for 0 < x < 2.

9. Inequalities

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________
Re: M20-33 &nbs [#permalink] 06 Nov 2017, 09:38
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# M20-33

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