GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2018, 13:44

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M20-33

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49251

### Show Tags

16 Sep 2014, 01:09
1
5
00:00

Difficulty:

65% (hard)

Question Stats:

57% (01:33) correct 43% (01:33) wrong based on 161 sessions

### HideShow timer Statistics

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

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

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49251

### Show Tags

16 Sep 2014, 01:09
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

### Show Tags

08 Jul 2015, 14:05
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: 49251

### Show Tags

09 Jul 2015, 02:03
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

### Show Tags

09 Jul 2015, 05:58
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
Director
Joined: 08 Jun 2015
Posts: 508
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

### Show Tags

28 Oct 2017, 01:10
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

### Show Tags

06 Nov 2017, 10:28

(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: 49251

### Show Tags

06 Nov 2017, 10:38
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, 10:38
Display posts from previous: Sort by

# M20-33

Moderators: chetan2u, Bunuel

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.