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

 It is currently 16 Oct 2019, 16:14

### 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

# M28-11

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58381

### Show Tags

16 Sep 2014, 01:28
1
10
00:00

Difficulty:

95% (hard)

Question Stats:

40% (02:15) correct 60% (03:02) wrong based on 147 sessions

### HideShow timer Statistics

If $$x^2 + 2x -15 = -m$$, where $$x$$ is an integer from -10 and 10, inclusive, what is the probability that $$m$$ is greater than zero?

A. $$\frac{2}{7}$$
B. $$\frac{1}{3}$$
C. $$\frac{7}{20}$$
D. $$\frac{2}{5}$$
E. $$\frac{3}{7}$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58381

### Show Tags

16 Sep 2014, 01:28
1
2
Official Solution:

If $$x^2 + 2x -15 = -m$$, where $$x$$ is an integer from -10 and 10, inclusive, what is the probability that $$m$$ is greater than zero?

A. $$\frac{2}{7}$$
B. $$\frac{1}{3}$$
C. $$\frac{7}{20}$$
D. $$\frac{2}{5}$$
E. $$\frac{3}{7}$$

Re-arrange the given equation: $$-x^2-2x+15=m$$.

Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15&gt;0$$.

Factorize: $$(x+5)(3-x) \gt 0$$. This equation holds true for $$-5 \lt x \lt 3$$.

Since $$x$$ is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is $$\frac{7}{21}=\frac{1}{3}$$.

_________________
Intern
Joined: 11 Aug 2014
Posts: 5

### Show Tags

08 Nov 2014, 12:24
1
Bunuel wrote:
Official Solution:

If $$x^2 + 2x -15 = -m$$, where $$x$$ is an integer from -10 and 10, inclusive, what is the probability that $$m$$ is greater than zero?

A. $$\frac{2}{7}$$
B. $$\frac{1}{3}$$
C. $$\frac{7}{20}$$
D. $$\frac{2}{5}$$
E. $$\frac{3}{7}$$

Re-arrange the given equation: $$-x^2-2x+15=m$$.

Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15&gt;0$$.

Factorize: $$(x+5)(3-x) \gt 0$$. This equation holds true for $$-5 \gt x \gt 3$$.

Since $$x$$ is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is $$\frac{7}{21}=\frac{1}{3}$$.

This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3.
Math Expert
Joined: 02 Sep 2009
Posts: 58381

### Show Tags

09 Nov 2014, 06:09
Bunuel wrote:
Official Solution:

If $$x^2 + 2x -15 = -m$$, where $$x$$ is an integer from -10 and 10, inclusive, what is the probability that $$m$$ is greater than zero?

A. $$\frac{2}{7}$$
B. $$\frac{1}{3}$$
C. $$\frac{7}{20}$$
D. $$\frac{2}{5}$$
E. $$\frac{3}{7}$$

Re-arrange the given equation: $$-x^2-2x+15=m$$.

Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15&gt;0$$.

Factorize: $$(x+5)(3-x) \gt 0$$. This equation holds true for $$-5 \gt x \gt 3$$.

Since $$x$$ is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is $$\frac{7}{21}=\frac{1}{3}$$.

This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3.

There was a typo. Should have been $$-5 \lt x \lt 3$$. Edited. Thank you.
_________________
Intern
Joined: 24 Jun 2015
Posts: 45

### Show Tags

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

If $$x^2 + 2x -15 = -m$$, where $$x$$ is an integer from -10 and 10, inclusive, what is the probability that $$m$$ is greater than zero?

A. $$\frac{2}{7}$$
B. $$\frac{1}{3}$$
C. $$\frac{7}{20}$$
D. $$\frac{2}{5}$$
E. $$\frac{3}{7}$$

Re-arrange the given equation: $$-x^2-2x+15=m$$.

Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15&gt;0$$.

Factorize: $$(x+5)(3-x) \gt 0$$. This equation holds true for $$-5 \gt x \gt 3$$.

Since $$x$$ is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is $$\frac{7}{21}=\frac{1}{3}$$.

This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3.

There was a typo. Should have been $$-5 \lt x \lt 3$$. Edited. Thank you.

Hi Bunuel,

Could you explain hoy do you factorize: (x+5)(3−x)>0. ? I have the doubt because -x^2 is negative in the original equation -x^2-2x+15.

*** Also I thought that when inequalitie is >0 (Like this exercise), the solution for x are OUT ranges (x<-5 and x>3), not IN ranges (-5<x<3) .

So I am confused in both things, how do you factorize the inequalitie and also why the solutions for the inequalitie are -5<x<3 (I already studied the inequalities section and I read what I mention in ***).

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
Math Expert
Joined: 02 Sep 2009
Posts: 58381

### Show Tags

06 Jul 2015, 07:53
1
1
luisnavarro wrote:
Hi Bunuel,

Could you explain hoy do you factorize: (x+5)(3−x)>0. ? I have the doubt because -x^2 is negative in the original equation -x^2-2x+15.

*** Also I thought that when inequalitie is >0 (Like this exercise), the solution for x are OUT ranges (x<-5 and x>3), not IN ranges (-5<x<3) .

So I am confused in both things, how do you factorize the inequalitie and also why the solutions for the inequalitie are -5<x<3 (I already studied the inequalities section and I read what I mention in ***).

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

My advice would be to brush up fundamentals before answering questions, especially such hard ones.

_________________
Intern
Joined: 24 Jun 2015
Posts: 45

### Show Tags

06 Jul 2015, 13:34
Bunuel wrote:
luisnavarro wrote:
Hi Bunuel,

Could you explain hoy do you factorize: (x+5)(3−x)>0. ? I have the doubt because -x^2 is negative in the original equation -x^2-2x+15.

*** Also I thought that when inequalitie is >0 (Like this exercise), the solution for x are OUT ranges (x<-5 and x>3), not IN ranges (-5<x<3) .

So I am confused in both things, how do you factorize the inequalitie and also why the solutions for the inequalitie are -5<x<3 (I already studied the inequalities section and I read what I mention in ***).

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

My advice would be to brush up fundamentals before answering questions, especially such hard ones.

Thanks a lot Bunuel, I appreciate your help.

I read the information and now it is totally clear to me.

Best regards.

Luis Navarro
Looking for 700.
Intern
Joined: 15 Apr 2015
Posts: 1

### Show Tags

24 Apr 2017, 00:01
Hi Bunuel, why X<-5 and X>3 is not considered?
Intern
Joined: 21 May 2017
Posts: 2

### Show Tags

16 Jul 2017, 12:13
I have the same question,
mkugan80 wrote:
Hi Bunuel, why X<-5 and X>3 is not considered?
Math Expert
Joined: 02 Sep 2009
Posts: 58381

### Show Tags

16 Jul 2017, 21:07
ajanpaul wrote:
I have the same question,
mkugan80 wrote:
Hi Bunuel, why X<-5 and X>3 is not considered?

Quote:
Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15&gt;0$$.

Factorize: $$(x+5)(3-x) \gt 0$$. This equation holds true for $$-5 \gt x \gt 3$$.

x < - 5 and x > 3 is not considered because x's from these ranges does not satisfy $$(x+5)(3-x) \gt 0$$.
_________________
Intern
Joined: 05 Oct 2018
Posts: 43
Location: United States
GMAT 1: 770 Q49 V47
GPA: 3.95
WE: General Management (Other)

### Show Tags

23 Oct 2018, 07:28
I tried to factor as written to (x+5)(x-3) = -m. I found 12 out of 21 possible values for making the left hand side positive and subtracted 12/21 from 1 and got answer E. I think my error was forgetting to subtract out the two cases for making m 0.

Thanks for the problem and explanation.
Director
Joined: 08 Jun 2013
Posts: 543
Location: France
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Consulting (Other)

### Show Tags

23 Oct 2018, 08:52
Bunuel wrote:
If $$x^2 + 2x -15 = -m$$, where $$x$$ is an integer from -10 and 10, inclusive, what is the probability that $$m$$ is greater than zero?

A. $$\frac{2}{7}$$
B. $$\frac{1}{3}$$
C. $$\frac{7}{20}$$
D. $$\frac{2}{5}$$
E. $$\frac{3}{7}$$

$$x^2 + 2x -15 = -m$$

i.e. $$x^2 + 2x +1 = 16-m$$

i.e $$m = 16 - (x+1)^2$$

For m>0 possible values x can take are -4, -3, -2, -1, 0, 1, 2, 3.

Total values x can take 21 (i.e. -10 to 10)

Required probability = 7/21 = 1/3
_________________
Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
It will all make sense.
Intern
Status: All our dreams can come true, if we have the courage to pursue them
Joined: 03 Jul 2015
Posts: 32
Location: India
Concentration: Technology, Finance
GMAT 1: 700 Q49 V35
WE: Information Technology (Computer Software)

### Show Tags

23 Oct 2018, 14:24
Hi Bunuel, chetan2u

I arrived at the correct answer but with a different approach/solution with the idea that we need to count the possible 'm' values but not 'x' values, as you did in OE . Not sure if this one is correct. Let me know if this is correct.

Rewriting the given solution, -(x+5)(x-3)=m

Now, substituting values of x from -10 to 10 gives m values as -65,-48,-33,-20,-9,0,7,12,15,16,15,12,7,0,-9,-20,-33,-36,-65,-84,-105 respectively.

Out of these, if we remove repetitions,
Count of +ve values of m is 4 (7,12,15,16)
Count of all m values is 12
So probability : 4/12 => 1/3

If we don't remove repetitions, we get 7/21=>1/3
Intern
Joined: 12 Feb 2019
Posts: 24
Location: India
Schools: Wharton '22, CBS, ISB, IIM
GPA: 4

### Show Tags

14 Aug 2019, 18:47
perfect!
Re: M28-11   [#permalink] 14 Aug 2019, 18:47
Display posts from previous: Sort by

# M28-11

Moderators: chetan2u, Bunuel