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M28-11

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Joined: 02 Sep 2009
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16 Sep 2014, 01:28
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95% (hard)

Question Stats:

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

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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. $$\frac{3}{7}$$

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16 Sep 2014, 01:28
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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 $$\frac{7}{21}=\frac{1}{3}$$.

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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.
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Joined: 02 Sep 2009
Posts: 58381

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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.
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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

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06 Jul 2015, 07:53
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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.

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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.
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Joined: 15 Apr 2015
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24 Apr 2017, 00:01
Hi Bunuel, why X<-5 and X>3 is not considered?
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16 Jul 2017, 12:13
I have the same question,
mkugan80 wrote:
Hi Bunuel, why X<-5 and X>3 is not considered?
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Posts: 58381

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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 $$(x+5)(3-x) \gt 0$$.
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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.
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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
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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
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14 Aug 2019, 18:47
perfect!
Re: M28-11   [#permalink] 14 Aug 2019, 18:47
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