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M28-11

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New post 16 Sep 2014, 01:28
1
10
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A
B
C
D
E

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  95% (hard)

Question Stats:

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

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New post 16 Sep 2014, 01:28
1
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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 \(-x^2-2x+15>0\).

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 \(\frac{7}{21}=\frac{1}{3}\).


Answer: B
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Re: M28-11  [#permalink]

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New post 08 Nov 2014, 12:24
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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 \(-x^2-2x+15>0\).

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 \(\frac{7}{21}=\frac{1}{3}\).


Answer: B


This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3.
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Re: M28-11  [#permalink]

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New post 09 Nov 2014, 06:09
parameswaranprasad 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 \(-x^2-2x+15>0\).

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 \(\frac{7}{21}=\frac{1}{3}\).


Answer: B


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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New post 06 Jul 2015, 05:08
Bunuel wrote:
parameswaranprasad 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 \(-x^2-2x+15>0\).

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 \(\frac{7}{21}=\frac{1}{3}\).


Answer: B


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 ***).

I appreciate your help.

Thanks a lot.

Regards.

Luis Navarro
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Re: M28-11  [#permalink]

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New post 06 Jul 2015, 07:53
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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 ***).

I appreciate your help.

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


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

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm



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Re: M28-11  [#permalink]

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New post 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 ***).

I appreciate your help.

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


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

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm






Thanks a lot Bunuel, I appreciate your help.

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

Best regards.

Luis Navarro
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Re: M28-11  [#permalink]

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New post 24 Apr 2017, 00:01
Hi Bunuel, why X<-5 and X>3 is not considered?
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Re: M28-11  [#permalink]

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New post 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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New post 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?


Please re-read the solution:
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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New post 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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New post 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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New post 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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New post 14 Aug 2019, 18:47
perfect!
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Re: M28-11   [#permalink] 14 Aug 2019, 18:47
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