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If p^2 – 13p + 40 = q, and p is a positive integer between 1

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If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post Updated on: 01 Feb 2012, 14:00
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If p^2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
E. 3/10

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Originally posted by AnkitK on 09 Mar 2011, 07:56.
Last edited by Bunuel on 01 Feb 2012, 14:00, edited 1 time in total.
Edited the question and added the OA
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Re: ___probability____  [#permalink]

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New post 09 Mar 2011, 08:37
7
5
Question moved to PS subfourm.

AnkitK wrote:
if p^2-13p+40=q ,and p is a positive integer between 1 and 10 ,inclusive what is the probability that q<0?

please suggest the answers guys.Also provide explanation.I would appreciate the efforts.

thnkx in advance.!
:?:


Welcome to Gmat Club.

Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
Provide answer choices for PS questions.

Also please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.

Question should read:

If p^2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
E. 3/10

\(p^2-13p+40=(p-5)(p-8)\) --> \(p^2-13p+40<0\) for \(5<p<8\):
Attachment:
MSP51419ef17g8chg8g5ch00002g8ag3h1ba7ibee0.gif
MSP51419ef17g8chg8g5ch00002g8ag3h1ba7ibee0.gif [ 3.69 KiB | Viewed 10496 times ]
Now, as \(p\) is an integer then \(p^2-13p+40=q<0\) for two values of \(p\) out of 10: 6 and 7, which means that: \(P=\frac{2}{10}=\frac{1}{5}\).

Answer: B.
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Re: ___probability____  [#permalink]

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New post 09 Mar 2011, 08:55
Thnkx a ton bunuel for quick solution.Excellent!!
Cheers:P
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Re: ___probability____  [#permalink]

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New post 10 Mar 2011, 01:32
1
(p-5)(p-8) = q

p = 6 or 7 for this to be true, so

2/10 = 1/5
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 08 Jul 2013, 19:23
1
P is any integer between 1-10, inclusive.
Plug in the each values. It took me about 1:30 min to substitute all the values.

I did this because I can't visualize the graph of the equation other than I know it will be a parabola because of the X^2.

When p = 1, 2, 3, 4, 5, 8, 9, 10 the value of q is not negative.
Only time q is negative is when p = 6 or 7.
Probability 2/10 = 1/5 = B
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 08 Jul 2013, 21:49
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jjack0310 wrote:
P is any integer between 1-10, inclusive.
Plug in the each values. It took me about 1:30 min to substitute all the values.

I did this because I can't visualize the graph of the equation other than I know it will be a parabola because of the X^2.

When p = 1, 2, 3, 4, 5, 8, 9, 10 the value of q is not negative.
Only time q is negative is when p = 6 or 7.
Probability 2/10 = 1/5 = B


Just a thought: Even if you can't visualize the graph, just to save up some precious time,

Once you factorize the quadratic as this : (p-5)(p-8); we have to find out the values of p, in the range [1,10] which will make it negative.

Or (p-5)(p-8)<0 . Now this is only possible if (p-5) and (p-8) have opposite signs,i.e.

I. p-5>0 AND p-8<0 \(\to\) p>5 AND p<8 \(\to\) 5<p<8

OR

II.p-5<0 and p-8>0\(\to\) p<5 AND p>8\(\to\) Invalid range.

Nonetheless, any logical way that gets you the valid answer is correct.
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 16 Sep 2013, 01:22
Is this question not just easy enough and factor and then plug in numbers??

if: \(p^2-13p+40\) Factors out to (p-8)(p-5)=q, which can easily be done in your head very quickly.

Then make a list of numbers that would give a negative result.

It can only be 2 numbers 6 and 7, since anything over 7 yields either 0 or a positive number, and anything less than 6 yields a positive number, there can only be two cases which would satisfy the solution. Is making a graph really necessary?
Making it 2/10 or 1/5
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 20 Aug 2014, 22:10
P(P-13) plus 40 = q

while testing the values q is negative only when p is 7 or 6

thus the answer is 2/10 or 1/5
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If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 20 Nov 2015, 01:46
1
AnkitK wrote:
If p^2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
E. 3/10


Given: \(p^2 – 13p + 40\) = q, 1< p < 10
Required: Probability of q < 0

\(p^2 – 13p + 40\) = (p -8)(p-5) = q

We need q < 0
Hence
(p -5)(p-8) < 0

Hence 5< p < 8.
Only two integral values lie in the range: 6 and 7

Probability = 2/10 = 1/5
Option B.

Solving an inequality with a less than sign:
The value of the variable will be greater than the smaller value and smaller than the greater value.
i.e. It will lie between the extremes.

Solving an inequality with a greater than sign:
The value of the variable will be smaller than the smaller value and greater than the greater value.
i.e. It can take all the values except the values in the range.
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 13 Jan 2017, 21:58
The quadratic equation could be expanded to (p-5)(p-8)=q
If p is a member of the set [1,10], p could be one of the 10 choices.

The equation (p-5)(p-8) will work out to < 0 for the range (5,8). The integers between 5 and 8 exclusive are 6 and 7.

The probability that q <0 = probability(p = 5 or p = 8 is selected) = 2/10 = 1/5
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 13 Mar 2017, 03:44
1
Factor the quadratic:
p^2 – 13p + 40 = q
(p – 8)(p – 5) = q
For p = 5 and p = 8, q = 0. Between p = 5 and p = 8, q has a negative sign, as (p – 8) is negative and (p – 5) is positive. With a total of 10 possible integer p values, only two (p = 6 and p = 7) fall in the range 5 < p < 8, so the probability is 2/10 or 1/5.
The correct answer is B.
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 15 Mar 2017, 16:45
AnkitK wrote:
If p^2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
E. 3/10


We can factor the given equation:

p^2 – 13p + 40 = q

(p - 5)(p - 8) = q

We see that in order for q to be negative, either (p - 5) is negative and (p - 8) is positive OR (p - 5) is positive and (p - 8) is negative.

Analyzing our expression a bit further, we see that it only produces a negative product when p = 6 and p = 7.

Thus, the probability that q < 0 is 2/10 = 1/5.

Answer: B
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 22 Jul 2018, 19:41
1
Bunuel wrote:
Question moved to PS subfourm.

Question should read:

If p^2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
E. 3/10

\(p^2-13p+40=(p-5)(p-8)\) --> \(p^2-13p+40<0\) for \(5<p<8\):
Attachment:
MSP51419ef17g8chg8g5ch00002g8ag3h1ba7ibee0.gif
Now, as \(p\) is an integer then \(p^2-13p+40=q<0\) for two values of \(p\) out of 10: 6 and 7, which means that: \(P=\frac{2}{10}=\frac{1}{5}\).

Answer: B.


Thanks for showing the explanation using the parabola Bunuel.

Was curious to understand how will it look like if for eg the question put the condition of q>0. In that case, the quadratic would turn out to be (p-8)(p-5) > 0. In such a case the parabola will open downward but then how do we make this out as the constant a (in a*p^2) is in any case +ve..
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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New post 09 Sep 2019, 03:46
AnkitK wrote:
If p^2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
E. 3/10


p² - 13p + 40 = 0
(p-5)(p-8) = 0

The CRITICAL POINTS are p=5 and p=8.
Only at these two critical points does p² - 13p + 40 = 0.
To determine the ranges in which p² - 13p + 40 < 0, test one value to the left and right of each critical point.

There are three ranges to consider:
p < 5, 5 < p < 8, p > 8
If p=0 or p=100, then p² - 13p + 40 is clearly GREATER than 0, implying that p<5 and p>8 are not viable ranges.
Implication:
The only viable range is 5 < p < 8.
Thus, of the 10 integers between 1 and 10, inclusive, only two -- 6 and 7 -- will yield a negative for p² - 13p + 40:
\(\frac{2}{10} = \frac{1}{5}\)


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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1   [#permalink] 09 Sep 2019, 03:46
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