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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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Updated on: 01 Feb 2012, 14:00
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Difficulty:

65% (hard)

Question Stats:

60% (01:44) correct 40% (01:51) wrong based on 581 sessions

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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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09 Mar 2011, 08:37
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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?

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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

$$p^2-13p+40=(p-5)(p-8)$$ --> $$p^2-13p+40<0$$ for $$5<p<8$$:
Attachment:

MSP51419ef17g8chg8g5ch00002g8ag3h1ba7ibee0.gif [ 3.69 KiB | Viewed 9043 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: $$P=\frac{2}{10}=\frac{1}{5}$$.

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09 Mar 2011, 08:55
Thnkx a ton bunuel for quick solution.Excellent!!
Cheers:P
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10 Mar 2011, 01:32
(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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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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08 Jul 2013, 21:49
2
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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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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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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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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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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13 Mar 2017, 03:44
1
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.
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Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1  [#permalink]

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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.

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

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

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: $$P=\frac{2}{10}=\frac{1}{5}$$.

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..
Re: If p^2 – 13p + 40 = q, and p is a positive integer between 1 &nbs [#permalink] 22 Jul 2018, 19:41
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# If p^2 – 13p + 40 = q, and p is a positive integer between 1

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