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

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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

The OA is B.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
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by Jay@ManhattanReview » Mon Dec 18, 2017 2:57 am
AAPL 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

The OA is B.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
We have p^2 - 13p + 40 = q

=> p^2 -5p -8p + 40 = q

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

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

Case 1: If p = 1, 2, 3, 4, 9, or 10, we have q > 0; there are six values of p out of 10 values that make q > 0.
Case 2: If p = 5 or 8, we have q = 0; there are two values of p out of 10 values that make q = 0.
Case 3: If p = 6 or 7, we have q < 0; there are two values of p out of 10 values that make q < 0.

So, if p = 1, 2, 3, 4, 5, 8, 9, or 10 (Eight values), then q is not less than 0 and if = p = 6 or 7 (Two values), then q < 0.

The probability that q < 0 = 2/10 = 1/5.

The correct answer: B

Hope this helps!

-Jay
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