AAPL wrote:Manhattan Prep
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:
2/10 = 1/5
The correct answer is
B.
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