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Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the valu

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Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the valu  [#permalink]

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New post 27 Nov 2019, 01:49
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  55% (hard)

Question Stats:

48% (01:15) correct 52% (02:27) wrong based on 23 sessions

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Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the value of n is chosen randomly from the set {-1, 0, 1, 2, 3, 4, 5}, what is the probability that point (-p, -q) is also on the graph?

A. \(\frac{1}{7}\)

B. \(\frac{3}{7}\)

C. \(\frac{4}{7}\)

D. \(\frac{5}{7}\)

E. \(\frac{6}{7}\)


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Re: Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the valu  [#permalink]

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New post 27 Nov 2019, 04:26
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Bunuel wrote:
Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the value of n is chosen randomly from the set {-1, 0, 1, 2, 3, 4, 5}, what is the probability that point (-p, -q) is also on the graph?

A. \(\frac{1}{7}\)

B. \(\frac{3}{7}\)

C. \(\frac{4}{7}\)

D. \(\frac{5}{7}\)

E. \(\frac{6}{7}\)


\(y = x^n\)
When \(x = p, y = p^n = q\)
When \(x = -p, y = (-p)^n = +p^n\) (or \(+q\)) when \(n\) is even; = \(-p^n\) (or \(-q\)) when \(n\) is odd

So, for \((-p, -q)\) to be on the graph, "\(n\)" must be odd
--> Favorable values of n = {-1, 1, 3, 5} = 4 values
--> Required Probability = \(\frac{4}{7}\)

IMO Option C
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Re: Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the valu   [#permalink] 27 Nov 2019, 04:26
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