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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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Joined: 02 Sep 2009
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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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27 Nov 2019, 00:49
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43% (01:35) correct 57% (02:26) wrong based on 35 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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27 Nov 2019, 03:26
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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
Re: Let (p, q) be a point on the graph of y = x^n where p ≠ 0. If the valu   [#permalink] 27 Nov 2019, 03:26
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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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