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# Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of

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Math Expert
Joined: 02 Sep 2009
Posts: 59721
Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of  [#permalink]

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28 Nov 2019, 00:44
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Difficulty:

45% (medium)

Question Stats:

75% (02:18) correct 25% (02:20) wrong based on 32 sessions

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Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of this set is 3, which of the following could be the value of x?

A. $$\frac{1}{2}$$
B. 1
C. $$\frac{3}{2}$$
D. 3
E. 4

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Concentration: General Management
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Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of  [#permalink]

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28 Nov 2019, 00:57
$$\frac{x^2+x+9+10x+x^2-16x}{4}=3$$

$$2x^2-5x+9=12$$

$$2x^2-5x-3=0$$

$$2x^2-6x+x-3=0$$

$$2x(x-3)+1(x-3)=0$$

$$(x-3)(2x+1)=0$$

$$x=3$$ or $$x=-\frac{1}{2}$$

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Joined: 22 Sep 2018
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Re: Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of  [#permalink]

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28 Nov 2019, 06:26
x^2+x, 9, 10x, x^2-16x
By adding the expressions and dividing it with 4

2x^2-5x+9 / 4 = 3
2x^2-5x-3=0
(2x+1)(x-3)=0
X= -1/2 or 3

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Re: Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of   [#permalink] 28 Nov 2019, 06:26
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