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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
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Joined: 02 Sep 2009
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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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New post 28 Nov 2019, 00:44
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

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

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Director
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Joined: 16 Jan 2019
Posts: 507
Location: India
Concentration: General Management
WE: Sales (Other)
Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of  [#permalink]

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New post 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 \(x=-\frac{1}{2}\)

Answer is (D)
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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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New post 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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Consider the set {x^2+x, 9, 10x, x^2-16x}. If the arithmetic mean of

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