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M31-04

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M31-04  [#permalink]

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New post 07 Jun 2015, 08:46
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A
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D
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  25% (medium)

Question Stats:

79% (01:12) correct 21% (02:52) wrong based on 33 sessions

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Re M31-04  [#permalink]

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New post 07 Jun 2015, 08:46
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Official Solution:

If \(1 + x^4 + x^3 + x^2 + x = 80\), then the average (arithmetic mean) of \(x\), \(x^2\), \(x^3\), \(x^4\) and \(x^5\) is equal to which of the following?

A. \(12x\)
B. \(13x\)
C. \(14x\)
D. \(16x\)
E. \(20x\)


The average (arithmetic mean) of \(x\), \(x^2\), \(x^3\), \(x^4\) and \(x^5\) is \(\frac{x + x^2 + x^3 + x^4 + x^5}{5} = \frac{x(1 + x + x^2 + x^3 + x^4)}{5} = \frac{80x}{5} = 16x\).


Answer: D
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Re: M31-04  [#permalink]

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New post 02 Feb 2016, 02:26
Hey Bunuel!

I do not see how x(1+x+x^2+x^3+x^4) = 80x. Can you please explain?

Many thanks!
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Re: M31-04  [#permalink]

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New post 02 Feb 2016, 03:02
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Re M31-04  [#permalink]

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New post 02 Aug 2019, 23:43
I think this is a high-quality question and I agree with explanation.
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M31-04  [#permalink]

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New post 29 Nov 2019, 04:43
This is how i approached this question:

Let say , If X=10, then what is the value of 5X? So you are given the value of X , and then you are asked about the value of 5X. These are two separate pieces of information and this is important to know.

Thus, If X=10 then 5X=5(10)= 50 is the the value of 5X.

Let's discuss the original question.

If 1+ x^4 +x^3 + x^2 + x = 80, then what is the average of x, x^2 +x^3 + x^4, x^5? so we have here two separate pieces of information as well.

We know that the formula for Average: Average=Sum of the terms/number of the terms.

Thus, Average= x, x^2 +x^3 + x^4, x^5 / 5(number of the terms) = but here is the trick, we can simplify the sum of the terms by pulling out the common factors, and if we do that, then we see that x, x^2 +x^3 + x^4, x^5 be comes X(1+x+x^2+x^3+x^4). Since from that question stem (1+x+x^2+x^3+x^4)= 80 , then simply substitute 80 for (1+x+x^2+x^3+x^4), then the avergae becomes X(80) / 5(number of the terms)= 16X

The correct answer is D

I hope it's clear.
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M31-04   [#permalink] 29 Nov 2019, 04:43
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