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# M31-04

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Math Expert
Joined: 02 Sep 2009
Posts: 59674

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07 Jun 2015, 08:46
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Difficulty:

25% (medium)

Question Stats:

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

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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. $$20x$$

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Joined: 02 Sep 2009
Posts: 59674

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07 Jun 2015, 08:46
1
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 $$\frac{x + x^2 + x^3 + x^4 + x^5}{5} = \frac{x(1 + x + x^2 + x^3 + x^4)}{5} = \frac{80x}{5} = 16x$$.

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Joined: 20 Aug 2015
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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!
Math Expert
Joined: 02 Sep 2009
Posts: 59674

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02 Feb 2016, 03:02
marcoelmanchi wrote:
Hey Bunuel!

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

Many thanks!

We are given that $$1 + x^4 + x^3 + x^2 + x = 80$$, thus $$x(1+x+x^2+x^3+x^4) = x*80=80x$$.
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Joined: 29 Jul 2017
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Location: India
Concentration: Operations, Finance
GMAT 1: 720 Q49 V39
GPA: 3.94
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02 Aug 2019, 23:43
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 14 Feb 2014
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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

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

Moderators: chetan2u, Bunuel