GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 11 Dec 2019, 12:34 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M31-04

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59674

### Show Tags 00:00

Difficulty:   25% (medium)

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

### HideShow timer Statistics

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. $$20x$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59674

### Show Tags

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 $$\frac{x + x^2 + x^3 + x^4 + x^5}{5} = \frac{x(1 + x + x^2 + x^3 + x^4)}{5} = \frac{80x}{5} = 16x$$.

_________________
Intern  Joined: 20 Aug 2015
Posts: 6

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 59674

### Show Tags

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 $$x(1+x+x^2+x^3+x^4) = x*80=80x$$.
_________________
Intern  B
Joined: 29 Jul 2017
Posts: 42
Location: India
Concentration: Operations, Finance
GMAT 1: 720 Q49 V39 GPA: 3.94
WE: Design (Manufacturing)

### Show Tags

I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 14 Feb 2014
Posts: 8

### Show Tags

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
Display posts from previous: Sort by

# M31-04

Moderators: chetan2u, Bunuel  