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If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of

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If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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New post 12 May 2015, 04:29
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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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Re: If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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New post 18 May 2015, 05:49
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Bunuel wrote:
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

Kudos for a correct solution.


OFFICIAL SOLUTION:

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

Answer: D.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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New post 12 May 2015, 10:01
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1 + x^4 + x^3 + x^2 + x = 81
i.e. 1 +x+ x^2+ x^3+ x^4 = 81
x+ x^2+ x^3+ x^4 = 80

x(1 +x+ x^2+ x^3) = 80
x(81-x^4) = 80

81x - x^5 = 80
x^5 = 81x -80

Now x+ x^2+ x^3+ x^4+ x^5 = 80 + 81x -80 = 81x

Average of {x, x^2, x^3, x^4, x^5} = 81x/5 ~ 16x

Answer D

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Re: If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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New post 12 May 2015, 10:09
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1 + x^4 + x^3 + x^2 + x = 81

multiplying by 'x' on both sides

x(1 + x^4 + x^3 + x^2 + x )= 81x

dividing by 5 we get

x+ x^2+ x^3+ x^4+x^5 = 81x/5 Approx 16x
But is approximation allowed? i don't see it mentioned in the question?
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Re: If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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New post 12 May 2015, 10:39
NavenRk wrote:
1 + x^4 + x^3 + x^2 + x = 81

multiplying by 'x' on both sides

x(1 + x^4 + x^3 + x^2 + x )= 81x

dividing by 5 we get

x+ x^2+ x^3+ x^4+x^5 = 81x/5 Approx 16x
But is approximation allowed? i don't see it mentioned in the question?



Ohh yeah. This is a much easier way. Kudos!
I think the question meant approximate value.
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Re: If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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New post 12 May 2015, 21:01
1
Bunuel wrote:
If 1 + x^4 + x^3 + x^2 + x = 81, 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

Kudos for a correct solution.


Ans: D

Solution: [x + x^2 + x^3 + x^4 + x^5]/5
Known 1 + x^4 + x^3 + x^2 + x = 81
taking x common from the first equation it becomes
=[1 + x^4 + x^3 + x^2 + x]*x/5
=81*x/5
=16x Answer
Ans.: D
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Re: If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of  [#permalink]

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