If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera

Question

A = \{x, 2x, 3x, 4x\} 
 
 B = \{x^2, 2x^2, 3x^2, 4x^2, 5x^2\}

If x ≠ 0 and the average (arithmetic mean) of set A is twice the average of set B, then x =

A. 1/3
B. 5/12
C. 2/3
D. 5/6
E. 5/3

BrentGMATPrepNow · Accepted Answer

There's a nice rule that says,   "In a set where the numbers are equally spaced, the mean will equal the median."  
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}

This means the average of set A is 2.5x
And the average of set B is 3x²

If the average (arithmetic mean) of set A is twice the average of set B, we can write: 2.5x = 2(3x²)
Simplify: 2.5x = 6x²
To make things easier on ourselves, let's multiply both sides of the equation by 2 to get: 5x = 12x²
Subtract 5x from both sides: 0 = 12x² - 5x
Factor: 0 = x(12x - 5)
So, either x = 0 or 12x - 5 = 0

Since we are told x ≠ 0, it must be the case that 12x - 5 = 0
Add 5 to both sides: 12x = 5
Divide both sides by 12 to get: x = 5/12

Answer: B

RamseyGooner · Answer

Highlighted portion doesn't look right, should be => 0= 12x^2 - 5x

which on solving will give us, x= 5/12, i.e., Option E

Hope this helps!

~R

Ved3 · Answer

We know that set A is a set of elements that are multiples of x with contiguous coefficients and similarly, set B has elements that are multiples of x^2.
So, The average will be the middle element if the number of elements is odd or the average of the two middle elements if the number of elements is even.
So average of set A -> 2.5x and
average of set B -> 3x^2
Now given is Avg(set A) = 2(Avg(set B))
Therefore 2.5x = 2*3x^2
Since x is not equal to zero
 2.5 = 6x
 x = 5/12

AbhaGanu · Answer

Bunuel  Is the answer 5/12?

Bunuel · Answer

____________________
Yes, the OA is 5/12 (B).

Kinshook · Answer

A = \{x, 2x, 3x, 4x\} 
 
 B = \{x^2, 2x^2, 3x^2, 4x^2, 5x^2\}

If x ≠ 0 and the average (arithmetic mean) of set A is twice the average of set B, then x =

Average of set A = (x+2x+3x+4x)/4 = 10x/4 = 2.5x
Average of set B = (x^2 + 2x^2 + 3x^2 + 4x^2 + 5x^2)/5 = 15x^2/5 = 3x^2

Since the average (arithmetic mean) of set A is twice the average of set B.
2.5x = 2*3x^2 
6x = 2.5
x = 2.5/6 = 25/60 = 5/12

IMO B

ProKiller · Answer

Doesn't need all the long explanations and formulae. Just think.

Given: Avg(A) = 2 * Avg(B) ==> (x+2x+3x+4x)/4 = 2 * Avg(B) ==> (x+2x+3x+4x)/4 = 2 * ((x^2 + 2x^2 + 3x^2 + 4x^2 + 5x^2)/5) ==> 10x/4 = 30x^2/5 Solve for x. You will get x = 5/12. Option B.

Archit3110 · Answer

for set A 
sum = 10x
and avg ; 10x/4
set B
sum = 15x^2
avg ; 15x^2/5 ; 3x^2
given that
average (arithmetic mean) of set A is twice the average of set B
10x/4 = 2*3x^2
x = 5/12
option B

If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera

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If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera

Prep Toolkit

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