Problem Solving

sanju09 wrote:If 0 is added to the number equal to the average (arithmetic mean) of the numbers x - 2, x - 1, x, x + 1, and x + 2 and if 10 is added to each number that is not equal to the average, then the average of the five resulting numbers is
(A) x + 4
(B) x + 8
(C) x + 10
(D) x + 40
(E) x + 50

First, there are 2 different ways to find the average of x-2, x-1, x, x+1, and x+2.

Method 1) If you recognize that the five numbers are consecutive numbers, then you will see that x must be the average. Why? There's a nice rule that says something like "Given a set of consecutive integers, the mean of the set is equal to the median." Since x is the the middlemost number in the set, it will be the median as well as the mean.

Aside: the rule can also be extended to include sets of numbers where the values are all equally spaced apart.

Method 2: Add the five values and divide by 5 to get: [x-2 + x-1 + x + x+1 + x+2]/5 = 5x/5 = x


Now that we know x is the average, we will add zero to x and add 10 to the remaining numbers to get:
x+8, x+9, x, x+11, and x+12

The new average is [x+8 + x+9 + x + x+11 + x+12]/5 = [5x+40]/5 = x+8 = B

Cheers,
Brent

sanju09 wrote:If 0 is added to the number equal to the average (arithmetic mean) of the numbers x - 2, x - 1, x, x + 1, and x + 2 and if 10 is added to each number that is not equal to the average, then the average of the five resulting numbers is
(A) x + 4
(B) x + 8
(C) x + 10
(D) x + 40
(E) x + 50

Alternatively, once we recognize that the average of the set is x, we know that [original sum of all five numbers] / 5 = x (this uses the average formula)

So, if we add 10 to four of the numbers in the set, the sum of all the numbers in the set will increase by 40

So, our new average = [original sum of all five numbers + 40] / 5

We can break the fraction into two pieces to get:
new average = [original sum of all five numbers]/5 + 40/5
new average = x + 8 = B

Cheers,
Brent