# Math Help - Probably very easy but i'm stupid

1. ## Probably very easy but i'm stupid

The average of three numbers is 57. The average of two of them is 45. What is the third number.

I tried defining each number as a random variable x, y, z and doing

(x+y+z)/3 = 57 and (x+y)/2 = 45 and solving simulataneously but couldn't get it to work.

any ideas?

2. Originally Posted by IndoorKid
The average of three numbers is 57. The average of two of them is 45. What is the third number.

I tried defining each number as a random variable x, y, z and doing

(x+y+z)/3 = 57 and (x+y)/2 = 45 and solving simulataneously but couldn't get it to work.

any ideas?
For starters one problem is that there is not a unique answer:
$x + y + z = 171$
and
$x + y = 90$

Solving the second equation for y gives
$y = 90 - x$

Inserting this into the first equation gives:
$x + (90 - x) + z = 171$

$90 + z = 171$

$z = 81$

So there are a range of possible values for y, subject to the restiction that y cannot be negative, and only one score we can predict with certainty: the 81.

-Dan

3. We assume:

Average of two numbers is 45, so the total of numbers is 45 * 2 = 90
Average of 3 numbers is 57 so the total is 57 * 3 = 171

x + y + z = 171 (1)
x + y = 90 (2)

Subtract (2) from (1):

x + y + z - x - y = 171 - 90
Then, z = 81