Probably very easy but i'm stupid

• Mar 25th 2008, 05:07 AM
IndoorKid
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?
• Mar 25th 2008, 05:32 AM
topsquark
Quote:

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:
\$\displaystyle x + y + z = 171\$
and
\$\displaystyle x + y = 90\$

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

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

\$\displaystyle 90 + z = 171\$

\$\displaystyle 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
• Mar 25th 2008, 05:39 AM
struck
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