Question

The average of five numbers is 10. If two of the five numbers are removed, the average of the remaining three numbers is 9. What is the sum of the two numbers that were removed?

Answer 1

@Mertsj

Answer 2

\[\frac{a+b+c+d+e}5=10\\\frac{a+b+c}3=9\]

Answer 3

a+b+c+d+e=50
a+b+c=27

Answer 4

I don't get the point that you are trying to say.

Answer 5

@hartnn

Answer 6

Subtract the two equations.

Answer 7

But how?

Answer 8

Let the numbers be a, b, c, d, e respectively.
Since their average is 10, we have:
a+b+c+d+e=50 ----(1)
Since the average of the first three is 9, we have:
a+b+c=27 ----(2)
Take (2) away from (1) and you get:
d+e=23

Answer 9

Is the answer 23?

Answer 10

yes

Answer 11

Can you explain why we have to subtract? It doesn't make any sense to me.

Sorry for my previous post. The auto-check on my iPad is very annoying.

Answer 12

The problem says you want the sum of the remaining numbers. The remaining numbers are d and e. So you have a system of two equations with 5 variables and you want to eliminate 3 of them. So study the two equations and tell me how you would solve for d+e?

Answer 13

Hey mertsj where do you live? Like what country.

Answer 14

USA