unknown averages: When each of three different numbers is...

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luissi

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When each of three different numbers is added to the average of two other numbers, the results are 105, 106 and 125. What is the average of the three original numbers?

I'm not exactly sure how to go about this problem... I'm pretty sure I'll need variables, but I'm just having a really hard time setting it up. I'd really appreciate some hlep.
 
I think there is a missing modifier: I think the exercise should say:

When each of three different numbers is added to the average of the two other numbers, the results are 105, 106, and 125.
Click to expand...
If so, then:

Let the three numbers be "x", "y", and "z". Then the relations are:

. . . . .(x + y)/2 + z = 105

. . . . .(x + z)/2 + y = 106

. . . . .(y + z)/2 + x = 125

Once you multiply through to clear the fractions and rearrange a bit, you get:

. . . . .x + y + 2z = 210

. . . . .x + 2y + z = 212

. . . . .2x + y + z = 250

Solve this system to find the values of the three original numbers, and then take their average. (Alternatively, since you don't actually care what x, y, or z might be, you could add the three equations, and manipulate the result to get (x + y + z)/3 on one side, with the average on the other side.)

If you get stuck, please reply showing how far you have gotten. Thank you! :D

Eliz.
 
I did it out and I got that x=124, y=254, and z=-84. They worked when I plugged them into the original equations, but I got 98 as the average, but according to the possbile answers my teacher gave my class to choose out of, it's wrong...
 
luissi said:
I did it out and I got that x=124, y=254, and z=-84. They worked when I plugged them into the original equations...
Click to expand...
They don't work in the second equation. I didn't check the third.

Since we cannot see your work, we cannot correct whatever errors were made. Try following the parenthetical advice in the earlier reply. :wink:

Eliz.
 
Parenthetical solution (copyright stapel)

x + y + 2z = 210

x + 2y + z = 212

2x + y + z = 250


Add all three

4 (x+y+z) = 672

Now finish it to find (x+y+z)/3.....
 
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