
Doug C. answered 09/03/23
Math Tutor with Reputation to make difficult concepts understandable
This does not seem like a problem for a middle schooler, but let's give it a shot.
Let x = #starting chips for girl who loses the 1st round (1st girl)
Let y = #starting chips for girl who loses the 2nd round (2nd girl)
Let z = #starting chips for girl who loses the 3rd round (3rd girl)
At the end of round 1: y and z will double their chips so they will have respectively 2y and 2z.
x will lose y and z chips (to help her opps double)
At the end of round 1 the chips look like this:
x - y - z , 2y, 2z
For round 2:
1st girl doubles: 2 (x - y - z) ; the expression in parentheses is part of what 2nd girl loses
3rd girl doubles: 2 (2z)
2nd girl takes away the amounts that double the other girls: 2y - (x - y - z) - 2z = -x +3y - z
At the end of round 2:
2x -2y - 2z, -x +3y - z, 4z
Ready for round 3:
1st girl doubles current chips: 2 ( 2x - 2y -2z)
2nd girl doubles: 2 (-x + 3y -z)
3rd girl loses the chips that double the others: 4z - (2x - 2y - 2z) - (-x +3y - z)
At the end of the final round:
1st girl: 4x - 4y - 4z = 40
or, x - y - z = 10
2nd girl: -2x +6y - 2z = 40
or, -x +3y -z = 20
3rd girl (let's simplify: 4z - 2x + 2y +2z + x - 3y + z = -x - y +7z
so:
-x - y +7z = 40
Now we have three equations and three unknowns:
1) x - y - z = 10
2) -x +3y -z =20
3) -x -y +7z =40
Eliminate x from 1 and 2 by adding those two equations:
2y - 2z = 30, y - z = 15
Eliminate x from 1 and 3 by adding those two equations:
-2y +6z = 50; -y +3z = 25
Down to two equations and two unknowns:
y - z = 15
-y + 3z = 25
Add to eliminate y:
2z = 40
z = 20
So, y - 20 = 15; y = 35
And:
x - 35 - 20 = 10
x = 65
Check:
Start: 65, 35, 20
Round 1 (65 loses 35 + 20)
10, 70 40
Round 2 (70 loses 10 + 40)
20, 20, 80
Round 3 (80 loses 20 + 20)
40, 40, 40
That does NOT seem to be a middle school problem!