Math
posted by George
A home owner goes to the hardware store to purchase the letters that make up a number for a sign he is making, (e.g., FOUR, NINE, etc.). When he arrives, other customers are doing the same thing. The first customer buys the letters to display ONE, and pays $40. The next customer buys the letters for the number TWO, and pays $60. The last customer buys the letters for the number ELEVEN, and pays $100. Our home owner wants to buy the letters to spell out TWELVE.
The question is: how much does the home owner pay?

Reiny
ok George, I'll play your game.
I will assume that the store owner is some mathnerd and sets the price of the numbers according to some cubic equation
cost = ax^2 + bx + c, where x is the number of letters and a,b,c are constants
so we have three ordered pairs:
(1,40), (2,60) and (11,100)
for (1,40) > a + b + c = 40
for (2,60) > 4a + 2b + c = 60
for (11,100) > 121a + 11b + c = 100
2nd  3rd: > 3a + b = 20
3rd  1st: > 120a + 10b = 60 or 12a + b = 6
subtracting those last two:
9a = 14
a = 14/9
from 12a+b=6
b = 612a = 612(14/9) = 74/3
back in 1st: a+b+c=40
14/9 + 74/3 + c = 40
c = 152/9
so cost = (14/9)x^2 + (74/3)x + 152/9
so when x = 12, Cost = $88.89 
Reiny
I meant to say:
...according to some quadratic equation
I would need 4 ordered pairs for a cubic.
My work reflects what is needed for a quadratic 
George
Teacher said answer was $120
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