There are 100-10 = 90 numbers from 11 to 100, inclusively.


Half of them, i.e. 45, are with even numbers.

They contribute this sequence of money in cents

    2*12, 2*14, 2*16, . . . , 2*100,


which sum up

    2*(12 + 14 + 16 + . . . + 100)  cents.      (1)


The sum in the parentheses is equal to the average  = 56 times 45, i.e. 2520.


When doubled, in accordance with the formula (1), it gives $50.40.




The other half of the tickets, i.e. 45, are with odd numbers.

They contribute this sequence of money in cents

    3*11, 3*13, 3*15, . . . , 3*99,


which sum up

    3*(11 + 13 + 15 + . . . + 99)  cents.      (2)


The sum in the parentheses is equal to the average  = 55 times 45, i.e. 2475.


When tripled, in accordance with the formula (2), it gives $74.25.



Thus the total is  $24.20 + $74.25 = $124.65.    ANSWER