SOLUTION: Tickets for a certain show were printed bearing numbers from 11 to 100. The odd number tickets were sold by receiving cents equal to thrice the number on the ticket while the even

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Question 1166212: Tickets for a certain show were printed bearing numbers from 11 to 100. The odd number tickets were sold by receiving cents equal to thrice the number on the ticket while the even number tickets were issued by receiving cents equal to twice the number on the ticket. How much was received by the issuing agency?
Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
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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 %2812%2B100%29%2F2 = 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 %2811%2B99%29%2F2 = 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

Solved.