Difference between revisions of "2009 AMC 10A Problems/Problem 8"
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| − | A senior ticket costs | + | A senior ticket costs <math>\$6.00</math>, so a regular ticket costs <math>6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8</math> dollars. Therefore children's tickets cost half that, or <math>\$4.00</math>, so we have: |
<math>2(6+8+4)\:=\:36</math> | <math>2(6+8+4)\:=\:36</math> | ||
Revision as of 15:23, 30 December 2016
Problem 8
Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a 
% discount as children. The two members of the oldest generation receive a 
discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs <dollar/>
, is paying for everyone. How many dollars must he pay?
Solution
A senior ticket costs 
, so a regular ticket costs 
dollars. Therefore children's tickets cost half that, or 
, so we have:
So Grandfather Wen pays 
, or 
.
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
