Difference between revisions of "2014 AMC 12B Problems/Problem 2"
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If every balloon costs <math>n</math> dollars, then Orvin has <math>30n</math> dollars. For every balloon he buys for <math>n</math> dollars, he can buy another for <math> \frac{2n}{3}</math> dollars. This means it costs him <math> \frac{5n}{3}</math> dollars to buy a bundle of <math>2</math> balloons. With <math>30n</math> dollars, he can buy <math>\frac{30n}{\frac{5n}{3}} = 18</math> sets of two balloons, so the total number of balloons he can buy is <math>18\times2 \implies \boxed{\textbf{(C)}\ 36 }</math> | If every balloon costs <math>n</math> dollars, then Orvin has <math>30n</math> dollars. For every balloon he buys for <math>n</math> dollars, he can buy another for <math> \frac{2n}{3}</math> dollars. This means it costs him <math> \frac{5n}{3}</math> dollars to buy a bundle of <math>2</math> balloons. With <math>30n</math> dollars, he can buy <math>\frac{30n}{\frac{5n}{3}} = 18</math> sets of two balloons, so the total number of balloons he can buy is <math>18\times2 \implies \boxed{\textbf{(C)}\ 36 }</math> | ||
| + | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=1|num-a=3}} | {{AMC12 box|year=2014|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:22, 22 February 2014
Problem
Orvin went to the store with just enough money to buy 
balloons. When he arrived he discovered that the store had a special sale on balloons: buy 
balloon at the regular price and get a second at 
off the regular price. What is the greatest number of balloons Orvin could buy?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}}\ 38\qquad\textbf{(E)}\ 39$ (Error compiling LaTeX. Unknown error_msg)
Solution
If every balloon costs 
dollars, then Orvin has 
dollars. For every balloon he buys for dollars, he can buy another for 
dollars. This means it costs him 
dollars to buy a bundle of 
balloons. With dollars, he can buy 
sets of two balloons, so the total number of balloons he can buy is 
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
