Difference between revisions of "2014 AMC 12B Problems/Problem 2"
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Orvin went to the store with just enough money to buy <math> 30 </math> balloons. When he arrived he discovered that the store had a special sale on balloons: buy <math> 1 </math> balloon at the regular price and get a second at <math> \frac{1}{3} </math> off the regular price. What is the greatest number of balloons Orvin could buy? | Orvin went to the store with just enough money to buy <math> 30 </math> balloons. When he arrived he discovered that the store had a special sale on balloons: buy <math> 1 </math> balloon at the regular price and get a second at <math> \frac{1}{3} </math> off the regular price. What is the greatest number of balloons Orvin could buy? | ||
| − | <math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 39 </math> |
| − | ==Solution== | + | ==Solution 1== |
| − | 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 | + | 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> |
| − | Solution | + | ==Solution 2== |
| + | |||
| + | Similar to solution 1, but quicker. | ||
| + | We see the fraction <math>\frac13,</math> so we let each balloon cost <math>\$3</math> WLOG. Thus, Orvin has <math>\$3\cdot30=\$90.</math> | ||
| + | Each pair of balloons (one full-priced and one a third off) costs <math>\$3+\left(1-\frac13\right)\cdot\$3=\$3+\$2=\$5.</math> Therefore, Orvin can buy <math>\frac{\$90}{\$5}=18</math> pairs of balloons, at maximum. <math>18\cdot2=36</math> balloons. | ||
| + | ~Technodoggo | ||
| + | |||
| + | ==Video Solution 1 (Quick and Easy)== | ||
| + | https://youtu.be/vkItt-jxrIs | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2014|ab=B|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:41, 30 May 2023
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?
Solution 1
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 
Solution 2
Similar to solution 1, but quicker.
We see the fraction 
so we let each balloon cost 
WLOG. Thus, Orvin has 
Each pair of balloons (one full-priced and one a third off) costs 
Therefore, Orvin can buy 
pairs of balloons, at maximum. 
balloons.
~Technodoggo
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
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. 
