Difference between revisions of "2014 AMC 12B Problems/Problem 2"
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Similar to solution 1, but quicker. | Similar to solution 1, but quicker. | ||
| − | We see the fraction <math>\frac13,</math> so we let each balloon cost <math>\$3 | + | 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. | 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 | ~Technodoggo | ||
Latest revision as of 18:41, 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. 
