Difference between revisions of "2014 AMC 10B Problems/Problem 6"
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<math> \textbf {(A) } 33 \qquad \textbf {(B) } 34 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 38 \qquad \textbf {(E) } 39</math> | <math> \textbf {(A) } 33 \qquad \textbf {(B) } 34 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 38 \qquad \textbf {(E) } 39</math> | ||
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==Solution== | ==Solution== | ||
Since he pays <math>\dfrac{2}{3}</math> the price for every second balloon, the price for two balloons is <math>\dfrac{5}{3}</math>. Thus, if he had enough money to buy <math>30</math> balloons before, he now has enough to buy <math>30 \cdot \dfrac{2}{\dfrac{5}{3}} = 30 \cdot \dfrac{6}{5} = \fbox{(C) 36}</math>. | Since he pays <math>\dfrac{2}{3}</math> the price for every second balloon, the price for two balloons is <math>\dfrac{5}{3}</math>. Thus, if he had enough money to buy <math>30</math> balloons before, he now has enough to buy <math>30 \cdot \dfrac{2}{\dfrac{5}{3}} = 30 \cdot \dfrac{6}{5} = \fbox{(C) 36}</math>. | ||
Revision as of 13:24, 20 February 2014
Problem 6
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
Since he pays 
the price for every second balloon, the price for two balloons is 
. Thus, if he had enough money to buy balloons before, he now has enough to buy 
.
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
