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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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 18:47, 27 December 2019
Contents
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 1
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 .
Solution 2
Suppose each balloon costs 3 dollars. Therefore, Orvin brought 90 dollars. Since every second balloon costs dollars, Orvin gets 2 balloons for 5 dollars. Therefore, Orvin gets
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 |
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