2014 AMC 10B Problems/Problem 6

Problem

Orvin went to the store with just enough money to buy $30$ balloons. When he arrived, he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?

$\textbf {(A) } 33 \qquad \textbf {(B) } 34 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 38 \qquad \textbf {(E) } 39$

Solution 1

Since he pays $\dfrac{2}{3}$ the price for every second balloon, the price for two balloons is $\dfrac{5}{3}$ . Thus, if he had enough money to buy $30$ balloons before, he now has enough to buy $30 \cdot \dfrac{2}{\dfrac{5}{3}} = 30 \cdot \dfrac{6}{5} = \fbox{(C) 36}$ .

Solution 2

Suppose each balloon costs $3$ dollars. Therefore, Orvin brought $90$ dollars. Since every second balloon costs $\frac{2}{3}\cdot3=2$ dollars, Orvin gets $2$ balloons for $5$ dollars. Therefore, Orvin gets $\dfrac{2\cdot90}{5}=\fbox{(C) 36}.$

Video Solution (CREATIVE THINKING)

https://youtu.be/Ivv9JXk5cK4

~Education, the Study of Everything

Video Solution

https://youtu.be/y1Q-_PBS7Zg

~savannahsolver

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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