2014 AMC 12B Problems/Problem 2

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

If every balloon costs $n$ dollars, then Orvin has $30n$ dollars. For every balloon he buys for $n$ dollars, he can buy another for $\frac{2n}{3}$ dollars. This means it costs him $\frac{5n}{3}$ dollars to buy a bundle of $2$ balloons. With $30n$ dollars, he can buy $\frac{30n}{\frac{5n}{3}} = 18$ sets of two balloons, so the total number of balloons he can buy is $18\times2 \implies \boxed{\textbf{(C)}\ 36 }$

Solution 2

Similar to solution 1, but quicker. We see the fraction $\frac13,$ so we let each balloon cost $$3$ WLOG. Thus, Orvin has $$3\cdot30=$90.$ Each pair of balloons (one full-priced and one a third off) costs $$3+\left(1-\frac13\right)\cdot$3=$3+$2=$5.$ Therefore, Orvin can buy $\frac{$90}{$5}=18$ pairs of balloons, at maximum. $18\cdot2=36$ balloons. ~Technodoggo

Video Solution 1 (Quick and Easy)

https://youtu.be/vkItt-jxrIs

~Education, the Study of Everything

2014 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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2014 AMC 12B Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

Video Solution 1 (Quick and Easy)

See also