2014 AMC 10B Problems/Problem 6

Revision as of 19:06, 31 December 2015 by Checkmatetang (talk | contribs) (Solution)

Problem 6

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

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 $\dfrac{2\cdot90}{5}=\fbox{(C) 36}.$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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