Difference between revisions of "2014 AMC 10B Problems/Problem 6"

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==Solution==
 
==Solution==
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==Solution==
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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{D) 36}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2014|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:13, 20 February 2014

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

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 $30 \cdot \dfrac{2}{\dfrac{5}{3}} = 30 \cdot \dfrac{6}{5} = \fbox{D) 36}$.

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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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