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Difference between revisions of "2014 AMC 10B Problems/Problem 6"
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==Solution 2==
 
==Solution 2==
 
Suppose each balloon costs <math>3</math> dollars. Therefore, Orvin brought <math>90</math> dollars. Since every second balloon costs <math>\frac{2}{3}\cdot3=2</math> dollars, Orvin gets <math>2</math> balloons for <math>5</math> dollars. Therefore, Orvin gets <math>\dfrac{2\cdot90}{5}=\fbox{(C) 36}.</math>
 
Suppose each balloon costs <math>3</math> dollars. Therefore, Orvin brought <math>90</math> dollars. Since every second balloon costs <math>\frac{2}{3}\cdot3=2</math> dollars, Orvin gets <math>2</math> balloons for <math>5</math> dollars. Therefore, Orvin gets <math>\dfrac{2\cdot90}{5}=\fbox{(C) 36}.</math>
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/Ivv9JXk5cK4
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~Education, the Study of Everything
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==Video Solution==
 
==Video Solution==

Latest revision as of 11:03, 2 July 2023

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

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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