Difference between revisions of "2014 AMC 10B Problems/Problem 6"
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| − | ==Problem | + | ==Problem== |
Orvin went to the store with just enough money to buy <math>30</math> balloons. When he arrived, he discovered that the store had a special sale on balloons: buy <math>1</math> balloon at the regular price and get a second at <math>\frac{1}{3}</math> off the regular price. What is the greatest number of balloons Orvin could buy? | Orvin went to the store with just enough money to buy <math>30</math> balloons. When he arrived, he discovered that the store had a special sale on balloons: buy <math>1</math> balloon at the regular price and get a second at <math>\frac{1}{3}</math> off the regular price. What is the greatest number of balloons Orvin could buy? | ||
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<math> \textbf {(A) } 33 \qquad \textbf {(B) } 34 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 38 \qquad \textbf {(E) } 39</math> | <math> \textbf {(A) } 33 \qquad \textbf {(B) } 34 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 38 \qquad \textbf {(E) } 39</math> | ||
| − | ==Solution | + | ==Solution 1== |
| − | + | 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{(C) 36}</math>. | |
| − | 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{C) 36}</math>. | + | |
| + | ==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> | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/Ivv9JXk5cK4 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/y1Q-_PBS7Zg | ||
| + | |||
| + | ~savannahsolver | ||
==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}} | ||
Latest revision as of 11:03, 2 July 2023
Contents
Problem
Orvin went to the store with just enough money to buy 
balloons. When he arrived, he discovered that the store had a special sale on balloons: buy 
balloon at the regular price and get a second at 
off the regular price. What is the greatest number of balloons Orvin could buy?
Solution 1
Since he pays 
the price for every second balloon, the price for two balloons is 
. Thus, if he had enough money to buy balloons before, he now has enough to buy 
.
Solution 2
Suppose each balloon costs 
dollars. Therefore, Orvin brought 
dollars. Since every second balloon costs 
dollars, Orvin gets 
balloons for 
dollars. Therefore, Orvin gets 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
