Difference between revisions of "2014 AMC 10B Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
| − | Suppose each balloon costs 3 dollars. Therefore, Orvin brought 90 dollars. Since every second balloon costs <math>\frac{2}{3}\cdot3=2</math> dollars, Orvin gets 2 balloons for 5 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> |
==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 18:48, 27 December 2019
Contents
Problem 6
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 
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. 
