Difference between revisions of "2014 AMC 12B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | Let <math>m</math> stand for the cost of a muffin, and let <math>b</math> stand for the value of a banana. | + | Let <math>m</math> stand for the cost of a muffin, and let <math>b</math> stand for the value of a banana. We need to find <math>\frac{m}{b}</math>, the ratio of the price of the muffins to that of the bananas. We have |
<cmath>2(4m + 3b) = 2m + 16b </cmath> | <cmath>2(4m + 3b) = 2m + 16b </cmath> | ||
<cmath>6m = 10b </cmath> | <cmath>6m = 10b </cmath> | ||
<cmath>\frac{m}{b} = \boxed{\textbf{(B)}\ \frac{5}{3}}</cmath> | <cmath>\frac{m}{b} = \boxed{\textbf{(B)}\ \frac{5}{3}}</cmath> | ||
| + | |||
| + | ==Video Solution 1 (Quick and Easy)== | ||
| + | https://youtu.be/K4HupKj78Yc | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2014|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:30, 17 October 2024
Problem
Susie pays for 
muffins and 
bananas. Calvin spends twice as much paying for 
muffins and 
bananas. A muffin is how many times as expensive as a banana?
Solution
Let 
stand for the cost of a muffin, and let 
stand for the value of a banana. We need to find 
, the ratio of the price of the muffins to that of the bananas. We have
![\[2(4m + 3b) = 2m + 16b\]](http://latex.artofproblemsolving.com/c/4/a/c4a833d68798cfa8c0721f1d738a16f24eedd2a7.png)
![\[6m = 10b\]](http://latex.artofproblemsolving.com/a/f/0/af0c19a2db3a0a83db506a7d8815b4843e953fc3.png)
![\[\frac{m}{b} = \boxed{\textbf{(B)}\ \frac{5}{3}}\]](http://latex.artofproblemsolving.com/0/a/f/0af7e28cf6bbd335cd902c6031b410b54dc11e03.png)
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
