Difference between revisions of "2011 AMC 12A Problems/Problem 1"
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<math>30.5-30=.5</math> hours <math>=30</math> minutes. | <math>30.5-30=.5</math> hours <math>=30</math> minutes. | ||
Since the price for phone calls is <math>10</math> cents per minute, the additional amount Michelle has to pay for phone calls is <math>30*10=300</math> cents, or <math>3</math> dollars. | Since the price for phone calls is <math>10</math> cents per minute, the additional amount Michelle has to pay for phone calls is <math>30*10=300</math> cents, or <math>3</math> dollars. | ||
| − | Adding <math>20+5+3</math> dollars <math>\boxed{\$28 \textbf{D}}</math>. | + | Adding <math>20+5+3</math> dollars <math>\boxed{\$28 \textbf{ (D)}}</math>. |
==Video Solution== | ==Video Solution== | ||
Latest revision as of 11:09, 19 October 2023
Contents
Problem
A cell phone plan costs 
dollars each month, plus 
cents per text message sent, plus 
cents for each minute used over 
hours. In January Michelle sent 
text messages and talked for 
hours. How much did she have to pay?
Solution
The base price of Michelle's cell phone plan is dollars.
If she sent text messages and it costs cents per text, then she must have spent 
cents for texting, or dollars. She talked for hours, but 
will give us the amount of time exceeded past 30 hours.
hours 
minutes.
Since the price for phone calls is cents per minute, the additional amount Michelle has to pay for phone calls is 
cents, or 
dollars.
Adding 
dollars 
.
Video Solution
~savannahsolver
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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 | |
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
