Difference between revisions of "2004 AMC 8 Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | + | Let's convert everything to minutes and add them together. On Monday he worked for <math>\frac54 \cdot 60 = 75</math> minutes. On Tuesday he worked <math>50</math> minutes. On Wednesday he worked for <math>2</math> hours <math>25</math> minutes, or <math>2(60)+25=145</math> minutes. On Friday he worked <math>\frac{60}{2}=30</math> minutes. This adds up to <math>75+50+145+30=300</math> minutes, or <math>300/60=5</math> hours and <math>5\cdot 3 = \boxed{\textbf{(E)}\ \textdollar 15}</math>. | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=9|num-a=11}} | {{AMC8 box|year=2004|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 01:09, 29 January 2023
Problem
Handy Aaron helped a neighbor 
hours on Monday, 
minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid 
per hour. How much did he earn for the week?
Solution
Let's convert everything to minutes and add them together. On Monday he worked for 
minutes. On Tuesday he worked minutes. On Wednesday he worked for 
hours 
minutes, or 
minutes. On Friday he worked 
minutes. This adds up to 
minutes, or 
hours and 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
