Difference between revisions of "2004 AMC 8 Problems/Problem 12"
m (→Problem) |
Megaboy6679 (talk | contribs) m |
||
Line 11: | Line 11: | ||
==Solution== | ==Solution== | ||
− | When not being used, the cell phone uses up <math>1 | + | When not being used, the cell phone uses up <math>\frac{1}{24}</math> of its battery per hour. When being used, the cell phone uses up <math>\frac{1}{3}</math> of its battery per hour. Since Niki's phone has been on for <math>9</math> hours, of those <math>8</math> simply on and <math>1</math> being used to talk, <math>8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}</math> of its battery has been used up. To drain the remaining <math>\frac{1}{3}</math> the phone can last for <math>\frac{\frac{1}{3}}{\frac{1}{24}}=\boxed{\textbf{(B)}\ 8}</math> more hours without being used. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=11|num-a=13}} | {{AMC8 box|year=2004|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:28, 29 January 2023
Problem
Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the battery will last for hours. If she is using it constantly, the battery will last for only hours. Since the last recharge, her phone has been on hours, and during that time she has used it for minutes. If she doesn’t use it any more but leaves the phone on, how many more hours will the battery last?
Solution
When not being used, the cell phone uses up of its battery per hour. When being used, the cell phone uses up of its battery per hour. Since Niki's phone has been on for hours, of those simply on and being used to talk, of its battery has been used up. To drain the remaining the phone can last for more hours without being used.
See Also
2004 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.