Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2004 AMC 8 Problems/Problem 12: Difference between revisions

← Older edit
Goldenn (talk | contribs)
43 edits
Delioa.j (talk | contribs)
53 edits
 
(One intermediate revision by one other user not shown)
Line 11: Line 11:


==Solution==
==Solution==
When not being used, the cell phone uses up <math>1/24</math> of its battery per hour. When being used, the cell phone uses up <math>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(1/24) + 1(1/3) = 2/3</math> of its battery has been used up. To drain the remaining <math>1/3</math> the phone can last for <math>\frac{1/3}{1/24}=\boxed{\textbf{(B)}\ 8}</math> more hours without being used.
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> with Niki using her phone, <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 07:17, 2 January 2025

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 $24$ hours. If she is using it constantly, the battery will last for only $3$ hours. Since the last recharge, her phone has been on $9$ hours, and during that time she has used it for $60$ minutes. If she doesn’t use it any more but leaves the phone on, how many more hours will the battery last?

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution

When not being used, the cell phone uses up $\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ with Niki using her phone, $8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}$ of its battery has been used up. To drain the remaining $\frac{1}{3}$ the phone can last for $\frac{\frac{1}{3}}{\frac{1}{24}}=\boxed{\textbf{(B)}\ 8}$ 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_12&oldid=238755"