Difference between revisions of "2022 AMC 8 Problems/Problem 7"
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<math>\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000</math> | <math>\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Notice that the number of kilobits in this song is <math>4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.</math> | Notice that the number of kilobits in this song is <math>4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.</math> | ||
| Line 12: | Line 12: | ||
Finally, we divide this number by <math>60</math> because this is the number of <i><b>seconds</b></i> to get the answer <math>\frac{600}{60}=\boxed{\textbf{(B) } 10}.</math> | Finally, we divide this number by <math>60</math> because this is the number of <i><b>seconds</b></i> to get the answer <math>\frac{600}{60}=\boxed{\textbf{(B) } 10}.</math> | ||
| − | + | ||
~wamofan | ~wamofan | ||
| + | |||
| + | ==Solution 2== | ||
| + | We seek a value of <math>x</math> that makes the following equation true, since every other quantity equals <math>1</math>. | ||
| + | |||
| + | <cmath>\frac{x\ \text{min}}{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \text{min}} = 1.</cmath> | ||
| + | Solving yields <math>x=\boxed{\textbf{(B) } 10}</math>. | ||
| + | |||
| + | -Benedict T (countmath1) | ||
| + | |||
| + | ==Video Solution by Math-X (First understand the problem!!!)== | ||
| + | https://youtu.be/oUEa7AjMF2A?si=ZRtIWQMPPOjOX9fB&t=822 | ||
| + | |||
| + | ~Math-X | ||
| + | |||
| + | ==Video Solution (THINKING CREATIVELY!!!)== | ||
| + | https://youtu.be/ZXMYmDPOypo | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=475 | ||
| + | |||
| + | ~Interstigation | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/1xspUFoKDnU?t=65 | ||
| + | |||
| + | ~STEMbreezy | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/scSofI5OqkQ | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/jmblFtkaKpc | ||
| + | |||
| + | ~harungurcan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=6|num-a=8}} | {{AMC8 box|year=2022|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:38, 23 November 2023
Contents
Problem
When the World Wide Web first became popular in the 
s, download speeds reached a maximum of about 
kilobits per second. Approximately how many minutes would the download of a 
-megabyte song have taken at that speed? (Note that there are 
kilobits in a megabyte.)
Solution 1
Notice that the number of kilobits in this song is 
We must divide this by in order to find out how many seconds this song would take to download: 
Finally, we divide this number by 
because this is the number of seconds to get the answer 
~wamofan
Solution 2
We seek a value of 
that makes the following equation true, since every other quantity equals 
.
![\[\frac{x\ \text{min}}{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \text{min}} = 1.\]](http://latex.artofproblemsolving.com/c/4/3/c4376075d7f61a88fc6bffbf587ddde57260792b.png)
Solving yields 
.
-Benedict T (countmath1)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=ZRtIWQMPPOjOX9fB&t=822
~Math-X
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=475
~Interstigation
Video Solution
https://youtu.be/1xspUFoKDnU?t=65
~STEMbreezy
Video Solution
~savannahsolver
Video Solution
~harungurcan
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
