Difference between revisions of "2022 AMC 8 Problems/Problem 7"
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==Solution== | ==Solution== | ||
| + | Notice that the amount of kilobits in this song is <math>4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.</math> | ||
| + | We must divide this by <math>56</math> in order to find out how many seconds this song would take to download. <math>\frac{\cancel{8}\cdot\cancel{7}\cdot6\cdot100}{\cancel{56}} = 600.</math> Finally, we divide this number by <math>60</math> because this is the number of <i>seconds</i>, not minutes in the song, to get an answer os <math>\boxed{\textbf{(B) } 10}</math> | ||
| + | |||
| + | ~wamofan | ||
==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}} | ||
Revision as of 11:47, 28 January 2022
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
Notice that the amount 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, not minutes in the song, to get an answer os 
~wamofan
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. 
