Difference between revisions of "2022 AMC 8 Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | Notice that the | + | Notice that the number 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 | + | |
| + | 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> to get an answer os <math>\boxed{\textbf{(B) } 10}.</math> | ||
~wamofan | ~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 11:02, 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 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 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. 
