Difference between revisions of "2022 AMC 8 Problems/Problem 7"

Revision as of 09:07, 9 December 2022

Problem

When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)

$\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000$

Solution 1

Notice that the number of kilobits in this song is $4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.$

We must divide this by $56$ in order to find out how many seconds this song would take to download: $\frac{\cancel{8}\cdot\cancel{7}\cdot6\cdot100}{\cancel{56}} = 600.$

Finally, we divide this number by $60$ because this is the number of seconds to get the answer $\frac{600}{60}=\boxed{\textbf{(B) } 10}.$

~wamofan

Solution 2

We seek a value of $x$ that makes the following equation true, since every other quantity equals $1$ .

$\frac{x\ \text{min}}{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \text{min}} = 1.$ Solving yields $x=\boxed{\textbf{(B) } 10}$ .

-Benedict T (countmath1)

Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=475

~Interstigation

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.

@@ Line 15: / Line 15: @@
 ~wamofan
-==Solution 2 (Dimensional Analysis)==
+==Solution 2==
-We seek a value of <math>x</math> that makes the following equation true, since evrery other quantity equals <math>1</math>.
+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>
+<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>.
-Solving yields <math>x=\boxed{\textbf{(B) } 10}</math>
 -Benedict T (countmath1)

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