# 2022 AMC 8 Problems/Problem 7

## 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)

~Interstigation

~STEMbreezy

## Video Solution

~savannahsolver

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