Question

A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD’s information at a constant rate of 1.4 megabits per second. How many minutes does the player take to read the entire CD?

Short answer

The CD player will take 74.59 min to read the entire CD.

Step-by-step solution

Step 1. Meaning of the rate of transfer 

The rate of transfer is equal to the amount of information being transferred each second.

Step 2. Given data and assumption

Size of the compact disc, $S=783.216\mathrm{megabytes}$.

The rate at which the CD player reads information, $r=1.4{\mathrm{megabitss}}^{-1}$.

Assume that the CD player takes $t_{\circ }$time to read the entire disc.

Step 3. Calculating the size of the disc in megabits

As it is mentioned in the question, 1 byte is equal to 8 bits. So, the total size of the disc in megabits will be

$\begin{array}{c}S=783.216\times 8\\ =6265.728\mathrm{megabits}\end{array}$

Step 4. Calculating the time taken by the CD player to read the CD

By the definition of rate, you can write the time taken by the CD player to read the drive as

$t_{\circ }=\frac{S}{r}$

Substituting the values in the above equation,

$\begin{array}{c}{t}_{\circ }=\frac{6265.728}{1.4}\\ =4475.52\mathrm{s}\end{array}$

Step 5. Converting seconds to minutes

As 1 minute is equal to 60 seconds, the time can be written in minutes as

$\begin{array}{c}{t}_{\circ }=\frac{4475.52}{60}\\ =74.59\min \end{array}$