Chapter 1: Q47. (page 20)

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?

Expert verified

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

Step by step solution

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

Size of the compact disc, $S = 783.216 megabytes$ .

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

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

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

$\begin{matrix} S = 783.216 \times 8 \\ = 6265.728 megabits \end{matrix}$

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,

$\begin{matrix} t_{\circ} = \frac{6265.728}{1.4} \\ = 4475.52 s \end{matrix}$

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

$\begin{matrix} t_{\circ} = \frac{4475.52}{60} \\ = 74.59 \min \end{matrix}$

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