Calculate the required bit rate of a channel if we need to download a text document at the rate of 100 pages per minute. Assume that a page has 24 lines with 80 characters in each line and one character is represented using 8 bits:

Question

Accepted Answer

The correct option is (1):
To calculate the required bit rate, we must first determine the total number of bits transferred per unit of time, which is typically measured in bits per second (bps). We start by finding the data size of a single page: a page has 24 lines, each with 80 characters, and one character is represented by 8 bits. Thus, one page contains $24 \times 80 \times 8 = 15,360$ bits. Since the required download speed is 100 pages per minute, the total data transferred per minute is $100 \times 15,360 = 1,536,000$ bits per minute. The standard bit rate unit is bits per second, so we must divide this by 60 seconds, which correctly yields a bit rate of $1,536,000 / 60 = 25,600$ bps, or $25.6$ Kbps. While this calculated result is $25.6 \text{ Kbps}$, the numerical value $1.536$ in the option is derived from expressing the bits-per-minute value in Megabits ($1,536,000 \text{ bits} = 1.536 \text{ Mbits}$). It is evident that the test question contains a unit and magnitude error, confusing the total data size or a raw figure from the calculation with the final rate in Kbps. We must select option 1 as it is the numerically intended but flawed answer.

The incorrect option is (2):
Option 2, 240 Kbps (or 240,000 bps), is mathematically incorrect and far exceeds the required rate calculated from the problem's constraints. The correct bit rate is calculated to be $25.6 \text{ Kbps}$, meaning that $240 \text{ Kbps}$ is nearly a ten-fold overestimation of the channel capacity needed for this download task. There is no logical miscalculation, such as missing a parameter or incorrect unit conversion, that would lead to a value of $240 \text{ Kbps}$. A channel operating at this bandwidth would be unnecessarily fast for the specified task, demonstrating a misunderstanding of the actual data rate requirement.

The incorrect option is (3):
The option 2.1 Mbps (or 2,100,000 bps) represents a very high data rate and is drastically wrong. This capacity is over 80 times the actual $25.6 \text{ Kbps}$ required to download $100$ pages per minute with the given character and line constraints. No combination of the factors in the question's formula, such as lines, characters, bits per character, or pages per minute, could reasonably result in a number near 2.1 million. Choosing this option indicates a fundamental error in the calculation of the data size or a major misapplication of the time conversion.

The incorrect option is (4):
Option 4, 1.636 Mbps, is incorrect both numerically and in unit magnitude. It is much higher than the actual required bit rate of $25.6 \text{ Kbps}$. This value represents approximately 64 times the necessary bandwidth. Although the numerical component of the correct option (1) hints at an original error involving $1.536 \text{ Mbits}$, $1.636 \text{ Mbps}$ is not derived from the components of the problem. This option is also clearly in the wrong unit, as it is over $1.5$ million bits per second, which does not match the actual requirement.