Discussion
Sixteen channels of \( 3 \ \mathrm{kHz} \) each are multiplexed, each coded to \( 7 \) bits and sampled at \( 8 \ \mathrm{kHz} \). The aggregate bit rate is
Answer: C
Rate per channel times the number of channels:
\[ \begin{aligned} 8000 \times 7 &= 56 \ \mathrm{kbit/s} \\ 16 \times 56 &= 896 \ \mathrm{kbit/s} \end{aligned} \]
giving 896 kbit/s before overhead. The 3 kHz figure is a distractor: the sampling rate is given directly and the signal bandwidth plays no further part. Spotting which of the quoted numbers the calculation does not need is half the work in a question of this shape. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.
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