= Solution
The <bit rate> counts data <bits> carried per second; the <baud rate> counts transmitted signal elements per second. A signal element need not carry one whole data <bit>. In <Manchester encoding>, each data <bit> occupies two half-bit signal elements, one high and one low. For example, choose low–high for zero and high–low for one. The compulsory transition in the middle provides a timing reference for <clock recovery>.
If one data <bit> lasts $T$ seconds, the <bit rate> is $1/T$, while each half-bit signal element lasts $T/2$. With this binary signal-element convention,
$$
\boxed{R_b=1/T\ \text{bits/s},\qquad R_s=2/T=2R_b\ \text{baud}.}
$$
Thus a 10 Mbit/s <Manchester encoding> signal uses 20 million half-bit signal elements per second. This is not a count of observed voltage transitions: some <bit> boundaries also contain a transition and others do not. Calling a whole two-element pattern a symbol gives a different symbol-counting convention; the physical fact is two half-bit elements per data <bit>.
\Image[/past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/networks-2015-line-codes.png]
{title=<Manchester encoding> and <differential Manchester encoding> for the same seven <bits>; dotted lines mark <bit> boundaries and dashed lines mark their midpoints}
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