Answer
The total length of the shaft is 12$\frac{9}{16}$in.
Work Step by Step
To find the total length of the shaft, all you need to do is add all the small sections the exercise has already given us.
So, total length = $\frac{15}{16}$in + 3$\frac{1}{4}$in + 2$\frac{1}{16}$in + 3$\frac{3}{8}$in + 1$\frac{13}{16}$in + 1$\frac{1}{8}$in
We can write "A$\frac{x}{y}$" as A + $\frac{x}{y}$, so we can do this for all the mixed numbers in our problem.
So,
total length = $\frac{15}{16}$in + 3 in + $\frac{1}{4}$in + 2 in + $\frac{1}{16}$in + 3 in + $\frac{3}{8}$in + 1 in + $\frac{13}{16}$in + 1 in + $\frac{1}{8}$in
Add the whole numbers and use a maximum common divisor to add all the fractions. In this case, the m.c.d. is 16.
total length = 10 in + $\frac{15+4(1)+1+2(3)+13+2(1)}{16}$in = 10 in + $\frac{41}{16}$in = $\frac{201}{16}$in
Written as a mixed number, the total length of the shaft is 12$\frac{9}{16}$in.
