Elementary Technical Mathematics

The total length of the shaft is 12$\frac{9}{16}$in.
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.