## Elementary Technical Mathematics

52$\frac{5}{32}$ in
Let us consider the mechanic needs tubing of length = 15$\frac{3}{8}$ in, 7$\frac{3}{4}$ in, 11$\frac{1}{2}$ in, 7$\frac{7}{32}$ in, 10$\frac{5}{16}$ in Total length = 15$\frac{3}{8}$ in + 7$\frac{3}{4}$ in + 11$\frac{1}{2}$ in + 7$\frac{7}{32}$ in+ 10$\frac{5}{16}$ in = $\frac{123}{8}$ + $\frac{31}{4}$ + $\frac{23}{2}$ + $\frac{231}{32}$ + $\frac{165}{16}$ =$\frac{123}{8}\times$ $\frac{4}{4}$ + $\frac{31}{4}\times$ $\frac{8}{8}$ + $\frac{23}{2}\times$ $\frac{16}{16}$ + $\frac{231}{32}$+ $\frac{165}{16}\times$ $\frac{2}{2}$ =$\frac{492}{32}$ + $\frac{248}{32}$ + $\frac{368}{32}$ + $\frac{231}{32}$+$\frac{330}{32}$ = $\frac{1669}{32}$ =52$\frac{5}{32}$ in