## Elementary Technical Mathematics

12$\frac{1}{16}$ in
Let the three distances between two holes are 3$\frac{3}{8}$ in, 5$\frac{5}{16}$ in and 3$\frac{3}{8}$ in total distance between the centers of the two end holes of the plate = 3$\frac{3}{8}$ in + 5$\frac{5}{16}$ in + 3$\frac{3}{8}$ in =$\frac{27}{8}$ + $\frac{85}{16}$ + $\frac{27}{8}$ =$\frac{54}{8}$ + $\frac{85}{16}$ =$\frac{54}{8}\times$ $\frac{2}{2}$ + $\frac{85}{16}$ = $\frac{108}{16}$ + $\frac{85}{16}$ = $\frac{193}{16}$ = 12$\frac{1}{16}$ in