## Elementary Technical Mathematics

Published by Brooks Cole

# Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise - Page 42: 78

#### Answer

$\frac{3}{16}$ in

#### Work Step by Step

Lets consider the thickness of the steel plates are $\frac{7}{16}$ in and $\frac{5}{8}$ in therefore the difference in thickness = $\frac{5}{8}$ - $\frac{7}{16}$ =$\frac{5}{8}\times$ $\frac{2}{2}$ - $\frac{7}{16}$ =$\frac{10}{16}$ - $\frac{7}{16}$ =$\frac{3}{16}$ in

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