Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise: 50

Answer

$\frac{1}{32}$ in

Work Step by Step

Diameter of 6011 welding rods = $\frac{1}{8}$ in. Diameter of super strength 100 rods = $\frac{3}{32}$ in. Difference of the diameters = $\frac{1}{8}$ in. - $\frac{3}{32}$ in =$\frac{1}{8}$x$\frac{4}{4}$ - $\frac{3}{32}$ = $\frac{4}{32}$ - $\frac{3}{32}$ = $\frac{1}{32}$ in
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