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 - Page 42: 73

Answer

a)$21\frac{13}{16}\ in$ b)$\frac{7}{8}\ in$

Work Step by Step

a) Sum the given lengths to find the total length. The denominators are all factors of 16, so the least common denominator is 16. $5\frac{1}{8}+5+7\frac{5}{8}+4\frac{1}{16}=$ $5\frac{1\times2}{8\times2}+5+7\frac{5\times2}{8\times2}+4\frac{1}{16}=$ $5\frac{2}{16}+5+7\frac{10}{16}+4\frac{1}{16}=$ $21\frac{13}{16}$ b) To find the diameter of the shaft, sum the given lengths and subtract from the total. 4 is a factor of 16, so the least common denominator is 16. $3\frac{3}{16}+3\frac{3}{16}=6\frac{6}{16}$ $7\frac{1}{4}-6\frac{6}{16}=7\frac{1\times4}{4\times4}-6\frac{6}{16}=7\frac{4}{16}-6\frac{6}{16}=6\frac{20}{16}-6\frac{6}{16}=$ $\frac{14}{16}=\frac{14\div2}{16\div2}=\frac{7}{8}$
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