## Elementary Technical Mathematics

Published by Brooks Cole

# Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise - Page 42: 72

#### Answer

$5\frac{3}{16}\ in$

#### Work Step by Step

Sum the lengths of each rod cut and the waste. The denominators are all factors of 16, so the least common denominator is 16. $1\frac{3}{8}+\frac{1}{16}+2\frac{5}{16}+\frac{1}{16}+4\frac{3}{4}+\frac{1}{16}=$ $1\frac{3\times2}{8\times2}+\frac{1}{16}+2\frac{5}{16}+\frac{1}{16}+4\frac{3\times4}{4\times4}+\frac{1}{16}=$ $1\frac{6}{16}+\frac{1}{16}+2\frac{5}{16}+\frac{1}{16}+4\frac{12}{16}+\frac{1}{16}=$ $7\frac{26}{16}=$ $8\frac{10}{16}$ Subtract from the original length to find the length of the remaining piece. $13\frac{13}{16}-8\frac{10}{16}=5\frac{3}{16}$

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