## Elementary Technical Mathematics

Published by Brooks Cole

# Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise - Page 42: 71

#### Answer

(a) $\frac{21}{2}$in (b) $\frac{3}{4}$in

#### Work Step by Step

For (a), you just need to simply add all 3 sides of the tool. So, 6$\frac{7}{8}$in + 1$\frac{3}{8}$in + 2$\frac{1}{4}$in. To add mixed numbers, we can separate the mixed number like so: 6 +$\frac{7}{8}$ + 1 + $\frac{3}{8}$ + 2 + $\frac{1}{4}$ Using the maximum common divisor, which in this case is 8, we just add all three fractions and add the 3 whole numbers. => 9 + $\frac{7+3+2(1)}{8}$ = 9 + $\frac{12}{8}$ Adding the whole number and the new fraction, we get = $\frac{21}{2}$in, and this is the length of the tool. For (b), to find the diameter A, we just need to subtract the two $\frac{7}{16}$in-sections from the total diameter of the tool, which is 1$\frac{5}{8}$in. So, A = 1$\frac{5}{8}$in - $\frac{7}{16}$in - $\frac{7}{16}$in. Using the same procedure as in (a), we get => A = $\frac{3}{4}$in

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