## Elementary Technical Mathematics

Published by Brooks Cole

# Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise - Page 40: 58

#### Answer

142$\frac{23}{24}$ oz

#### Work Step by Step

Let the weight of the different parts of the product are 3$\frac{1}{2}$ oz, 33$\frac{1}{8}$ oz, 6 lb, 10$\frac{1}{3}$ oz given 1 lb =16 oz So 6 lb = 96 oz total product weight = 3$\frac{1}{2}$ oz + 33$\frac{1}{8}$ oz + 96 oz + 10$\frac{1}{3}$ oz =$\frac{7}{2}$ + $\frac{265}{8}$ + 96 + $\frac{31}{3}$ = $\frac{7}{2} \times$ $\frac{12}{12}$ + $\frac{265}{8} \times$ $\frac{3}{3}$ + 96 $\times$ $\frac{24}{24}$ + $\frac{31}{3} \times$ $\frac{8}{8}$ = $\frac{84}{24}$ + $\frac{795}{24}$ + $\frac{2304}{24}$ + $\frac{248}{24}$ =$\frac{3431}{24}$ =142$\frac{23}{24}$ oz

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