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 43: 84


The total length of the shaft is 12$\frac{9}{16}$in.

Work Step by Step

To find the total length of the shaft, all you need to do is add all the small sections the exercise has already given us. So, total length = $\frac{15}{16}$in + 3$\frac{1}{4}$in + 2$\frac{1}{16}$in + 3$\frac{3}{8}$in + 1$\frac{13}{16}$in + 1$\frac{1}{8}$in We can write "A$\frac{x}{y}$" as A + $\frac{x}{y}$, so we can do this for all the mixed numbers in our problem. So, total length = $\frac{15}{16}$in + 3 in + $\frac{1}{4}$in + 2 in + $\frac{1}{16}$in + 3 in + $\frac{3}{8}$in + 1 in + $\frac{13}{16}$in + 1 in + $\frac{1}{8}$in Add the whole numbers and use a maximum common divisor to add all the fractions. In this case, the m.c.d. is 16. total length = 10 in + $\frac{15+4(1)+1+2(3)+13+2(1)}{16}$in = 10 in + $\frac{41}{16}$in = $\frac{201}{16}$in Written as a mixed number, the total length of the shaft is 12$\frac{9}{16}$in.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.