Elementary Technical Mathematics

Published by Brooks Cole

Chapter 1 - Section 1.14 - Rate, Base, and Part - Exercises - Page 84: 39

Answer

$53.\overline{3}\%$

Work Step by Step

First, find the increase in pressure: $=115-75 \\=40\text{ lb/in}^2$ RECALL: $\text{percent increase}=\dfrac{\text{increase in value}}{\text{original value}}$ The given problem has: original value =$75 \text{ lb/in}^2$ increase in value = $40 \text{ lb/in}^2$ Solve for the percent increase using the formula above to obtain: percent increase $= \dfrac{40}{75} \\=0.5\overline{3} \\=0.5\overline{3} \times 100\% \\=43.\overline{3}\%$ Thus, there was an $53.\overline{3}\%$ increase in pressure.

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