Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 1 - Section 1.14 - Rate, Base, and Part - Exercises - Page 84: 37



Work Step by Step

RECALL: $R=\dfrac{P}{B}$ where $B$=base, $P$=part, and $R$= rate or percent The given problem has: $B=0.600$ grain $P=0.150\%$ Solve for $R$ using the formula above to obtain: $R = \dfrac{P}{B} \\R = \dfrac{0.150}{0.600} \\P=0.25 \\P=0.25 \times 100\% \\P=25\$$ Thus, $0.150$ grain is $25\%$ of $0.600$ grain.
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