Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 426: 58


$\$12,500$ is invested at the $7\%$ rate and $\$7500$ is invested at the $9\%$ rate.

Work Step by Step

We are given that $\$20,000$ is invested in two accounts, at the rate of $7\%$ and $9\%$. Total yearly interest earned is $\$1550$. Let the amount x be invested at a rate of $9\%$. This means that the amount of $\left( 20,000-x \right)$ is invested at a rate of 7%. Now, it is given that the interest earned in one year is 1550. This implies: $\left\{ \left( 9 \right)\times \left( x \right) \right\}+\left\{ \left( 7 \right)\times \left( 20,000-x \right) \right\}=\$1550$ This means: $0.09x+0.07\left( 20,000-x \right)=1550$ Using distributive property: $0.09x+1400-0.07x=1550$ Now, $0.02x+1400=1550$ $0.02x=150$ $x=7500$ Thus, amount invested at the rate of $9\%$ is $\$7500$ and the amount invested at $7\%$ is $20,000-x=20000-7500=12500$.
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