Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 2 - Section 2.6 - Percent and Mixture Problem Solving - Practice: 7

Answer

2 liters of the 20% dye solution. 4 liters of the 40% dye solution.

Work Step by Step

number of liters$\times$dye solution strength=amount of dye solution x=# of liters Liters of 20% solution = x Liters of 50% solution = 6-x Liters of 40% solution needed = 6 $0.20x + 0.50(6-x)=0.40(6)$ $0.20x + 3 - 0.50x=2.4$ $-0.30x+3=2.4$ $-0.30x=2.4-3$ $-0.30x=-0.6$ $x=-0.6\div-0.30$ $x=2$ so we need 2 liters of the 20% solution 50% solution = 6-x with x=2 50% solution =6-2=4 so we'll need 4 liters of the 50% solution.
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