Math Word Problems Demystified - A Self Teaching Guide

The equation is

Alloy 1 þ Alloy 2 ¼ Alloy 3

70 %xþ 50 %ð 12 
xÞ¼ 65 %ð 12 Þ

IMPLEMENTATION: Solve the equation:

70 %xþ 50 %ð 12 
xÞ¼ 65 %ð 12 Þ

0 : 70 xþ 0 : 50 ð 12 
xÞ¼ 0 : 65 ð 12 Þ

0 : 70 xþ 6 
 0 : 50 x¼ 7 : 8

0 : 20 xþ 6 
 6 ¼ 7 : 8 
 6

0 : 2 x¼ 1 : 8

0 : 21 x

0 : 21

¼

1 : 8

0 : 2

x¼9 ounces of Alloy 1

12 
x¼ 12 
 9 ¼3 ounces of Alloy 2

Hence 9 ounces of the 70% alloy should be mixed with 3 ounces of the

50% alloy to get 12 ounces of an alloy which is 65% silver.

EVALUATION: Check the equation:

70 %xþ 50 %ð 12 
xÞ¼ 65 %ð 12 Þ

70 %ð 9 Þþ 50 %ð 3 Þ¼ 65 %ð 12 Þ

6 : 3 þ 1 : 5 ¼ 7 : 8

7 : 8 ¼ 7 : 8

Another type of percent mixture requires that a mixture bedilutedto make

aweakerconcentration of the mixture. If the amounts are the same, then a

portion of the higher concentration mixture must be removed, and the same

amount of the weaker mixture must be added. In this case, the equation

would look like this:

Mixture 1
Amount to beþAmount of weaker mixture¼Mixture 2

removed to be added

150 LESSON 14 Solving Mixture Problems