Question 159796
Let X be the amount of 60% pure alloy and Y be the amount of 70% pure alloy.
First, the total weight is
1.{{{X+Y=100}}}
Next, the percentage of alloy is,
2.{{{0.60*Y+0.70*Y=0.66*(100)}}}
Let's use eq. 1 to solve for X in terms of Y,
1.{{{X+Y=100}}}
{{{X=100-Y}}}
Now substitute that into eq. 2 and solve for y.
2.{{{0.60*X+0.70*Y=0.66*(100)}}}
{{{0.60*(100-Y)+0.70*Y=0.66*(100)}}}
{{{60-0.60Y+0.70Y=66}}}
{{{0.10Y=6}}}
{{{Y=60}}}
From eq. 1,
{{{X=100-Y}}}
{{{X=40}}}
Let's check the answer using eq. 2,
2.{{{0.60*X+0.70*Y=0.66*(100)}}}
{{{0.60*(40)+0.70*(60)=66}}}
{{{24+42=66}}}
{{{66=66}}}
True statement.
Good answer.
Use 40 ounces of 60% pure alloy and 60 ounces of 70% pure alloy.