SOLUTION: Mr. Lead want to mix pencils costing $0.15 each with pencils costing $0.35 each. The mixture will cost $3.45. The number of $0.35 pencils is 7 more then the number of $0.15 pencils

Algebra ->  Customizable Word Problem Solvers  -> Evaluation -> SOLUTION: Mr. Lead want to mix pencils costing $0.15 each with pencils costing $0.35 each. The mixture will cost $3.45. The number of $0.35 pencils is 7 more then the number of $0.15 pencils      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 25801: Mr. Lead want to mix pencils costing $0.15 each with pencils costing $0.35 each. The mixture will cost $3.45. The number of $0.35 pencils is 7 more then the number of $0.15 pencils. How many of each type of pencil should be included in the package??

Answer by elima(1433) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=pencils that cost .15
so; x+7=pencils that cost .35
.15x+.35(x+7)=3.45
.15x+.35x+2.45=3.45
.50x+2.45=3.45
.50x=1
x=2
So there were 2 pencils at .15;
and there were 9 pencils at .35.
Hope you understand
=)