# SOLUTION: Lane invested \$10,000, part at 15% and part at 14%. If the total interest at the end of the year is \$1,430, how much did she invest at 15%?

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Finance -> SOLUTION: Lane invested \$10,000, part at 15% and part at 14%. If the total interest at the end of the year is \$1,430, how much did she invest at 15%?       Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Word Problems: Money, Business and Interest Solvers Lessons Answers archive Quiz In Depth

 Question 41616: Lane invested \$10,000, part at 15% and part at 14%. If the total interest at the end of the year is \$1,430, how much did she invest at 15%? Answer by tutorcecilia(2152)   (Show Source): You can put this solution on YOUR website!This is a type of mixture problem. The formula to remember is: (rate of A)(amount of A) + (rate of B)(amount of B) = Total Value Rate A = 15% = .15 Amount A = "how much" = x Rate B = 14% = .14 Total Value = "the total amount of interest" = \$1430 (this wording is confusing because this is really the "Total Value of interest of \$10,000) . Total Amount invested = \$10,000 =(Amount of A + Amount B) So, 10,000 - A = B or 10,000 - B = A . Putting it all together: (.15)(x) + (.14)(10,000 - x) = 1430 Simplifying: .15x + (1400 -.14x) = 1430 .15x + 1400 - .14x = 1430 .01x + 1400 - 1400 = 1430 - 1400 .01x = .30 .01x/.01 = 30/.01 x = 3000 . Plugging x=3000 back into the original equation: (.15)(x) + (.14)(10,000 - x) = 1430 (.15)(3000) + (.14)(10,000 - 3000) = 1430 450 + 980 = 1430 1430 = 1430