SOLUTION: Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1083955: Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much did he invest at each rate?
Found 2 solutions by Fombitz, MathTherapy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let A be the amount at 6%, B the amount at 8%.
1.A%2BB=2100
.
.
.
0.08%2AB=0.06%2AA%2B28
%284%2F3%29B=A%2B1400%2F3
2.A=%284%2F3%29B-1400%2F3
Substituting from 2 into 1,
%284%2F3%29B-1400%2F3%2BB=2100
%287%2F3%29B=%286300%2B1400%29%2F3
%287%2F3%29B=7700%2F3
Solve for B, then use either equation to solve for A.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much did he invest at each rate?
Let amount invested in the 8% fund be E

Then amount invested in the 6% fund = 2,100 - E

We then get the following INTEREST equation: .08E = 2(.06)(2,100 - E) + 28

Solve this for E, the amount invested in the 8% fund. You should get $1,400.

Subtract value of E from $2,100 to get amount invested in 6% fund.