SOLUTION: HAROLD INVESTED $8000 FOR ONE YEAR, PART AT 11% SIMPLE INTREST AND PART AT 10% INTREST. IF HE EARNED A TOTAL OF $850 IN INTREST AT EACH RATE? FOR 11% I GOT 5000 FOR 10% I GOT 3

Algebra.Com
Question 1093099: HAROLD INVESTED $8000 FOR ONE YEAR, PART AT 11% SIMPLE INTREST AND PART AT 10% INTREST. IF HE EARNED A TOTAL OF $850 IN INTREST AT EACH RATE?
FOR 11% I GOT 5000
FOR 10% I GOT 3000
IS THIS Correct?

Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
x at 11%
8000-x at 10%
.11x+.10(8000-x)=850
.11x+800-.10x=850
.01x=50
x=$5000
8000-x=$3000
yes, you are correct.

RELATED QUESTIONS

a teacher invested $11,000 in two accounts. part of the money was invested at 8% annual... (answered by solver91311)
A man invested $6,000, part of it at 5% simple intrest and the rest at 7% simple intrest. (answered by math problem solving)
two investments were made totaling $15,000. For a certain year these investments yielded (answered by ptaylor)
Mitch invested $7500 for 1 year, part at 12% annual intrest and the rest at 10% annual... (answered by bam878s)
a total of $8000 is invested. part at 7% and the remainder at 15%. how much is invested... (answered by Fombitz)
two investments were made totaling 8800 dollars. for a certain year these investments... (answered by Maths68)
The Gibsons invested $7000, part at 6% and the rest at 10% simple annual intrest. If they (answered by stanbon)
Jeremy inherits $18,000 and invests part of the money in bonds with an intrest rate of... (answered by checkley71)
a college student earned $8900 during the summer. the student invested part of the money... (answered by addingup)