SOLUTION: A total of $3600 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $385, how much was invested at each rate? 6 %+ 11%=

Algebra.Com
Question 1108863: A total of $3600 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $385, how much was invested at each rate?
6 %+
11%=

Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
Let "part of it" be x. So, x was invested at 6% or 0.06x
The remainder is invested at 11%, and that amount = 3600 - x
Put it all together:
0.06x + 0.11(3600 - x) = 385
0.06x + 396 - 0.11x = 385
-0.05x = -11 divide, and remember that -/- = + so when you divide you can ignore the - sign, just divide 11/0.05
x = 220 this is how much is invested at 6%
3600 - 220 = 3380 this is the amount invested at 11%
-------------------------------------------
Check:
220 * 0.06 = 13.2
3380 * 0.11 = 371.80
13.2 + 371.80 = 385 Correct.

RELATED QUESTIONS

A total of $3700 was invested, part of it at 6% interest and the remainder at 11%. If the (answered by mananth)
A total of $3500 was invested, part of it at 6% interest and the remainder at 11%. If the (answered by richwmiller)
A total of $4500 was invested, part of it at 6% interest and the remainder at 12%. If the (answered by subudear)
A total of $1500 was invested, part of it at 10% per year and the remainder at 6% per... (answered by oscargut)
A total of $5000 was invested, part of it at 4% per year and the remainder at 6% per... (answered by lwsshak3)
A total of $3000 was invested, part of it at 4% per year and the remainder at 6% per... (answered by josgarithmetic)
a total of 3500 was invested, part of it at 8% interest and the remainder at 13%. if the... (answered by richwmiller)
A total of $4000 was invested, part of it at 8% interest and the remainder at 9%. If the... (answered by checkley77)
A total of $4300 was invested, part of it at 5% interest and the remainder at 10%. If the (answered by mananth)