Math Problem?
I have no idea how to set up/solve this math problem. please help! Thanks!
A retired couple invested part of $11,800 at 6% interest and the rest at 7.5%. If their annual income from these investments is $786, how much was invested at each rate?
Comments
Lets consider the couple invested x amount at 6%
so balance (11800-x) will be at 7.5%
Interest on 6% investment = 0.06x
Interest on 7.5% investment = (11800-x)0.075
Total interest = 0.06x + (11800-x)0.075 = 786
0.06x + 885 - 0.075x = 786
0.015x = 99
x = 6600
Investment at 6% is for $6,600
and investment at 7.5% for (11800-6600) = $5,200
Let the amount invested at 7.5% be $ 100X
THEN, the amount invested at 6% WILL BE $ ( 11800 - 100X )
Interest earned = 7.5X + 0.06 ( 11800 - 100X ) = 786
7.5X + 708 - 6X = 786
1.5X = 78
X = 52
ANSWER $5200 at 7.5% + $6600 at 6%
Ash solution is the right answer.