Eva invested a certain amount of money at 9% interest and $1600 more than that amount at 11%. Her total yearly interest was $760. How much did she invest at each rate?
Let x = amount at 9% and y = amount at 11%
y = x + 1600
.09x + .11y = 760
Substitute x + 1600 for y in the second equation and solve for y. Put that value in first equation to find x. To check, put both values in second equation.
2920
To solve this problem, we can set up a system of equations to represent the given information.
Let's assume that Eva invested x dollars at 9% interest. Then, she invested (x + $1600) at 11% interest.
The interest earned from the first investment can be calculated as 0.09x, and the interest earned from the second investment is 0.11(x + $1600).
According to the problem, the total yearly interest earned was $760. So, we can set up the equation: 0.09x + 0.11(x + $1600) = $760.
Now, let's solve this equation step-by-step:
0.09x + 0.11(x + $1600) = $760
0.09x + 0.11x + 0.11($1600) = $760
0.09x + 0.11x + $176 = $760
Combining like terms:
0.20x + $176 = $760
Next, subtract $176 from both sides of the equation:
0.20x = $760 - $176
0.20x = $584
Now, divide both sides of the equation by 0.20 to isolate x:
x = $584 / 0.20
x = $2920
Therefore, Eva invested $2920 at 9% interest. To find the amount she invested at 11%, we can substitute this value back into our assumption:
Amount invested at 11% = $2920 + $1600 = $4520.
So, Eva invested $2920 at 9% interest and $4520 at 11% interest.