mary had 25000 dollars to invest. she invested part of that amount at 3% annual interest and part at 5% annual interest for one year. the amount of interest she earned for both investments was 1100. how much was invested at each rate
To solve this problem, let's assume that Mary invested x dollars at 3% annual interest and the remaining amount (25000 - x) dollars at 5% annual interest. We will set up an equation based on the given information and solve for x.
The interest earned from the 3% investment can be calculated as (x * 0.03) since 3% is equivalent to 0.03 in decimal form.
Similarly, the interest earned from the 5% investment can be calculated as ((25000 - x) * 0.05) since 5% is equivalent to 0.05 in decimal form.
According to the problem, the total interest earned from both investments is $1100. So, we can write the equation as:
(x * 0.03) + ((25000 - x) * 0.05) = 1100
Now, let's solve the equation for x to find the amount invested at 3% interest:
0.03x + 0.05(25000 - x) = 1100
0.03x + 1250 - 0.05x = 1100
0.03x - 0.05x = 1100 - 1250
-0.02x = -150
x = -150 / -0.02
x = 7500
According to the calculation, Mary invested $7500 at 3% annual interest. Therefore, she invested the remaining amount of (25000 - 7500) = $17500 at 5% annual interest.
If x was invested at 3%, then the rest (25000-x) was invested at 5%. So, just add up the interest:
.03x + .05(25000-x) = 1100
Continued from steeve...
0.03x + 1250 - 0.05x = 1500.
0.03x - 0.05x = 1500 - 1250
-0.02x = 250
Divide both sides by 0.02.
X = -12500