Question

Mary had $25,000 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?

Answer 1

Mary invested $7,500 at a 3% interest rate and $17,500 at a 5% interest rate. We found these amounts by setting up a system of linear equations based on the total investment and total interest earned, and solving the system for both x and y, which represent the amounts invested at each interest rate.

Step-by-step explanation:

To solve the problem of how much Mary invested at each rate, we can use a system of linear equations. Let's define x as the amount invested at 3% interest and y as the amount invested at 5% interest. Since the total investment is $25,000, our first equation is x + y = 25,000. The second equation comes from the total interest earned, which is $1,100. The interest from each part of the investment is 0.03x + 0.05y = 1,100. To solve the system, we can use substitution or elimination.

Step 1: Rewrite the first equation in terms of x:

x = 25,000 - y

Step 2: Replace x in the second equation with 25,000 - y:

0.03(25,000 - y) + 0.05y = 1,100

Step 3: Simplify and solve for y:

750 - 0.03y + 0.05y = 1,100

0.02y = 1,100 - 750

0.02y = 350

y = 350 / 0.02

y = 17,500

Step 4: Use the value of y to find x:

x = 25,000 - 17,500

x = 7,500

Mary invested $7,500 at 3% interest rate and $17,500 at 5% interest rate.

Answer 2

(25000-f)(.03)+.05f=1100

750-.03f+.05f=1100

750+.02f=1100

.02f=350

f=$17500

So Mary invested $17500 @ 5% and 7500 at 3% to earn $1100 in interest.