Question

Mary invested a certain amount of money at a 10% interest rate and $2000 more than that at a 12% interest rate. Her total yearly interest earned is $1340. How much money did she invest at each rate?

a) $5000 at 10%, $7000 at 12%
b) $6000 at 10%, $8000 at 12%
c) $7000 at 10%, $9000 at 12%
d) $8000 at 10%, $10000 at 12%

Answer 1

Mary invested $5000 at a 10% interest rate and $7000 at a 12% interest rate. We found this by solving the system of equations based on the total interest earned, which was $1340.

Step-by-step explanation:

The student is asking how much money Mary invested at two different interest rates given that she earned a total of $1340 in yearly interest. To solve this, we use the system of equations:

Let x be the amount invested at 10% and x + $2000 be the amount at 12%. The total interest earned from these investments is $1340, which can be expressed as:

• 0.10x + 0.12(x + $2000) = $1340

By solving the system of equations, we will be able to find the values of x and x + $2000, determining the amount invested at each rate.

1. 0.10x + 0.12x + $240 = $1340
2. Combine like terms: 0.10x + 0.12x = 0.22x
3. Subtract $240 from both sides: 0.22x = $1100
4. Divide both sides by 0.22 to find x: x = $5000
5. To find the amount at 12%, add $2000 to x: $5000 + $2000 = $7000

Therefore, Mary invested $5000 at 10% and $7000 at 12%.

Answer: a) $5000 at 10%, $7000 at 12%