Larry Mitchel invested part of his $32,000 advance at 7% annual simple interest and the rest at 6% annual simple interest. If this total yearly interest from both accounts was $2,050, find the amount invested at each.

Question

Larry Mitchel invested part of his $32,000 advance at 7% annual simple interest and the rest at 6% annual simple interest. If this total yearly interest from both accounts was $2,050, find the amount invested at each.

The amount invested at 7% is....
The amount invested at 6% is....

Thank You for the Help

Answer 1

Investment @ 7% = Po

Investment @ 6% = 32,000-Po

Po*r*t + (32000-Po)*r*t = 2,050
Po*0.07*1 + (32000-Po)*0.06*1 = 2050
0.07Po + 1920 - 0.06Po = 2050
0.01Po = 2050-1920 = 130
Po = $13000 @ 7%.

32,000-13000 = $19,000 @ 6%.

Answer 2

To solve this problem, we can use a system of equations. Let's assign variables to the unknowns:

Let x represent the amount invested at 7% annual simple interest.
Let y represent the amount invested at 6% annual simple interest.

According to the problem, the total amount of money invested is $32,000, so we can set up the first equation:

x + y = 32,000 .......... Equation (1)

Next, we know that the total yearly interest from both accounts is $2,050. The interest earned from an investment can be calculated using the formula:

Interest = Principal * Rate * Time

For the investment at 7%, the interest earned would be:

x * 0.07

For the investment at 6%, the interest earned would be:

y * 0.06

Since the total yearly interest is $2,050, we can set up the second equation:

0.07x + 0.06y = 2,050 .......... Equation (2)

Now, we have a system of two equations with two unknowns. We can solve this system using various methods such as substitution or elimination.

Let's solve it using the elimination method. We'll multiply equation (2) by 100 to eliminate the decimals:

7x + 6y = 205,000 .......... Equation (3)

Now, we can solve the system of equations (1) and (3) simultaneously. Subtract equation (1) from equation (3):

(7x + 6y) - (x + y) = 205,000 - 32,000

Simplifying the equation:

6x + 5y = 173,000 .......... Equation (4)

From equations (1) and (4), we have a new system of equations:

6x + 5y = 173,000 .......... Equation (4)
x + y = 32,000 .......... Equation (1)

We can solve this system of equations to find the values of x and y.

Multiplying equation (1) by 6:

6(x + y) = 6(32,000)
6x + 6y = 192,000

Now, subtract equation (4) from this new equation:

(6x + 6y) - (6x + 5y) = 192,000 - 173,000

Simplifying the equation:

6y - 5y = 19,000
y = 19,000

Now, substitute the value of y back into equation (1):

x + 19,000 = 32,000
x = 32,000 - 19,000
x = 13,000

Therefore, the amount invested at 7% is $13,000, and the amount invested at 6% is $19,000.

Answer 3

Let's solve this step-by-step:

Step 1: Assign variables
Let x represent the amount invested at 7% interest.
Since the total amount invested is $32,000, the amount invested at 6% interest can be represented as (32,000 - x).

Step 2: Calculate the interest for each investment
The interest earned from the investment at 7% interest is 0.07x.
The interest earned from the investment at 6% interest is 0.06(32,000 - x).

Step 3: Set up and solve the equation
The total interest from both investments is $2,050.
So we can write the equation:
0.07x + 0.06(32,000 - x) = 2,050.

Step 4: Solve the equation
Let's simplify and solve the equation for x:

0.07x + 0.06(32,000 - x) = 2,050
0.07x + 1,920 - 0.06x = 2,050
0.01x + 1,920 = 2,050
0.01x = 2,050 - 1,920
0.01x = 130
x = 130 / 0.01
x = 13,000

Step 5: Calculate the amounts invested at each interest rate
The amount invested at 7% interest is $13,000.
The amount invested at 6% interest is $32,000 - $13,000 = $19,000.

So, Larry Mitchel invested $13,000 at 7% interest and $19,000 at 6% interest.