Larry Mitchell invested part of his $27,000 advance at 5% annual simple interest and the rest at 2% annual simple interest. If his total yearly interest from both accounts was $720​, find the amount invested at each rate
Larry Mitchell invested part of his $31,000 advance at 3% annual simple interest and the rest at 2% annual simple interest. If his total yearly interest from bothaccounts was $730, find the amount invested at each rate.
Let's denote the amount invested at 5% annual interest as x (in dollars).
Since the total amount invested is $27,000, the amount invested at 2% annual interest will be 27,000 - x (in dollars).
The formula for calculating simple interest is I = P * r * t, where I is the interest, P is the principal (amount invested), r is the rate, and t is the time.
For the amount invested at 5% annual interest:
Interest obtained = P * r * t
720 = x * 0.05 * 1 (since the time is 1 year)
For the amount invested at 2% annual interest:
Interest obtained = P * r * t
720 = (27,000 - x) * 0.02 * 1 (since the time is 1 year)
Now, we have the following system of equations:
0.05x = 720
0.02(27,000 - x) = 720
Simplifying these equations, we get:
0.05x = 720
0.02 * 27,000 - 0.02x = 720
Multiplying the second equation by 100 to eliminate decimals, we get:
540 - 2x = 72
Solving for x, we have:
-2x = 72 - 540
-2x = -468
x = (-468) / (-2)
x = 234
Therefore, Larry Mitchell invested $234 at 5% annual interest and $27,000 - $234 = $26,766 at 2% annual interest.
To find the amount invested at each rate, let's assign variables to represent the unknown quantities.
Let's say Larry invested x dollars at 5% annual interest rate. Hence, the amount invested at 2% would be (27,000 - x) dollars.
Now, we can set up the equation based on the interest earned:
Interest from the first account (5%): x * 5% = 0.05x
Interest from the second account (2%): (27,000 - x) * 2% = 0.02(27,000 - x)
According to the problem, the total annual interest earned is $720:
0.05x + 0.02(27,000 - x) = 720
Now, let's solve for x:
0.05x + 0.02(27,000 - x) = 720
0.05x + 0.54 - 0.02x = 720
0.03x + 0.54 = 720
0.03x = 720 - 0.54
0.03x = 719.46
x = 719.46 / 0.03
x ≈ 23,982
Therefore, Larry invested approximately $23,982 at 5% annual interest, and the remaining amount ($27,000 - $23,982 = $3,018) was invested at 2% annual interest.