Different Interest Rates. Mrs. Brighton invested $30,000 and received a total of $2300 in interest. If she invested part of the money at 10% and the remainder at 5%, then how much did she invest at each rate?
$30,000 Total
She invested $16,000 x .10 = $1600
She invested $14,000 x .05% = $700
$1600 +$700 = $2300
To solve this problem, we can use a system of equations. Let's denote the amount invested at 10% as "x" and the amount invested at 5% as "y".
According to the problem, the total investment amount is $30,000, so we have the equation:
x + y = 30,000 -- (Equation 1)
Now, let's calculate the interest earned at each rate. The amount invested at 10% gives an interest of 10% of x, which can be written as 0.10x. Similarly, the amount invested at 5% gives an interest of 5% of y, which can be written as 0.05y.
The problem states that the total interest earned is $2,300, so we have another equation:
0.10x + 0.05y = 2,300 -- (Equation 2)
Now, we have a system of equations:
x + y = 30,000 -- (Equation 1)
0.10x + 0.05y = 2,300 -- (Equation 2)
We can solve this system of equations to find the values of x and y, representing the amounts invested at each rate.
There are several methods to solve this system, such as substitution, elimination, or using matrices. Let's use the elimination method in this case:
Multiplying Equation 1 by 0.10 (to match the coefficients of x in both equations) gives:
0.10x + 0.10y = 3,000 -- (Equation 3)
Next, we subtract Equation 3 from Equation 2:
(0.10x + 0.05y) - (0.10x + 0.10y) = 2,300 - 3,000
0.10x - 0.10x + 0.05y - 0.10y = -700
-0.05y = -700
Dividing both sides by -0.05:
y = (-700) / (-0.05)
y = 14,000
Substituting the value of y back into Equation 1:
x + 14,000 = 30,000
x = 30,000 - 14,000
x = 16,000
Therefore, Mrs. Brighton invested $16,000 at 10% and $14,000 at 5%.