a mother wants to invest 7000 for her son's future education.she invest in two account: a money market fund that pays 4% and a certificate of deposit that pays 7%. if the total interest earned after one year is $420, how much money was invested in each account?
this is just a typical mixture problem. If $x is at 4%, then the rest (7000-x) is at 7%. So, adding up the interest, you have
.04x + .07(7000-x) = 420
23333
To determine how much money was invested in each account, we can use a system of equations.
Let's assume the amount invested in the money market fund is 'x' and the amount invested in the certificate of deposit is 'y'.
According to the problem:
- The total amount invested is $7000, so we have the equation: x + y = 7000.
- The interest earned from the money market fund will be: 4% of x, which is 0.04x.
- The interest earned from the certificate of deposit will be: 7% of y, which is 0.07y.
From the given information, the total interest earned after one year is $420, so we have the equation: 0.04x + 0.07y = 420.
We now have a system of equations:
Equation 1: x + y = 7000
Equation 2: 0.04x + 0.07y = 420
To solve this system, we can use the substitution or elimination method.
Let's solve the system using the substitution method:
1. Solve Equation 1 for x: x = 7000 - y.
2. Substitute this value of x into Equation 2: 0.04(7000 - y) + 0.07y = 420.
3. Simplify the equation: 280 - 0.04y + 0.07y = 420.
4. Combine like terms: 0.03y = 140.
5. Divide both sides by 0.03: y = 140 / 0.03
6. Calculate y: y ≈ $4666.67.
Now that we know the value of y, we can substitute it back into Equation 1 to find x:
x = 7000 - y
x = 7000 - 4666.67
x ≈ $2333.33
Therefore, approximately $2333.33 was invested in the money market fund, and approximately $4666.67 was invested in the certificate of deposit.