Mr. Wilson invested money in two accounts. His total investment was $30,000. If one account pays 3% in interest and the other pays 7% in interest, how much does he have in each account if he earned a total of $1,300 in interest in 1 year?
He invested $ in the 3% account and $in the 7% account.
Let's solve this problem step by step:
Step 1: Assign variables
Let's assign variables to represent the unknowns:
Let's say Mr. Wilson invested x dollars in the 3% account and (30000 - x) dollars in the 7% account.
Step 2: Calculate interest earned
The interest earned from the 3% account would be x * (3/100) = 0.03x dollars.
The interest earned from the 7% account would be (30000 - x) * (7/100) = 0.07(30000 - x) dollars.
Step 3: Set up an equation
The total interest earned from both accounts is $1,300. So, we can set up the equation:
0.03x + 0.07(30000 - x) = 1300.
Step 4: Solve the equation
Let's simplify and solve the equation:
0.03x + 0.07(30000 - x) = 1300
0.03x + 2100 - 0.07x = 1300
-0.04x + 2100 = 1300
-0.04x = 1300 - 2100
-0.04x = -800
x = (-800)/(-0.04)
x = 20000
Step 5: Calculate the amounts in each account
Now that we have the value of x, we can calculate the amounts in each account:
Amount in the 3% account = $20,000.
Amount in the 7% account = $30,000 - $20,000 = $10,000.
Therefore, Mr. Wilson has $20,000 in the 3% account and $10,000 in the 7% account.
t = amount at three % , s = amount at seven %
t + s = 30000 ... 3 t + 3 s = 90000
.03 t + .07 s = 1300 ... 3 t + 7 s = 130000
subtracting equations ... 4 s = 40000
find s , then substitute back to find t