Last year, Susan had 30,000 to invest. She invested some of it in an account that paid 8% simple interest per year, and she invested the rest in an account that paid 7% simple interest per year. After one year, she received a total of $2150 in interest. How much did she invest in each account?
Well, it seems like Susan had quite the financial adventure! Let me do some clown calculations for you.
Let's suppose Susan invested x dollars in the account that paid 8% interest, and the remaining (30,000 - x) dollars in the account that paid 7% interest.
Now, we need to figure out the interest generated by each investment.
For the first account, the interest can be calculated as 0.08x.
For the second account, the interest can be calculated as 0.07(30,000 - x).
Since the total interest received was $2150, we can set up the equation:
0.08x + 0.07(30,000 - x) = 2150
Now, let me simplify this equation while juggling some numbers...
0.08x + 2100 - 0.07x = 2150
0.01x = 50
x = 50 / 0.01
x = 5000
Ta-da! Susan invested $5000 in the account that paid 8% interest and the remaining $25,000 in the account that paid 7% interest.
To solve this problem, we can use a system of equations. Let's assume that Susan invested x dollars at 8% and y dollars at 7%.
First, we know that the total amount invested is $30,000, so we have the equation:
x + y = 30,000
Second, we know that the interest earned after one year is $2,150. The interest earned from the 8% account is 0.08x, and the interest earned from the 7% account is 0.07y. So, we have the equation:
0.08x + 0.07y = 2,150
Now we can solve the system of equations. Here's one way to do it:
Step 1: Solve the first equation for x:
x = 30,000 - y
Step 2: Substitute the expression for x into the second equation:
0.08(30,000 - y) + 0.07y = 2,150
Step 3: Simplify and solve for y:
2,400 - 0.08y + 0.07y = 2,150
0.01y = 2,150 - 2,400
0.01y = -250
y = -250 / 0.01
y = 25,000
Step 4: Substitute the value of y back into the first equation to solve for x:
x + 25,000 = 30,000
x = 30,000 - 25,000
x = 5,000
Therefore, Susan invested $5,000 at 8% and $25,000 at 7%.
To solve this problem, we can set up a system of equations. Let's say Susan invested x dollars in the account that paid 8% interest and y dollars in the account that paid 7% interest.
We know that the total amount Susan invested was $30,000, so the first equation is:
x + y = 30,000
Next, we can calculate the interest earned in each account using the simple interest formula (I = P * r * t), where I is the interest, P is the principal, r is the interest rate, and t is the time in years.
The interest earned in the 8% account is (x * 0.08) and the interest earned in the 7% account is (y * 0.07). Since Susan received a total of $2,150 in interest, we can write the second equation as:
0.08x + 0.07y = 2,150
Now we have a system of equations we can solve:
1) x + y = 30,000
2) 0.08x + 0.07y = 2,150
To solve this system, we can use the method of substitution or elimination.
Let's use the method of elimination to eliminate one variable. We can multiply equation 1) by 0.08 and equation 2) by 100 to make the coefficients of x the same:
0.08(x + y) = 0.08(30,000)
0.08x + 0.08y = 2,400
8x + 7y = 215,000
Now we have:
1) 8x + 7y = 215,000
2) 0.08x + 0.07y = 2,150
To eliminate the decimals, we can multiply equation 2) by 100:
8x + 7y = 215,000
8x + 7y = 215,000
Now we have two equations with the same coefficients of x, so we can subtract equation 1) from equation 2):
8x + 7y - (8x + 7y) = 215,000 - 215,000
0 = 0
Since we ended up with 0 = 0, this means the two equations are equivalent, and there are infinitely many solutions.
Therefore, there is not enough information given to determine the exact amounts Susan invested in each account.
.08 x + .07(30000-x) = 2150
.08 x + 2100 -.07 x = 2150
.01 x = 50
x = 5,000
30,000 - x = 25,000