Let \( x \) represent the amount Marie invested at 3% for 2 years. Therefore, the amount invested at 5.5% for 3 years would be \( 12000 - x \).
Step 1: Calculate the interest for each account
Interest from the first account:
\[ \text{Interest}_1 = x \times \frac{3}{100} \times 2 = 0.06x \]
Interest from the second account:
\[ \text{Interest}_2 = (12000 - x) \times \frac{5.5}{100} \times 3 = 0.165(12000 - x) \]
Step 2: Set up the equation
According to the problem, the total interest from both accounts after their terms is K850:
\[ 0.06x + 0.165(12000 - x) = 850 \]
Step 3: Simplify and solve the equation
Now, let's simplify the equation:
\[ 0.06x + 1980 - 0.165x = 850 \]
Combine the \( x \) terms:
\[ -0.105x + 1980 = 850 \]
Step 4: Isolate \( x \)
Subtract 1980 from both sides:
\[ -0.105x = 850 - 1980 \]
\[ -0.105x = -1130 \]
Then divide by -0.105:
\[ x = \frac{-1130}{-0.105} \approx 10762.86 \]
Step 5: Calculate the investment in each account
So, Marie invested approximately K10,762.86 in the first account.
Now, to find the investment in the second account:
\[ 12000 - x = 12000 - 10762.86 \approx 1237.14 \]
Step 6: Summary
Marie invested approximately:
- K10,762.86 at 3% for 2 years
- K1,237.14 at 5.5% for 3 years
To verify:
-
Interest from the first account: \[ 0.06 \times 10762.86 = 645.77 \]
-
Interest from the second account: \[ 0.165 \times 1237.14 = 204.23 \]
Total interest: \[ 645.77 + 204.23 = 850 \]
The calculations confirm that the amounts are accurate.