Marie invested K12000 in two different accounts at simple interest. The first one at 3% for 2 years and the second one for 3 years at 5.5%. Both accounts earned K850 as their accumulated interest after their respective terms has matured.

Question

How much did Marie invested in each account?

Answer 1

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.