An amount was invested at 11% simple interest and another amount was invested at 10% simple interest. The amount invested at 11% was 140% of the amount invested at 10%. If the total interest earned af

Let $P_1$ be the amount invested at $11\%$ and $P_2$ be the amount invested at $10\%$.
The interest rate for the first account is $R_1 = 11\% = 0.11$.
The interest rate for the second account is $R_2 = 10\% = 0.10$.
The time period for both investments is $T = 3\frac{1}{2} \text{ years} = 3.5 \text{ years}$.
The total interest earned is $I_{total} = \text{GH¢ } 22,580.60$.
The formula for simple interest is $I = P \times R \times T$.

The problem states that the amount invested at $11\%$ ($P_1$) is $140\%$ of the amount invested at $10\%$ ($P_2$).

The total interest is the sum of the interest from each account:

Substitute the known values:

Calculate the products:
$1.4 \times 0.11 \times 3.5 = 0.154 \times 3.5 = 0.539$
$0.10 \times 3.5 = 0.35$
So the equation becomes:

Now, solve for $P_2$:

So, $P_2 = \text{GH¢ } 25,400$.

Using the relationship $P_1 = 1.4 P_2$:

The amount invested at $11\%$ is $\boxed{\text{GH¢ 35,560}}$.