Given:
- Two equal sums of money were lent at simple interest.
- The rate of interest is 11% per annum.
- The time periods are 3½ years and 4½ years respectively.

To Find:
The difference in interest on a certain sum is 0.125 times the principal at 10% per annum.

Solution:

Step 1: Understand the Problem
- We have two equal sums of money.
- The interest is calculated at a rate of 11% per annum.
- The time periods are 3½ years and 4½ years respectively.
- We need to find the difference in interest on a certain sum, which is 0.125 times the principal at 10% per annum.

Step 2: Apply the Simple Interest Formula
The formula for simple interest is:
\[I = P \cdot R \cdot T\]
where:
- I is the interest
- P is the principal (the initial sum of money)
- R is the rate of interest per annum
- T is the time period in years

Step 3: Calculate the Interest for the Given Time Periods
Let's assume the principal amount is P.

For the first time period (3½ years):
\[I_1 = P \cdot 0.11 \cdot 3.5\]

For the second time period (4½ years):
\[I_2 = P \cdot 0.11 \cdot 4.5\]

Step 4: Calculate the Difference in Interest
The difference in interest is given as 0.125 times the principal at 10% per annum.
\[0.125 \cdot P \cdot 0.1 = 0.0125 \cdot P\]

Step 5: Set up the Equation
According to the given problem, the difference in interest is equal to 0.0125 times the principal at 10% per annum.
\[I_2 - I_1 = 0.0125 \cdot P\]

Step 6: Solve the Equation
Substituting the values of \(I_1\) and \(I_2\) from Step 3 into the equation:
\[P \cdot 0.11 \cdot 4.5 - P \cdot 0.11 \cdot 3.5 = 0.0125 \cdot P\]

Simplifying the equation:
\[0.11 \cdot P \cdot (4.5 - 3.5) = 0.0125 \cdot P\]

\[0.11 \cdot P \cdot 1 = 0.0125 \cdot P\]

Cancelling out the P's:
\[0.11 \cdot 1 = 0.0125\]

\[0.11 = 0.0125\]

The equation is not satisfied, which means that there is no solution to this problem.