I was completely confused by that same explanation for a while. Don't even think of it like that. Think of it like this.
Definition: The effective rate of interest during the $n$th time period is
$$i_n = \frac{A(n) - A(n-1)}{A(n-1)}$$
Definition: The effective rate of discount during the $n$th time period is
$$d_n = \frac{A(n) - A(n-1)}{A(n)}$$
where $A(n)$ is the amount function (as defined in Kellison), the amount of money you have at time $n$. So, all you really need to understand here is that the rate of interest is a rate based on what you start with during the period. The rate of discount is a rate based on what you end up with. It's just two ways of looking at the same situation. There aren't a whole lot of real world situations where you borrow a bunch of money and immediately give some of it back. You would just borrow less.
By the way, that formula is all you need to calculate the effective rate of discount during period $n$ no matter what your $A(n)$ function is. So, in particular, it would work for your specific question of simple discount.
Question: Given a rate of 10% simple discount, calculate the effective rate of discount during period 5.
Answer: If we have 10% simple discount, then we know our accumulation function is $a(t) = \frac{1}{1 - 0.1t}$ for $0 \leq t < \frac{1}{d} = 10$. This is basically the definition of simple discount. If you have simple discount, this is your accumulation function. Memorize that. Then use it.
Therefore
$$d_5 = \frac{a(5) - a(4)}{a(5)} = \frac{2-10/6}{2} = \frac{1}{6} = 16.666666... \%$$
If you wanted to calculate the effective rate of interest when you are given the effective rate of simple discount, you can do that too. For example, in this same example,
$$i_5 = \frac{a(5) - a(4)}{a(4)} = \frac{2-10/6}{10/6} = \frac{1}{5} = 20 \%$$
Nothing changed. We're just looking at the same problem differently. In the discount case, how much money did we earn that period relative to how much we had at the end? In the interest case, how much money did we earn that period relative to how much we had at the beginning.
I do not think that conceptually there is a shorter way. I would probably let $Q_A$ be the amount in fund A after $10$ years, and similarly for $Q_B$. Then we want $Q_A+Q_B$.
After $10$ years we have $10000$, so
$$Q_A(1.03)^{10}+Q_B(1.025)^{10}=10000.\tag{$1$}$$
After $21$ years, fund A is twice as big as fund B, so
$$Q_A(1.03)^{21}=2Q_B(1.025)^{21}.\tag{2}$$
These are your equations, simplified a bit.
Now instead of keeping everything in symbols, I think it is time to compute say $Q_B$. Using Equation $(2)$, we find that $Q_A(1.03)^{10}=\dfrac{2Q_B(1.025)^{21}}{(1.03)^{11}}$.
Shove this into the calculator. Now we can use Equation $`$ to find $Q_B$, and then $Q_A$.
Best Answer
Answered by @mathguy that both are correct.