The difference between the Compound Interest and Simple Interest on a certain sum at the same rate of interest for the second year is Rs. 32 and for the third year is Rs. 66.56. Find the rate of interest (in %).
Solution:-
The difference between SI and CI in the third year 2 * The difference between SI and CI in the second year + Interest on the difference between the interests for the second year
66.56 = 2 x 32 + D, where D is the interest on the difference between the interests for the second year.
D = 2.56
2.56 is 8% of 32
I am not able to understand the bolded part in the solution given
Here is my algebraic description:
Let $B$ be the (unknown) starting balance, and let $r$ be the (unknown) interest rate.
Let $S_i$ be the balance with simple interest in year $i$, So $$S_i=B\times (1+i\times r)$$
Let $C_i$ be the balance with compound interest in year $i$. So $$C_i=B\times (1+r)^i$$
Then the interest paid to the simple account in year $i$ is $S_i-S_{i-1}$, with a similar result for the interest paid to the compound account.
The balances coincide at the end of year $1$ (both are $(1+r)B$. In year $2$, the simple account of course receives $rB$. The compound account receive $rC_1=r(1+r)B$ Thus the difference in interest over year $2$ is $$\Delta_2=r(1+r)B-rB=r^2B$$
Now we move to year $3$. The simple account receives $rB$ as always. The compound account receives $rC_2=r(1+r)^2B$. Thus $$\Delta_3=r(1+r)^2B-rB=(r+2r^2+r^3)B-rB=(2r^2+r^3)B$$
We remark that we have shown that $$\Delta_3=2\Delta_2+r\Delta_2$$ which is the bolded part of the official solution.
In any case, solving the pair of equations, $\Delta_2=32, \Delta_3=66.56$ yields $$\boxed {r=.08\quad \&\quad B=5000}$$