John invested $100 in each of the funds A and B. After one year, the

Since there is a difference of 10$ on an amount of 100$ in first year, the difference in interest is 10%.

Now the easiest way would be to make use of options and solve backwards, but let me solve it for every one to understand the method.

Let the rate of interest on A be a, so on B it becomes a-10

So, at the end of 2 years:-
A becomes \( 100(1+\frac{a}{100})^2\)
B becomes \( 100(1+\frac{a-10}{100})^2\)

The difference in values after two years is given as 25$.

Thus,
\(100(1+\frac{a}{100})^2-100(1+\frac{a-10}{100})^2=25\)

We can use the formula \(a^2-b^2=(a-b)(a+b)\)
\(100(1+\frac{a}{100}+1+\frac{a-10}{100})(1+\frac{a}{100}-1-\frac{a-10}{100})=25\)

\(100(2+\frac{2a-10}{100})(\frac{10}{100})=25\)

\(\frac{200+2a-10}{100}*10=25\)

\(190+2a=250\)

\(2a=60\)

\(a=30\)


B