SI = (P × R × t)/100  

$(A\; = \;P\;{\left( {\;1\; + \frac{R}{{100\;}}\;} \right)^t})$ ; CI = A - P

Where,

CI = Compound interest

SI = Simple interest

A = Amount on compound interest

P = Principal

R = rate %

t = time in years

Given, rate is same for calculating both SI and CI

Let the rate be ‘R’.

SI is calculated for 1 year

SI = (P × R × 1)/100 = PR/100

Given, SI = Rs. 260

⇒ PR/100 = 260 -------------eq (1)

CI is calculated for 2 year

$(\therefore A\; = \;P\;{\left( {\;1\; + \frac{R}{{100\;}}\;} \right)^2}\;)$

As, CI = A – P

$(\Rightarrow C.I = \;P\;{\left( {\;1\; + \frac{R}{{100\;}}\;} \right)^2} - P)$

$(\Rightarrow C.I = P\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right))$

Given, CI = Rs. 540.80

$(\therefore P\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right) = 540.80)$ -----------eq (2)

Dividing eq 2 by eq 1

$(\frac{{P\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right)}}{{\frac{{PR}}{{100}}}} = \frac{{540.80}}{{260}})$

$(\Rightarrow \frac{{\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right)}}{{\frac{R}{{100}}}} = 2.08)$

Using a² – b² = (a + b)(a – b)

$(\Rightarrow \frac{{\left( {1 + \frac{R}{{100}} + 1} \right)\left( {1 + \frac{R}{{100}} - 1} \right)}}{{\frac{R}{{100}}}} = 2.08)$

⇒ 2 + R/100 = 2.08