If compound interest on a certain sum for 3 years at 11\(\frac{1}{9}\)% p.a. is Rs. 2710, then find at the same rate of interest how much the compound interest of three years is greater than the simple interest of two years?

Rs. 1090

11\(\frac{1}{9}\)% = \(\frac{1}{9}\)

For 1 year: Ratio b/w Amount & Principal = \(\frac{10}{9}\)

For 3 year: Ratio b/w Amount & Principal = (\(\frac{10}{9}\))³

Here,

\(\frac{Amount}{Principal}\) = \(\frac{1000}{729}\)

Now, C.I.= Amount - Principal = 1000 - 729 = 271

ATQ,

271R = 2710

1R = 10

So, Principal = 7290

S.I. of two years @11\(\frac{1}{9}\)% = 7290 × 11\(\frac{1}{9}\)% × 2

                                                         = 7290 × \(\frac{1}{9}\) × 2 = 1620

Difference in C.I. for 3years and S.I. for 2years = 2710 - 1620 = 1090