Correct Answer: $21 \frac{9}{11} \%$

Solution : Let the rate of interest be $r\%$ per annum.
Amount to be paid = $10 + \frac{10 \times r \times\frac{11}{12}}{100}= 10 + \frac{11r}{120}$
Total effective payment = (Re. 1 + interest on Re. 1 for 10 months) + (Re. 1 + interest on Re. 1 for 9 months) + ... +(Re. 1 + interest on Re. 1 for 1 months) + Re. 1
= $(1 + \frac{1 × r × \frac{10}{12}}{100}) + (1 + \frac{1 × r × \frac{9}{12}}{100}) + .... + (1 + \frac{1 × r × \frac{1}{12}}{100})+ 1$
= $(1 + \frac{10r}{1200}) + (1 + \frac{9r}{1200}) + .... + (1 + \frac{r}{1200}) + 1$
= $11 + \frac{(1 + 2 + .... + 10)r}{1200}$
= $11 + \frac{(10 × \frac{11}{2})r}{1200}$
= $11 + \frac{11r}{240}$
Now we have,
$10 + \frac{11r}{120} = 11 + \frac{11r}{240}$
⇒ $\frac{11r}{240} = 1$
⇒ $r = \frac{240}{11} = 21\frac{9}{11}\%$
Hence, the correct answer is $21\frac{9}{11}\%$.