Question

48 number question.please solve it..

Answer 1

Hello!

Let the rate of interest be $\textbf{R\: \%}$ per annum

$\boxed{\sf AMOUNT \:TO \:BE\: PAID}$ ( if paid at the end of 11 months ).

$\sf 10 + \dfrac{\sf 10\:x\:R\:x\: \dfrac{11}{12}}{\sf 100} = \boxed{\sf 10 + \dfrac{\sf 11R}{\sf 120}}$

Then,

$\boxed{\sf TOTAL\:EFFECTIVE \:PAYMENT}$

( Rs. 1 + for 10 months ) + ... + ( Rs. 1 + for 1 month ) + Rs. 1

⇒ $\sf 1 + \dfrac{\sf 10R}{\sf 1200} + ... + \sf 1 + \dfrac{\sf R}{\sf 1200} + 1$

⇒ $\sf 11 + \dfrac{\sf 1 + 2 + ... + 10}{\sf 1200}$ ⇒ $\boxed{\sf 11 + \dfrac{\sf 11R}{\sf 240}}$

$\textbf{Now}$ :

$\sf 10 + \dfrac{\sf 11R}{\sf 120} = \sf 11 + \dfrac{\sf 11R}{\sf 240}$

⇒ $\dfrac{\sf 11R}{\sf 240} = \sf 1$

⇒ $\sf R = \dfrac{\sf 240}{\sf 11} = \boxed{\sf 21 \dfrac{\sf 9}{\sf 11}\%}$

Hence,
Option $\textbf{D.}$ : $\sf 21 \dfrac{\sf 9}{\sf 11}\%$ is Correct.

Cheers