( i )  Let the deposit per month  = Rs. P

Number of months = ( n ) = 36

Rate of interest ( r ) = 7.5 % p.a.

( ii ) Maturity value = P x n + S.I. 

                            = Rs. ( 2000 x 36 + 8325 )

                            = Rs. 80,325. 

( i )  P = Rs. 5000,   T = 1 year,   A = Rs. 5325

   I = A - P

$$\begin{gathered} \Rightarrow \mathrm{I} = 5325 - 5000 \\ \Rightarrow \mathrm{I} = 325 \end{gathered}$$

So, the interest at the end of first year is Rs. 325

$$\begin{gathered} \mathrm{I} = \frac{\mathrm{PRT}}{100} \\ \Rightarrow \mathrm{R} = \frac{\mathrm{I} \mathrm{X} 100}{\mathrm{P} \mathrm{X} \mathrm{T}} \\ \Rightarrow \mathrm{R} = \frac{325 \mathrm{X} 100}{5000 \mathrm{X} 1} \\ \Rightarrow \mathrm{R} = \frac{32500 }{5000 } = 6.5 \% \end{gathered}$$

So, the rate of interest at the end of the first year is  6.5 %.

( ii ) The amount at the end of the first year will be the principal for the second year.

P = Rs. 5325,  T = 1 year,   R = 6.5 %

$$\begin{gathered} \mathrm{I} = \frac{\mathrm{PRT}}{100} \\ \Rightarrow \mathrm{I} = \frac{5325 \mathrm{X} 6.5 \mathrm{X} 1}{100} \\ \Rightarrow \mathrm{I} = 346.125 \\ \mathrm{A} = \mathrm{P} + \mathrm{I} \\ \Rightarrow \mathrm{A} = 5325 + 346.125 \\ \Rightarrow \mathrm{A} = 5671.125 \\ \Rightarrow \mathrm{A} \approx \mathrm{RS}. 5671 \end{gathered}$$

So, the amount at the end of the second year is  Rs. 5671.

For  1^st  year :

P = Rs. 50,000;    R = 12 %   and    T = 1 year

$$\begin{gathered} \therefore \mathrm{Interest} = \mathrm{Rs}. \frac{50,000 \mathrm{x} 12 \mathrm{x}1}{100} \\ = \mathrm{Rs}. 6,000 \end{gathered}$$

And,  Amount  =  Rs. 50,000 + Rs. 6,000 = Rs. 56,000

Since  Money repaid  =  Rs. 33,000 

$\therefore$ Balance =  Rs.  56,000 - Rs.  33,000

                 = Rs.  23,000

For  2^nd  year :

P = Rs.  23,000;    R = 15 %     and    T = 1 year

$$\begin{gathered} \therefore \mathrm{Interest} = \mathrm{Rs}. \frac{23,000 \mathrm{x} 15 \mathrm{x} 1}{100} \\ = \mathrm{Rs}. 3,450 \end{gathered}$$

And,  Amount  =  Rs. 23,000 +  Rs. 3,450

                      = Rs. 26,450

Thus, Jaya must pay  Rs. 26,450  at the end of  2^nd  year to clear her debt.  

Principal for the month of Jan  =  Rs.  5600

Principal for the month of Feb  =  Rs.  4100

Principal for the month of Mar  =  Rs.  4100

Principal for the month of Apr  =  Rs.  2000

Principal for the month of May  =  Rs.  8500

Principal for the month of June =  Rs.  10000

Total Principal for one month    =  Rs.  34300

Rate of interest = 6 % pa

$$\begin{gathered} ( \mathrm{i} ) \mathrm{Simple} \mathrm{interest} = \frac{\mathrm{PRT}}{100} \\ = \frac{34300 \mathrm{X} 6 \mathrm{X} 1}{100 \mathrm{X} 12} \\ = \mathrm{Rs}. 171.50 \end{gathered}$$

( ii ) Total amount = Rs. 10000 + Rs. 171.50

                           = Rs. 10171.50

Given:  P = Rs. 1000,   r = 10 %   and   I = Rs. 5550

$$\begin{gathered} \mathrm{I} = \mathrm{P} \times \frac{\mathrm{n} \times \left( \mathrm{n} + 1 \right)}{2 \times 12} \times \frac{\mathrm{r}}{100} \\ \Rightarrow 5550 = 1000 \times \frac{\mathrm{n} \times \left( \mathrm{n} + 1 \right)}{24} \times \frac{10}{100} \\ \Rightarrow 1332 = \mathrm{n} \times \left( \mathrm{n} + 1 \right) \\ \Rightarrow {\mathrm{n}}^2 + \mathrm{n} - 1332 = 0 \\ \Rightarrow {\mathrm{n}}^2 + 37 \mathrm{n} - 36 \mathrm{n} - 1332 = 0 \\ \Rightarrow \mathrm{n} \left( \mathrm{n} + 37 \right) - 36 \left( \mathrm{n} + 37 \right) = 0 \\ \Rightarrow \left( \mathrm{n} + 37 \right) \left( \mathrm{n} - 36 \right) = 0 \\ \Rightarrow \mathrm{n} = - 37 \mathrm{or} \mathrm{n} = 36 \end{gathered}$$   

Since number of months cannot be negative, we reject  n = - 37

$\Rightarrow$ n = 36

Thus, total time is  36  months.