Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

  P × T × ( R + 2 ) 100 - P × T × R 100 = 108 ? 2 P T = 10800 . . . . . . ( 1 )

 And, 

  P × R × ( T + 2 ) 100 - P × R × T 100 = 180 ? 2 P R = 18000 . . . . . . ( 2 )

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

Let the Time be 'N' and Rate be 'R' 

J = (K x NR)/100 K = (L x NR)/100 

J/K = NR/100 K/L = NR/100 

J/K = K/L 

K x K = JL

 Let sum be Rs.x then, S.I. =Rs. [ x * (27/2) * 4 * (1/100) ] = Rs. 27x/50 

Amount =Rs [ x + (27x/50)] = Rs.77x/50 

=> 77X/50 = 2502.50  

=>X= (2502.50 * 50)/77 = 1625 

Hence Sum = Rs.1625

Cash price = $21 000

Deposit = 10% × $21 000 = $2100 

Loan amount = $21000 ? $2100 = $18900

I=p x r x t/100 

I=11340

Total amount = 18900 + 11340 = $30240

Regular payment = total amount /number of payments

We know, 

S.I = PTR/100 where P = principal amount, T = time, R = rate of interest

Here in the given data,

Interest for two years S.I = 924 - 812 = Rs. 112

Now, Principal amount P = 812 - 112 = Rs. 700 

Now,

R = S.I x 100/PT

R = 112 x 100/700 x 2

R = 11200/1400

R = 8%

Hence, the rate of interest R = 8%.

Let the two different rates of interests be r1 and r2 respectively.

From the given data,

$$\begin{gathered} 2600 x 4 x r 1 - r 2 100 = 402 . 80 \\ r 1 - r 2 = 402 . 80 104 = 3.87 \end{gathered}$$

Hence, the difference in the interest rates = 3.87

Given,
S.I = 3 Principal Amount
=> 3A = A x 16 x R/100
By solving, we get
=> R = 18.75%

I = (600x11x5)/100 = 330 Paise = Rs. 3.30

P = Rs. 900, R = 3 ½ % = 7/2 %, T = 2 years. 

Therefore, 

S.I. = PTR/100

S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63

Now, P = Rs. 200, S.I. = Rs. 63, R = 5 %

Time = ((100 x 63) / (200 x 5) ) years = 6.3 years.