Let the two different rates of interests be r1 and r2 respectively.
From the given data,
Hence, the difference in the interest rates = 3.87
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%.
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
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
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 the sum be Rs. p, rate be R% p.a. and time be T years.
Then,
And,
Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.
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.
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