Job Roles

Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the numberof years for which it is invested, r is the rate of interest per annum

In this case, Rs. 1250 has become Rs.10,000.

Therefore, the interest earned = 10,000 – 1250 = 8750.

8750 = [(1250 x n x 12.5)/100]

=> n = 700 / 12.5 = 56 years.

From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of  interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.

Let sum = P and original rate = R. Then

[(P * (R+2) * 3)/100] - [ (P * R * 3)/100] = 360

3P*(R+2) - 3PR = 36000

3PR + 6P - 3PR = 36000

6P = 36000

P = 6000

Let principal = P, Then, S.I.= P and Time = 8 years

We know that S.I. = PTR/100

Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.

 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

(24 + 31 + 18) days = 73 days = 1/5 year .

P = Rs 3000 and R = 18 % p.a.

 S.I=3000*18*15100

P = Rs. 6250, R = 14 %  & T = (146/365) years = 2/5 years .

 S.I=Rs.6250*14*25100=Rs.350

P = 68000, R =503%  & T = 9 months (3/4 years)  

 S.I=p×r×t100=(6800×503×34)100=Rs.8500