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General Concepts :

 Principal or Sum : The money borrowed or lent out for a certain period is called the principal or the sum.

Interest: Extra money paid for using other's money is called interest.

Simple Interest: If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

FORMULAE: Let Principal = P, Rate = R% per annum

And Time = T years. Then,

(l) S.I. = [P x R x T/100]

(ii) P = [(100 x S.I.) / (R x T)],    R= [(100 x S.I.)/ (P x T)]   and  T= [(100 x S.I)./ (P x R)]

                            Solved Problems

Ex.l Find:

(i)                  S.l. on Rs 68000 at 16 ^2/3  % per annum for  9 months.

(ii)               S.l. on Rs 6250 at 14% per annum for 146 days.

(iii)             S.l. on Rs 3000 at 18% per annum for the period from 4th

Sol. (i) P = 68000, R = 50/3 % p.a.     &  T = 9/12 year = 3/4 years.

            .. S.I. = ( PxRxT/100)

                     = Rs. [6800 x 50/3 x ¾ x 1/100]   = Rs. 8500.

 (ii) P = Rs. 6250,    R = 14 % p.a.   & T = (146/365) year = 2/5 year .

:. S.l. = Rs (6250 x 14 x 2/5 x 1/100)= Rs 350.

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

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

:. S.I. = Rs (3000 x 18 x 1/5 x 1/100)= Rs 108 .

Remark: The day on which money is deposited is not counted while the day on which money is withdrawn is counted.

 Ex. 2. A sum at simple interest at 13 i % per annum amounts to Rs. 2502.50 after 4 years. Find the sum.

Sol.  Let sum be x. Then

       S.I. = (X x 27/2 x 4 x 1/100)    = 27X/50

       .. Amount = ( X + 27X/50)  = 77X/50

       .. 77X/50  = 2502.50    or X= (2502.50 X 50)/77 = 1625

Hence Sum = Rs.1625

 Ex. 3. A certain sum of money amounts to Rs 1008 in 2 years and to Rs 1164 in 3 i years. Find the sum and (be rate of interest.

Sol. S.I. for 1 ½  years = Rs (1164 - 1008) = Rs 156 .

S.I. for 2 years = Rs (156 x 2/3  x 2)= Rs 208.

:. Principal = Rs (1008 - 208) = Rs 800.

Now, P = 800, T= 2 and S.I. = 208.

        .. Rate = (100 x S.I.) / (P x T)  =[ (100x208)/(800x2)]% =13%

 Ex.4. At what rate percent per annum will a sum of money double in 8 years

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

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

Ex. 5. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum.

 Sol. Let, sum = P and original rate = R. Then,

Px (R+2}x 3      P xRx 3   = 360

100                   100

or 3PR + 6P - 3PR = 36000  or 6P = 36000   or P = 6000 . Hence, sum = Rs 6000 .

Ex. 6. Simple interest on a certain sum is 16/25 of  the sum. Find the rate percent and time, If both are numerically equal.

        Sol. Let sum = x. Then, S.I. = 16x / 25

Let rate = R % and time = R years.

.. ( X x R x R)/(100) = 16x/25    or R² =1600/25    R= 40/5 =8

:. Rate = 8% and Time = 8 years.

 Ex. 7. A man borrowed Rs 24000 from two money lenders. For one loan, he paid 15% per annum and for the other 18% per annum. At the end of one year, he paid Rs 4050. How much did he borrow at each rate ?

 Sol. Let the sum at 15% be Rs x  and that at 18% be Rs (24000 - x).

          .. {(X x 15 x 1)/100  + {[(2400 – x) X 18 X 1]/100} = 4050

or 15 x + 432000 - 18x = 405000    or    x = 9000.

:. Money borrowed at 15% = Rs 9000 .

Money borrowed at 18% = Rs 15000.

 Ex. 8. What annual instalment wiU discharge a debt of Rs 1092 due in 3 years at 12% simple interest?

 Sol. Let each instalment be Rs. x . then,

                 { [x + (X x 12 x 2)/100]   + [ X + (X + 12 x 2)/100]  + x} =1092

              or 28x/25 + 31x/25 +x=1092 or (28x+ 31x+25x) = (1092x25)

or X= {(1092 x 25)/84} = 325

:. Each instalment = Rs. 325