Simple Intrest


Important Facts and Formulae: 

Principal or Sum:- The money borrowed or lent out for a 
certain period is  called Principal or the Sum. 
 
Interest:- Extra money paid for using others 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:
          Principal = P 
          Rate   = R% per annum 
          Time  = T years. Then, 
 
(i)Simple Interest(S.I)=  (P*T*R)/100 
 
(ii) Principal(P) = (100*S.I)/(R*T) 
        Rate(R) = (100*S.I)/(P*T) 
        Time(T) = (100*S.I)/(P*R) 

Simple Problems

1.Find S.I on Rs68000 at 16 2/3% per annum for 9months. 

Sol:-           P=68000 
                R=50/3% p.a 
                T=9/12 years=4/3 years 
                S.I=(P*R*T)/100 
                   =(68000*(50/3)*(3/4)*(1/100)) 
                   =Rs 8500 
 
Note:If months are given we have to converted into 
years by dividing 12 ie., no.of months/12=years 
 
2.Find S.I on Rs3000 at 18% per annum for the period from
4th Feb to 18th  April 1995 

Sol:-          Time=(24+31+18)days 
                   =73 days 
                   =73/365=1/5 years 
               P= Rs 3000 
               R= 18% p.a 
               S.I = (P*R*T)/100 
                   =(3000*18*1/5*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counted while the day on  which money is withdrawn is 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. 
 
3. In how many years will a sum of money becomes triple
at 10% per annum. 

Sol:-     Let principal =P 
                    S.I = 2P 
                    S.I = (P*T*R)/100 
                     2P = (P*T*10)/100 
                      T = 20 years 
Note:
 (1) Total amount = Principal + S.I 
 (2) If sum of money becomes double means Total amount 
     or Sum 
                 = Principal + S.I 
                 = P + P = 2P 
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Medium Problems

1.A sum at Simple interest at 13 1/2% per annum amounts 
to Rs 2502.50 after 4 years.Find the sum. 

Sol:-     Let Sum be x. then, 
              S.I = (P*T*R)/100 
                  = ((x*4*27)/(100*2)) 
                  = 27x/100 
           Amount = (x+(27x)/100) 
                  = 77x/50 
           77x/50 = 2502.50 
                x = (2502.50*50)/77 
                  = 1625 
              Sum = 1625 
 
2. A some of money becomes double of itself in 4 years 
in 12 years it will become how many times at the same
rate. 

Sol:-         4 yrs  - - - - - - - - -  P 
              12 yrs - - - - - - - - -   ? 
                  (12/4)* P =3P 
              Amount or Sum = P+3P = 4 times 
 
3. A Sum was put at S.I 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 
          original rate = R 
                      T =  3 years 
 If 2% is more than the original rate ,it would have 
  fetched 360 more ie., R+2 
        (P*(R+2)*3/100) - (P*R*3)/100 = 360 
                          3PR+ 6P-3PR = 36000 
                                   6P = 36000 
                                    P = 6000 
                                  Sum = 6000. 
 
4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest
rate is increased by 3%,  it would amount to how much? 

Sol:-             S.I = 920 - 800 = 120 
                 Rate = (100*120)/(800*3) = 5% 
             New Rate = 5 + 3 = 8%   
            Principal = 800   
                 Time = 3 yrs 
                  S.I = (800*8*3)/100 = 192 
           New Amount = 800 + 192 
                      = 992 
 
5. Prabhat took a certain amount as a loan from bank at 
the rate of 8% p.a S.I and gave the same amount to Ashish 
as a loan at the rate of 12% p.a . If at the end of 12 yrs,
he made a profit of Rs. 320 in the deal,What was the 
original amount?        

Sol:-      Let the original amount be Rs x.    
                             T = 12 
                            R1 = 8%      
                            R2 = 12%    
                        Profit = 320 
                             P = x 
         (P*T*R2)/100 - (P*T*R1)/100 =320 
              (x*12*12)/100 - (x*8*12)/100 = 320 
                        x = 2000/3 
                        x = Rs.666.67 
  
6. Simple Interest on a certail sum at a certain rate is
9/16 of the sum . if the number representing rate percent
and time in years be equal ,then the rate is. 

Sol:-           Let Sum = x .Then, 
                    S.I = 9x/16 
               Let time = n years & rate = n% 
                      n = 100 * 9x/16 * 1/x * 1/n 
                  n * n = 900/16 
                     n = 30/4 = 7 1/2% 

Complex Problems

1. A certain sum of money amounts t 1680 in 3yrs & it 
becomes 1920 in 7 yrs .What is the sum. 

Sol:-       3 yrs - - - - - - - - - - - - - 1680 
            7 yrs - - - - - - - - - - - - - 1920 
  then,     4 yrs - - - - - - - - - - - - - 240 
            1 yr  - - - - - - - - - - - - -   ? 
                   (1/4) * 240 = 60 
           S.I in 3 yrs = 3*60 = 18012 
                           Sum = Amount - S.I 
                               = 1680 - 180 
                               = 1500 
       we get the same amount if we take S.I in 7 yrs 
                    I.e., 7*60 =420 
                           Sum = Amount - S.I 
                                = 1920 - 420 
                                = 1500     

2. A Person takes a loan of Rs 200 at 5% simple Interest.
He returns Rs.100 at  the end of 1 yr. In order to clear
his dues at the end of 2yrs ,he would pay: 

Sol:-              Amount to be paid 
                      = Rs(100 + (200*5*1)/100 + (100*5*1)/100) 
                      = Rs 115 
 
3. A Man borrowed Rs 24000 from two money lenders.For one
loan, he paid 15% per annum and for other 18% per annum.
At the end of one year,he paid Rs.4050.How much did he 
borrowed at each rate? 

Sol:-            Let the Sum at 15% be Rs.x 
                   & then at 18% be Rs (24000-x) 
                P1 = x                    R1 = 15 
                P2 = (24000-x)            R2 = 18 
                At the end of ine year T = 1 
              (P1*T*R1)/100 + (P2*T*R2)/100 = 4050 
             (x*1*15)/100 + ((24000-x)*1*18)/100 = 4050 
                   15x + 432000 - 18x = 405000 
                             x = 9000 
              Money borrowed at 15% = 9000 
              Money borrowed at 18% = (24000 - 9000) 
                                    = 15000 
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4.What annual instalment will discharge a debt of Rs. 1092 
due in 3 years at 12% Simple Interest ? 

Sol:-             Let each instalment be Rs x  
(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092 
                   28x/25 + 31x/25 + x =1092 
                   (28x +31x + 25x) = (1092 * 25) 
                                84x = 1092 * 25 
                                  x = (1092*25)/84 = 325 
                  Each instalement  =  325 


5.If x,y,z are three sums of money such that y is the simple 
interest on x,z is the simple interest on y for the same 
time and at the same rate of interest ,then we have: 

Sol:-      y  is simple interest on x, means 
                    y = (x*R*T)/100 
                   RT = 100y/x 
           z   is simple interest on y, 
                    z = (y*R*T)/100 
                   RT = 100z/y 
               100y/x = 100z/y 
                y * y = xz 
 
6.A Sum of Rs.1550 was lent partly at 5% and partly at 5% 
and partly at 8% p.a  Simple interest .The total interest 
received after 3 years was Rs.300.The ratio of the money 
lent at 5% to that lent at 8% is: 

Sol:-      Let the Sum at 5% be Rs x 
           at 8% be Rs(1550-x) 
      (x*5*3)/100 + ((1500-x)*8*3)/100 = 300 
          15x + 1500 * 24 - 24x = 30000 
                              x = 800  
      Money at 5%/ Money at 8% = 800/(1550 - 800) 
                               = 800/750 = 16/15 
 
7. A Man invests a certain sum of money at 6% p.a Simple 
interest and another  sum at 7% p.a Simple interest. His
income from interest after 2 years was  Rs 354 .one 
fourth of the first sum is equal to one fifth of the 
second sum.The total sum invested was: 

Sol:-       Let the sums be x & y 
                  R1 = 6     R2 = 7 
                  T = 2 
           (P1*R1*T)/100 + (P2*R2*T)/100 = 354 
     (x * 6 * 2)/100  + (y * 7 * 2)/100 =  354 
            6x + 7y = 17700 ———(1) 
  also one fourth of the first sum is equal to one 
            fifth of the second sum 
     x/4 = y/5 => 5x - 4y = 0 —— (2) 
    By solving 1 & 2 we get, 
       x = 1200       y = 1500 
      Total sum = 1200 +1500 
                = 2700      
 
8. Rs 2189 are divided into three parts such that their       
amounts after 1,2& 3 years respectively may be equal,
the rate of S.I being 4% p.a in all cases. The Smallest 
part is: 

Sol:-   Let these parts be x,y and[2189-(x+y)] then, 
    (x*1*4)/100 = (y*2*4)/100 =  (2189-(x+y))*3*4/100 
                       4x/100 = 8y/100 
                            x = 2y 
              By substituting values 
         (2y*1*4)/100 = (2189-3y)*3*4/100 
                  44y = 2189 *12 
                    y = 597 
        Smallest Part = 597 
 
9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and 
the remainder at 10%.If his annual income is Rs.561. The 
capital is: 

Sol:-        Let  the capital be Rs.x 
    Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)
                + (5x/12 * 10/100 * 1) = 561 
            7x/300 + x/50 + x/24 = 561 
                       51x = 561 * 600 
                         x = 6600


    

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