8th Compound Interest Solved problems

Ex.1. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.

  Sol.  Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112.            

            Therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

Ex. 2. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.

   Sol.         Time = 2  years  4  months = 2(4/12) years = 2(1/3) years.
                    Amount = Rs'. [8000 X (1+­(15/100))2 X (1+((1/3)*15)/100)]

                         =Rs. [8000 * (23/20) * (23/20) * (21/20)]     = Rs. 11109.                             .

                   :. C.I. = Rs. (11109 - 8000) = Rs. 3109.

    Ex. 3. Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the   interest being compounded half-yearly
    Sol.         Principal = Rs. 10000; Rate = 2%  per half-year;  Time = 2 years = 4 half-years.

Amount =  Rs [10000 * (1+(2/100))4] = Rs(10000 * (51/50) * (51/50) * (51/50) * (51/50))  

= Rs. 10824.32.

        :. C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

   Ex. 4. Find the compound interest on Rs. 16,000 at 20% per annum for 9 months,    compounded quarterly.                

   Sol.           Principal = Rs. 16000; Time = 9 months =3 quarters;

                  Rate = 20% per annum = 5% per quarter. 

                  Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.   

      CJ. = Rs. (18522 - 16000) = Rs. 2522.

Ex. 5. If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.

Sol.  Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200.          . .

So principal=RS [100*1200]/3*5=RS 8000

Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261.

.. C.I. = Rs. (9261 - 8000) = Rs. 1261.
Ex. 6. In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?  
Sol.  Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a.     Let the time be n years. Then,

[ 1000 (1+ (10/100))n  ] = 1331 or (11/10)n =  (1331/1000) = (11/10)3          n = 3 years. 

 

Ex. 7. If Rs. 600 amounts to Rs. 683.20 in two years compounded annually, find the rate of interest per annum.
      Sol. Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.
               Let the rate be R% per annum.. 'Then,

      [ 500 (1+(R/100)2 ] = 583.20 or [ 1+ (R/100)]2  =  5832/5000 = 11664/10000 

      [ 1+ (R/100)]2 = (108/100)2  or  1 + (R/100) = 108/100  or  R = 8           

               So, rate = 8% p.a.
Ex. 8. If the compound interest on a certain sum at 16 (2/3)% to  3 years is Rs.1270,  find the simple interest on the same sum at the same rate and f or the same period.
Sol. Let the sum be Rs. x. Then,

    C.I. = [ x * (1 + (( 50/(3*100))3- x ] = ((343x / 216) - x) = 127x / 216

127x /216 = 1270  or  x = (1270 * 216) / 127  =  2160.
Thus, the sum is Rs. 2160

S.I. = Rs  ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.             

Ex. 9. The difference between the compound interest and simple interest on a  certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.

 Sol. Let the sum be Rs. x. Then,

               C.I. =  x ( 1 + ( 10 /100 ))2 - x  =  21x / 100 ,  

              S.I.  = (( x * 10 * 2) / 100) =  x / 5

               (C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) =  x / 100

               ( x / 100 )  =  632  ó  x  =  63100.

Hence, the sum is Rs.63,100. 

Ex. 10. The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a. ?                 
Sol. Let the rate be R% p.a. then,

[ 18000 ( 1 + ( R / 100 )2 ) - 18000 ] - ((18000 * R *  2) / 100 ) = 405

18000 [ ( 100 + (R / 100 )2  / 10000)  -  1 - (2R / 100 ) ]  =  405

 18000[( (100 + R ) 2 - 10000 - 200R) / 10000 ]  =  405

 9R2 / 5  =  405   R2ó  =((405 * 5 ) / 9) = 225    R = 15.         Rate = 15%.

 
Ex. 11. Divide Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per  annum.

Sol. Let the two parts be Rs. x and Rs. (1301 - x).

x(1+4/100)7 =(1301-x)(1+4/100)9

x/(1301-x)=(1+4/100)2=(26/25*26/25)

625x=676(1301-x)

1301x=676*1301

x=676.
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625.

 Ex.12. a certain sum amounts to rs.7350 in 2 years and to rs.8575 in 3 years.find the sum and rate percent.

S.I on rs.7350 for 1 year=rs.(8575-7350)=rs.1225.

Rate=(100*1225/7350*1)%=16 2/3%
Let the sum be rs.x.then,

X(1+50/3*100)2=7350

X*7/6*7/6=7350

X=(7350*36/49)=5400.

Sum=rs.5400.

Ex.13.a sum of money amounts to rs.6690 after 3 years and to rs.10,035 after 6 years on compound interest.find the sum.

Sol. Let the sum be rs.P.then

P(1+R/100)3=6690…(i) and   P(1+R/100)6=10035…(ii)

On dividing,we get (1+R/100)3=10025/6690=3/2.

Substituting this value in (i),we get:

P*3/2=6690 or P=(6690*2/3)=4460

Hence,the sum is rs.4460.

Ex.14. a sum of money doubles itself at compound interest in 15 years.in how many years will it beco,e eight times?

P(1+R/100)15=2P

(1+R/100)15=2P/P=2

LET P(1+R/100)n=8P

(1+R/100)n=8=23={(1+R/100)15}3[USING (I)]

(1+R/100)N=(1+R/100)45

n=45.

Thus,the required time=45 years.

Ex.15.What annual payment will discharge a debt of Rs.7620 due in 3years at  16and  2/3% per annum interest?

 Sol. Let each installment beRs.x.

     Then,(P.W. of Rs.x due 1 year hence)+(P>W of Rs.x due 2 years hence)+(P.W of Rs. X due 3    

     years hence)=7620.\ x/(1+(50/3*100))+ x/(1+(50/3*100))2 + x/(1+(50/3*100))3=7620

 (6x/7)+(936x/49)+(216x/343)=7620. 

294x+252x+216x=7620*343.

x=(7620*343/762)=3430.

Amount of each installment=Rs.3430.