COMPOUND INTEREST

Ans .

612

1. Explanation :

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

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

Ans .

3109

1. Explanation :

   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.

Ans .

824.32

1. Explanation :

   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.

Ans .

2522

1. Explanation :

   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

Ans .

1261

1. Explanation :

   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.

Ans .

3 yrs

1. Explanation :

   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.

Ans .

8% p.a.

1. Explanation :

   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.

Ans .

1080

1. Explanation :

   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.

Ans .

Rs.63,100.

1. Explanation :

   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

   Hence, the sum is Rs.63,100.

Ans .

15%.

1. Explanation :

   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 9\(R^2\) / 5 = 405 \(R^2\) =((405 * 5 ) / 9) = 225

   R = 15. Rate = 15%.

Ans .

Rs.676 and Rs.625.

1. Explanation :

   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.

Ans .

Rs.5400

1. Explanation :

   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

Ans .

Rs.4460.

1. Explanation :

   . 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.

Ans .

45 years

1. Explanation :

   \(P(1+R/100)^15\)=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

Ans .

Rs.3430

1. Explanation :

   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 3years 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.