# 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.