Question

A certain sum of money at compound interest becomes 7396in 2years and 7950.70in 3years find the rate of interest

Answer 1

 Let the principal be P
Case i:
 CI=7396
CI=p((1+r/100)^t
7396=P( 1+r/100)^2
P=7396x10000/ (100+r)^2   ..................1
case ii:
CI= 7950.70
P=7950.70x1000000/(100+r)^3 ..............2
from 1 and 2 we get
 7396x10000/ (100+r)^2   =7950.70x1000000/(100+r)^3 
7396=795070/(100+r)
100+r =795070 / 7396
r= (795070 / 7396)-100
r=(795070 - 739600)/ 7396
r=55470/7396
r= 7.5%
Hope this Helps you.

Answer 2

Step-by-step explanation:

Let the principal be P

Case i:

 CI=7396

CI=p((1+r/100)^t

7396=P( 1+r/100)^2

P=7396x10000/ (100+r)^2   ..................1

case ii:

CI= 7950.70

P=7950.70x1000000/(100+r)^3 ..............2

from 1 and 2 we get

 7396x10000/ (100+r)^2   =7950.70x1000000/(100+r)^3 

7396=795070/(100+r)

100+r =795070 / 7396

r= (795070 / 7396)-100

r=(795070 - 739600)/ 7396

r=55470/7396

r= 7.5%

Hope this Helps you.

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