If the difference between the compound interest and
simple interest for 2 years at 12%p.a. compounded annually is
Rs108, find the sum borrowed.
Answers
Rs. 7500
Step-by-step explanation:
Let the sum borrowed was P
Here Time = 2 years and Rate = 12% per annum compounded annually
Lets Find Compound Interest
We know that
CI=P[1+\frac{12}{100}]^2-PCI=P[1+
100
12
]
2
−P
CI=P[\frac{112}{100}]^2-PCI=P[
100
112
]
2
−P
CI=P[\frac{28}{25}]^2-PCI=P[
25
28
]
2
−P
CI=P[\frac{784}{625}]-PCI=P[
625
784
]−P
CI=[\frac{159P}{625}]CI=[
625
159P
] ------- ( i )
Now Simple Interest
SI = (P × R × n)/100
SI = ( P × 12 × 2)/100
SI = 6P/25 ---- (ii )
According to Question CI - SI = 108
Putting values of CI and SI from ( i ) and ( ii )
[\frac{159P}{625}]-\frac{6P}{25}=108[tex] < /p > < p > [tex]\frac{9P}{625} = 108[
625
159P
]−
25
6P
=108[tex]</p><p>[tex]
625
9P
=108
P=\frac{108\times625}{9}P=
9
108×625
P = 7500
∴ Sum borrowed was Rs. 7500