Question

a certain sum of money is invested at the rate of 10% per annum compound interest,the interest compounded annually is the difference between the interest of third year and first year is rupees 1105 find the sum invested

Answer 1

For annual compound interest amount A at the end of the year is given by A=p(1+r)n where r is the interest expressed as % , rate of an amount c=c*r, Let invested amount be P rate of interest (r=10/100=0.1) second year r=A2=p(1+0.1)2 third yr A2*0.1
different of 3rd year to 1st year is 1105
So,0.1*p*(1.1)2
0.1p=1105
0.1p=[(1.1)2 - 1]=1105
0.1p(0.21)=1105
P=52619