If the difference of the compound interest and the simple interest on a sum of money for 2 years is 1% of the sum of money then find the annual rate of interest.
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CI = P[1+r/100]^n-P
SI = pnr/100
From the problem,
N = 2
CI – SI = P*1/100
We need to find r.
CI – SI = P[1+r/100]^n-P-Pnr/100
=P[(1+r/100)^n-1-nr/100]
Substituting the values of n and r, we get
CI – SI = P * 1/100 = P[1+r/100)^2-1-2*r/100]
Since P is common on both sides of the equation, they get cancelled.
Let us consider the expression within the square brackets.
Expanding and subtracting the remaining values, we get
1+2r/100+r^2/10000-1-2r/100 = r^2/10000
Therefore, 1/100 = r^2/10000.
R^2 = 10000/100 = 100
Therefore, r = 10%
Thus, the rate of interest is 10%.
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