interest being compounded annually

Question

What sum would amount to 31,104 in three years at 20% p.a Compound interest, interest being compounded annually ? (Ans: 18,900)

Solution: \begin{align*} C I &= A – P \\[1ex] P &= 31104 – P\left( 1 + \frac{20}{100}\right)^3 \\[1ex] P &= 31104 – P\left(1 + \frac{1}{5}\right)^3 \\[1ex] P + P\left(1 + \frac{1}{5}\right)^3 &= 31104 \\[1ex] P + P\left(1+\frac{1}{25}\right) &= 31104 \\[1ex] P + P\left(\frac{16}{25}\right) &= 31104 \\[1ex] P \left(\frac{41}{25}\right) &= 31104 \\[1ex] P &= \frac{31104 \times 25}{41} \\[1ex] P &= 18,965 \end{align*} So answer doesn't match actual answer. please help me.

Answer 1

The formula used isn't right.

For a principal $P$, rate of interest $r$%, time in years $n$, the amount $A$ when compounded annually is given by $$A=P(1+\frac{r}{100})^n$$ Thus $$31104=P(1+\frac{20}{100})^3$$ $$31104=P(\frac{6}{5})^3$$ $$P=\frac{31104*125}{216}=18000$$

The principal must be $18000$.

Answer 2

If interest is being compounded annually use the formula-

$$A=p(1+\frac{r}{100})^t$$ instead of calculation for each year.

Note-Here $A$ is the final amount,$p$=initial principle,$r$ is rate of compound interest,$t$ is the number of years.

Answer 3

$A(3)=A(0)(1+0.20)^{3}$

$A(0)=\frac{A(3)}{1.20^3}=\frac{31104}{1.20^3}=18000$