Maths 5 Mark Question

ii. Here, $P=$ Rs $5000, R=8$ p.c.p.a. and $N=3$ years
$
\begin{aligned}
A & =P\left[1+\frac{R}{100}\right]^{ N } \\
& =5000\left[1+\frac{8}{100}\right]^3 \\
& =5000\left[\frac{100+8}{100}\right]^3 \\
& =5000\left[\frac{108}{100}\right]^3=5000\left[\frac{27 \times 4}{25 \times 4}\right]^3 \\
& =5000\left[\frac{27}{25}\right]^3 \\
& =5000 \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25} \\
& =\frac{8}{25} \times 27 \times 27 \times 27=0.32 \times 19683 \\
\therefore & A=\text { Rs } 6298.56 \\
I = & \text { Amount }(A)-\text { Principal }(P) \\
= & 6298.56-5000 \\
= & \text { Rs } 1298.56
\end{aligned}
$
$\therefore$ The amount is Rs 6298.56 and the compound interest is Rs 1298.56.

iii. Here, $P=R s 4000, R=7.5$ p.c.p.a. and $N=2$ years
$
\begin{aligned}
A & =P\left[1+\frac{R}{100}\right]^N \\
& =4000\left[1+\frac{7.5}{100}\right]^2 \\
& =4000\left[1+\frac{75}{1000}\right]^2 \\
& =4000\left[\frac{1000+75}{1000}\right]^2 \\
& =4000\left[\frac{1075}{1000}\right]^2=4000\left[\frac{43 \times 25}{40 \times 25}\right]^2 \\
& =4000\left[\frac{43}{40}\right]^2 \\
& =4000 \times \frac{43}{40} \times \frac{43}{40}=2.5 \times 43 \times 43 \\
\therefore A & =\text { Rs } 4622.50 \\
I & =\text { Amount (A) }- \text { Principal (P) } \\
= & 4622.50-4000 \\
= & \text { Rs } 622.50
\end{aligned}
$
$\therefore$ The amount is Rs 4622.50 and the compound interest is Rs 622.50