Question
Find the difference between simple and compound interest on $Rs. 5000$ invested for $3$ years at $6\%\ \ p.a.$, interest payable yearly.
✓
Answer
Case $1:$
Here $P_1=R s .5000$ and $r=6 \%$
So, Amount after $1$ year
$=P\left(1+\frac{r}{100}\right)$
$=5000\left(1+\frac{6}{100}\right)$
$=5000 \times \frac{106}{100}$
$=5300$
Amount after $2$ year
$=P\left(1+\frac{r}{100}\right)$
$=5300\left(1+\frac{6}{100}\right)$
$=5300 \times \frac{106}{100}$
$=5618$
Thus, $P_3=\text { Rs. } 5618$ and $ r=6 $
Amount after $3$ year
$=P\left(1+\frac{r}{100}\right)$
$=5618\left(1+\frac{6}{100}\right)$
$=5618 \times \frac{106}{100}$
$=5955.08$
Hence, Amount $=\text { Rs. } 5955.08$
Also, $\text {C.I. }$
$=A \cdot P$
$=\text { Rs. } 5955.08-R s .5000$
$=\text { Rs. } 955.08$
Case$ II:$
Simple interest $=\frac{5000 \times 6 \times 3}{100}$
$=900$
Difference between $C.I.$ and $S.I.$
$=\text { Rs. } 955.08 \text { - Rs. } 900$
$=\text { Rs. } 55.08 \text {. }$
Here $P_1=R s .5000$ and $r=6 \%$
So, Amount after $1$ year
$=P\left(1+\frac{r}{100}\right)$
$=5000\left(1+\frac{6}{100}\right)$
$=5000 \times \frac{106}{100}$
$=5300$
Amount after $2$ year
$=P\left(1+\frac{r}{100}\right)$
$=5300\left(1+\frac{6}{100}\right)$
$=5300 \times \frac{106}{100}$
$=5618$
Thus, $P_3=\text { Rs. } 5618$ and $ r=6 $
Amount after $3$ year
$=P\left(1+\frac{r}{100}\right)$
$=5618\left(1+\frac{6}{100}\right)$
$=5618 \times \frac{106}{100}$
$=5955.08$
Hence, Amount $=\text { Rs. } 5955.08$
Also, $\text {C.I. }$
$=A \cdot P$
$=\text { Rs. } 5955.08-R s .5000$
$=\text { Rs. } 955.08$
Case$ II:$
Simple interest $=\frac{5000 \times 6 \times 3}{100}$
$=900$
Difference between $C.I.$ and $S.I.$
$=\text { Rs. } 955.08 \text { - Rs. } 900$
$=\text { Rs. } 55.08 \text {. }$
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