Question
Find the sum on which the difference between the simple interest and compound interest at the rate of $8\ \%\ $ per annum compounded annually would be $Rs. 64$ in $2$ years.
✓
Answer
Let $Rs, X$ be the sum.
Simple Interest$(I)=\frac{X \times 8 \times 2}{100}=0.16 X$
Compound interest
For $1^{\text {st }}$ year :
$P=\text { Rs. } X_i R=8 \%\ $ and $T=1$
$ \Rightarrow$ Interest $(I)=\frac{X \times 8 \times 1}{100}=0.08 X$
And amount $=\text{₹}(x+0.08 x)$
$\text { = ₹ } 1.08 \mathrm{x}$
For $2^{\text {nd }}$ year :
$P=\text { Rs. } X+\text { Rs. } 0.08 X=\text { Rs. } 1.08 X$
$ \Rightarrow$ Interest $(I)=\frac{1.08 \mathrm{X} \times 8 \times 1}{100}=0.0864 X$
And, amount $=\text{₹}(1.08 x+0.0864 x)$
$=\text{₹} 1.1664 \mathrm{x}$
So,
$C.I =$ Amount $-p$
$=\text{₹}(1.1664 x-x)$
$=\text{₹} 0.1664 x$
The difference between the simple interest and compound interest at the rate of $8\ \%\ $ per annum compounded annually should be $Rs. 64$ in $2$ years.
$\text{₹} 0.1664 x-\text{₹} 0.16 x=\text{₹} 64$
$\text{₹} 0.0064 \mathrm{x}=\text{₹} 64$
$x=\text{₹} 10000$
Therefore, the sum is $\text{₹ 10,000}$.
Simple Interest$(I)=\frac{X \times 8 \times 2}{100}=0.16 X$
Compound interest
For $1^{\text {st }}$ year :
$P=\text { Rs. } X_i R=8 \%\ $ and $T=1$
$ \Rightarrow$ Interest $(I)=\frac{X \times 8 \times 1}{100}=0.08 X$
And amount $=\text{₹}(x+0.08 x)$
$\text { = ₹ } 1.08 \mathrm{x}$
For $2^{\text {nd }}$ year :
$P=\text { Rs. } X+\text { Rs. } 0.08 X=\text { Rs. } 1.08 X$
$ \Rightarrow$ Interest $(I)=\frac{1.08 \mathrm{X} \times 8 \times 1}{100}=0.0864 X$
And, amount $=\text{₹}(1.08 x+0.0864 x)$
$=\text{₹} 1.1664 \mathrm{x}$
So,
$C.I =$ Amount $-p$
$=\text{₹}(1.1664 x-x)$
$=\text{₹} 0.1664 x$
The difference between the simple interest and compound interest at the rate of $8\ \%\ $ per annum compounded annually should be $Rs. 64$ in $2$ years.
$\text{₹} 0.1664 x-\text{₹} 0.16 x=\text{₹} 64$
$\text{₹} 0.0064 \mathrm{x}=\text{₹} 64$
$x=\text{₹} 10000$
Therefore, the sum is $\text{₹ 10,000}$.
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