Question
Find the difference between the compound interest compounded yearly and half$-$yearly on $Rs. 10,000$ for $18$ months at $10\%$ per annum.
✓
Answer
$(i)$ When interest is compounded yearly:
Given : $P=R s .10,000 ; n=18$ months $=1 \frac{1}{2}$ year and $r=10 \% p.a.$
For $1$year
$A=P\left(1+\frac{r}{100}\right)^n\left(1+\frac{r}{200}\right)^{\frac{1}{ Z }} \times Z$
$=10,000 \times \frac{11}{10} \times \frac{21}{20}$
$=1020 \times 11$
$=11,550$
When interest is compounded half$-$yearly
$A=P\left(1+\frac{r}{200}\right)^{n \times 2}$
$=10,000\left(1+\frac{10}{200}\right)^{\frac{3}{\nexists} \times \not 2}$
$=10,000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$=\frac{105 \times 441}{4}$
$=\frac{46305}{4}$
$=11576.25$
$=\frac{46305}{4}$
$=11576.25$
Therefore, the difference between both $C.I$
$₹11,576.25 - ₹11,550$
$= ₹26.25$
Therefore, the difference between both $C.I$
Given : $P=R s .10,000 ; n=18$ months $=1 \frac{1}{2}$ year and $r=10 \% p.a.$
For $1$year
$A=P\left(1+\frac{r}{100}\right)^n\left(1+\frac{r}{200}\right)^{\frac{1}{ Z }} \times Z$
$=10,000 \times \frac{11}{10} \times \frac{21}{20}$
$=1020 \times 11$
$=11,550$
When interest is compounded half$-$yearly
$A=P\left(1+\frac{r}{200}\right)^{n \times 2}$
$=10,000\left(1+\frac{10}{200}\right)^{\frac{3}{\nexists} \times \not 2}$
$=10,000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$=\frac{105 \times 441}{4}$
$=\frac{46305}{4}$
$=11576.25$
$=\frac{46305}{4}$
$=11576.25$
Therefore, the difference between both $C.I$
$₹11,576.25 - ₹11,550$
$= ₹26.25$
Therefore, the difference between both $C.I$
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