Question

Find the amount and the compound interest on $Rs.10,000$ for $1\frac{1}{2}$ years at $10\%$ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Answer 1

Here,

Principal $\left(P\right)=Rs.10000$

Rate of interest $\left(R\right)=10\%$ per annum compounded half yearly.

$\Rightarrow R=5\%$ per half yearly
Time $\left(n\right)=1\frac{1}{2}$ years $=3$ half years
Amount $\left(A\right)=P{(1+\frac{R}{100})}^n$

$=10000{(1+\frac{5}{100})}^3$

$=10000{(1+\frac{1}{20})}^3$
$=10000{\left(\frac{21}{20}\right)}^3$

$=10000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}$

$=Rs11,576.25$
Compound Interest $(C.I)=A-P$

$=Rs11,576.25-Rs10,000$

$\Rightarrow C.P=Rs.1576.25$

So, Compound interest half yearly is $Rs.1576.25$---(1)

Rate of interest $\left(R\right)=10\%$

Time $\left(n\right)=1$ years
Amount $\left(A\right)$ for $1$ year

$A=P{(1+\frac{R}{100})}^n$

$=10000{(1+\frac{10}{100})}^1$

$=10000{(1+\frac{1}{10})}^1$
$=10000{\left(\frac{11}{10}\right)}^1$

$=10000\times \frac{11}{10}$
$=Rs.11,000$
Interest for half year $\frac{1}{2}$ year,

$=\frac{11000\times 1\times 10}{2\times 100}$ [ $\because S.I=\frac{PTR}{100}$]

$=Rs.550$
Therefore, total amount $=Rs.11,000+Rs.550=Rs.11,550$
Now,

$C.I=A-P$

$=Rs.11,550-Rs.10000$

$=Rs.1550$

So, Compound interest per annum is $Rs.1550$. ---(2)
From (1) and (2),

the interest $Rs.1576.25$ is more than $Rs.1550$.

Hence, the answer is 'Yes'.