Solution: (a)
A = \(\displaystyle p{{(1+\frac{r}{{100}})}^{2}}\)
or, 15800 + 7716.72 = 15800 \(\displaystyle {{(1+\frac{r}{{100}})}^{2}}\)
\(\displaystyle \Rightarrow \) \(\displaystyle {{(1+\frac{r}{{100}})}^{2}}\)
\(\displaystyle \frac{{2351.72}}{{15800}}={{(1+\frac{r}{{100}})}^{2}}\)
Taking square root of both the sides,
\(\displaystyle 1.22=1+\frac{r}{{100}}\)
or, r = 22%

Solution: (a)
C.I. when interest compounded yearly
\(\displaystyle \begin{array}{l}=Rs.\left[ {5000\times \left( {1+\frac{4}{{100}}} \right)\times \left( {1+\frac{{\frac{1}{2}\times 4}}{{100}}} \right)} \right]\\=Rs.\left( {5000\times \frac{{26}}{{25}}\times \frac{{51}}{{50}}} \right)\\=Rs.5304\end{array}\)
C.I. when interest compounded half-yearly
\(\displaystyle \begin{array}{l}=Rs.\left[ {5000\times {{{\left( {1+\frac{2}{{100}}} \right)}}^{3}}} \right]\\=Rs.\left( {5000\times \frac{{51}}{{50}}\times \frac{{51}}{{50}}\times \frac{{51}}{{50}}} \right)\\=Rs.5306.4\end{array}\)
Therefore, difference – Rs. (5306.40-5304)=Rs. 2.04

Solution: (d)
Simple interest = \(\displaystyle \frac{{principal\times time\times rate}}{{100}}\)
\(\displaystyle \frac{{12000\times 3\times 12}}{{100}}=4320\)
C.I. = \(\displaystyle p[{{(1+\frac{{rate}}{{100}})}^{{time}}}-1]\)
= \(\displaystyle 12000[{{(1+\frac{{12}}{{100}})}^{3}}-1]\)
= \(\displaystyle 12000[{{(\frac{{28}}{{25}})}^{3}}-1]\)
= \(\displaystyle 12000[\frac{{21952}}{{15625}}-1]\) = \(\displaystyle 12000\times \frac{{6327}}{{15625}}\)
= ₹ 4859.136
Therefore, required difference = 4859.136 – 4320
= ₹ 539.136

Solution: (c)
Compound Interest
=\(\displaystyle \begin{array}{l}=Rs.\left[ {4000\times {{{\left( {1+\frac{{10}}{{100}}} \right)}}^{2}}-4000} \right]\\=Rs.\left( {4000\times \frac{{11}}{{10}}\times \frac{{11}}{{10}}-4000} \right)\\=Rs.840\end{array}\)
Therefore, Sum is
=\(\displaystyle \begin{array}{l}=Rs.\left( {\frac{{420\times 100}}{{3\times 8}}} \right)\\=Rs.1750\end{array}\)

Solution: (a)
Let the principal be ₹ x, then
\(\displaystyle x{{(1+\frac{{15}}{{100}})}^{3}}-x=\frac{{650052}}{{100}}\)
\(\displaystyle x\times \frac{{23}}{{20}}\times \frac{{23}}{{20}}\times \frac{{23}}{{20}}-x=\frac{{650052}}{{100}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle 12167x-8000x=\frac{{650052\times 8000}}{{100}}\)
x = \(\displaystyle \frac{{650052\times 8000}}{{100\times 4167}}=156\times 80=12480\)

Given that, Time = 3 years, Rate = R%

Let the principal be ₹ 1 and Amount = ₹ 8

Amount= \(\displaystyle =P\times {{\left( {1+\frac{r}{{100}}} \right)}^{n}}\)⇒ 8= 1×

\(\displaystyle \begin{array}{l}\Rightarrow {{\left( 2 \right)}^{3}}={{\left( {1+\frac{r}{{100}}} \right)}^{3}}\\\Rightarrow 2=\left( {1+\frac{r}{{100}}} \right)\end{array}\)

Compounded amount has to become sixteen times. So, 2 can be 16 if it is multiplied 4 times

\(\displaystyle \Rightarrow {{\left( 2 \right)}^{4}}={{\left( {1+\frac{r}{{100}}} \right)}^{4}}\)

Therefore time T=4 years

Alternate method

Let principal = P

From the given data,

In Case(I)Time = 3 years,

Amount = 8P

\(\displaystyle \begin{array}{l}\Rightarrow 8P=P{{\left( {1+\frac{r}{{100}}} \right)}^{3}}\\\Rightarrow {{\left( 2 \right)}^{3}}={{\left( {1+\frac{r}{{100}}} \right)}^{3}}\end{array}\)

Taking cube root of both sides,

\(\displaystyle \Rightarrow 2=\left( {1+\frac{r}{{100}}} \right)\)

\(\displaystyle \Rightarrow r=100\%\)

In Case(II)

Let after ‘t’ years it will be 16 times

\(\displaystyle \Rightarrow 16P=P{{\left( {1+\frac{r}{{100}}} \right)}^{t}}\)

\(\displaystyle \Rightarrow 16={{\left( 2 \right)}^{t}}\)

\(\displaystyle \Rightarrow {{\left( 2 \right)}^{4}}={{\left( 2 \right)}^{t}}\)

⇒t=4

Hence required time (t) = 4 years

One more alternate method

Here, p = 8, \(\displaystyle {{n}_{1}}\)= 3 and q = 16, \(\displaystyle {{n}_{2}}\)= ?

Using,

Given that , Time = 3 years , Rate = R%

Let the principal be ₹ 1 and Amount = ₹ 8

Amount= \(\displaystyle =P\times {{\left( {1+\frac{r}{{100}}} \right)}^{n}}\)⇒ 8= 1×

\(\displaystyle {{P}^{{\frac{1}{{{{n}_{1}}}}}}}={{Q}^{{\frac{1}{{{{n}_{2}}}}}}}\)

\(\displaystyle {{8}^{{\frac{1}{{{{n}_{1}}}}}}}={{16}^{{\frac{1}{{{{n}_{2}}}}}}}\)

\(\displaystyle {{\left( {{{2}^{3}}} \right)}^{{\frac{1}{{{{n}_{1}}}}}}}={{\left( {{{2}^{4}}} \right)}^{{\frac{1}{{{{n}_{2}}}}}}}\)

\(\displaystyle {{2}^{1}}={{2}^{{\frac{4}{{{{n}_{2}}}}}}}\)

Bases are same so equating powers,

\(\displaystyle 1=\frac{4}{{{{n}_{2}}}}\) Therefore time T=4 years