Solution: (c)
Principal (P)=Rs. 8000
Rate of interest (R)=5% per annum
Time (n)=2 years
Total amount on compounded annually,
Amount= \(\displaystyle =P\times {{\left( {1+\frac{r}{{100}}} \right)}^{n}}\)
\(\displaystyle \begin{array}{l}=8000\times {{\left( {1+\frac{5}{{100}}} \right)}^{2}}\\=8000\times {{\left( {\frac{{21}}{{20}}} \right)}^{2}}\\=8000\times \frac{{21}}{{20}}\times \frac{{21}}{{20}}\\=Rs.820\end{array}\)
\(\displaystyle \Rightarrow \)C.I=A−P
=Rs. 8820−Rs. 8000
=Rs. 820
Therefore, the amount is Rs. 8820 and compound interest is Rs. 820.

Solution: (d)
\(\displaystyle \begin{array}{l}CI=P\left[ {{{{\left( {1+\frac{R}{{100}}} \right)}}^{T}}-1} \right]\\\Rightarrow 3041.28=P\left[ {{{{\left( {1+\frac{{16}}{{100}}} \right)}}^{2}}-1} \right]\\\Rightarrow 3041.28=P\left[ {{{{\left( {\frac{{29}}{{25}}} \right)}}^{2}}-1} \right]\\\Rightarrow 3041.28=P\left[ {\frac{{941}}{{625}}-1} \right]\\\Rightarrow 3041.28=P\times \frac{{216}}{{625}}\\\Rightarrow P=\frac{{3041.28\times 625}}{{216}}=8800\end{array}\)

Solution: (b)
Effective rate of CI for 2 years
\(\displaystyle =5+5+\frac{{5\times 5}}{{100}}\)
=10.25 %
Effective rate of SI for 3 years= 6×3 = 18%
Required SI=\(\displaystyle \frac{{246}}{{10.25}}\times 18\)
=Rs. 432
Alternate method,
CI = Rs. 246, R = 5%, T = 2 years
CI = A – P
\(\displaystyle \begin{array}{l}246=P{{\left( {1+\frac{5}{{100}}} \right)}^{2}}-P\\246=P{{\left( {\frac{{21}}{{20}}} \right)}^{2}}-P\\246=P\left[ {{{{\left( {\frac{{21}}{{20}}} \right)}}^{2}}-1} \right]\\246=P\frac{{41}}{{400}}\\P=\frac{{246\times 400}}{{41}}\\=Rs.2400\end{array}\)
Now, P = Rs. 2400, R = 6%, T = 3 years
SI= \(\displaystyle \frac{{2400\times 6\times 3}}{{100}}\)
=432

Solution: (d)
Amount after three years = Rs. 2662
Amount after two years = Rs. 2420
Net interest earned in the 3rd year = 2662−2420
= Rs. 242
Rate of interest (r)= \(\displaystyle \frac{{242}}{{2420}}\times 100\)
=10%
Alternate method
\(\displaystyle Let\text{ }the\text{ }sum\text{ }of\text{ }money\text{ }invested\text{ }be\text{ }Rs.\text{ }x\text{ }and\text{ }interest\text{ }rate\text{ }per\text{ }annum\text{ }=\text{ }r\%\)
\(\displaystyle Amount\text{ }=\text{ }Principal\text{ }\times \text{ }{{\left[ {1+\frac{r}{{100}}} \right]}^{n}}where\text{ }n\text{ }is\text{ }time\)
\(\displaystyle {Then\text{ }x\text{ }\times \text{ }{{{\left[ {1+\frac{r}{{100}}} \right]}}^{2}}=\text{ }Rs.\text{ }2420\text{ }….\text{ }\left( i \right)}\)
\(\displaystyle {and\text{ }x\text{ }\times \text{ }{{{\left[ {1+\frac{r}{{100}}} \right]}}^{3}}=\text{ }Rs.\text{ }2662\text{ }….\text{ }\left( {ii} \right)}\)
\(\displaystyle Dividing\text{ }equation\text{ }\left( {ii} \right)\text{ }by\text{ }\left( i \right),\text{ }we\text{ }get\)
\(\displaystyle {\left( {1\text{ }+\text{ }\frac{r}{{100}}} \right)\text{ }=\text{ }\frac{{2662}}{{2420}}}\)
\(\displaystyle {\Rightarrow \text{ }\frac{r}{{100}}\text{ }=\text{ }\left( {\frac{{2662}}{{2420}}} \right)-1}\)
\(\displaystyle {\Rightarrow \text{ }r\text{ }=\text{ }\frac{1}{{10}}\text{ }=\text{ }10\%}\)

Solution: (b)
Let each installment be x
P=13360
Rate o interest=8\(\displaystyle \frac{3}{4}\)
=\(\displaystyle \frac{35}{4}\)%
\(\displaystyle \begin{array}{l}\Rightarrow \frac{x}{{{{{\left( {1+\frac{{35}}{{100}}} \right)}}^{2}}}}+\frac{x}{{\left( {1+\frac{{35}}{{100}}} \right)}}=13360\\\Rightarrow \frac{{6400x+6960x}}{{7569}}=13360\\\Rightarrow 13360x=13360\times 7569\\x=7569\end{array}\)