Solution: (c)
Amount invested in Scheme B = X
Amount invested in Scheme A = 5200 – x
\(\displaystyle x\times 10\times \frac{4}{{100}}=(5200-x)\times \frac{{21}}{{100}}\)
\(\displaystyle [(1-\frac{{10}}{{100}})2-1]=\frac{{21}}{{100}}\)
\(\displaystyle \frac{{40x}}{{100}}=(5200-x)\times \frac{{21}}{{100}}\)
\(\displaystyle \frac{{2x}}{5}=(5200-x)\times \frac{{21}}{{100}}\)
200x = 5200×21×5 – x×5×21
200x = 546000 – 105x
305x = 546000
x = 1790
Scheme A = 5200 – 1790 = 3410

Solution: (a)

S.I. on Rs. 4624 for 1 year

= Rs. (4913−4624)

= Rs. 289

Therefore,

Rate=\(\displaystyle =\left( {\frac{{100\times 289}}{{4624\times 1}}} \right)\%\)

\(\displaystyle =6\frac{1}{4}\%\)

So,

\(\displaystyle \begin{array}{l}x{{\left( {1+\frac{{25}}{{400}}} \right)}^{2}}=4624\\\Rightarrow x\times \frac{{17}}{{16}}\times \frac{{17}}{{16}}=4624\\\Rightarrow x=4624\times \frac{{16}}{{17}}\times \frac{{16}}{{17}}\\\Rightarrow x=4096\end{array}\)

Therefore, the sum of money is Rs. 4096

Alternate method,

Given,

In 2 years = Rs. 4624

In 3 years = Rs. 4913

If the year difference in CI is 1 so we can directly calculate the rate %

\(\displaystyle \Rightarrow \)Rate % = \(\displaystyle (4913-4624)\times \frac{{100}}{{4624}}\)

\(\displaystyle \Rightarrow \)Rate % = 6.25 %

\(\displaystyle \Rightarrow \)6.25 % = 1/16

Let, P = 16 , A = 17

For two years

\(\displaystyle \begin{array}{*{20}{c}} P \\ {16} \\ {16} \\ {256} \end{array}\begin{array}{*{20}{c}} A \\ {17} \\ {17} \\ {289} \end{array}\)

\(\displaystyle \Rightarrow \)289 unit = 4624

\(\displaystyle \Rightarrow \)1 unit = 16

\(\displaystyle \Rightarrow \)P = 256 × 16

\(\displaystyle \Rightarrow \)P = 4096

Therefore, the sum of money is Rs. 4096

Solution: (b)
Rate = 5%
Time = 3 years
Compound Interest Rs. 252.20
Effective rate% of CI for 3 years = 15.7625%
Effective rate% of SI for 3 years = 5×3 = 15%
Required SI= \(\displaystyle \frac{{252.50}}{{15.7625}}\times 15\)
=Rs. 240
Alternate method,
Suppose principal be P .
Here , Principal ( P ) = ? , Compound Interest ( CI ) = ₹ 252.20 , Rate ( R ) = 5% , Time = 3 years
\(\displaystyle \begin{array}{l}\Rightarrow P\left[ {{{{\left( {1+\frac{5}{{100}}} \right)}}^{3}}-1} \right]=252.20\\\Rightarrow P\left[ {{{{\left( {\frac{{21}}{{20}}} \right)}}^{3}}-1} \right]=252.20\\\Rightarrow P\left[ {\frac{{21\times 21\times 21-20\times 20\times 20}}{{20\times 20\times 20}}} \right]=252.20\\\Rightarrow P\left[ {\frac{{1261}}{{8000}}} \right]=252.20\\\Rightarrow P=1600\\SimpleInterest=\frac{{PRT}}{{100}}\\\Rightarrow SI=\frac{{1600\times 5\times 3}}{{100}}\\=Rs.240\end{array}\)

Solution: (c)
\(\displaystyle \begin{array}{l}\Pr incipal=Rs.\left( {\frac{{100\times 1000}}{{5\times 4}}} \right)\\=Rs.5000\\Now,P=Rs.10000,\\T=2years,\\R=5\%\\Amount=Rs.\left[ {10000\times {{{\left( {1+\frac{5}{{100}}} \right)}}^{2}}} \right]\\=Rs.\left( {10000\times \frac{{21}}{{20}}\times \frac{{21}}{{20}}} \right)\\=Rs.11025\\\Rightarrow CI=11025-10000\\=Rs.1025\end{array}\)
Alternate method,
Let the principal amount be ‘x’
Therefore, According to the 1st given condition,
\(\displaystyle \Rightarrow \)1000 = (x × 4 × 5)/100
\(\displaystyle \Rightarrow \) x = 5000
Therefore the required amount= \(\displaystyle 2\times 5000\times {{\left( {1+\frac{{0.05}}{1}} \right)}^{{1\times 2}}}-\left( {2\times 5000} \right)\)
=Rs. 1025

Solution: (d)
Simple Interest=Rs.\(\displaystyle \frac{{P\times R\times 2}}{{100}}=\frac{{2PR}}{{100}}\)
Compound Interest=Rs. \(\displaystyle \left[ {P\times {{{\left( {1+\frac{R}{{100}}} \right)}}^{2}}-P} \right]\)
=Rs. \(\displaystyle \left[ {\frac{{P{{R}^{2}}}}{{{{{\left( {100} \right)}}^{2}}}}+\frac{{2PR}}{{100}}} \right]\)
Difference=Rs. \(\displaystyle \left\{ {\left[ {\frac{{P{{R}^{2}}}}{{{{{\left( {100} \right)}}^{2}}}}+\frac{{2PR}}{{100}}} \right]-\frac{{2PR}}{{100}}} \right\}\)
=Rs. \(\displaystyle {\frac{{P{{R}^{2}}}}{{{{{\left( {100} \right)}}^{2}}}}}\)

Solution: (a)

\(\displaystyle \begin{array}{l}P{{\left( {1+\frac{R}{{100}}} \right)}^{3}}=8P\\\Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{3}}=8—–(i)\end{array}\)

Let

\(\displaystyle P{{\left( {1+\frac{R}{{100}}} \right)}^{n}}=16P\)

\(\displaystyle \Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}=16\)
\(\displaystyle \Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}={{2}^{4}}\)

\(\displaystyle \Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}={{\left( {{{2}^{3}}} \right)}^{{\frac{4}{3}}}}\)

\(\displaystyle \Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}={{\left( 8 \right)}^{{\frac{4}{3}}}}——(ii)\)

Substituting the value of 8 from equation (i) in equation (ii)

\(\displaystyle \Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}={{\left[ {{{{\left( {1+\frac{R}{{100}}} \right)}}^{3}}} \right]}^{{\frac{4}{3}}}}\)

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

\(\displaystyle \Rightarrow n=4\)

Therefore, the required time is 4 years

Alternate method

Let the principal =x in Rs.

In the first case Time (n)=3 yrs, amount =8x, rate =R.

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

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

In the second case the amount =16x, Time =n.

\(\displaystyle \begin{array}{l}x{{\left( {1+\frac{R}{{100}}} \right)}^{n}}=16x\\\Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}=16\\\Rightarrow {{\left( {1+\frac{R}{{100}}} \right)}^{n}}={{2}^{4}}\\\Rightarrow n=4\end{array}\)

So, after 4 years, the amount will become 16 times.