Solution: (c)
The sum of money doubles itself in 20 years,
So,
\(\displaystyle \begin{array}{l}2P=P\times {{\left( {1+\frac{r}{{100}}} \right)}^{{20}}}\\{{\left( {1+\frac{r}{{100}}} \right)}^{{20}}}=2——-(i)\end{array}\)
And let it treble itself in n years. so,
\(\displaystyle 3={{\left( {1+\frac{r}{{100}}} \right)}^{n}}——-(ii)\)
\(\displaystyle \Rightarrow 1+\frac{r}{{100}}={{3}^{{\frac{1}{n}}}}——-(iii)\)
Substituting equation (iii) in (i) we get
\(\displaystyle {{3}^{{\frac{{20}}{n}\ }}}=2\)
Taking log on both sides
\(\displaystyle \begin{array}{l}\frac{{20}}{n}\log 3=\log 2\\\frac{{20}}{n}=\frac{{\log 2}}{{\log 3}}\\n=\frac{{20\log 3}}{{\log 2}}\\n=\frac{{20\times 0.477}}{{0.3010}}\\n=31.74\end{array}\)
It will treble itself in 31.7 years

Solution: (a)
CI = \(\displaystyle p[{{(1+\frac{r}{{100}})}^{t}}-1]\)
\(\displaystyle 594.5=5800[{{(1+\frac{r}{{100}})}^{2}}-1]\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{594.5}}{{5800}}+1={{(1+\frac{r}{{100}})}^{2}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{6394.5}}{{5800}}={{(1+\frac{r}{{100}})}^{2}}\)
\(\displaystyle 1.05=(1+\frac{r}{{100}})\)
\(\displaystyle \Rightarrow \)\(\displaystyle 1.05-1=\frac{r}{{100}}\) r = 0.05 = 5%

Solution: (b)

N=1year

R=10 %

We have SI\(\displaystyle =\frac{{PRT}}{{100}}\)=\(\displaystyle =\frac{{P\times 1\times 10}}{{100}}\)=0.10P

When interest being compounded for half yearly, for 1 year

We have, N=2

And R=10/2 =5 %

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

\(\displaystyle \begin{array}{l}=P\times {{\left( {1+\frac{5}{{100}}} \right)}^{2}}\\=P\times {{\left( {1+0.05} \right)}^{2}}\\=P\times {{\left( {1.05} \right)}^{2}}\end{array}\)

Amount=1.1025P

And C.I.=A−P=1.1025P−P=0.1025P

Given, C.I.−S.I.=Rs. 180

\(\displaystyle \Rightarrow \)0.1025P −0.10P=180

\(\displaystyle \Rightarrow \)0.0025P=180

\(\displaystyle \Rightarrow \)P=180/0.0025=Rs. 72000

Alternate method

Rate % = 10%,

Time = 1 year

Case (I) : When interest is calculated yearly, Rate = 10%

Case (II) : When interest is calculated half yearly

⇒New rate %=10/2=5%

Time = 1×2=2 years

Effective rate% = \(\displaystyle 5+5+\frac{{5\times 5}}{{100}}\)=10.25%

Difference in rates = (10.25−10)%=0.25%

As per the question,

0.25% of sum = Rs 180

Sum = \(\displaystyle \frac{{180}}{{0.25}}\times 100\)=Rs. 72000

One more method:

When the money is compounded half yearly the effective rate of interest for 6 months =\(\displaystyle \frac{10}{2}\) =5%

=\(\displaystyle \frac{5}{100}\)= \(\displaystyle \frac{1}{20}\)

Let principal = \(\displaystyle {{\left( {20} \right)}^{2}}\)= 400

\(\displaystyle {\Rightarrow 1units\to 180}\)

\(\displaystyle {\Rightarrow \Pr incipal=180\times 400}\)

\(\displaystyle {=Rs.72000}\)

Solution: (d)

Effective Rate of CI for 3 years = 15.7625%
Effective Rate of SI for 3 years = 5 × 3 = 15%
According to the question
(15.7625−15)%of sum = Rs. 36.600.7625% of sum = Rs. 36.60
Sum = \(\displaystyle =\frac{{36.60}}{{0.7625}}\times 10=4800\)
Alternate method,
Simple Interest= \(\displaystyle =\frac{{PRT}}{{100}}\)
\(\displaystyle \begin{array}{l}=\frac{{P\times 5\times 10}}{{100}}\\=0.15P\end{array}\)
Compound Interest= \(\displaystyle =P{{\left( {1+\frac{5}{{100}}} \right)}^{2}}-P\)
$\(\displaystyle \begin{array}{l}=P{{\left( {1.05} \right)}^{2}}-P\\=1.15763P-P\\=0.15763P\end{array}\)
As per question,
CI – SI = 36.60
\(\displaystyle \Rightarrow \)0.15763 P – 0.15 P = 36.60
\(\displaystyle \Rightarrow \)0.007625 P = 36.60
\(\displaystyle \Rightarrow \)P = 4800.

Solution: (a)

SI=\(\displaystyle =\frac{{PRT}}{{100}}\)

\(\displaystyle =\frac{{5000\times 4\times 3}}{{100}}=600\)

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

\(\displaystyle =5000\times {{\left( {1+\frac{4}{{100}}} \right)}^{3}}-5000\)

\(\displaystyle =5000\times {{\left( {\frac{{104}}{{100}}} \right)}^{3}}-5000\)

\(\displaystyle =5624.32-5000\)

\(\displaystyle =624.32\)

Difference=624.32-600=24.32

Alternate method:

Difference in rate of interest at 4%

\(\displaystyle =\left( {4+4+4+\frac{{16+16+16}}{{100}}+\frac{{4\times 4\times 4}}{{{{{(100)}}^{2}}}}} \right)-3\times 4\)

\(\displaystyle =\frac{{48}}{{100}}+\frac{{64}}{{{{{(100)}}^{2}}}}=0.48+0.0064=0.4864\%\)

Difference in amount = \(\displaystyle \frac{{5000\times 0.4864}}{{100\times 10000}}=24.32\)

Solution: (e)
For 2 years, difference in CI and SI = \(\displaystyle \frac{{{{{\Pr }}^{2}}}}{{{{{100}}^{3}}}}\)
For 3 years difference is \(\displaystyle \frac{{{{{\Pr }}^{2}}(r+300)}}{{{{{100}}^{3}}}}\)
\(\displaystyle \frac{{\frac{{{{{\Pr }}^{2}}}}{{{{{100}}^{3}}}}}}{{\frac{{{{{\Pr }}^{2}}(r+300)}}{{{{{100}}^{3}}}}}}=\frac{{16}}{{49}}\)
So \(\displaystyle \frac{{100}}{{r+300}}=\frac{{16}}{{49}}\)
Solve, \(\displaystyle r=6\frac{1}{4}\)