Question

☹️☹️Rs 4800 become 7500 at a certain rate of interest compounded annually in 2 years 8 months. then in how many years rs 19200 will amounts to 37500 at the same rate of interest compounded annually?☹️☹️​@ aditi

Answer 1

Given that Rs.4800 becomes Rs.7500 at a certain rate of interest (say $\sf{r}$) compounded annually in 2 years 8 months.

If an amount $\sf{P}$ becomes $\sf{P'}$ at a rate of interest $\sf{r}$ compounded annually in $\sf{y}$ years and $\sf{m}$ months, then,

$\longrightarrow\sf{P'=P\left[1+\dfrac{r}{100}\right]^y\left[1+\dfrac{m}{12}\times\dfrac{r}{100}\right]\quad\quad\dots(1)}$

Since the rate of interest is smaller than 100, the value of $\sf{\dfrac{r}{100}}$ will be smaller than 1. If rate is considered to be very smaller than 100, then $\sf{\dfrac{r}{100}}$ can be neglected.

$\longrightarrow\sf{r\ \textless\textless\ 100\quad\implies\quad\dfrac{r}{100}\ \textless\textless\ 1}$

We know that, for $\sf{|x|\ \textless\textless\ 1,}$

$\longrightarrow\sf{(1+x)^n\approx1+nx\quad \forall\,x,\,n\in\mathbb{R}.}$

Thus we have,

$\longrightarrow\sf{1+\dfrac{m}{12}\times\dfrac{r}{100}=\left[1+\dfrac{r}{100}\right]^{\frac{m}{12}}\quad\quad\left[\,\because\dfrac{r}{100}\ \textless\textless\ 1\,\right]}$

Hence (1) becomes,

$\longrightarrow\sf{P'=P\left[1+\dfrac{r}{100}\right]^y\left[1+\dfrac{r}{100}\right]^{\frac{m}{12}}}$

$\longrightarrow\sf{P'=P\left[1+\dfrac{r}{100}\right]^{y+\frac{m}{12}}\quad\quad\dots(2)}$

Here,

• $\sf{P=4800}$

• $\sf{P'=7500}$

• $\sf{y=2}$

• $\sf{m=8}$

By (2),

$\longrightarrow\sf{4800\left[1+\dfrac{r}{100}\right]^{2+\frac{8}{12}}=7500}$

$\longrightarrow\sf{\left[1+\dfrac{r}{100}\right]^{2+\frac{2}{3}}=\dfrac{7500}{4800}}$

$\longrightarrow\sf{\left[1+\dfrac{r}{100}\right]^{\frac{8}{3}}=\dfrac{25}{16}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{25}{16}\right)^{\frac{3}{8}}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{5^2}{4^2}\right)^{\frac{3}{8}}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{5}{4}\right)^{2\times\frac{3}{8}}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{5}{4}\right)^{\frac{3}{4}}\quad\quad\dots(3)}$

Given that Rs.19200 becomes Rs.37500 at the same rate of interest $\sf{r}$ compounded annually in $\sf{n}$ years (say). We have to find this $\sf{n.}$

Then,

$\longrightarrow\sf{19200\left[1+\dfrac{r}{100}\right]^n=37500}$

$\longrightarrow\sf{\left[1+\dfrac{r}{100}\right]^n=\dfrac{37500}{19200}}$

$\longrightarrow\sf{\left[1+\dfrac{r}{100}\right]^n=\dfrac{125}{64}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{125}{64}\right)^{\frac{1}{n}}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{5^3}{4^3}\right)^{\frac{1}{n}}}$

$\longrightarrow\sf{1+\dfrac{r}{100}=\left(\dfrac{5}{4}\right)^{\frac{3}{n}}\quad\quad\dots(4)}$

Since left hand sides of (3) and (4) are same, we can equate their right hand sides.

$\longrightarrow\sf{\left(\dfrac{5}{4}\right)^{\frac{3}{4}}=\left(\dfrac{5}{4}\right)^{\frac{3}{n}}}$

The powers can be equated since the bases are same.

$\longrightarrow\sf{\dfrac{3}{4}=\dfrac{3}{n}}$

Since the numerators are same, the denominators can be equated. Hence,

$\longrightarrow\sf{\underline{\underline{n=4}}}$

Hence the answer is 4 years.

Answer 2

Answer:

Given :-

Rs 4800 become Rs 7500 at a certain rate of interest compound annually in 2 years 8 months.

Rs 19200 will amount to Rs 37500 at the same rate of interest compound annually.

To Find :-

How many years Rs 19200 will amounts to Rs 37500 at the same rate of interest compounded annually.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{Amount =\: P\bigg(1 + \dfrac{r}{100}\bigg)^{n} \bigg(1 + \dfrac{r}{100}\bigg)^{F}}}}$

where,

P = Principal

r = Rate of Interest

n = Time

F = Fraction Time

Solution :-

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {1}^{{st}}\: case\: :-}}}}}$

$\mapsto$ Rs 4800 become Rs 7500 at a certain rate of interest compound annually in 2 years 8 months.

Given :

Time = 2 years 8 months

Then,

$\implies \sf Time =\: 2\: years + 8\: months$

2 years = 12 + 12 = 24

$\implies \sf Time =\: 24\: months + 8\: months$

$\implies \sf\bold{\green{Time =\: 32\: months}}$

Again, convert it into years we get,

$\implies \sf \dfrac{\cancel{32}}{\cancel{12}}$

$\implies\sf\dfrac{\cancel{16}}{\cancel{6}}$

$\implies \sf\dfrac{\cancel{8}}{\cancel{3}}$

$\implies \sf\bold{\green{2\dfrac{2}{3}}}$

Now, we can say that :

Time (n) = 2

Fraction Time (F) = ⅔

According to the question by using the formula we get,

Given :

Principal = Rs 4800

Amount = Rs 7500

Time = 2

Fraction Time = ⅔

Then,

$\leadsto\sf 7500 =\: 4800\bigg(1 + \dfrac{r}{100}\bigg)^{2} \bigg(1 + \dfrac{r}{100}\bigg)^{\frac{2}{3}}$

$\leadsto \sf \dfrac{75\cancel{00}}{48\cancel{00}} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{2 + \frac{2}{3}}$

$\leadsto \sf \dfrac{75}{48} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{\frac{8}{3}}$

$\leadsto \sf {\bigg(\dfrac{5}{4}\bigg)}^{2} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{\frac{8}{3}}$

$\leadsto \sf \bigg(\dfrac{5}{4}\bigg)^{\cancel{2} \times \frac{3}{\cancel{8}}} =\: \bigg(1 + \dfrac{r}{100}\bigg)$

$\leadsto \sf \bold{\green{\bigg(\dfrac{5}{4}\bigg)^{\frac{3}{4}} =\: \bigg(1 + \dfrac{r}{100}\bigg)\: ----\: (Equation\: No\: 1)}}$

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {2}^{{nd}}\: case\: }}}}}$

$\mapsto$ Rs 19200 will amounts to Rs 37500 at the same rate of interest compound annually.

Given :

Principal = Rs 19200

Amount = Rs 37500

As we know that,

$\longmapsto \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^{n}}}}$

Then,

$\leadsto \sf 37500 =\: 19200\bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto \sf \dfrac{374\cancel{00}}{192\cancel{00}} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto \sf \bigg(\dfrac{375}{192}\bigg) =\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto \sf {\bigg(\dfrac{5}{4}\bigg)}^{3} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto\sf\green{\bold{{\bigg(\dfrac{5}{4}\bigg)}^{\frac{3}{n}} =\: \bigg(1 + \dfrac{r}{100}\bigg)\: ----\: (Equation\: No\: 2)}}$

Now, by putting the equation no 1 in the equation no 2 we get,

$\leadsto \sf \bigg(\dfrac{5}{4}\bigg)^{\frac{3}{4}} =\: \bigg(\dfrac{5}{4}\bigg)^{\frac{3}{n}}$

$\leadsto \sf {\cancel{\bigg(\dfrac{5}{4}\bigg)}}^{\frac{3}{n}} =\: {\cancel{\bigg(\dfrac{5}{4}\bigg)}}^{\frac{3}{4}}$

$\leadsto \sf \dfrac{3}{n} =\: \dfrac{3}{4}$

$\leadsto \sf\bold{\red{n =\: 4\: years}}$

$\therefore$ The time is 4 years.