☹️☹️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. Created by foreverbesties1. Math

Given that Rs.4800 becomes Rs.7500 at a certain rate of interest (say ) compounded annually in 2 years 8 months. If an amount becomes at a rate of interest compounded annually in years and months, then,Since the rate of interest is smaller than 100, the value of will be smaller than 1. If rate is considered to be very smaller than 100, then can be neglected. We know that, for Thus we have,Hence (1) becomes,Here,By (2),Given that Rs.19200 becomes Rs.37500 at the same rate of interest compounded annually in years (say). We have to find this Then,Since left hand sides of (3) and (4) are same, we can equate their right hand sides.The powers can be equated since the bases are same.Since the numerators are same, the denominators can be equated. Hence,Hence the answer is 4 years.

☹️☹️Rs 4800 become 7500 at a certain rate of interest compounded annually in 2 y...

shadowsabers03

Verified answer

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.

shiny12014

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 :-

$>where,P = Principalr = Rate of Interestn = TimeF = Fraction TimeSolution :-<img src=$

$\mapsto$ 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

$><img src=$

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,

$><img src=$

$\mapsto$ 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,

$><img src=$

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

$\therefore$ The time is 4 years.

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