Question

in what time will 8000 yield 2648 as compound interest at 10% p.a. compounded annually.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A sum of money Rs 8000 yield Rs 2648 as compound interest at the rate of 10 % per annum compounded annually.

So,

➢ Principal, P = Rs 8000

➢ Compound Interest, CI = Rs 2648

➢ Rate of interest, r = 10 % per annum compounded annually.

➢ Let assume that the time period be n years.

So, We know that,

➢ Compound interest (CI) on a certain sum of money Rs p invested at the rate of r % per annum compounded annually for n years is

$\red{\boxed{ \rm{ \: \: \: C \: I = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - p \: \: \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:2648 = 8000 {\bigg[1 + \dfrac{10}{100} \bigg]}^{n} - 8000$

$\rm :\longmapsto\:2648 + 8000 = 8000 {\bigg[1 + \dfrac{1}{10} \bigg]}^{n}$

$\rm :\longmapsto\:10648 = 8000 {\bigg[\dfrac{10 + 1}{10} \bigg]}^{n}$

$\rm :\longmapsto\:\dfrac{10648}{8000} = {\bigg[\dfrac{11}{10} \bigg]}^{n}$

$\rm :\longmapsto\:\dfrac{1331}{1000} = {\bigg[\dfrac{11}{10} \bigg]}^{n}$

$\rm :\longmapsto\:\dfrac{11 \times 11 \times 11}{10 \times 10 \times 10} = {\bigg[\dfrac{11}{10} \bigg]}^{n}$

$\rm :\longmapsto\: {\bigg[\dfrac{11}{10} \bigg]}^{3} = {\bigg[\dfrac{11}{10} \bigg]}^{n}$

So, on comparing, we get

$\red{\boxed{\bf :\longmapsto\: \bf{ \: \: \: n \: \: = \: \: 3 \: \: years \: \: \: }}}$

Additional Information :-

1. Amount on a certain sum of money Rs p invested at the rate of r % per annum compounded annually for n years is

$\red{\boxed{ \rm{ \: \: \: Amount = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: \: }}}$

2. Amount on a certain sum of money Rs p invested at the rate of r % per annum compounded semi - annually for n years is

$\red{\boxed{ \rm{ \: \: \: Amount = p {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: \: }}}$

3. Amount on a certain sum of money Rs p invested at the rate of r % per annum compounded quarterly for n years is

$\red{\boxed{ \rm{ \: \: \: Amount = p {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: \: }}}$

Answer 2

Answer:

P=8000, SI=2648, A= 8000+2648 ie : 10648

R = 20%, halfyearly R = 10% , n =?  

A=p(1+r/100)^n  

A/P =(1+r/100)^n  

(10648/8000) = (110/100)^n  

(1331/1000) = (11/10)^n  

[11/10]^3= [11/10]^n

By the laws of indices when bases are equal, then the powers are equal .  

n=time  

n =3  

Mode of interest : yearly  

n = 3  years.  

Step-by-step explanation:

thanks.