Question

At compound interest, if a certain sum of money doubles in 'n ' years, then the amount will be four fold in..
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Answer 1

hey... here's ur solution,
let P = Rs x
then, amount = Rs 2x
time = n years


( Use Attachment )


Hence,

N = 2n years.


hope it might be helpful !!

Answer 2

Given :

A certain sum of money doubles in n years

To Find :

The Amount will be four fold in how many years

Solution :

Let The principal = Rs p

Amount = Rs A

Time period = n

Rate of interest = r %

From Compound interest

Amount = Principal × $(1+\dfrac{rate}{100})^{time}$

A/Q

2 p = p × $(1+\dfrac{r}{100})^{n}$

Or, $(1+\dfrac{r}{100})^{n}$ = $\dfrac{2p}{p}$

i.e  $(1+\dfrac{r}{100})^{n}$ = 2

taking power $\dfrac{1}{n}$ both side

So,  1 + $\dfrac{r}{100}$  = $2^{\dfrac{1}{n} }$        ..........1

Again

Let The Amount will be four fold in T years

i.e   A = 4 p

So, Amount = Principal × $(1+\dfrac{rate}{100})^{time}$

4 p = p × $(1+\dfrac{r}{100})^{T}$

Or, $(1+\dfrac{r}{100})^{T}$ = $\dfrac{4p}{p}$

i.e  $(1+\dfrac{r}{100})^{T}$ = 4

Taking power $\dfrac{1}{T}$ both side

So, 1 + $\dfrac{r}{100}$  = $4^{\dfrac{1}{T} }$

Now, From eq 1

    $2^{\dfrac{1}{n} }$  = $4^{\dfrac{1}{T} }$

Or,  $2^{\dfrac{1}{n} }$  = $(2^{2})^{\dfrac{1}{T} }$

i.e $2^{\dfrac{1}{n} }$  = $(2^{})^{\dfrac{2}{T} }$

As common base both side

So,   $\dfrac{1}{n} = \dfrac{2}{T}$

i.e   T = 2 n

Or, The Time period = 2 n

Hence, The Amount will be four fold in 2 n years . Answer