Question

In how many years will Rs.6750 amount to Rs.$8192$ at $6\dfrac{2}{3}\% $ p.a. compounded annually?

Answer 1

Hint: Note that Rs.$6750$ is the principal ( $P$), Rs.$8192$ is the amount ( $A$) and $6\dfrac{2}{3}\% $ p.a. is the rate of interest ( $r$). The number of years could be calculated using the formula of compound interest, i.e.,
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$
Where $n$ is the number of years. Substituting the values in the formula, $n$ comes out to be $3$.

Complete step by step answer:
We are given
Rate $ = 6\dfrac{2}{3}\% = \dfrac{{20}}{3}\% $ per annum ………………...(1)
Principal amount = Rs.$6750$ ……………….. (2)
Let the number of years be $n$
The amount at the end of $n$ years = Rs.$8192$ ………………………….(3)
The easiest way to calculate n is by applying the formula of compound interest which is;
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$ ………………….(4)
You will find that you have the value of amount ($A$), principal ($P$), and rate ($r$). n is the number of years.
By substituting the values of these variables in the formula, n can be easily found out.
Substituting the values from equation (1), (2), (3) in (4), we get
$\Rightarrow 8192 = 6750{\left( {1 + \dfrac{{20}}{{3 \times 100}}} \right)^n}$
$\Rightarrow \dfrac{{8192}}{{6750}} = {\left( {1 + \dfrac{1}{{15}}} \right)^n} $
$\Rightarrow \dfrac{{4096}}{{3375}} = {\left( {\dfrac{{16}}{{15}}} \right)^n} $
Breaking the numerator and denominator on LHS and RHS in prime factors,
$\Rightarrow \left(\dfrac{2^4}{5 \times 3} \right)^n= \left(\dfrac{2^4}{5 \times 3} \right)^3 $
Comparing the above equation, ee get
$n = 3$

Therefore, the number of years $ = 3$

Note:
The easiest way to solve the equations formed in these type of questions is to break the number into prime factors, so that on comparing, we may get the value of n. Also, do not confuse the amount and principal. Students often do that. The amount is the total amount after n years whereas the principal is the initial amount.