Question

In how many years \[Rs.7000\] amount to \[Rs.9317\] as compound interest at \[10\% \] per annum compound annually.

Answer 1

Hint: Here we solve for the time period by substituting the values of Amount, Compound interest and rate of interest in the formula for compound interest.
* Compound interest is the interest earned on money that was previously earned as an interest. This leads to increasing the interest.

Complete step-by-step answer:
Given, the value of Amount \[(A)\] is \[Rs.7000\]
The value of Compound interest \[(C.I)\] yielded is \[Rs.9317\]
The value of Rate of Interest \[(R.I)\] is \[10\% \]
Let us assume the time period in which \[Rs.7000\] yields \[Rs.9317\] as compound interest at \[10\% \] per annum compound annually be denoted by \[t\].
Since, we know the formula of compound interest
\[C.I = A{\left( {1 + \dfrac{{R.I}}{{100}}} \right)^t}\]
Where \[C.I\] stands for compound interest, \[A\] stands for amount, \[R.I\] stands for rate of interest and \[t\] is the time.
Substitute the values of \[A,C.I,R.I\] in the formula.
\[9317 = 7000{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Divide both sides by \[7000\]
 \[\dfrac{{9317}}{{7000}} = \dfrac{{7000}}{{7000}}{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Cancel out both denominator and numerator from RHS as they are equal.
\[\dfrac{{9317}}{{7000}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Since, we can write \[9317 = 7 \times 1331,7000 = 7 \times 1000\]
Substitute both the values in numerator and denominator.
\[\dfrac{{7 \times 1331}}{{7 \times 1000}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Cancel out common factors from both numerator and denominator in LHS
\[\dfrac{{1331}}{{1000}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Now we can write \[1331 = (11 \times 11 \times 11)\]
And \[1000 = 10 \times 10 \times 10\]
Therefore LHS becomes \[\dfrac{{1331}}{{1000}} = \dfrac{{{{(11)}^3}}}{{{{(10)}^3}}}\]
Cancel out same terms from denominator and numerator
LHS becomes \[\dfrac{{{{(11)}^3}}}{{{{(10)}^3}}} = {\left( {\dfrac{{11}}{{10}}} \right)^3}\]
Now we substitute the value of LHS back in equation \[\dfrac{{1331}}{{1000}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
\[{\left( {\dfrac{{11}}{{10}}} \right)^3} = {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Solve RHS by taking LCM
RHS becomes \[{\left( {\dfrac{{100 + 10}}{{100}}} \right)^t} = {\left( {\dfrac{{110}}{{100}}} \right)^t}\]
Cancel out ‘0’ from both numerator and denominator.
RHS is \[{\left( {\dfrac{{11}}{{10}}} \right)^t}\]
Now substitute the value of RHS back into the equation \[{\left( {\dfrac{{11}}{{10}}} \right)^3} = {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
\[{\left( {\dfrac{{11}}{{10}}} \right)^3} = {\left( {\dfrac{{11}}{{10}}} \right)^t}\]
Since both sides of the equations have same base, therefore their powers can be equated,
So, \[t = 3\] years.

Note: Students are advised to write the final answer along with the S.I unit. Always try to convert the part on LHS of the equation as a number with some power, this makes it easier to compare the value of time at the end.