Question

In what time will Rs.1500 yield Rs.1996.50 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 \[Rs.1500\]
The value of Compound interest \[(C.I)\] yielded is \[Rs.1996.50\]
The value of Rate of Interest \[(R.I)\] is \[10\% \]
Let us assume the time period in which \[Rs.1500\] yields \[Rs.1996.50\] 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 in years.
Substitute the values of \[A,C.I,R.I\] in the formula.
\[1996.50 = 1500{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Divide both sides by \[1500\]
 \[\dfrac{{1996.50}}{{1500}} = \dfrac{{1500}}{{1500}}{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Remove the decimal from the value \[1996.50\] by dividing it from \[100\]
\[\dfrac{{199650}}{{1500 \times 100}} = \dfrac{{1500}}{{1500}}{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Cancel out both denominator and numerator from RHS as they are equal.
\[\dfrac{{199650}}{{1500 \times 100}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Solve LHS by cancelling out \[10\] as exists in both numerator and denominator.
\[\dfrac{{19965}}{{1500 \times 10}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Now we can write \[19965 = 5 \times (3993) = 5 \times \left[ {3 \times (1331)} \right] = 5 \times 3 \times (11 \times 11 \times 11)\]
And \[1500 \times 10 = 5 \times 3000 = 5 \times 3 \times 1000\]
Therefore LHS becomes \[\dfrac{{19965}}{{1500 \times 10}} = \dfrac{{5 \times 3 \times {{(11)}^3}}}{{5 \times 3 \times {{(10)}^3}}}\]
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 \[\dfrac{{19965}}{{1500 \times 10}} = 1 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
\[{\left( {\dfrac{{11}}{{10}}} \right)^3} = {\left( {1 + \dfrac{{10}}{{100}}} \right)^t}\]
Solve RHS by taking LCM
RHS becomes \[{\left( {\dfrac{{100 + 10}}{{100}}} \right)^t} = {\left( {\dfrac{{110}}{{100}}} \right)^t}\]
Cancel out 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,

Therefore, the required answer is \[t = 3\] 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.