What is the difference between the simple interest and compound interest on the principal of 10000?

Question

Vedantu Content Team · Accepted Answer

Hint: For simplifying the problem of simple interest and compound interest on a particular amount of money, we must have a ‘rate of interest’ for which the amount is deposited and a ‘time period’ for which the amount is held. Hence, we will assume the values of the rate of interest and time period to be the same for calculating both simple and compound interest in the problem.Formula used:Formula used for computing simple interest is $Simple\,Interest = \dfrac{{ Principal\,Amount \times Rate\,of\,Interest \times time\,period}}{{100}}$Complete step by step solution: 
Principal amount of money is given as: $(P)$ = $Rs.10000$Let us consider the value of Rate of Interest $(r)$ be $5\% $ and the value of time period $(t)$ be $3years$Now, we know, by the formula of Simple Interest: -$SI = \dfrac{{P \times r \times t}}{{100}}$ where, SI = Simple InterestP = Principal Amountr = Rate of Interestt = Time PeriodNow, Apply the formula for calculating the simple interest on principal of $Rs.10000$ $SI = \dfrac{{10000 \times 5 \times 3}}{{100}} = Rs.1500$i.e., The Simple Interest on Principal Amount of $Rs.10000$ is $Rs.1500$for given conditions.Now, to find the compound interest, we need to calculate the interest added to the principal amount of each year as: -For the first-year$P = Rs.10000$$t = 1$Simple Interest of 1st year, $S{I_1} = \dfrac{{10000 \times 5 \times 1}}{{100}} = Rs.500$Amount at the end of 1 year, ${A_1} = Rs.10000 + Rs.500 = Rs.10500$For second-year $P = {A_1} = Rs.10500$$t = 1$Simple Interest of 2nd year, $S{I_2} = \dfrac{{10500 \times 5 \times 1}}{{100}} = Rs.525$Amount at the end of 2 years, ${A_2} = Rs.10500 + Rs.525 = Rs.11025$For third-year$P = {A_2} = Rs.11025$$t = 1$Simple Interest of 3rd year, $S{I_3} = \dfrac{{11025 \times 5 \times 1}}{{100}} = Rs.551.25$Amount at the end of 3 years, ${A_3} = Rs.11025 + Rs.551.25 = Rs.11576.25$Then, the Compound Interest (CI) will be calculated as: -Compound Interest = Amount at the end of 3rd year- Principal Amount$CI = {A_3} - P = 11576.25 - 10000 = Rs.1576.25$That means, the Compound Interest on the Principal Amount of $Rs.10000$ is $Rs.1576.25$for given conditions.Clearly, it is shown by the above computations that both Compound and Simple Interest are different for the same amount of principal having the same rate of interest and time period i.e., $CI > SI$Hence, the difference between the simple interest and compound interest on the principal of $Rs.10000$ is: -$difference = CI - SI = 1576.25 - 1500 = Rs.76.25$Note: The term derivative is the root of the first order differential equation. A solid understanding of derivatives can make learning differential equations easier and more digestible. A derivative is a mathematical tool for measuring the rate of change of values in a function at a specific point in the function.

For the first-year$P = Rs.10000$$t = 1$	Simple Interest of 1^st year, $S{I_1} = \dfrac{{10000 \times 5 \times 1}}{{100}} = Rs.500$	Amount at the end of 1 year, ${A_1} = Rs.10000 + Rs.500 = Rs.10500$
For second-year $P = {A_1} = Rs.10500$$t = 1$	Simple Interest of 2^nd year, $S{I_2} = \dfrac{{10500 \times 5 \times 1}}{{100}} = Rs.525$	Amount at the end of 2 years, ${A_2} = Rs.10500 + Rs.525 = Rs.11025$
For third-year$P = {A_2} = Rs.11025$$t = 1$	Simple Interest of 3^rd year, $S{I_3} = \dfrac{{11025 \times 5 \times 1}}{{100}} = Rs.551.25$	Amount at the end of 3 years, ${A_3} = Rs.11025 + Rs.551.25 = Rs.11576.25$