Often, students face problems on simple interest (SI) and compound interest (CI), wherein they are asked the difference between compound interest and compound interest over certain period or years. Generally, questions are asked to find the difference for up to 3 years. In competitive exams solving the problem in general method to find the difference between simple interest and compound interest is time taking which is not acceptable for aspirants preparing for competitive exams. So, candidates have to know how to find the difference between compound interest and simple interest within one minute of time duration. Here we will discuss the shortcut tricks for difference between si and ci questions and also provide you few examples for practice.

**Case 1**

Firstly, if it is asked to find the difference between CI and SI for 1 year, then the answer is simple and short, there is no difference between the interest occurring between SI and CI for the first year. We all know that the interest for a certain amount say ‘P’ at rate of interest ‘R’ for year is calculated as same. So, there is no difference between the SI and CI, so it is zero.

**Case 2**

As, in the earlier case we have seen that there is no difference in the interest, as the principal amount is same in SI and CI. But when we are calculating the difference between the compound interest and simple interest for two years, then the principal amount is different in both the case. For compound interest, new principle for second year is the original principal of first year plus the CI of year one. The result will have a difference in the amount of interest generated in case of CI and SI for second year. Let us see how we can find the difference between simple and compound interest for 2 years.

**General method to calculate the difference between SI and CI for 2 years**

We are aware that,

SI = \(\displaystyle \frac{{PRT}}{{100}}\)

CI = A – P

Where,

\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{t}}\)

Where: A = amount

P = the principal

r = the annual interest rate

t = the number of years (here t = 2 years)

Difference = CI – SI

\(\displaystyle \Rightarrow \) Difference = \(\displaystyle P{{\left( {1+\frac{R}{{100}}} \right)}^{t}}-P-\frac{{PRT}}{{100}}\)

\(\displaystyle \Rightarrow \) Difference = \(\displaystyle P{{\left( {1+\frac{R}{{100}}} \right)}^{2}}-P-\frac{{P\times R\times 2}}{{100}}\)

Substituting all the values and subsequent calculations will lead us to the answer. But, we can see that this is time consuming and lengthy process. So, we need to have a formula to find the difference between compound interest and simple interest for two years. In the successive paragraphs we have elaborated difference between compound interest and simple interest with example. On solving the above difference, we can come up with a formula that is given below.

To determine the two years difference in CI and SI, formula is

Difference = \(\displaystyle P{{\left( {\frac{R}{{100}}} \right)}^{2}}\)

or can also be written as

Difference = \(\displaystyle \frac{{SI\times R}}{{100}}\)

Where,

P=principal, R = rate of interest, and SI=Simple Interest (in case, SI is given)

**Case 3**

**Difference between CI and SI for 3 years**

Let us see how we can find the difference between simple and compound interest for 3 years. If we start solving the problem in general method the process is same as in case of calculating for two years, which is explained above. So, let’s directly get to the shortcut trick to find the difference between compound interest and simple interest for three years, is given by

Difference = \(\displaystyle 3P{{\left( {\frac{R}{{100}}} \right)}^{2}}+P{{\left( {\frac{R}{{100}}} \right)}^{3}}\)

or this can also be written as

Difference = \(\displaystyle P{{\left( {\frac{R}{{100}}} \right)}^{2}}\left( {\frac{R}{{100}}+3} \right)\)

or can also be written as

Difference = \(\displaystyle \frac{{P{{R}^{2}}(300+R)}}{{{{{100}}^{3}}}}\)

Where,

P=principal, R = rate of interest

From the above formula we can see that the solving the problems using shortcut tricks is very useful to save time and efforts in any competitive exam. Let us check whether we can save time in finding the difference between CI and SI, by solving some example questions.

**Difference between compound interest and simple interest questions**

**Example 1:**

Bharath had invested same amount of sums at simple as well as compound interest. The time period of both the sums was 2 years and rate of interest too was same 4% per annum. At the end, he found a difference of Rs. 50 in both the interests received. What were the sums invested?

Solution:

We know that,

SI = \(\displaystyle \frac{{PRT}}{{100}}\)

CI = A – P

Where,

A = \(\displaystyle P{{\left( {1+\frac{R}{{100}}} \right)}^{t}}\)

Where: A = amount

P = the principal

r = the annual interest rate

t = the number of years

CI – SI = \(\displaystyle P{{\left( {1+\frac{R}{{100}}} \right)}^{t}}-P-\frac{{P\times R\times T}}{{100}}\)

\(\displaystyle \Rightarrow 50=P{{\left( {1+\frac{4}{{100}}} \right)}^{2}}-P-\frac{{P\times 4\times 2}}{{100}}\)

\(\displaystyle \Rightarrow 50=P\left[ {{{{\left( {\frac{{104}}{{100}}} \right)}}^{2}}-1-\frac{8}{{100}}} \right]\)

\(\displaystyle \Rightarrow 50=P\left[ {\frac{{104}}{{100}}\times \frac{{104}}{{100}}-1-\frac{8}{{100}}} \right]\)

\(\displaystyle \Rightarrow 50=P\left[ {\frac{{104\times 104-10000-800}}{{10000}}} \right]\)

\(\displaystyle \Rightarrow 50=P\left[ {\frac{{16}}{{10000}}} \right]\)

\(\displaystyle \Rightarrow \) P = Rs. 31250

**Alternate method**

Difference between compound interest and simple interest **for two years** is given by the formulae

Difference = \(\displaystyle P{{\left( {\frac{R}{{100}}} \right)}^{2}}\)

or can also be written as

Difference = \(\displaystyle \frac{{SI\times R}}{{100}}\)

Where,

P=principal, R = rate of interest, and SI=Simple Interest

In this case we know rate of interest, time and difference, so

\(\displaystyle 50=P{{\left( {\frac{4}{{100}}} \right)}^{2}}\)

\(\displaystyle \Rightarrow P=\frac{{100\times 100\times 50}}{{16}}\)

P=Rs. 31250

We can see that there is a large scale difference in the amount of time consumed in solving the problem by general method.

**Example 2**

The difference between simple interest and compound interest on a certain amount at 5% per annum for three years is Rs. 122. Find the sum?

Solution:

Given,

Rate = 5%

Time = 3 years

SI – CI = 122

Difference = \(\displaystyle 3P{{\left( {\frac{R}{{100}}} \right)}^{2}}+P{{\left( {\frac{R}{{100}}} \right)}^{3}}\)

\(\displaystyle \Rightarrow 122=3P{{\left( {\frac{5}{{100}}} \right)}^{2}}+P{{\left( {\frac{5}{{100}}} \right)}^{3}}\)

\(\displaystyle \Rightarrow 122=P{{\left( {\frac{5}{{100}}} \right)}^{2}}\left[ {3+\frac{5}{{100}}} \right]\)

\(\displaystyle \Rightarrow 122=P{{\left( {\frac{1}{{20}}} \right)}^{2}}\left[ {3+\frac{1}{{20}}} \right]\)

\(\displaystyle \Rightarrow 122=P{{\left( {\frac{1}{{20}}} \right)}^{2}}\left[ {\frac{{61}}{{20}}} \right]\)

\(\displaystyle \Rightarrow P=\frac{{122\times 20\times 20\times 20}}{{61}}\)

\(\displaystyle \Rightarrow \) P = 16000

Or use another formula

Difference = \(\displaystyle \frac{{P{{R}^{2}}(300+R)}}{{{{{100}}^{3}}}}\)

\(\displaystyle \Rightarrow 122=\frac{{P{{5}^{2}}(300+5)}}{{{{{100}}^{3}}}}\)

\(\displaystyle \Rightarrow 122=\frac{{P\times 25\times 305}}{{{{{100}}^{3}}}}\)

\(\displaystyle \Rightarrow P=\frac{{122\times 1000000}}{{25\times 305}}\)

P=16000

We can see that, by this method finding difference between ci and si for 3 years is very simple and easy.

Aspirants are suggested to practice more number of questions on how to find the difference between compound interest and simple interest for 2 years and 3 years to excel in exams.

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