The difference between the compound interest and simple interest on a certain

Simple interest is only based on the principal amount of a loan, while compound interest is based on the principal amount and the accumulated interest.

We can explain the difference between simple interest and compound interest in simple words as given below.

Simple Interest

Principal is always constant and it will never  change.

Compound Interest

Principal is a variable and it will be changing for each term.

Example :

$1000 is invested at 10% simple interest and 10% compound interest where the interest is compounded annually.

The picture given below explains the difference between simple interest and compound interest for the above investment.

Important Note :

When we look at the above picture, it is clear that interest earned in simple interest and compound interest is same ($100) for the 1st year when interest is compounded annually in compound interest.

Formula for Difference between Simple and Compound Interest (2 years and 3 years)

The above two formulas are applicable only in the following conditions.

1. The principal in simple interest and compound interest must be same.

2. Rate of interest must be same in simple interest and compound interest.

3. In compound interest, interest has to be compounded annually.

Solved Questions

Question 1 :

The difference between the compound interest and simple interest on a certain investment at 10% per year for 2 years is $631. Find the value of the investment.

Answer :

The difference between compound interest and simple interest for 2 years is 631.

P(R/100)2 = 631

Substitute R = 10.

P(10/100)2 = 631

P(1/10)2 = 631

P(1/100) = 631

Multiply each side by 100.

P = 631 x 100

P = 63100

So, the value of the investment is $63100.

Question 2 :

The difference between the compound interest and simple interest on a certain principal is at 10% per year for 3 years is $31. Find the principal.

Answer :

The difference between compound interest and simple interest for three years is 31.

P(R/100)2(R/100 + 3) = 31

Substitute R = 10.

P(10/100)2(10/100 + 3) = 31

P(1/10)2(1/10 + 3) = 31

P(1/10)2(31/10) = 31

Multiply each side by 1000/31.

P = 31 x (1000/31)

P = 1000

So, the principal is $1000.

Question 3 :

The compound interest and simple interest on a certain sum for 2 years is $ 1230 and $ 1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principal ?

Answer :

To find the principal, we need rate of interest. So, let us find the rate of interest first.

Step 1 :

Simple interest for two years is $1200. So interest per year in simple interest is $600.

So, C.I for 1st year is $600 and for 2nd year is $630.

(Since it is compounded annually, S.I and C.I for 1st year would be same)

Step 2 :

When we compare the C.I for 1st year and 2nd year, it is clear that the interest earned in 2nd year is 30 more than the first year.

Given,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + R/100)^n - 1] – (PTR)/100 = 360

P [(1 + 7.5/100)^2 - 1] – (P(2)(7.5))/100 = 360

P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

∴ The sum is Rs 64000

Hint: Let the sum be \[x\] rupees. We know the compound interest \[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\] , we need to find P. we know simple interest formula \[S.I = P \times \dfrac{r}{{100}} \times T\] . We know compound interest is the difference between amount and principal amount. Since the difference between compound and simple interest is given we can find the value of \[x\] .

Complete step-by-step answer:
We know,
 \[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\] , where A is amount, R is rate of interest, n is number of times the interest is compounded per year.
 \[P = x\] , \[n = 2\] \[r = 10\] , substituting we get,
 \[ \Rightarrow A = x{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\]
 \[ \Rightarrow A = x{\left( {1 + \dfrac{1}{{10}}} \right)^2}\]
Taking L.C.M and simplifying we get,
 \[ \Rightarrow A = x{\left( {\dfrac{{10 + 1}}{{10}}} \right)^2}\]
 \[ \Rightarrow A = x{\left( {\dfrac{{11}}{{10}}} \right)^2}\]
We know that compound interest is the difference between the amount of money accumulated after n years and the principal amount.
 \[C.I = A - P\]
 \[ \Rightarrow C.I = x{\left( {\dfrac{{11}}{{10}}} \right)^2} - x\] .
Now to find the simple interest we have, \[S.I = P \times \dfrac{r}{{100}} \times T\]
Substituting the known values,
 \[ \Rightarrow S.I = x \times \dfrac{{10}}{{100}} \times 2\]
 \[ \Rightarrow S.I = x \times \dfrac{1}{{10}} \times 2\]
 \[ \Rightarrow S.I = \dfrac{x}{5}\]
Given the difference between compound and simple interest is 500
 \[ \Rightarrow C.I - S,I = 500\]
Substituting C.I and S.I we get
 \[ \Rightarrow x{\left( {\dfrac{{11}}{{10}}} \right)^2} - x - \dfrac{x}{5} = 500\]
Simple division \[\dfrac{{11}}{{10}} = 1.1\] and \[\dfrac{1}{5} = 0.2\] we get,
 \[ \Rightarrow x{(1.1)^2} - x - 0.2x = 500\]
 \[ \Rightarrow 1.21x - 1x - 0.20x = 500\]
 \[ \Rightarrow 0.21x - 0.20x = 500\]
Taking x as common,
 \[ \Rightarrow (0.21 - 0.20)x = 500\]
 \[ \Rightarrow 0.01x = 500\]
 \[ \Rightarrow x = \dfrac{{500}}{{0.01}}\]
Multiply numerator and denominator by 100.
 \[ \Rightarrow x = 50,000\]
That is \[P = 50,000\] .
 \[50,000\] Rupees is the sum when the interest is compounded annually.
So, the correct answer is “\[P = 50,000\]”.

Note: Here we used three formulas. Remember the formula for simple interest, compound interest and amount formula. We can also take P as P as it is, and solve for P. Above all we did is substituting the given data in the formula and simplifying. Principal amount is the initial amount you borrow or deposit.

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