What is the formula to calculate the difference between S.I and CI for 2 years?

The formula given below can be used to find the difference between compound interest and simple interest for two years.

The above formula is 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.

Example 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.

Solution :

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

Then we have,

P(R/100)2 = 631

Substitute R = 10.

P(10/100)2 = 631

P(1/10)2 = 631

P(1/100) = 631

Multiply both sides by 100.

P = 631 x 100

P = 63100

So, the value of the investment is $63100.

Example 2 :

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 ?

Solution :

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.

Because, in C.I, interest $600 earned in 1st year earned this $30 in 2nd year.

It can be considered as simple interest for one year.

That is, principle = 600, interest = 30

I = PRT/100

30 = (600 x R x 1)/100

30 = 6R

Divide both sides by 6.

5 = R

So, R = 5%.

Step 3 :

The difference between compound interest and simple interest for two years is

= 1230 - 1200

= 30

Then we have,

P(R/100)2 = 30

Substitute R = 5.

P(5/100)2 = 30

P(1/20)2 = 30

P(1/400) = 30

Multiply both sides by 400.

P = 30 x 400

P = 12000

So, the principal is $12000.

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We will discuss here how to find the difference of compound interest and simple interest.

If the rate of interest per annum is the same under both simple interest and compound interest  then for 2 years, compound interest (CI) - simple interest (SI) = Simple interest for 1 year on “Simple interest for one year”.

Compound interest for 2 years – simple interest for two years

= P{(1 + \(\frac{r}{100}\))\(^{2}\) - 1} - \(\frac{P × r × 2}{100}\)

= P × \(\frac{r}{100}\) × \(\frac{r}{100}\)

= \(\frac{(P × \frac{r}{100}) × r × 1}{100}\)

= Simple interest for 1 year on “Simple interest for 1 year”.

Solve examples on difference of compound interest and simple interest:

1. Find the difference of the compound interest and simple interest on $ 15,000 at the same interest rate of 12\(\frac{1}{2}\) % per annum for 2 years.

Solution:

In case of Simple Interest:

Here,

P = principal amount (the initial amount) = $ 15,000

Rate of interest (r) = 12\(\frac{1}{2}\) % per annum = \(\frac{25}{2}\) % per annum = 12.5 % per annum

Number of years the amount is deposited or borrowed for (t) = 2 year

Using the simple interest formula, we have that

Interest = \(\frac{P × r × 2}{100}\)

           = $ \(\frac{15,000 × 12.5  × 2}{100}\)

           = $ 3,750

Therefore, the simple interest for 2 years = $ 3,750

In case of Compound Interest:

Here,

P = principal amount (the initial amount) = $ 15,000

Rate of interest (r) = 12\(\frac{1}{2}\) % per annum = \(\frac{25}{2}\) % per annum = 12.5 % per annum

Number of years the amount is deposited or borrowed for (n) = 2 year

Using the compound interest when interest is compounded annually formula, we have that

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

A = $ 15,000 (1 + \(\frac{12.5}{100}\))\(^{2}\)

   = $ 15,000 (1 + 0.125)\(^{2}\)

   = $ 15,000 (1.125)\(^{2}\)

   = $ 15,000 × 1.265625

   = $ 18984.375

Therefore, the compound interest for 2 years = $ (18984.375 - 15,000)

                                                             = $ 3,984.375

Thus, the required difference of the compound interest and simple interest = $ 3,984.375 - $ 3,750 = $ 234.375.

2. What is the sum of money on which the difference between simple and compound interest in 2 years is $ 80 at the interest rate of 4% per annum?

Solution:

In case of Simple Interest:

Here,

Let P = principal amount (the initial amount) = $ z

Rate of interest (r) = 4 % per annum

Number of years the amount is deposited or borrowed for (t) = 2 year

Using the simple interest formula, we have that

Interest = \(\frac{P × r × 2}{100}\)

           = $ \(\frac{z × 4  × 2}{100}\)

           = $ \(\frac{8z}{100}\)

           = $ \(\frac{2z}{25}\)

Therefore, the simple interest for 2 years = $ \(\frac{2z}{25}\)

In case of Compound Interest:

Here,

P = principal amount (the initial amount) = $ x

Rate of interest (r) = 4 % per annum

Number of years the amount is deposited or borrowed for (n) = 2 year

Using the compound interest when interest is compounded annually formula, we have that

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

A = $ z (1 + \(\frac{4}{100}\))\(^{2}\)

   = $ z (1 + \(\frac{1}{25}\))\(^{2}\)

   = $ z (\(\frac{26}{25}\))\(^{2}\)

   = $ z × (\(\frac{26}{25}\)) × (\(\frac{26}{25}\))

   = $ (\(\frac{676z}{625}\))

So, the compound interest for 2 years = Amount – Principal

                                                    = $ (\(\frac{676z}{625}\)) - $ z

                                                    = $ (\(\frac{51z}{625}\))

Now, according to the problem, the difference between simple and compound interest in 2 years is $ 80

Therefore,

    (\(\frac{51z}{625}\)) - $ \(\frac{2z}{25}\) = 80

⟹ z(\(\frac{51}{625}\) - \(\frac{2}{25}\)) = 80

⟹ \(\frac{z}{625}\) = 80

⟹ z = 80 × 625

⟹ z = 50000

Therefore, the required sum of money is $ 50000

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