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

**● Compound Interest**

**Compound Interest with Growing Principal**

**Compound Interest with Periodic Deductions**

**Compound Interest by Using Formula**

**Compound Interest when Interest is Compounded Yearly**

**Compound Interest when Interest is Compounded Half-Yearly**

**Compound Interest when Interest is Compounded Quarterly**

**Variable Rate of Compound Interest**

**Practice Test on Compound Interest**

**● Compound Interest - Worksheet**

**Worksheet on Compound Interest**

**Worksheet on Compound Interest with Growing Principal**

**8th Grade Math Practice****From Difference of Compound Interest and Simple Interest to HOME PAGE**

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