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 HalfYearly
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
Worksheet on Compound Interest with Periodic Deductions8th Grade Math Practice
From Difference of Compound Interest and Simple Interest to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.