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 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 Problems on Compound Interest 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
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Discussion :: Compound Interest - General Questions (Q.No.14)
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