Find the amount and the compound interest on 25,000 for 3 years at 6% per annum compounded annually
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We will learn how to use the formula for calculating the compound interest when interest is compounded quarterly. Show Computation of compound interest by using growing principal becomes lengthy and complicated when the period is long. If the rate of interest is annual and the interest is compounded quarterly (i.e., 3 months or, 4 times in a year) then the number of years (n) is 4 times (i.e., made 4n) and the rate of annual interest (r) is one-fourth (i.e., made \(\frac{r}{4}\)). In such cases we use the following formula for compound interest when the interest is calculated quarterly. If the principal = P, rate of interest per unit time = \(\frac{r}{4}\)%, number of units of time = 4n, the amount = A and the compound interest = CI Then A = P(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) Here, the rate percent is divided by 4 and the number of years is multiplied by 4. Therefore, CI = A - P = P{(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) - 1} Note: A = P(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) is the relation among the four quantities P, r, n and A. Given any three of these, the fourth can be found from this formula. CI = A - P = P{(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) - 1} is the relation among the four quantities P, r, n and CI. Given any three of these, the fourth can be found from this formula. Word problems on compound interest when interest is compounded quarterly: 1. Find the compound interest when $1,25,000 is invested for 9 months at 8% per annum, compounded quarterly. Solution: Here, P = principal amount (the initial amount) = $ 1,25,000 Rate of interest (r) = 8 % per annum Number of years the amount is deposited or borrowed for (n) = \(\frac{9}{12}\) year = \(\frac{3}{4}\) year. Therefore, The amount of money accumulated after n years (A) = P(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) = $ 1,25,000 (1 + \(\frac{\frac{8}{4}}{100}\))\(^{4 ∙ \frac{3}{4}}\) = $ 1,25,000 (1 + \(\frac{2}{100}\))\(^{3}\) = $ 1,25,000 (1 + \(\frac{1}{50}\))\(^{3}\) = $ 1,25,000 × (\(\frac{51}{50}\))\(^{3}\) = $ 1,25,000 × \(\frac{51}{50}\) × \(\frac{51}{50}\) × \(\frac{51}{50}\) = $ 1,32,651 Therefore, compound interest $ (1,32,651 - 1,25,000) = $ 7,651. 2. Find the compound interest on $10,000 if Ron took loan from a bank for 1 year at 8 % per annum, compounded quarterly. Solution: Here, P = principal amount (the initial amount) = $ 10,000 Rate of interest (r) = 8 % per annum Number of years the amount is deposited or borrowed for (n) = 1 year Using the compound interest when interest is compounded quarterly formula, we have that A = P(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) = $ 10,000 (1 + \(\frac{\frac{8}{4}}{100}\))\(^{4 ∙ 1}\) = $ 10,000 (1 + \(\frac{2}{100}\))\(^{4}\) = $ 10,000 (1 + \(\frac{1}{50}\))\(^{4}\) = $ 10,000 × (\(\frac{51}{50}\))\(^{4}\) = $ 10,000 × \(\frac{51}{50}\) × \(\frac{51}{50}\) × \(\frac{51}{50}\) × \(\frac{51}{50}\) = $ 10824.3216 = $ 10824.32 (Approx.) Therefore, compound interest $ (10824.32 - $ 10,000) = $ 824.32 3. Find the amount and the compound interest on $ 1,00,000 compounded quarterly for 9 months at the rate of 4% per annum. Solution: Here, P = principal amount (the initial amount) = $ 1,00,000 Rate of interest (r) = 4 % per annum Number of years the amount is deposited or borrowed for (n) = \(\frac{9}{12}\) year = \(\frac{3}{4}\) year. Therefore, The amount of money accumulated after n years (A) = P(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) = $ 1,00,000 (1 + \(\frac{\frac{4}{4}}{100}\))\(^{4 ∙ \frac{3}{4}}\) = $ 1,00,000 (1 + \(\frac{1}{100}\))\(^{3}\) = $ 1,00,000 × (\(\frac{101}{100}\))\(^{3}\) = $ 1,00,000 × \(\frac{101}{100}\) × \(\frac{101}{100}\) × \(\frac{101}{100}\) = $ 103030.10 Therefore, the required amount = $ 103030.10 and compound interest $ ($ 103030.10 - $ 1,00,000) = $ 3030.10 4. If $1,500.00 is invested at a compound interest rate 4.3% per annum compounded quarterly for 72 months, find the compound interest. Solution: Here, P = principal amount (the initial amount) = $1,500.00 Rate of interest (r) = 4.3 % per annum Number of years the amount is deposited or borrowed for (n) = \(\frac{72}{12}\) years = 6 years. A = amount of money accumulated after n years Using the compound interest when interest is compounded quarterly formula, we have that A = P(1 + \(\frac{\frac{r}{4}}{100}\))\(^{4n}\) = $1,500.00 (1 + \(\frac{\frac{4.3}{4}}{100}\))\(^{4 ∙ 6}\) = $1,500.00 (1 + \(\frac{1.075}{100}\))\(^{24}\) = $1,500.00 × (1 + 0.01075)\(^{24}\) = $1,500.00 × (1.01075)\(^{24}\) = $ 1938.83682213 = $ 1938.84 (Approx.) Therefore, the compound interest after 6 years is approximately $ (1,938.84 - 1,500.00) = $ 438.84. ● 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 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 Deductions 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. What is the compound interest of 25000 for 3 years?(1+12100)3=25000(2825×2825×2825)=35123.20. Q.
What does 6% compounded annually mean?Imagine you put $100 in a savings account with a yearly interest rate of 6% . After one year, you have 100+6=$106 . After two years, if the interest is simple , you will have 106+6=$112 (adding 6% of the original principal amount each year.)
What will be the compound interest of Rs 25000 for 3 years at 8% per annum compounded annually?Amount = ₹ 33264.
What is the compound interest for 5000 rupees at 6% for 3 years?Compounded monthly, is the same as he got at 6% interest per annum for 3 years. Compounded annually. Therefore, compound interest is 955.08.
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