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https://byjus.com/cbse-sample-papers-for-class-8-maths/ We can see that interest increases for successive years.
For detailed discussion on compound interest, download BYJU’S -The learning app. Compound interest finds its usage in most of the transactions in the banking and finance sectors and also in other areas as well. This is called the future value of the investment and is calculated with the following formula. Required fields are marked *, \(P\left(1~+~\frac{R}{100}\right) \left(1~+~\frac{R}{100}\right)\), \(P \left[\left(1~+~ \frac{R}{100}\right)^n~ –~ 1\right]\), \(P_2\left(1~+~\frac{R}{2~×~100}\right)\), \(P \left(1~+~\frac{R}{2~×~100}\right)^2\), \(P\left(1~+~\frac{R}{2~×~100}\right)^2\), \(P\left(1~+~\frac{\frac{R}{2}}{100}\right)^{2~×~1}\), \(P\left(1~+~\frac{R’}{100}\right)^{T’}\), \(= 10000 \left ( 1 + \frac{10}{100} \right )^{2} = 10000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right )= Rs.12100\), \(A = P \left ( 1 + \frac{R}{100} \right )^{T}\), \(P\times \frac{R}{100} = 5000 \times \frac{10}{100} =500\), \(P\times \frac{R}{100}\left (1 + \frac{R}{100} \right ) = 5000 \times \frac{10}{100}\left ( 1 + \frac{10}{100} \right ) = 550\), \(2000~×~\left(\frac{21}{20}\right)^3 = Rs.2315.25\). Let, Principal amount = \(P\), Time = \(n\) years, Rate = \(R\). Example, 6% interest with " monthly compounding " does not mean 6% per month, it means 0.5% per month (6% divided by 12 months), and is worked out like this: FV = PV × (1+r/n)n = $1,000 × (1 + 6%/12)12 = $1,000 × (1 + 0.5%)12 Interest (I1) = \(P\times \frac{R}{100} = 5000 \times \frac{10}{100} =500\), Interest (I2) = \(P\times \frac{R}{100}\left (1 + \frac{R}{100} \right ) = 5000 \times \frac{10}{100}\left ( 1 + \frac{10}{100} \right ) = 550\), Total Interest = I1+ I2 = 500 + 550 = Rs. Now we have the final amount at the end of 1 year: \(A\) = \(P\left(1~+~\frac{R}{2~×~100}\right)^2\), \(A\) = \(P\left(1~+~\frac{\frac{R}{2}}{100}\right)^{2~×~1}\), Let \(\frac{R}{2}\) = \(R'\); \(2T\) = \(T’\), the above equation can be written as, [for the above case \(T\) = \(1\) year], \(A\) = \(P\left(1~+~\frac{R’}{100}\right)^{T’}\). Its population declines at a rate of 10% per annum. The price of a radio is Rs 1400 and it depreciates by 8% per month.
This interest varies with each year for the same principal amount.
Let us solve various examples to understand the concepts in a better manner.
Solution: Principal, \(P\) = \(Rs.2000\), Time, \(T’\) = \(2~×~\frac{3}{2}\) years = 3 years, Rate, \(R’\) = \(\frac{10%}{2}\) = \(5%\), amount, \(A\) can be given as: \(A = P ~\left(1~+~\frac{R}{100}\right)^n\), \(A = 2000~×~\left(1~+~\frac{5}{100}\right)^3\), = \(2000~×~\left(\frac{21}{20}\right)^3 = Rs.2315.25\), \(CI = A – P = Rs.2315.25~ –~ Rs.2000\) = \(Rs.315.25\). Other than the first year, the interest compounded annually is always greater than that in case of simple interest. The count of a certain breed of bacteria was found to increase at the rate of 2% per hour. Compound Interest Formula The formula for the Compound Interest is, This is the total compound interest which is just the interest generated minus the principal amount. In the formula, A represents the final amount in the account after t years compounded 'n' times at interest rate 'r' with starting amount 'p'. First, we will look at the simplest case where we are using the compound interest formula to calculate the value of an investment after some set amount of time. P is the principal; that's the amount you start with. Calculate the compound interest and amount he has to pay at the end of 2 years. The formula for compound interest is P (1 + r/n)^ (nt), where P is the initial principal balance, r is the interest rate, n is the number of times interest is compounded per time period and t is the number of time periods. Since interest is compounded half-yearly, the principal amount will change at the end of first 6 months. Hence, we can conclude that the interest charged by the bank is not simple interest, this interest is known as compound interest.
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For the total accumulated wealth (or amount), the formula is given as: Students can also use a compound interest calculator, to solve compound interest problems in an easier way. 1050.
Illustration 1: A sum of Rs.10000 is borrowed by Akshit for 2 years at an interest of 10% compounded annually. 10000, Rate = 10%, and Time = 2 years, From the table shown above it is easy to calculate the amount and interest for the second year, which is given by-, Amount(\(A_{2}\)) = \(P\left (1+\frac{R}{100} \right )^{2}\), \(A_{2}\)= \(= 10000 \left ( 1 + \frac{10}{100} \right )^{2} = 10000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right )= Rs.12100\), Compound Interest (for 2nd year) = \(A_{2} – P \) = 12100 – 10000 = Rs. Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. Compound interest basics have been explained here along with solved examples.
We can also reduce the formula of compound interest of yearly compounded for quarterly as given below: \(CI =P(1+\frac{\frac{R}{4}}{100})^{4T}-P\). Let us calculate the compound interest on a principal, P kept for 1 year at interest rate R % compounded quarterly. Directly we can use the formula for calculating the interest for second year, which will give us the same result.
The interest for the next six months will be calculated on the amount remaining after the first six months. Illustration 3: Calculate the compound interest to be paid on a loan of Rs.2000 for 3/2 years at 10% per annum compounded half-yearly? When we observe our bank statements, we generally notice that some interest amount is credited to our account every year. nice questions , but some hard questions must be added. Similarly if we proceed further to \(n\) years, we can deduce: \(A\) = \(P\left(1~+~\frac{R}{100}\right)^n\), \(CI\) = \(A~–~P\) = \(P \left[\left(1~+~ \frac{R}{100}\right)^n~ –~ 1\right]\). Also, interest for the third quarter will be calculated on the amount remaining after the first 6 months and for the last quarter on the remaining after the first 9 months. Principal (P) = Rs.5000 , Time (T)= 2 year, Rate (R) = 10 %, We have, Amount, \(A = P \left ( 1 + \frac{R}{100} \right )^{T}\), \(A = 5000 \left ( 1 + \frac{10}{100} \right )^{2} = 5000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right ) = 50 \times 121 = Rs. Compound Interest Formula Example #3 Case of Compounded Quarterly Fin International Ltd makes an initial investment of $ 10,000 for a period of 2 years. The population of the town decreases by 10% every year. A stands for the amount of money that has accumulated. Find the value of the investment after the two years if the investment earns the return of 2 % compounded quarterly. Simple interest at the end of first six months, \(A_1\) = \(P~ + ~SI_1\) = \(P ~+~ \frac{P~×~R~×~1}{2~×~100}\) = \(P \left(1~+~\frac{R}{2~×~100}\right)\) = \(P_2\), Simple interest for next six months, now the principal amount has changed to \(P_2\), \(SI_2\) = \(\frac{P_2~×~R~×~1}{100~×~2}\), \(A_2\) = \(P_2~ +~ SI_2\) = \(P_2 ~+~ \frac{P_2~×~R~×~1}{2~×~100}\) = \(P_2\left(1~+~\frac{R}{2~×~100}\right)\) = P(1 + R/ 2×100)(1 + R/2×100) = \(P \left(1~+~\frac{R}{2~×~100}\right)^2\). This data will be helpful in determining the interest and amount in case of compound interest easily.
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