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Quantitative Methods Session 8– 31.07.12 Chapter 5 – Simple Interest & Compound Interest Pranjoy Arup Das.

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Presentation on theme: "Quantitative Methods Session 8– 31.07.12 Chapter 5 – Simple Interest & Compound Interest Pranjoy Arup Das."— Presentation transcript:

1 Quantitative Methods Session 8– Chapter 5 – Simple Interest & Compound Interest Pranjoy Arup Das

2 Interest is the payment to be made for the use of money. Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A will have to repay not only the sum of Rs. X but also some extra money (say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. Bs money. Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using Mr.Bs money & Rs. (X + Y) is termed as the Amount(A) which Mr. A will have to repay to Mr. B after a certain period of time (T). The interest to be paid by the borrower Mr. A to the lender Mr. B is fixed before or while lending the money. Interest is fixed as a percentage of the principal and is called Rate of interest (R). Money is usually lent out for a fixed period of time (T). Interest is calculated on the basis of this time period at the agreed rate, which is charged either yearly, half-yearly or quarterly. T is always calculated in Years. So a loan period of 6 months will have to be taken as 0.5 years.

3 Simple Interest: Original Principal Rs Interest at 1 st year end = 10%of 2000 = Rs. 200 Interest at 2 nd year end = 10%of 2000 = Rs. 200 So at the end of 2 years Mr. A has to pay Rs. ( )= Rs. 2400/- Compound Interest: Original principal Rs Interest at 1 st year end = 10% of 2000 = Rs. 200 New principal=Rs = 2200 Interest at 2 nd year end = 10% of 2200 = Rs. 220 So at the end of 2 years Mr. A has to pay Rs. ( ) = Rs. 2420/- There are two kinds of interest Simple interest & Compound Interest Simple interest is always calculated on the actual principal originally borrowed. Throughout the entire time period, a fixed amount of interest is paid whenever due which is either yearly, half yearly of quarterly For Eg. Mr. A borrows Rs from Mr. B for a period of 2 years at an interest rate of 10% payable yearly. How much will A pay B after 2 years? Compound interest is calculated on the increasing principal. The interest is calculated on the original principal and then added to the original principal which becomes the new principal for calculating the interest on next due date.

4 Part 1 Simple Interest

5 If a sum of money (principal) Rs. P is borrowed at the rate of R% simple interest for T years, then Simple interest amount for T years = 1 st years int. + 2 nd yrs int. +….+ Tth years int = (R% of P) + (R% of P) + (R% of P) + …T times = (R% of P) ( ……………upto T) = (R% of P) * T = I = Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years => P + I = A => P + (R*P*T)/100 = A

6 Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that : a) The loan will be repaid after 3 years b) The borrower will pay a simple interest of 8% per annum (per year) 1) How much interest will Mr. A have to pay at the end of 3 years? 2) What is the amount that Mr. A will have to pay at the end of 3years? Solution: Here the Principal = Rs Time = 3years Rate of interest= 8% Interest at 1 st year end = 8% of 5400 = Rs. 432 Interest at 2 nd year end = 8% of 5400 = Rs. 432 Interest at 3 nd year end = 8% of 5400 = Rs. 432 Total interest at the end of the 3 year loan period = Rs. ( ) = Rs So at the end of 3 years Mr. A has to pay Rs. ( ) = Rs. 6696

7 Alternative Solution: USING FORMULA OF SIMPLE INTEREST Here Principal (P) = Rs.5400, Time (T) = 3 years, Rate of int.(R) =8%pa So the interest to be paid at the end of the loan period of 3 years : Simple Interest (I) = (R*P*T) / 100 = (8 * 5400* 3) / 100 I = Rs The amount to be paid to Mr. B at the end of 3 years : Amount = P + I = = Rs. 6696

8 Page 388 (RSA) Ex 1 SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS Here the Principal = Rs. 5400Time = 10/12 years Rate of int. = 8%pa Interest for 1 year = 8% of 5400 = Rs. 432 That means Interest for 12 months = Rs. 432 Interest for 10 months will be: (10/12) * 432 = Rs. 360 Total interest at the end of the 10 month loan period = Rs. 360 SOLUTION 2 : Here the Principal (P) = 5400Time (T) = 10/12 yearsRate of int.(R) =8%pa So the interest to be paid at the end of the loan period of 3 years : Simple Interest (I) = (R*P*T) / 100 = (8 * 5400* ) / 100 = Rs.__________

9 Page 390 Exercise 21A, Pr no. 4 Solution: Let the sum of money be Rs. x which is the principal (P) It is given that interest on Rs. p.a. in 6 years = Rs Here principal is Rs.x, Rate of interest (R ) is 8% p.a. and the time (T) is 6 years Interest on Rs. 8% for 6 years (I) = (R*P*T)/100 = (8 * x * 6) /100 = Rs.12x/25 So that means, 12x/25 = 8376 => x = (8376 * 25) /12 => x = Rs. ___________

10 Page 388, Ex. 4 Solution: Let the sum of money (P) be Rs. x And let the rate of interest (R ) be y% p.a In the first case Rs. x becomes Rs. 854 in 2 y% p.a y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100 = (2xy)/100 = Rs. xy/50 So, x + (xy/50) = 854 => 50x + xy = 42700…………(i) Similarly, in the 2 nd case Interest on Rs. y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200) So, x + (7xy/200) = => 200x + 7xy = …….(ii) Eliminating xy from (i) & (ii) we can get the value of x and then y.

11 Page 389, Ex. 5 Solution: Let the sum of money (P) be Rs. x So after 8 years, this sum of money becomes Rs. 2x. Since, Principal + Interest = Amount x + (Interest on Rs. x for 8 years) = 2x => (Interest on Rs. x for 8 years) = 2x – x = Rs. x Let the rate of interest be R% p.a. Here principal (P) = Rs. x, Interest (I)= Rs. x and Time (T)= 8 years We know that I = (R * P* T) /100 R = (I * 100) / (P*T) = ( x* 100) / (x * 8) = (100x) /(8x) = 100/8 = 25/2 = 12.5 %

12 Page 389, Ex. 6 Solution: Let the 8% p.a. be Rs. x 10% p.a. = Rs. (8000 – x) Given that total annual interest = Rs. 714 That means, in 1 year: Interest earned on Rs. 8% + Interest 10% on Rs. (8000-x) = 714 => {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714 = > (8x/100) + {(8000-x) / 10} = 714 Solve for x.

13 Session 9; Chapter 5; Part 2 Compound Interest

14 Page 401 Ex 1. SOLUTION 1 : Without using formula. Here Principal (P)= Rs. 8000, Rate ( R) = 10% p.a., Time (T) = 3 years Interest at 1 st year 10% on 8000 for 1 year= (8000 * 10 * 1) /100 = Rs. 800 New principal for 2 nd year = Rs = 8800 Interest at 2 nd year 10% on 8800 for 1 year = (8800 * 10 * 1) /100 = Rs. 880 New principal for 3 rd year = Rs = Rs Interest at 3 rd year 10% on 9680 for 1 year = (9680 * 10 * 1) /100 = Rs. 968 So at the end of 3 years Rs will become Rs. ( ) = Rs /- So the compound interest (CI) = A – P = – 8000 = Rs ALTERNATIVELY : CI = Rs. ( ) = Rs. 2648

15 FORMULAS OF COMPOUND INTEREST: If a sum of money (principal) Rs. P is borrowed /lent at the rate of R% compound interest for T years compounded yearly, then : Compound Interest (CI) on Rs. R% for T years = If Rs. P is borrowed / lent at the rate of R% compound interest for T years compounded half yearly, then : Compound Interest (CI) on Rs. R% for T years = If Rs. P is borrowed /lent at the rate of R% compound interest for T years compounded quarterly, then : Compound Interest (CI) on Rs. R% for T years =

16 SOLUTION 2 : Using formula. Here Principal (P)= Rs. 8000Rate ( R) = 10% p.aTime (T) = 3 years Compound interest (CI) on Rs at the end of 3 10% = CI = = = = = = = Rs. _____________

17 Page 402 Ex 4. SOLUTION 1 : Without using formula. Here Principal (P)= Rs , Time (T) = 1 year, Yearly rate of interest =12% The Interest is compounded half yearly So the half yearly rate of interest = 12 * ½ = 6% per half year Interest at 1 st half year end = 6% of = Rs New principal for 2 nd half year = Rs = Rs Interest at 2 nd half year end = 6% of = Rs So at the end of 1 year Rs will become Rs. ( ) = Rs /- So the compound interest (CI) = – = Rs Or, CI = Rs. ( )= Rs. 3090

18 SOLUTION 2 : Using formula. Here Principal (P)= Rs Rate ( R) = 12% p.a Time (T) = 1 year Compound interest (CI) on Rs at the end of 1 12% pa compounded half yearly = CI = = = = = = = Rs. _____________

19 Page 402 Ex 5. SOLUTION : Without using formula. Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs, Note: 9 months = 3 quarters Yearly rate of interest = 20% pa The Interest is compounded quarterly. In a year there are 4 quarters So the per quarter rate of interest = 20/4 = 5% per quarter. Principal for 1 st quarter = Rs Interest at 1 st quarter end = 5% of 8000 = Rs. 400 New principal for 2 nd quarter = Rs = Rs Interest at 2 nd quarter end = 5% of 8400 = Rs. 420 New principal for 3 rd quarter = Rs = Rs Interest at 3 rd quarter end = 5% of 8820 = Rs. 441 So at the end of 3 quarters or 9 months, Rs will become Rs. ( ) = Rs. 9261/- So the compound interest (CI) = 9261 – 8000 = Rs Or, CI = Rs. ( )= Rs. 1261

20 Page 402 Ex 3. SOLUTION : Here Principal (P)= Rs. 5000, Time (T) = 3years, Rate of interest for 1 st year (R1) = 5% pa Principal for 1 st year = Rs Interest at 1 st year end = 5% of 5000 = Rs. 250 New principal for 2 nd year = Rs = Rs Rate of interest for 2 nd year (R2) = 8% pa Interest at 2 nd year end = 8% of 5250 = Rs. 420 New principal for 3 rd year = Rs = Rs Rate of interest for 3 rd year (R3) = 10% pa Interest at 3 rd year end = 10% of 5670 = Rs. 567 So the amount at the end of 3 years = Rs. ( ) = Rs. 6237/- This means, at the end of 3 years, Rs will become Rs. 6237/- Compound interest (CI) on Rs after 3 5%,8% &10%= 6237 – 5000 = Rs Or, CI on Rs after 3 5%,8% &10%= ( )= Rs. 1237

21 Page 402 Ex 6. SOLUTION : Here Principal (P) = a certain sum of money, CI = Rs. 3783, Time (T) = 3years, Rate of interest (R) = 5% pa Let the principal (P) be Rs. x Assuming interest is being charged annually, We know that, CI = => CI on Rs. 5% for 3 years = => 3783 = Find x then find SI on Rs. 5% for 3 years = (x*5*3)/100


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