# Quantitative Methods Session 8–

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Quantitative Methods Session 8– 31. 07
Quantitative Methods Session 8– Chapter 5 – Simple Interest & Compound Interest Pranjoy Arup Das

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. B’s money. Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using Mr.B’s 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.

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. Simple Interest: Original Principal Rs. 2000 Interest at 1st year end = 10%of 2000 = Rs. 200 Interest at 2nd year end = 10%of 2000 So at the end of 2 years Mr. A has to pay Rs. ( )= Rs. 2400/- Compound Interest: Original principal Rs. 2000 Interest at 1st year end = 10% of 2000 = Rs. 200 New principal=Rs = 2200 Interest at 2nd year end = 10% of 2200 = Rs. 220 So at the end of 2 years Mr. A has to pay Rs. ( ) = Rs. 2420/-

Part 1 Simple Interest

= (R% of P) + (R% of P) + (R% of P) + …T times
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 = 1st years int. + 2nd 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

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 1st year end = 8% of 5400 = Rs. 432 Interest at 2nd year end = 8% of 5400 Interest at 3nd year end = 8% of 5400 Total interest at the end of the 3 year loan period = Rs. ( ) = Rs. 1296 So at the end of 3 years Mr. A has to pay Rs. ( ) = Rs. 6696

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. 1296 The amount to be paid to Mr. B at the end of 3 years : Amount = P + I = = Rs. 6696

Page 388 (RSA) Ex 1 SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS Here the Principal = Rs Time = 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) = 5400 Time (T) = 10/12 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* ) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4 Solution: Let the sum of money be Rs
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. ___________

Page 388, Ex. 4 Solution: Let the sum of money (P) be Rs
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 2nd 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.

Page 389, Ex. 5 Solution: Let the sum of money (P) be Rs
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 %

Page 389, Ex. 6 Solution: Let the loan @ 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} = Solve for x.

Session 9; Chapter 5; Part 2 Compound Interest

Page 401 Ex 1. SOLUTION 1 : Without using formula. Here Principal (P)= Rs , Rate ( R) = 10% p.a ., Time (T) = 3 years Interest at 1st year 10% on 8000 for 1 year= (8000 * 10 * 1) /100 = Rs. 800 New principal for 2nd year = Rs = 8800 Interest at 2nd year 10% on 8800 for 1 year = (8800 * 10 * 1) /100 = Rs. 880 New principal for 3rd year = Rs = Rs. 9680 Interest at 3rd 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. 2648 ALTERNATIVELY : CI = Rs. ( ) = Rs. 2648

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 : If Rs. P is borrowed /lent at the rate of R% compound interest for T years compounded quarterly, then :

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

SOLUTION 1 : Without using formula.
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 1st half year end = 6% of 25000 = Rs. 1500 New principal for 2nd half year = Rs = Rs Interest at 2nd half year end = 6% of 26500 = Rs. 1590 So at the end of 1 year Rs will become Rs. ( ) = Rs /- So the compound interest (CI) = – = Rs. 3090 Or , CI = Rs. ( )= Rs. 3090

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. _____________

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 1st quarter = Rs. 8000 Interest at 1st quarter end = 5% of 8000 = Rs. 400 New principal for 2nd quarter = Rs = Rs. 8400 Interest at 2nd quarter end = 5% of 8400 = Rs. 420 New principal for 3rd quarter = Rs = Rs. 8820 Interest at 3rd 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. 1261 Or, CI = Rs. ( )= Rs. 1261

Page 402 Ex 3. SOLUTION : Here Principal (P)= Rs. 5000, Time (T) = 3years, Rate of interest for 1st year (R1) = 5% pa Principal for 1st year = Rs. 5000 Interest at 1st year end = 5% of 5000 = Rs. 250 New principal for 2nd year = Rs = Rs. 5250 Rate of interest for 2nd year (R2) = 8% pa Interest at 2nd year end = 8% of 5250 = Rs. 420 New principal for 3rd year = Rs = Rs. 5670 Rate of interest for 3rd year (R3) = 10% pa Interest at 3rd 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. 1237 Or, CI on Rs after 3 5%,8% &10%= ( )= Rs. 1237

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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