Mathematics of Finance - newagepublishers.comnewagepublishers.com/samplechapter/002031.pdf · CHAPTER 1 Mathematics of Finance 1.1. INTRODUCTION In this chapter we will discuss mathematical

CHAPTER 1

Mathematics of Finance

1.1. INTRODUCTION

In this chapter we will discuss mathematical methods and formulae which are helpful in businessand personal finance. One of the fundamental concepts in the mathematics of finance is the ‘timevalue of money’, i.e., the value of a particular sum of money at different points of time. Forexample, if you have Rs. 100 today, what will it be worth at the end of one year?

1.1.1. Simple Interest (S.I.)

When we take loan from Bank for a certain period we pay off the loan and a certain sum of moneyto the Bank for the use of the money lent. The loan amount is called the “Principal” and theadditional amount paid for the use of the loan is called “Interest”. The sum of the principal andthe interest due at the end of the period is called the “Amount”, i.e., Amount = Principal +Interest. The borrower is called the “Debtor” and the lender is called the “Creditor”. If Rs. 10 ispaid as interest on Rs. 100 for 1 year, the rate of interest is said to be 10% p.a. Simple Interest(S.I.) is the interest on the principal alone for the time for which it is used.

S.I. on the principal P for N years at the rate of R% p.a. is given by:

S.I. = P.N. R

100, where i =

R

100 or R = 100i; and the amount (A) is given by:

A = P + S.I. = P + P.N. × i = P (1 + Ni) or P = A

Ni1 +Unless otherwise stated the interest is always calculated yearly.

Example 1.1. (i) Find the simple interest on Rs. 1,000 for 10 years at 10% p.a.

(ii) At what rate percent will Rs. 1,000 amount to Rs. 1,500 in 2 years?

(iii) What principal will amount to Rs. 3,000 in 5 years at 20% p.a. S.I.?

(iv) In what time will Rs. 1,200 amount to Rs. 3,600 at 10% p.a. S.I.?Answer: (i) Interest on Rs. 100 for 1 year = Rs. 10

Interest on Rs. 1 for 1 year = Rs. 10

100

Interest on Rs. 1,000 for 1 year = Rs. 10 1 000

100

× ,

2 Financial Mathematics

Interest on Rs. 1,000 for 10 years = Rs. 10 1 000

10010

× ×,= Rs. 1,000

(ii) Here P = Rs. 1,000, A = Rs. 1,500, N = 2 years

Now, A = P (1 + Ni) or, 1,500 = 1,000 (1 + 2i) or, 1+ 2i = 1.5 or i = 0 5

20 25

..= .

Again, i = R

100∴ R = 100 × 0.25 = 25

(iii) Here, A = Rs. 3,000, N = 5 years, i = 20

1000 2= .

Now, A = P (1+ Ni), or, 3,000 = P (1+ 5 × 0.2), or, P = 3 000

2

, = Rs. 1,500

(iv) Here, P = Rs. 1,200, A = Rs. 3,600, i = 10% = 0.1

Now, A = P (1+ Ni), or, 3,600 = 1,200 (1+ N × 0.1) or, 3 = 1+ 0.1N or N = 2

0 1.= 20 years.

1.2. COMPOUND INTEREST (C.I.)

If the interest, as and when it becomes due, is added to the principal and the whole amountproduces interest for the subsequent period, then it is called “compound interest”. The periodafter which the interest becomes due is known as interest period. Interest is compounded monthly,quarterly, half-yearly, yearly etc., if it is specifically mentioned. If this is not given in the problemwe assume that the interest is payable yearly.

Example 1.2. (i) Find the difference between Simple and Compound interests on Rs. 3,000 in-vested for 3 years at 6% p.a., interest payable annually.

(ii) What is the present value of Rs. 1,000 due in 2 years at 6% p.a., compound interest payablehalf-yearly?

(iii) What rate of interest p.a., does a man get who is paid @ 6% compound interest payable1/2 yearly.

(iv) Find the compound interest on Rs. 1,000 @ 4% for the first year, 5% for the second yearand 6% for the third year.

Answer: (i) Simple Interest = PNi = = × × =PNR

100

3 000 3 6

100

, Rs. 540

Compound Interest:

Principal (Original) = Rs. 3,000.00

Interest for 1st year = 180.00

Principal for 2nd year = 3,180.00

Interest for 2nd year = 190.80

Principal for 3rd year = 3,370.80

Interest for 3rd year = 202.25–––––––––––

Amount in 3 years = Rs. 3,573.05–––––––––––

∴ Compound Interest = Rs. 3573.05 − 3000 = 573.05

Compound Interest – Simple Interest = Rs. (573.05 − 540)

= Rs. 33.05

Mathematics of Finance 3

(ii) Let, Principal (original) = Rs. 100.00

Interest for 1st 1/2 year = 3.00–––––––

Principal = 103.00

Interest for 2nd 1/2 year = 3.09–––––––

Principal = 106.09

Interest for 3rd 1/2 year = 3.18

Principal = 109.27

Interest for 4th 1/2 year = 3.28–––––––––

Amount in 2 years = Rs. 112.55–––––––––

∴ The required principal or present value = Rs. ,

.

100 1 000

112 55

¥

= Rs. 888.49––––––––––

(iii) Let, Principal (original) = Rs. 100.00

Interest for 1/2 year = 3.00––––––

Principal = 103.00

Interest for 1/2 year = 3.09––––––––––

Rs. 106.09––––––––––

∴ Compound interest = Rs. (106.09 − 100.00) = Rs. 6.09

∴ The required rate of interest = 6.09%

(iv) Principal (original) = Rs. 1,000.00

Interest for 1st year @ 4% 40.00–––––––––––

Principal for 2nd year 1,040.00

Interest for 2nd year @ 5% 52.00–––––––––––

Principal for 3rd year 1,092.00

Interest for 3rd year @ 6% 65.52–––––––––––

Amount in 3 years 1,157.52–––––––––––

∴ The required compound interest = Rs. (1,157.52 − 1,000) = Rs. 157.52

1.3. FORMULA FOR COMPOUND INTEREST

Let, P = Principal, N = No. of years or interest periods, i = Interest on unit sum for 1 interestperiod or 1 year, A = Amount and I = Total interest.

Now, Principal = P

Interest in the 1st period = Pi

∴ Amount in 1st period, or Principal for 2nd period = P (1+ i)

Interest for 2nd period = P (1+ i) i


Amount in the 2nd period or Principal for 3rd period = P (1+ i) + P (1+ i) i

= P (1+ i) {1+ i} = P (1+ i)2

Interest for 3rd period = P (1+ i)2i

Amount in the 3rd period = P (1+ i)2+ P (1+ i)2 i

= P (1+ i)2 {1+ i} = P (1+ i)3 and so on.

∴ A = Amount in N periods = P (1+ i)N ...(1)

If R is the rate per cent, then A = PR N

1100

+FHIK , i.e., in compound interest the amounts at the

end of different years are in G.P.

In formula (1), P is called the present value of the sum A due in N periods.

∴ =+( )

= ⋅ +( )−PA

iA iN

N

11 ...(2)

I = Compound Interest = A − P = P (1+ i)N − P = P {(1+ i)N − 1} ...(3)

Cor. 1. If (1+ i) = I1, i.e., I1 = Amount of principal 1 in 1 period, then A = P I IN N⋅ −1 1 1and I = P{ }

Hence, for the compound interest, the amount increases in geometric progression.

Cor. 2. From equation (1), log A = log P+ N log (1+ i). If we know any three of the four unknownsA, P, N, i, we can find the other.

Cor 3. In case of uniform decrease, we use ‘−i’ instead of ‘i’ in all the above formulae for com-pound interest. Hence, for depreciation at compound rate, we apply the formula,

A = P (1 − i)N

Similarly, if P = Present population of a country and R% = Rate of decrease of population p.a.,

then population after N years = PR N

1100

−FH

IK

Cor. 4. We use the following formulae:

(i) A = Pi N

12

2

+FHIK , when interest is paid half-yearly.

(ii) A = Pi N

14

4

+FHIK , when interest is paid quarterly.

(iii) A = Pi N

112

12

+FHIK , when interest is paid monthly.

(iv) A = Pi N

1365

365

+FHIK , when interest is paid daily.


Cor. 5. If ‘i’ is the normal rate of interest on Re. 1 for 1 year and the interest is convertible ‘b’ timesa year, then the effective rate of interest is

100 1 1+FHIK −

RST

UVW

i

b

b

Note 1: In compound interest calculations, the principal and interest vary for each unit of time. The interest for any unitof time again earns interest at the same rate over subsequent units of time. Hence principal for any unit of timeis the amount due at the end of previous unit of time. Unit of time is known as the ‘conversion period’.

Note 2: The rate of interest R% p.a. compounded at given number of times per year is known as ‘nominal rate’. Therate of interest R% p.a. which if compounded yearly would yield the same amount of interest as r% rate com-pounded m times per year, then R% is called the “effective rate”. For example, if a man borrows Rs. 100 at

10% p.a. compounded half-yearly, then 10% p.a. is the nominal rate and 100 110

200100

2

+FH

IK

−L

NMM

O

QPP

= 10.25% is

the effective rate (here N = 1). If the nominal rate is compounded annually, then the nominal rate of interestbecomes equal to the effective rate.

Cor. 6. If the rate of interest is different for each year, e.g., r1, r2, r3 for the first, second and thirdyears respectively, then the amount after 3 years is given by:

A Pr r r

= +FHIK +FH

IK +FH

IK1

1001

1001

1001 2 3 .

Note 3: The compound interest law A = PR N

1100

+FH

IK applies to any quantity which increases or decreases so that

amount at the end of each period of constant length bears a constant ratio to the amount at the starting of thatperiod. This ratio is known as ‘growth factor’ if it is more than 1, and ‘decay factor’ if less than 1. For example,if the population of a city steadily increases by 3% p.a. of the population at the beginning of each year, then theyearly growth factor will be 1.03 and the population after N years will be (1.03)N times the population at thebeginning of the period.

Again, if the value of the machinery depreciates steadily by 12% p.a. of its value at the beginning of each year,then the yearly decay factor will be (1 − 0.12) = 0.88 and the value after N years will be (0.88)N times its valuewhen new.

Example 1.3. (i) The simple interest on a sum of money in one year is Rs. 50 and compoundinterest in two years is Rs. 102. Find the principal and the rate of interest.

(ii) A machine is depreciated at the rate of 10% on reducing balance. The original cost wasRs. 10,000 and the ultimate scrap value was Rs. 3,750. Find the effective life of the machine.

[B.Com. (C.U.), 1966, B.Com. (B.U.), Hons. 1990]

(iii) Find in what time a sum of money will double itself at 5% interest compounded annually[log 2 = 0.3010; log 105 = 2.0212]

(iv) The machinery in a factory is valued at Rs. 24,537 and it is decided to reduce the estimatedvalue at the end of each year by 18% of the value at the beginning of that year. When willthe value be Rs. 20,000?

(v) A man left Rs. 18,000 with the direction that it should be divided in such a way that his3 sons aged 9, 12, 15 years should each receive the same amount when they would reach the


age of 25 years. If the rate of compound interest is 3.5% p.a., what should each son receivewhen he is 25 years old?

(vi) When a boy is born, Rs. 500 is placed to his credit in an account that pays @ 6% com-pounded annually, (b) 6% compounded quarterly, (c) 6% compounded monthly. If the ac-count is not distributed, what amount will there be to his credit on his twentieth birthday?

(ICWAI, Dec. 1979)

(vii) A truck purchased by a transport company at Rs. 60,000 depreciates at the rate of 10% p.a.and its maintenance cost for the first year is Rs. 2,000 which increases by 2% every year. Ifthe scrap value realised when sold is Rs. 35,429.40, find the minimum average annualreturn from the truck the company should get so as not to sustain any loss.

[ICWAI, Jan. 1978]

(viii) A man borrowed Rs. 20,000 from a money-lender but he could not repay any amount for aperiod of 4 years. Accordingly, the money lender’s demand showed Rs. 26,500 due fromhim. At what rate per cent p.a. C.I. did the money lender lend his money?

[Given: log 2.65 = 0.4232, log 2 = 0.3010, log 1073 = 3.0306] [C.A. (Ent), Nov. 1991]

Answer: (i) S.I. = PNi, or, PNR

100, where P = Principal, N = Number of Years,

i = Interest on Re. 1 for one yearCompound Interest for 2 years = P{(1+ i)2 − 1} = P {i2 + 2i} = Pi (2 + i)Now, Pi = 50 ...(1) and Pi (i + 2) = 102 ...(2)

From (2), 50 (i + 2) = 102, or, i + 2 =102

50= 2.04 ∴ i = 2.04 − 2 = 0.04 =

R

100 ∴ R = 4%

From (1), P = 50

0 04.= Rs. 1,250

∴ The required principal and the rate of interest are Rs. 1,250 and 4% respectively.

(ii) Let the effective life of the machine be N years. For depreciation at constant rate,

A = P (1 − i)N ...(1)

Here, P = Rs. 10,000, A = Rs. 3,750, i = 10

100

From (1), 3,750 = 10,000 110

100−FH

IK

N

or, 0.375 = (0.9)N

or, log 0.375 = N log 0.9

or, 1 5740. = N × 1 9542.

or, (−1 + 0.5740) = N × (−1 + 0.9542)

or, −0.426 = N × −0.0458

or, N =0

0 0458

.426

.= 9.3 years


(iii) If P = Rs. 100, then A = Rs. 200, i =5

100

Now, A = P (1 + i)N or, 200 = 100 (1 + 0.05)N or, 2 = (1.05)N, or, log 2 = N log 1.05

or, log 2 = N (log 105 − log 100) or, 0.3010 = N (2.0212 − 2), or, N = 0 3010

0 021214 2

.

..= years.

(iv) Here, P = Rs. 24,537, A = Rs. 20,000, i = 0.18

Now, A = P (1 − i)N or, 20,000 = 24,537 (1 − 0.18)N or, 0.8151 = 0.82N, or, log 0.8151 = N log 0.82

or, 1 9113 1 9138. .= ×N or, (−1 + 0.9113) = N (−1 + 0.9138) or, N = 0 0887

0 0862

.

.= 1.03 years

(v) Let the sons aged 9, 12 and 15 years receive Rs. P1, Rs. P2 and Rs. P3 respectively.

Again, each of them receive Rs. P at the age of 25 years.

∴ P1+ P2+ P3 = 18,000 ...(1)

For the son aged 9 years, P = P P1

25 9

1161

3 5

1001 035+FH

IK = ( )

−..

Similarly, " " 12 years, P = P2 (1.035)13

" " 15 years, P = P3 (1.035)10

∴ From (1), P(1.035)−16 + P(1.035)−13 + P(1.035)−10 = 18,000 ...(2)

Now, (1.035)−16 = x (say) or,−16 log 1.035 = log x or, −16 × 0.0149 = log x or, −0.2384 = log x

or, 1 7616. = log x

or, x = 0.5776Similarly, using logarithm we get,

(1.035)−13 = 0.6401

(1.035)−10 = 0.7096

∴ From (2) P(0.5776 + 0.6401 + 0.7096) = 18,000

or, P =18 000

1 9273

,

.= Rs. 9,339.49.

(vi) (a) Here P = Rs. 500, i = 0.06, N = 20

Now A = P (1 + i)N

or, A = 500 (1.06)20

or, log A = log 500 + 20 log 1.06

= 2.6990 + 20 × 0.0253

= 3.2050

∴ A = Antilog 3.2050 = Rs. 1,603.

(b) Now, A = Pi N

14

4

+FHIK

= 500 (1 + 0.015)80

or, log A = log 500 + 80 log 1.015

= 2.6990 + 80 × 0.0064

= 3.2110

∴ A = Antilog 3.2110 = Rs. 1,626.


(c) Now A = Pi N

112

12

+FHIK

= 500 (1 + 0.005)240

or, log A = log 500 + 240 log 1.005

= 2.6990 + 240 × 0.0021

= 3.2030

∴ A = Antilog 3.2030 = Rs. 1,596.

(vii) Here, P = Rs. 60,000, i = 0.1, A = Rs. 35,429.40, the truck is sold after N years.

Now, A = P (1−i)N

or, 35,429.40 = 60,000 (1−0.1)N

or, 0.5905 = 0.9N

or, log 0.5905 = N log 0.9

or, 1 7713 1 9542. .= ×N

or, (−1+ 0.7713) = N × (−1 + 0.9542)

or, −0.2287 = N × (−0.0458)

or, N = 4.99 ≈ 5 years

The maintenance cost for 5 years: = Rs. [2,000 + 2,040 + 2,080.8 + 2,122.42 + 2,164.86]

= Rs. 10,408.08

Minimum return required for no loss: = Rs. (60,000 + 10,408.08) − Rs. 35,429.40 = Rs. 34,978.68.

∴ Average minimum yearly return required for no loss = =

RsRs

. , .. , , .

34 978 68

56 995 74

(viii) If r% p.a. is the rate at which the money-lender lends his money, then 26,500 = 20,000 (1 + r/100)4; or log 26,500

= log 20,000 + 4 log (1 + r/100) or, 4.4232 = 4.3010 + 4 log (1 + r/100); or, 4 log 1100

+FH

IK

r = 0.1222; or, log

(1 + r/100) = 0.0306; or (1 + r/100) = 1.073; or r/100 = 0.073, or r = 7.3%.

Example 1.4. (i) A sum of money put out at simple interest amounts to Rs. 690 in two years andto Rs. 757.50 in 3½ years. Find the sum invested and the rate of simple interest.

[ICWAI (Prel.), June 1992, Dec. 1993]

(ii) Two equal sums are lent at 6.75% and 4.5% simple interest per annum respectively. If theformer is recovered two years earlier than the later and the amount in each case is Rs. 1,905.Find the sum lent in each case. [B.Com. (Bangalore), Nov. 1992]

(iii) A pressure cooker is available for Rs. 250 cash or Rs. 100 cash down payment followed byRs. 165 after six months. Find the rate of interest charged under the instalment plan.

[ICWAI (Prel.), Dec. 1992, Dec. 1993]

(iv) Ram deposited a sum of Rs. 10,000 in a bank. After 2 years, he withdrew Rs. 4,000 and atthe end of 5 years he received an amount of Rs. 7,520. Find the rate of simple interest.

[ICWAI (Prel.), June 1990]

(v) If I ask you for a loan and agree to repay you Rs. 300 after nine months from to-day, howmuch should you loan me if you are willing to make the loan at the rate of 6% p.a.?



(vi) A sum of Rs. 1,200 becomes Rs. 1,323 in 2 years at compound interest compounded annu-ally. Find the rate per cent. [ICWAI (Prel.), Dec. 1990, June 1993]

(vii) If the population of a town increases every year by 2 per cent of the population at thebeginning of that year, in how many years will the total increase of population be 40%?

[B.Com. (C.U.), Hons. 1990]

(viii) A sum of Rs. 1,000 is invested for 5 years at 12% interest per year. What is the simpleinterest? If the same amount had been invested for the same period at 10% p.a. compoundinterest compounded per year, how much more interest would he get?


(ix) On what sum the difference between simple and compound interest for 3 years at the rate of20% is Rs. 1,600? [ICWAI (Prel.), Dec. 1993]

(x) A man deposits Rs. 5,000 in a Savings Bank which pays compound interest at the rate of4½ % for first two years and then at the rate of 5% p.a. for next three years. Find his amountafter 5 years. [B.Com. (C.U.), 1981]

(xi) A machine depreciates at the rate of 7% of its value at the beginning of a year. If themachine was purchased for Rs. 8,500, what is the minimum number of complete years atthe end of which the worth of the machine will be less than or equal to half of its originalcost price? [ICWAI, Dec. 1976]

(xii) A man wishes to have Rs. 2,500 available in a bank account when his daughter’s first yearcollege expenses begin. How much must he deposit now at 3.5% compounded annually, ifthe girl is to start in college six years from now? [ICWAI, Dec. 1982]

(xiii) A machine, the life of which is estimated to be 10 years, costs Rs. 10,000. Calculate thescrap value at the end of its life, depreciation on the reducing instalment system beingcharged at 10% p.a.[Given: log 30 = 1.4771 and log 3.483 = 0.5420] [CA (Ent.), May 1991]

Answer: (i) Let, P = Principal and R = Rate of simple interest

∴ Rs. 690 = P (1 + 2i) ...(1)

Rs. 757.50 = P (1 + 3.5i) ...(2)

From (1) and (2), 690

1 2

757 50

1 3 5+=

+i i

.

. or, 690 + 2,415i = 757.50 + 1,515i or, 900i = 67.5, or, i = 0.075, i.e., 7.5%.

From (1), 690 = P (1 + 2 × 0.075) or, P = 690

115600

.. .= Rs

(ii) Let Rs. P = Two equal sums, and sums at 6.75% and 4.5% S.I. p.a. be recovered after n years and (n + 2) yearsrespectively.

∴ P n16 75

100+ ×F

HIK

.= Rs. 1,905 ...(1)

P n Rs1 24 5

1001 905+ + ¥

L

NM

O

QP =b g

.. , ...(2)

From (1) and (2), 16 75

1001

4 5 9

100+ = + +. .n n

or, 100 + 6.75n = 100 + 4.5n + 9 or, 2.25n = 9 ∴ n = 4 years


From (1), P 14 6 75

1001 905+ ×F

HIK =.

, or, P Rs= =1 905

1 271500

,

.. , .

(iii) Interest on Rs. 150 [i.e., Rs. (250 − 100)] for 6 months = Rs. (165 − 150) = Rs. 15.

S.I. on Rs. 150 for 1 year = Rs. (15 × 2) = Rs. 30.

∴ % of interest = ¥ =

Rs

Rs

.

.

30

150100 20%.

(iv) Total interest received by Ram = Rs. (7,520 − 6,000) = Rs. 1,520. If R% p.a. = Rate of interest, then S.I. on

Rs. 10,000 for 2 years =¥ ¥

=

10 000 2

100200

,.

RRs R

Again, principal after 2 years = Rs. (10,000 − 4,000) = Rs. 6,000

∴ S.I. on Rs. 6,000 for 3 years =¥ ¥

=

6 000 3

100180

,.

RRs R

∴ Total interest = Rs. (200 + 180) R = Rs. 1,520 or, R = 4%.

(v) Let Rs. P be the loan.

∴ Amount = Rs P Rs. .13

4

6

100300+ ¥

FH

IK =

or, PRs

Rs= × =.. . .

300 400

418287 08 ∵ N = =L

NM

O

QP

9

12

3

4year

(vi) Here, P = Rs. 1,200, A = Rs. 1,323, N = 2, i = ?

∴ 1,323 = 1,200 (1+ i)2 or, 11 323

1 200110252+( ) = =i

,

,.

Taking logarithms of both sides, we get

2 log (1 + i) = log 1.1025 or, 2 log (1 + i) = 0.0426

or, log (1 + i) = 0.0213 or, (1 + i) = Antilog 0.0213 = 1.051

or, i = 0.051 and R = 100 × 0.051 = 5.1%.

(vii) Let P = Original population = 100

∴ A = 100 + 0.4 × 100 = 140

Here, i N= =2

100, ? ∴ 140 100 1

2

100= +FH

IK

N

or, 12

1001+FH

IK =

N

.40 or, N log 1.02 = log 1.40 or, N × 0.0086 = 0.1461

or, N years= = ≈0 1461

0 008616 99 17

.

.. .

(viii) S.I. on Rs. 1,000 for 5 years @ 12% p.a. = × × =Rs Rs.,

. .1 000 5 12

100600

Compound interest on Rs. 1,000 for 5 years @ 10% p.a.

= +FHIK −

RST

UVW

= ( ) −1 000 110

1001 1 000 11 1

55, , .n s


[Let, x = 1,000 (1.1)5 or, log x = log 1,000 + 5 log 1.1

or, log x = 3 log 10 + 5 × 0.0414 = 3 + 0.2070 or, x = Rs. 1,611]

∴ Compound interest = Rs. {1,000 (1.1)5 − 1,000} = Rs. (1,611 − 1,000) = Rs. 611.

∴ Difference = Rs. (611 − 600) = Rs. 11.

(ix) Let, Rs. 100 = Sum of money ∴ = × ×FH

IK =S I Rs Rs. . . .100 3

20

10060

Compound Interest (C.I.) = P i RsN1 1 100 120

1001

3

+( ) − = +FHIK −

RST

UVW

n s .

= Rs Rs. .1006

51 100

216

1251

3FHIK −

RST

UVW

= −RST

UVW

= Rs Rs.,

. .9 100

12572 80=

∴ C.I. − S.I. = Rs. (72.80 − 60) = Rs. 12.80

Difference is Rs. 12.80 when sum of money is Rs. 100

" Re. 1 " " 100

12 80.

" Rs. 1,600 " " Rs..

,100

12 801 600×

= Rs. 12,500.

(x) Amount after 5 years = +FHG

IKJ

L

NMM

O

QPP

× +FHG

IKJ

Rs. ,.

5 000 14 5

1001

5

100

2 3

= Rs. [5,000 (1.045)2] × (1.05)3.

= Rs. [5,000 × 1.092025] × 1.157625 = Rs. (5,460.125 × 1.157625) = Rs. 6,320.

(xi) For depreciation at compound rate: A = P (1 − i)N

Here P = Rs. 8,500, i = 7% = 0.07, A ≤ FHIK

× =Rs N. ,500, ?1

28

∴ A = 8,500 (1 − 0.07)N = 8,500 (0.93)N. Again, 8 0 931

28,500 . ,500( ) ≤ ×N

or, 0 931

2.( ) ≤N or, N log . log .0 93 0 5≤ or, N × ⋅ ≤ ⋅1 9685 1 6990 or, N ( ) ( . )− + ⋅ ≤ − +1 0 9685 1 0 6990

or, N ( ) .− ⋅ ≤ −0 0315 0 3010 or, N years≤ = ≈0 3010

0 03159 56 10

.

.. .

∴ Minimum number of complete years = 10.

(xii) Let P be the money deposited now.

Here, A = Rs. 2,500, i = 3.5% = 0.035, N = 6, P = ?

∴ 2,500 = P (1+ 0.035)6

or, log 2,500 = log P + 6 log 1.035 or, log P = −6 log 1.035 + log 2,500 or, log P = −6 (0.0149) + 3.3979

or, log P = 3.3085 or, P = Antilog 3.3085 = Rs. 2,034.


(xiii) A = Scrap value = P (1 − i)N = 10,000 (1 − 0.1)10 = 10,000 (0.9)10

Let x = 0.910; ∴ log x = 10 [log 9 − log 10] = 10 [2 log 3 − 1] = 10 [2 × 0.4771 − 1]

= 10 [0.9542 − 1] = − 0.458 = −1 + 1 − 0.458 = T.542 = log 0.3483

∴ x = 0.3483; ∴ A = Rs. (10,000 × 0.3483) = Rs. 3,483.

1.4. SOME RELATED TERMS

(i) Exact Time, Exact Interest, Ordinary Interest: In many transactions, the time may be givenin months, weeks or days.

But in the simple interest formula, N must be expressed in years. Thus, months, weeks or daysare to be converted into years, e.g.,

4 months = = =4

12

1

3

13

52

1

4year year, 13 . weeks =

When an annual simple interest rate and the time ‘d’ in days are given, the following methodsare used to convert days into years:

(1) If N (in years) = d (,

in days)

365 then the interest is said to be exact.

(2) If N (in years) = d (,

in days)

360 then the interest is said to be ordinary.

The exact number of days between the date of deposit and date of interest calculation isreferred to as exact time. For example, the exact time in days from February 1, 2001 to March 10,2001 is the exact number of days between the two dates. Since 2001 is not a leap year, the exacttime will be as follows:

No. of remaining days in February 28 − 1= 27

No. of days in March = 10––––––––

Exact time 37 days–––––––––

Illustration 1.1. A borrows Rs. 2,000 on June 1, 2001, for 60 days. The simple interest rate is5½%.

(1) Calculate the exact simple interest.

(2) Calculate the ordinary simple interest.

Answer: (1) Here, P = Rs. 2,000, i = 0.055, N year= 60

365. ∴ Exact S.I. = Rs. 2,000 × 0.055 ×

60

36518 08= Rs. . .

(2) Here N year= =60

360

1

6 ∴ Ordinary S.I. = Rs. 2,000 × 0.055 ×

1

618 33= Rs. . .

(ii) Equations of Value, Time Value of Money and S.I.: Consider a case when A borrows Rs. 100from B at 5% S.I. and agrees to pay Rs. 50 on the loan in 6 months. What payment 1 year fromnow will settle the debt?


Set up the information on a time diagram as follows:

Rs. 100 (Money borrowed)

0 6 months 12 months

Rs. 50 x (Payment in 12 months)

For this type of problem where payments are made at different dates, we need the followingfundamental principle of mathematics of finance:

Equation of Value; i.e.,

Value of loan at focal date = Value of payments at focal date

Focal date is the particular date at which amounts of money payable at different times can only becompared. Focal date is fixed by the lender and the borrower.

(a) In this problem, if the focal date is chosen 1 year from now, then the value of each sum ofmoney must be calculated at the focal date as follows:

Value of loan at focal date = Value of payments at focal date.

Value of Rs. 100 at focal date Value of Rs. 50 at focal date Value of x at focal dateRs. 100 1+0.05 1

Rs. 50 1+0.05f time)× = F

HGIKJ

L

NM

O

QP=

+= +

Rs. .. .

(105 00 1

251 25

� ��

Rsx No shift o

or, x = Rs. 53.75 [Focal date at 12 months].

(b) In this problem, if the focal date is chosen 6 months from now, then the equation of value willbe as follows:

Value of Rs. 100 at focal date Value of Rs. 50 at focal date Value of x at focal date

A=P (1+Ni), Rs. 100 1+0.051

2

Rs. 50 (No shift of time)FHGIKJ

L

NM

O

QP= +

+ FHGIKJ

= +

Rsx

i ex

. ..

, . ..

102 51 0 05

1

2

1 025

� ��

i e PA

Ni. .

( )=

+1 or,

xor x Rs

1 02552 5 53 8125

.. , . .= =

The time diagram will be as follows:

Focal date

100 102.5

0 6 (months) 12 (months)

50 x

x

1 025.

Thus, different focal dates give different values of x. This difference will exist in simple interesttransactions and hence it is important for the parties concerned to agree on the focal date.


Illustration 1.2. A borrows Rs. 200 now and agrees to pay Rs. 50 after 2 months and Rs. 70after 6 months. What final payment should A make 18 months from now to settle this debt if theS.I. rate is 10% and the focal date is now?Answer: Let Rs. x be the final payment in the following time diagram. The values of the loan and payments at the focaldate are shown in Table 1.1.

Value of money at focal date Focal date

200 Rs. 200

0 2 6 12 18 (months)

49.02 Rs. 50 Rs. 70 Rs. x

66.67

x

115.(Time diagram)

Table 1.1

Rs. Formula to use Value at focal date (Rs.)

200 No shift in time 200

50 PA

Ni=

+1

50

1 012

12

49 02+ FHGIKJ

=.

.

70 PA

Ni=

+1

70

1 016

12

66 67+ FHGIKJ

=.

.

x PA

Ni=

+1

x x

1 011812

115+ FHGIKJ

=.

.

∴ The equation of value at the focal date is

200 49 02 66 67115

= + +. ..

x or,

xor x Rs

11584 31 96 96

.. , . . .= =

(iii) Equations of Value and C.I.: Let us again consider transactions in which one or more debtsare repaid with one or more payments due at various points of time.

Illustration 1.3. A borrows Rs. 200 and agrees to pay Rs. 50 after 2 months and Rs. 70 after6 months. What final payment should A make 18 months from now to settle the debt if interest is6% compounded monthly?Answer: This illustration is similar to illustration 1.2. The only difference is that C.I. is used here. With C.I., the focaldate may be any date at which interest is compounded, and the resulting equations of value will give identical result forthe quantity to be determined.

(a) Let 18 months be the focal date and x be the amount of final payment.


Value of money at focal date

Focal date

Rs. 200 200 (1.005)18

0 2 6 18 (months)

Rs. 50 Rs. 70 x x

70 (1.005)12

50 (1.005)16

(Time Diagram)

Table 1.2 shows the value of each amount of money at the focal date.

Table 1.2

Rs. Formula used Value at focal date (Rs.)

200 A Pi N

= +FHIK1

12

12

200 106

12200 1 005

1218

12 18+FH

IK

=×.

.a f

50 " 50 106

1250 1 005

1216

12 16+FH

IK

=×.

.a f

70 " 70 106

1270 1 005

1212

12 12+FH

IK

=×.

.a f

x No shift in time x


200 (1.005)18 = 50 (1.005)16 + 70 (1.005)12 + x

or, 200 (1.09392894) = 50 (1.083071151) + 70 (1.061677812) + x or, 218.79 = 54.15 + 74.32 + x

or, x = Rs. 90.32 [Using Calculator]

(b) Let us solve this illustration by selecting focal date now.

Table 1.3 shows the values of each amount of money at focal date (i.e., t = 0). The time diagram is shown below:

Value of money at focal date Focal date

200 200

0 2 6 18 (months)

50 (1.005)−2 50 70 x

70 (1.005)−6

x (1.005)−18


Table 1.3

Rs. Formula used Value at focal date

200 No shift in time 200

50 P A= +FH

IK

− ×

106

12

122

12.50 (1.005)−2

70 P A= +FH

IK

− ×

106

12

126

12.70 (1.005)−6

x P A= +FH

IK

− ×

106

12

1218

12.x (1.005)−18

∴ The equation of value is:


200 = 50 (1.005)−2 + 70 (1.005)−6 + x (1.005)−18

200 50 0 990074503 70 0 970518078 0 914136159= + +( . ) ( . ) ( . )x

or, 200 = 49.50 + 67.94 + 0.914136159 x or, 0.914136159x = 82.56 or, x = Rs. 90.32 [using calculator]

Thus values of x in (a) and (b) are equal. Also if we multiply both sides of equation of value of (b) by (1.005)18, we getthe equation of value of (a), which shows that the two equations (for two different focal dates) are algebraically equivalent.

(iv) Continuous Compounding: The compound amount A for a deposit of Rs. P at interest rate Rper year compounded continuously for N years is given by:

A = P . eRN. (R in decimal form)

To get Rs. A at the end of N years, an initial investment of P = A . e−RN is required. The value ofe is 2.7182818. The values of ex and e−x can be obtained from tables or by using calculators.

For a constant principal, time period and annual interest rate, the more frequent the compounding,the larger is the return on the investment. But there is a theoretical upper limit on the return thatcan be obtained in this way. If we imagine the number of yearly conversions to increase indefi-nitely, we arrive at a situation when interest is compounded “continuously”, i.e., at each instant oftime the investment grows in proportion to its current value. This is known as continuous com-pounding.

EXERCISE 1(a)

1. Mr. Rama invested equal amounts—one at 6% simple interest and the other at 5% compoundinterest. If the former earns Rs. 486.56 more as interest at the end of two years, find the totalamount invested. [ICWAI (Prel.), Dec. 1985]

2. Mr. Ram borrowed Rs. 25,000 from a moneylender but he could not repay any amount in aperiod of 5 years. Accordingly the moneylender demands now Rs. 35,880 from him. At whatrate per cent per annum compound interest did the latter lend his money?

[ICWAI, June 1987]


3. In how many years will the population of a village change from 2,500 to 2,601, if the rate ofincrease is 2% per annum? [ICWAI (Prel.), Dec. 1992]

4. Two partners A and B together lend Rs. 12,615 at 5% compounded annually. The amount Agets in 2 years is the same as B gets at the end of 4 years. Determine the share of each in theprincipal. [ICWAI (Prel.), June 1991]

[Hints: If A lends Rs. P, then B lends Rs. (12,615 − P).

∴ P (1+ 0.05)2 = (12,615 − P) (1 + 0.05)4]

5. A sum of money invested at compound interest amounts to Rs. 10,816 at the end of secondyear and to Rs. 11,248.64 at the end of the third year. Find the rate of interest and the suminvested. [B.Com. (C.U.), 1983]

6. The population of a town is 1,25,000. If the annual birth rate is 3.3% and annual death rate is1.3%, calculate the population of the town after 3 years. [ICWAI, June 1992]

[Hints: Population increases each year by (3.3 − 1.3) = 2%

∴ A = 1,25,000 12

100

3

+FHIK ]

7. A man left for his three sons aged 10, 12 and 14 years Rs. 10,000, Rs. 8,000 and Rs. 6,000respectively. The money is invested in 3%, 6% and 10% compound interest respectively. Theywill receive the amount when each of them attains the age of 21 years. Find, using a five-figurelog-table, how much each would receive. [B.Com. (C.U.), 1967]

8. A borrower pays interest on his loan at the rate of 4% in quarterly instalments. He wishes topay monthly in the future. What should be the new nominal rate so that the lender will receivean equivalent amount? [ICWAI (Prel.), Dec. 1989]

9. A machine depreciates 10% p.a. for first two years and then 7% p.a. for the next three years,depreciation being calculated on the diminishing value. If the value of the machine beRs. 10,000 initially, find the average rate of depreciation and the depreciated value of themachine at the end of the fifth year. [B.Com. (C.U.), 1974]

10. The difference between the simple and the compound interests at the same rate for 2 years ona certain amount is 1/400 of amount. Find the rate of interest. [ICWAI (Prel.), June 1986]

11. The compound interest on a certain sum of money invested for two years at 5% is Rs. 238.What will be the simple interest on it at the same rate and for the same period?

[ICWAI (Prel.), Dec. 1986]

12. A sum of money invested now at x% per annum compound interest quadruples in 18 years.Find x. [ICWAI (Prel.), June 1984]

13. In how many years will the population of a village change from 15,625 to 17,576 if the rate ofincrease is 4% per year? [Given: 17576 = 263 and 15625 = 253] [ICWAI (Prel.), Dec. 1984]

14. A machine depreciates at the rate of 10 p.c. of its value at the beginning of a year. The machinewas purchased for Rs. 44,000 and the scrap value realised when sold was Rs. 25,981.56. Findthe number of years the machine was used. [ICWAI, Dec. 1983]


15. A machine depreciates each year by 10% of its value at the beginning of the year. At the endof fourth year its value is Rs. 1,31,200. Find its original value. [Given: log 1312 = 3.1179,log 90 = 1.9542, Antilog 5.3011 = 20,000] [B.Com. (C.U.), 1982]

16. Mr. Needy borrowed money from the short-term loan company and promised to pay Rs. 200 atthe end of the year. The interest rate is 2½% per month on the first Rs. 100 and 2% on thesecond Rs. 100. How much does he receive? [ICWAI, Dec. 1979]

17. What is the effective rate of interest corresponding to a nominal rate of 5% p.a. if interest iscompounded quarterly? [ICWAI, June 1991]

18. A lends B Rs. 320. B is to pay interest on whatever amount he has not paid back at the rateof 5% per annum for the first year, 6% for the second year and 7% for the third year. B paysRs. 100 at the end of first year, Rs. 100 at the end of the second year and enough to pay offcompletely the debt and the interest at the end of the third year. How much is the last payment?

[ICWAI, 1971]

19. The population of a country increases every year by 2.4% of the population at the beginning ofthat year. In what time will the population double itself? Answer to the nearest year.

[ICWAI, June 1977]

20. The difference between simple and compound interest on a sum of money put out for 4 yearsat 5% p.a. is Rs. 150. Find the sum. [ICWAI, June 1989]

21. A machine depreciates in value each year at the rate of 10% of its value at the beginning of ayear. The machine was purchased for Rs. 10,000. Obtain, to the nearest rupee, its value at theend of the tenth year. [ICWAI, June 1975]

22. A sum of money invested at compound interest, payable yearly, amounts to Rs. 2,704 at theend of the second year and to Rs. 2,812.16 at the end of the third year. Find the rate of interestand the sum. [B.Com. (C.U.), 1983]

23. In a certain population the annual birth and death rates per 1,000 are 39.4 and 19.4 respec-tively. Find the number of years in which the population will be doubled assuming that there isno immigration. [ICWAI, Dec. 1974]

24. Find the amount that Rs. 100 will become after 20 years at compound interest @ 5% calcu-lated annually. [B.Com. (C.U.), 1966]

25. A man wants to invest Rs. 5,000 for four years. He may invest the amount at 10% p.a. com-pound interest, interest accruing at the end of each quarter of the year, or, he may invest it at10½% p.a. compound interest, interest accruing at the end of each year. Which investment willgive him slightly better return? [ICWAI, June 1976]

26. The interest on a sum of money invested at compound interest are Rs. 832 for the second yearand Rs. 865.28 for the third year. Find the rate of interest and the sum invested.

[ICWAI, June 1986 (old)]

27. Mr. Brown was given the choice of two payment plans on a piece of property. He may payRs. 10,000 at the end of 4 years, or, Rs. 12,000 at the end of 9 years. Assuming money can beinvested annually at 4% p.a. converted annually, what plan should Mr. Brown choose?

[ICWAI, June 1983]


ANSWERS

(1) Rs. 27,800; (2) 7.5%; (3) 2 years; (4) Rs. 6,615, Rs. 6,000; (5) 4%, Rs. 10,000;(6) 1,32,651; (7) Rs. 13,843, Rs. 13,517; Rs. 11,691; (8) 3.6%; (9) 8.2%, Rs. 6,515;(10) 5% p.a.; (11) Rs. 232.20; (12) 8%; (13) 3 years; (14) 5 years; (15) Rs. 2,00,000;(16) Rs. 156.45; (17) 5.0945%; (18) Rs. 160.67; (19) 29 years; (20) Rs. 9,673.52;(21) Rs. 3,483; (22) 4% p.a., Rs. 2,500; (23) 35 years; (24) Rs. 265.50; (25) Second invest-ment will give him better return; (26) 4%, Rs. 20,000; (27) Second plan.

1.5. ANNUITIES

An annuity is a fixed amount paid at regular intervals e.g., monthly, quarterly, yearly, etc., undercertain conditions. When the interval is not given, we take it as one year.

An annuity payable for a fixed number of periods, or, years is defined as “Annuity Certain”.When an annuity is to continue for ever, it is said to be a “Perpetual Annuity or Perpetuity”. Ifthe payment of an annuity commences or ceases at the occurrence of a contingent event, it is saidto be an “Annuity Contingent”. If the payments are made at the end of each period, the annuityis called “Immediate or Ordinary Annuity”. When the payments are made at the beginning ofeach period, the annuity is defined as “Annuity Due”. An annuity is taken as immediate unlessotherwise stated.

If the payments of an annuity are deferred or delayed for certain periods, or years, it is calleda “Deferred Annuity”. If an annuity is deferred for ‘n’ years, its first instalment will be paid at theend of (n + 1) years. For a “Deferred Perpetuity”, the payments commence after the deferredperiod and thereafter continue for ever. If we add the present values of all the payments of anannuity, we get its “Present Value”.

A “Free-hold Estate” is that which generates a perpetual annuity, i.e., rent. A “Lease-holdEstate” generates rent for the fixed lease period. The value of a freehold estate is equal to thepresent value of the perpetuity (or rent). If an annuity remains unpaid for a certain period, it is saidto be “Unpaid Annuity” for that period. The amount of the unpaid annuity is obtained by addingthe instalments and the compound interest on each for the period during which it remained unpaid.

1.6. FORMULAE

(1) Amount of an annuity:

AP

ii n= +( ) −1 1

where, A = Amount of an annuity or (immediate annuity)

P = Annuity; n = Unpaid years

i = Interest on unit sum for 1 year or period

Proof: The first instalment P due at the end of the first year will earn compound interest for(n − 1) years and the amount will be P (1+ i)n−1. Similarly, the second instalment P will earncompound interest for (n − 2) years and amount to P (1+ i)n−2 and so on. The last instalment P willnot earn any interest.


∴ A P i P i Pn n= + + + + +− −

1 11 2

b g b g .......

[This is a G.P. with C.R. 1/(1 + i)]

= +( ) ⋅−

+FH

IK

RST

UVW

−+

−P ii

i

n

n

1

11

1

11

1

1 = +( ) ⋅ +( ) −+( )

RST

UVW

× +( )+ −

−P ii

i

i

in

n

n11 1

1

1

1 11

= +( ) ⋅ +( ) −+( )

RST

UVW

×−−P i

i

i in

n

n11 1

1

111 = +( ) −P

ii n1 1n s

Cor. If t = Times of payments made per year.

Rs. P = Rent per year, Rs.P

t= Rent per period

i = Interest per unit sum per year, then

A = Amount of the annuity P for n years

= +FHIK −

RST

UVW

= +FHIK −

RST

UVW

P t

i t

i

t

P

i

i

t

nt nt

1 1 1 1

(2) Present value of an annuity:

VP

ii n= − +( )−1 1n s

where, P = Annuity to continue for n years.

V= Present value of the annuity (immediate).

i = Interest on unit sum for 1 year.

Proof: The present values of the first, second, third, ......., nth payments are:

P

i

P

i

P

i

P

i n1 1 1 12 3+ + + +a f a f a f a f, , , . .. . . .. . ., respectively.

∴ V = Sum of the present values of all the payments.

=+

++

++

+ ++

P

i

P

i

P

i

P

i n1 1 1 12 3a f a f a f a f…… =

+−

+FHG

IKJ

RS|

T|

UV|

W|

−+

Pi i

i

n

11

11

11

1

b g

= − + −P

ii

n1 1b g{ }


Cor. If t = Times of payments made per year

Rs. P = Rent per year Rs.P

t= Rent per period

i = Interest on unit sum for 1 year, then

VP t

i t

i

t

P

i

i

t

nt nt

= ⋅ − +FHIK

RST

UVW

= − +FHIK

RST

UVW

1 1 1 1

(3) Present value of a perpetuity:

VP

i=

Proof: For perpetuity, n is indefinitely large. Hence the value of 1

1 +( )i n in (2) may be taken

as zero.

∴ = −+( )

RST

UVW

= −( ) =From VP

i i

P

i

P

in( ),2 11

11 0

(4) Present value of a deferred annuity:

VP

i

i

i

n

m n= +( ) −+( )

RST

UVW

+1 1

1= −

+( )RST

UVW

− −+( )

RST

UVW

+P

i i

P

i im n m11

11

1

1

Proof: Let P = Annuity, V = Present value, i = Interest on unit sum for 1 year. Payment startsat the end of m years and thereafter continues for n years. Hence, the first, second, third,...., lastpayments are due at the end of (m + 1), (m + 2), (m + 3),......, (m + n) years respectively.

∴ The present values of the successive payments are:

P

i

P

i

P

im m m n1 1 11 2+ + ++ + +a f a f a f

, , ..........., respectively.

∴ =+

++

+ +++ + +V

P

i

P

i

P

im m m n1 1 11 2a f a f a f

. . . .. . .. . . . .

= +( ) −+( )

RST

UVW

+P

i

i

i

n

m n

1 1

1 [Infinite geometric series]

=+

−+

L

NMM

O

QPP

+P

i i im m n

1

1

1

1a f a f= −

+( )L

NM

O

QP − −

+( )L

NM

O

QP+

P

i i

P

i im n m11

11

1

1

= [Present value of an immediate annuity to continue for (m + n) years] − [Presentvalue of an immediate annuity to continue for m years.]


(5) The present value of a deferred perpetuity:

VP

i i m=+( )

L

NM

O

QP

1

1

where, V = Present value of a deferred perpetuity P to begin after m years.

Proof: In (4), putting 1

10

+( )= +( )+i

as m nm n , is indefinitely large, we get

VP

i

P

i

P

i i

P

i im m= − + ×+( )

= ⋅+( )

1

1

1

1.

(6) Sinking fund:

It is a fund which is created by investing annually a fixed amount at compound interest to pay offa loan or a debenture stock or bond on a given date or to redeem certain liabilities or to provide forreplacement of assets (i.e., capital expenditure), e.g., plant and machinery, etc.

If A is the amount of loan to be paid off at the end of n years and P is the accumulations of theannual sum, then

AP

ii n= +( ) −1 1

where, i = Interest on unit sum for 1 year [Same as formula (1)]

(7) Present value of uneven cash inflows:

If P1, P2 ,....., Pn be the unequal cash inflows received at the end of year 1, 2,...... n respectively,then the present value of these sums at interest rate ‘i’ is given by:

VP

i

P

i

P

in

n=+( )

++( )

+ ++( )

1 221 1 1

......... .

Example 1.5. (i) A machine costs Rs. 97,000 and its effective life is estimated to be 12 years. Afund is created for replacing the machine at the end of its effective life time. If the scrap realisesRs. 2,000, what amount should be retained out of profits at the end of each year to accumulate atcompound interest at 5% p.a.? [Given (1.05)12 = 1.797].

(ii) S. Roy borrows Rs. 20,000 at 4% compound interest and agrees to pay both the principal andthe interest in 10 equal annual instalments at the end of each year. Find the amount of theseinstalments. [C.U., 1972]

(iii) The annual rent of a free-hold estate is Rs. 1,000. What is its present value, if the compoundinterest rate is 4% p.a.?

(iv) Which is better, an annuity of Rs. 100 to last for 12 years , or, the reversion of a free-holdestate of Rs. 80 p.a. to commence 6 years hence, the rate of interest being 6%?

(v) A man buys an old piano for Rs. 500, agreeing to pay Rs. 100 down and the balance in equalmonthly instalment of Rs. 20 with interest at 6%. How long will it take him to completepayment? [ICWAI, Dec. 1979]


(vi) The accumulations in a Provident Fund are invested at the end of every year to earn 10% p.a.A person contributes 12½% of his salary to which his employer adds 10% every month. Findhow much the accumulations will amount to at the end of 30 years of his service, for every100 rupees of his monthly salary [Give the answer to the nearest rupee]. [ICWAI, June 1975]

(vii) A wagon is purchased on instalment basis such that Rs. 5,000 is to be paid on the signing ofthe contract and four-yearly instalments of Rs. 3,000 each payable at the end of the first,second, third and fourth years. If interest is charged at 5% p.a., what would be the cash downprice? [B.Com. (C.U.), C.A. (Ent.), Nov. 1991]

(viii) For endowing an annual scholarship of Rs. 12,000 a man wishes to make three equal contri-butions. The first award of the scholarship is to be made 3 years after the last of his threecontributions. What would be the value of each contribution, assuming interest at 2.5% p.a.compounded annually? (Assume that the first contribution is to be made now and the othertwo at an interval of one year thereafter.) [ICWAI, Dec. 1980]

(ix) A sinking fund is created for redemption of debentures of Rs. 2,00,000 at the end of 20 years.How much money should be provided out of profit each year for the sinking fund if theinvestment can earn interest @ 4% p.a.?

Answer: (i) Here, A = Cost price of the machine − Scrap value

= Rs. (97,000 − 2,000) = Rs. 95,000

n = 12 years, i = 0.05

Now, Amount of an annuity = AP

ii n= +( ) −1 1n s

where, P = Amount to be retained out of profits at the end of each year = Annuity

∴ = −95 0000 05

1 05 112,.

.Pn s

[Let x = 1.0512 ∴ log x = 12 log 1.05 = 12 × 0.0212 = 0.2544

∴ x = Antilog 0.2544 = 1.797]

or, 4,750 = P (1.797 − 1) or, P Rs= =4 750

0 7975 960

,

.. , .

(ii) Here, V = Rs. 20,000 = P.V. of an annuity of Rs. P to continue for 10 years at 4% p.a., n = 10, i = 0.04, P = Amountof each instalment or annuity.

Now, Present value of annuity P

= = − +( )−VP

ii n1 1n s or, 20 000

0 041

1

1 04 10,. .

= −( )

L

NM

O

QP

P

[Let, x = (1.04)10 ∴ log x= 10 log 1.04 = 10 × 0.0170 = 0.1700, ∴ x = 1.479]

or, 800 11

10 3239= −L

NMOQP

= ( )P P.479

. ∴ P = Rs. 2,469.90 ≈ Rs. 2,470

(iii) A freehold estate yields a perpetual annuity P.

If V = P.V. of the freehold estate, then VP

iRs= = =1 000

0 0425 000

,

.. , .


(iv) First case: VP

ii n= = − +( ) =−1 1 P.V. of an annuity P to continue for n years.

Here, P = 100, i = 0.06, n = 12 or, V = - +LNM

OQP

-100

0 061 1 06

12

..b g

[Let x = 1.0612, ∴ log x = 12 log 1.06 = 12 × 0.0253 = 0.3036

∴ x = Antilog 0.3036 = 2.0120] or, V = 1,666.67 (1 − 0.4970) = Rs. 838.34

Second case: It is a deferred perpetuity which commences after 6 years.

Now, VP

ii m= +( )−1 = P.V. of a deferred perpetuity P to commence after m years.

Here, P = Rs. 80, i = 0.06, m = 6 ∴ = ( ) = ( )− −V80

0 061 06 1 333 33 1 066 6

.. , . . .

[Let, x = (1.06)6, ∴ log x = 6 log 1.06 = 6 × 0.0253 = 0.1518

∴ x = Antilog 0.1518 = 1.419] or, V = 1,333.33 × 0.7047 = Rs. 939.60

Since P.V. of the second case, i.e., Rs. 939.60 is greater than the P.V. of the first case, i.e., Rs. 838.34, the second caseis better.

(v) Amount paid in cash = Rs. 100.

∴ V = Rs. (500 − 100) = Rs. 400 = The balance amount.

If n be the number of years, then VP

i

i n

= − +FHIK

L

NM

O

QP

−1 1

12

12

Here, P = Rs. 20 × 12 = Rs. 240, i = 0.06 ∴ = − +FHIK

LNM

OQP

−400

240

0 061 1

0 06

12

12

.

. n

or, 400 = 4,000 {1 − (1.005)−12n} or, 1

101 1 005 12= − ( )−. n

or, 1 005 11

10

9

1012.( ) = − =− n or, 1 005

10

912.( ) =n

or, 12n log 1.005 = log 1.1111 or, 12n × 0.0021 = 0.0457 or, 12n = 21.76

Here, the required time = n years = 12n months = 21.76 ≈ 22 months

(vi) Total monthly contributions to P.F. = 12.5 + 10 = Rs. 22.5; if we assume the monthly salary of the person as Rs. 100.

∴ Total annual contribution to P.F. = 12 × 22.5 = Rs. 270.

If A = Total accumulation at the end of 30 years, then

AP

ii n= +( ) −1 1n s . Here, P = Rs. 270, i = 0.1, n = 30 = ( ) − = −270

0 111 1 2 700 11 130 30

.. , .n s d h

[Let x = 1.130, or log x = 30 log 1.1 = 30 × 0.0414 = 1.242

∴ x = Antilog 1.2420 = 17.46] ∴ A = 2,700 × 16.46 = Rs. 44,442.

(vii) If V = P.V. of the annuity of Rs. 3,000 for 4 years at 5% compound interest, then cash down price of the wagon willbe Rs. (V + 5,000).

Now, VP

ii

n= - +

-

1 1b g{ } = − ( )−3 000

0 051 1 05 4,

..n s Here P = Rs. 3,000, n = 4, i = 0.05


[Let, x = 1.05−4, ∴ log x = − 4 log 1.05 = − 4 × 0.0212 = −0.0848 = − 1+ 1 − 0.0848 = 1 9152.

∴ x = Antilog 1 9152 0 8226. . ]= or V = 60,000 (1 − 0.8226) = Rs. 10,644

∴ The required cash down price = Rs. (10,644 + 5,000) = Rs. 15,644.

(viii) The first scholarship is to be paid at the end of the 5th year and thereafter it will continue for ever. Hence we havea perpetuity of Rs. 12,000 deferred by 4 years.

∴ V = P.V. of the endowments = + − +( )−xx

ii n1 1n s

where, x = The annual contribution

Here, n = 2, as the first contribution is made now, i = 0.025.

∴ V = + − −xx

0 0251 1 025 2

..n s

= + × = +FHIKx

xx

0 0250 0481 1

0 0481

0 025..

.

. = 2 924. x ...(1)

Again, for the P.V. of deferred perpetuity

VP

i i m= ×+( )1

1, Here, P = Rs. 12,000, i = 0.025, m = 4

= ( )−12 000

0 0251 025 4,

..

[Let, x = 1.025−4, or log x = −4 log 1.025 = −4 × 0.0107 = −0.0428 = −1+ 1 − 0.0428 = 1 9572.

∴ x = =Antilog 1 9572 0 9061. . ] ∴ V = 4,80,000 × 0.9061 = Rs. 4,34,928 ...(2)

From (1) and (2), 2.924x = Rs. 4,34,928 ∴ x = Rs. 1,48,744.18

(ix) Amount of annuity, AP

ii or

Pn= + − = −1 1 2 00 0000 04

1 04 120a fo t a fo t; , ,.

.

or, P Rs=( ) −

=−( )

=8 000

1 04 1

8 000

2 188 16 734

20

,

.

,

.. , .

n s

[Let x = (1.04)20; log x = 20 log 1.04 = 20 × 0.0170 = 0.34, ∴ x = 2.188]

Example 1.6. (i) A government constructed housing flats costing Rs. 1,36,000; 40% to be paid atthe time of possession and the balance reckoning C.I. @ 9% p.a. is to be paid in 12 equal annual

instalments. Find the amount of each such instalment. Given: 1

1 090 355812.

.( )

=L

NM

O

QP

[B.Com. (C.U.), 1984]

(ii) A loan of Rs. 10,000 is to be repaid in 30 equal annual instalments of Rs. P. Find P if thecompound interest charged is at the rate of 4% p.a. (Annuity is an immediate annuity, i.e.,first payment is made at the end of first year). [Given: (1.04)30 = 3.2434]

[B.Com. (C.U.), 1982]

(iii) A machine costs a company Rs. 52,000 and its effective life is estimated to be 25 years. Asinking fund is created for replacing the machine by a new model at the end of its life time,when its scrap realises a sum of Rs. 2,500 only. The price of the new model is estimated to


be 25% higher than the price of the present one. Find what amount be set aside each yearout of the profits for the sinking fund, if it accumulates at 3½% per annum compound?[Given: log 10.35 = 1.01494, log 236.32 = 2.3735] [C.A. (Ent.), May 1992]

(iv) An investor has a capital of Rs. 20,000 on which he earns interest @ 5% p.a. If he spendsRs. 1,800 per year, show that he will be ruined of his capital before the end of 17th year.

(v) Determine the present value of a perpetual annuity of Rs. 100 payable at the end of 1st year,Rs. 200 at the end of 2nd year, and Rs. 300 at the end of 3rd year, and so on, increasingRs. 100 payable at the end of each subsequent year. Assume a time preference rate of 5% p.a.,compounded annually.

Answer: (i) Amount to be paid at the time of possession = Rs. (1,36,000 × 0.4) = Rs. 54,400

Balance to be paid in 12 instalments along with interest = Rs. (1,36,000 − 54,400) = Rs. 81,600.

If Rs. P is the annual instalment, then 81 6000 09

11

1 09 12,. .

= −( )

L

NM

O

QP

P

[Let x = (1.09)12; ∴ log x = 12 log 1.09 = 12 × .0374 = 0.4488

∴ x = Antilog 0.4488 = 2.810]

or, 81 6000 09

1 0 35590 09

0 6441,.

..

.= − = ×P P or, P = =7344

0 644111 402

.. , .Rs

(ii) 10 0000 04

11

1 0 04 30,. .

= −+( )

L

NM

O

QP

P

[Let x = 1.0430; ∴ log x = 30 log 1.04 = 30 × 0.017 = 0.51

∴ x = Antilog 0.51 = 3.236]

or, 10 0000 04

1 0 3090 400 0 691,.

. , .= −( ) = ×Por P ∴ = =P Rs

400

0 691578 87

.. . .

(iii) A = Cost of the machine − Scrap value

= Rs. (1.25 × 52,000 − 2,500) = Rs. (65,000 − 2,500) = Rs. 62,500

Amount of an annuity = = +( ) −AP

ii n1 1n s

where, A = Amount to be retained out of profits at the end of each year

∴ = ( ) −620 035

1 035 125,500.

.Pn s

[Let x = 1.03525; ∴ log x = 25 log 1.035 = 25 × 0.01494 = 0.3735 = log 2.3632

∴ x = 2.3632]

or, P (2.3632 − 1) = 2,187.5 or, P = Rs Rs., .

.. , . .

2 187 5

136321 604 68=

(iv) VP

ii n= − +( )−1 1n s or, 20 000

1 800

0 051 1 05,

,

..= − ( )−nn s

or, 1 000

1 8001 1 05

,

,.= − ( )−n or, 1 05 1

10

18

8

18

4

9.( ) = − = =−n or, −n log 1.05 = log 4 − log 9


or, −n × 0.0212 = 0.6021 − 0.9542 or, −n × 0.0212 = −0.3521 ∴ = = ≈n th year0 3521

0 021216 61 17

.

..

(v) The present value of the perpetual annuity can be obtained as follows:

VP

i

P

i

P

i=

++

+( )+

+( )+ + ∞1 2

23

21 1 1( )............. = +

( )+

( )+100

1 05

200

1 05

300

1 052 3. . ................

or, V

100

1

1 05

2

1 05

3

1 052 3= +( )

+( )

+. . .

...............

or, A = x+ 2x2+ 3x3 +..........+ ∞ Put x andV

A1

1 05 100.= =L

NMO

QP

Multiplying both sides by x, we get

Ax = x2+ 2x3+ 3x4+ ......

Now, A − Ax = x+ x2+ x3 +........ or, A (1 − x) = x (1+ x+ x2 +.......) = x (1 − x)−1

or, A x xV= ⋅ −( ) =−1

1002 or, V = 100x (1 − x)−2

=( )

− ( )=

FH

IK

−

−

100 1 05

1 1 05

100

1 050 051 05

1

1 2 2

.

. ...

n s=

×( )

= =100 1 05

0 05

105

0 002542 0002

.

. .. , .Rs

1.7. AMORTIZATION

A loan is amortized if both the principal and interest are paid by a sequence of equal periodicpayments. Amortization means removal of loan. Each payment consists of:

(i) The interest on the loan outstanding at the beginning of the payment period.(ii) A part repayment of the loan or principal.

To facilitate accounting, it is often required to split up each instalment paid into repayment ofloan and payment of interest on the outstanding balances. With each payment, the actual loan orprincipal decreases, interest part in each payment successively decreases and loan repayment (i.e.,amortization) increases.

1.7.1. To calculate the amortizations for the 1st, 2nd, 3rd, etc., years, of a loanrepaid by fixed annual payments in n years, compound interest being allowed

Let A be the amount of each annual payment paid at the end of each year for n years. Hence theloan P is the present value of the n annual payments. Let i be the rate of interest per Re. p.a.

∴ The loan is given by:

P P P P Pn= + + + +1 2 3 .. . .. . .. . . .. . ..

= + + + + + + + +− − − −A i A i A i A i n1 1 1 11 2 3a f a f a f a f. . . .. . .. . . .

= + + + + + + + +− − − − −A i i i i n1 1 1 1 11 1 2 1a f a f a f a f{ }a f. . .. . .. . . ..


= +( ) ⋅ − +( )− +( )

=+( )

⋅−

+( )

−+( )

−−

−A ii

i

A

ii

i

n n

11 1

1 1 1

11

1

11

1

11

=+

⋅+ −

+×

++ −

=+ −

+

RS|

T|

UV|

W|

A

i

i

i

i

i

A

i

i

i

n

n

n

n1

1 1

1

1

1 1

1 1

1a f

a f

a f

a f

a f

a f

a f Hence, A

Pi i

i

n

n= +( )+( ) −

1

1 1...(1)

Now, the annual payment A paid at the end of the 1st year contains the interest on P for 1 year,which is Pi and the rest of A is amortization.

∴ Amortization at the end of the 1st year

= − = +( )+( ) −

−A PiPi i

iPi

n

n

1

1 1...[From (1)]

= +( ) − +( ) ++( ) −

=+( ) −

Pi i Pi i Pi

i

Pi

i

n n

n n

1 1

1 1 1 1

After amortization the loan amount at the end of the 1st year = −+ −

PPi

i n( )1 1...(2)

Interest on (2) in 1 year = −+ −

PiPi

i n

2

1 1( )

Hence amortization at the end of the 2nd year

= − −+ −

RST

UVW

=+

+ −− +

+ −A Pi

Pi

i

Pi i

iPi

Pi

in

n

n n

2 2

1 1

1

1 1 1 1( ) ( ) ( )

a f[From (1)]

=+( ) − +( ) − +

+( ) −=

++( ) −

Pi i Pi i Pi

i

Pi Pi

i

n n

n n

1 1 1

1 1 1 1

2 2n s = +( )+( ) −

Pi i

i n

1

1 1.

Similarly, amortization at the end of the 3rd year = +( )+( ) −

Pi i

i n

1

1 1

2

and so on

Amortization at the end of the nth year = ++ −

−Pi i

i

n

n

1

1 1

1a f

a f

∴ Total amortization in n years

=+ −

++

+ −+

++ −

+ ++

+ −

−Pi

i

Pi i

i

Pi i

i

Pi i

in n n

n

n( ) ( ) ( ).........

( )1 1

1

1 1

1

1 1

1

1 1

2 1a f a f a f


=+ −

+ + + + + + + −Pi

ii i in

n

( )( ) ( ) ............. ( )

1 11 1 1 12 1n s

=+ −

⋅+ −+ −

= =Pi

i

i

iPn

n

( )

( )

( )1 1

1 1

1 1 Loan amount.

Example 1.7. A loan of Rs. 1,000 is to be paid in 5 equal annual payments, interest being at 6%p.a. compound interest and first payment being made after a year. Analyse the payments into thoseon account of interest and on account of amortization of the principal. [ICWAI, Jan. 1970]

Answer: Present value of an annuity,

VP

ii n= − + −1 1( ) ,n s

where P is the annuity (i.e., yearly payment) in rupees.

Here, V = 1,000, i = 0.06, n = 5, P = ? ∴ = − + −1 0000 06

1 1 06 5,.

( . ) ,Po t

or, 60 = P 1 1 06 5− −( . )o t ...(1)

[Let x = (1.06)−5, log x = −5 log 1.06 = −5 × .0253 = −0.1265 = − 1 + 1 −0.1265 = 1.8735

∴ x = Antilog ( . )1 8735 = 0.7473]

From (1), 60 = P P1 0 7473 0 2527−{ } = ×. . or, P Rs= ≈60

0 2527237

.. .40

Amortization Table

End of Yearly Interest Amortization Principalyear payment due (Rs) (Rs) earning interest

P (Rs) (Rs)

1st 237.40 60.00 177.40 1,000.00

2nd 237.40 49.36 188.04 822.60

3rd 237.40 38.08 199.32 634.56

4th 237.40 26.12 211.28 435.24

5th 237.40 13.44 223.96 223.96

Total 1,187 187 1,000 –

1.8. PRESENT VALUE (P.V.) IN CAPITAL EXPENDITURE

The expenditures made for buying fixed assets like land, building, plant and machinery projects,etc. are called capital expenditure. We have already seen how present values of such assets can becalculated. P.V. concept in capital expenditure helps us to select the best out of alternatives. Forexample, suppose a pipe line is due for repairs. It will cost Rs. 10,000 and lasts for 3 years.Alternately, a new pipe line can be laid at a cost of Rs. 30,000 which will last for 10 years. Assumecost of capital is 10% and ignore salvage value.


The present values of both the cases are to be calculated and compared. The project havinglesser P.V. is to be selected.

1.8.1. Free-hold and Lease-hold Estates

1. A free-hold estate yields perpetual annuity, i.e., rent whereas a lease-hold estate held for afixed period yields rent for that fixed period only.

2. The value of free-hold estate can be taken as the P.V. of the perpetuity having each paymentequal to rent. The value of certain lease hold property is the P.V. of an annuity certain havingeach payment equal to rent, the status being the no. of years to the time the lease terminates.

3. Value of a free-hold estate = P.V. of the perpetuity of P (Annual rent) = = =P

iPx say x

i( ), ,

1

is the no. of year’s purchase.4. If a man having purchased a lease yielding a rent of Rs. P yearly and lasting for n years, wants,

after ‘x’ years (n > x), to renew it for another period of ‘y’ years, he must pay fine for therenewal. The fine is the P.V. of an annuity of P for y years deferred (n − x) years.

5. To obtain the P.V. of a lease hold property with a rent P, we are to find the P.V. of the annuityof P for the remaining period of the lease.

6. A lease-hold property reverts to the original holder after termination of the lease. An estate

worth X to be reverted after n years has the reversion value = P.V of XX

i n.( )

.=+1

If the yearly rent is P, then from (3), XP

i= . ∴ The value of reversion =

+P

i i n( ).

1

7. A company increases its capital by issuing loans, which are called “Debentures”. The deben-ture-holders get their fixed rate of interest whether the company earns profit or not. It is alsoissued by State or Central Govts., or, other public bodies, e.g., Improvement Trusts, Corpora-tions, etc. Like debenture bond it is also a written promise to pay or do something.

1.9. PRESENT VALUE AND DISCOUNT

The P.V. of a given sum A due after a given period is that sum of money [P (say)] which becomesthe sum A after the given period.

(a) P.V PA

Ni. = =

+1 (Simple interest)

True Discount (T.D.) = − = −+

= −+

FH

IKA P A

A

NiA

Ni11

1

1 =

+ANi

Ni1

(b) P.V. = =+( )

PA

i N1 (Compound interest)

T.D AA

iA iN

N. = −+( )

= − +( )−

11 1n s


1.9.1. Banker’s Discount and Present Value

If a sum of money is due after a given period, then the interest (simple or compound) on the sumof money for that period is called the ‘Banker’s Discount’. It is the interest on the face value ofthe bill. [Face value = P.V. + T.D.].

True discount (T.D.) is the interest (simple or compound) on the present value of the sum dueor present value of the amount of the bill. Banker’s P.V. is the difference between the sum ofmoney due and the Banker’s Discount.

Banker’s Gain = Banker’s Discount − True Discount

= Interest on Bill value − Interest on Present value

= Interest on (Bill value − Present value)

= Interest on True Discount

Legal due date = Nominal due date + Three days of grace.

Example 1.8. (i) The difference between true discount and banker’s discount on a bill due after 6months at 4% p.a. is Rs. 24. Find out true discount, banker’s discount and bill amount.

[B.Com. (Bangalore), 1991]

(ii) If the difference between true discount and banker’s discount on a sum due in 6 months at5% p.a. is Rs. 100, find the amount of the bill. [B.Com. (Bangalore), 1990]

(iii) The banker’s gain for a sum due 10 months hence at 6% p.a. is Rs. 25. Find the sum due.[ICWAI, June 1993]

(iv) A bill for Rs. 12,900 was drawn on 3rd February, 1988 at six months date and discounted on13th March at the rate of 8% p.a. For what sum was the bill discounted and how much didthe banker gain on this? [B. Com. (Bangalore), 1991]

Answer: (i) (Banker’s discount − True discount) for 6 months = Interest on true discount for 6 months

∴ True discount × 0 041

224. × = or, True discount = =24

0 021 200

.. ,Rs

If Rs. x is the P.V. of the bill, then x × × =0 041

21 200. , ∴ x = Rs. 60,000

∴ Amount of the bill = Rs. (60,000 + 1,200) = Rs. 61,200.

True discount = Rs. 1,200; Banker’s discount = Rs. 1,224.

(ii) Let the amount of the bill be Rs. x.

Present value of the bill = =+

=+ ×

= ×PA

ni

xx

1 15

10012

200

205

T.D xx

.= × × =200

205

5

100

1

2 41Banker’s discount = × × =x

x0 05

1

2 40. .

Banker’s discount − True discount = −FH

IK = =x x x

Rs40 41 1640

100. (given)

∴ Amount of bill = Rs. 1,64,000.


(iii) Let Rs. x be the P.V. of the sum due

∴ Sum due = + × ×FH

IK = +FH

IKRs x x Rs x

x. . .0 06

10

12 20T.D x

xx

x.= + − =

20 20

Banker’s gain = Interest on T.Dx

. .= × × =20

0 0610

1225 (given) ∴ x = Rs. 10,000

(iv) Due date of the bill = 6th August (Includes 3 days of grace)

No. of days for which Banker’s discount is paid.

= 18 + 30 + 31 + 30 + 31 + 6 = 146 days.

Banker’s discount = × × =Rs Rs. , . . .12 900146

3650 08 412 8

Present value of the bill on the day of discounting:

Present value + Present value ×146

3650 08 12 900× =. , or, Present value 1

16

50012 900+L

NMOQP

= ,

or, Present value = × =Rs.,

,500.12 900 500

51612 ∴ T.D. = Rs. (12,900 − 12,500) = Rs. 400

Banker’s gain = Rs. (412.8 − 400) = Rs. 12.8.

1.10. TYPICAL EXAMPLES

Example 1.9: (i) Machine A costs Rs. 10,000 and has useful life of 8 years. Machine B costsRs. 8,000 and has useful life of 6 years. Suppose machine A generates an annual savings ofRs. 2,000 while machine B generates an annual saving of Rs. 1,800. Assuming the time value ofmoney is 10% p.a., which machine is preferable? [D.U., B.Com. (Hons.), 1997]

(ii) How much is needed to be saved each year in a savings account paying 6% p.a. com-pounded continuously in order to accumulate Rs. 6,000 in three years?

[D.U., B.Com. (Hons.), 1992]

(iii) A man retires at the age of 60 years and his employer gives him a pension of Rs. 1,200 ayear for the rest of his life. Reckoning his expectations of life to be 13 years and that interestis at 4% p.a., what single sum is equivalent to his pension?

[Given: log 104 = 2.0170 and log 6012 = 3.7790]. [C.A. (Ent.) May, 1991]

(iv) A fixed royalty of Rs. 20,000 p.a. for 15 years is granted to an author by a publishingcompany. The right of receiving the royalty is sold after 10 years. Find to the nearest rupeethe price at which it is sold, assuming money is worth 12% p.a. compounded annually.

(v) Find the present value of Rs. 500 due 10 years hence when interest of 10% is compounded(i) half-yearly, (ii) continuously. [D.U., Eco. (Hons.), 1989]

(vi) A money-lender charges ‘interest’ at the rate of 10 paise per rupee per month, payable inadvance. What effective rate of interest does he charge p.a.? [D.U., B.Com. (Hons.), 1996]

(vii) Rs. 3,000 is invested at annual rate of interest of 10%. What is the amount after 3 years ifthe compounding is done:

(a) Annually? (b) Semi-annually? (c) Monthly? (d) Daily?


(viii) An investor wants to buy a 3-year Rs. 1,000 per value bond having nominal interest rate of10%. At what price the bond be purchased now if it matures at par and the investor requiresa rate of return of 14%?

(ix) What annual rate of interest compounded annually doubles an investment in 5 years?

(x) A man opened an account on April, 2004 with a deposit of Rs. 900. The account paid 5%interest compounded quarterly. On October 1, 2004, he closed the account and added enoughadditional money to invest in a 6-month Time Deposit for Rs. 1,200 earning 5% com-pounded monthly.

(a) How much extra amount did the man invest on October 1, 2004?

(b) What was the maturity value of his Time Deposit on April 1, 2005?

(c) How much total interest was earned?

(xi) Rs. 10,000 is invested in a Term Deposit Scheme which yields interest 5% p.a. compoundedquarterly. What will be the interest after 1 year? Work-out ab initio. What is the effectiverate of interest?

(xii) Mr. Y has made real estate investment for Rs. 15,000 which he expects will have a maturityvalue equivalent to interest at 12% compounded monthly for 5 years. If most of the savingsinstitutions currently pay 6% compounded quarterly on a 5-year term, what is the leastamount for which Mr. Y should sell his property?

(xiii) Mr. X plans to receive an annuity of Rs. 8,000 semi-annually for 10 years after he retires in15 years. Money is worth 10% compounded semi-annually.

(a) How much amount is needed to finance the annuity?

(b) What amount of single deposit made now would provide the funds for the annuity?

(c) What amount will Mr. Y receive from the annuity?

(xiv) Find the effective rate equivalent to nominal rate of 10% compounded (a) half-yearly;(b) quarterly; (c) monthly; (d) continuously.

(xv) An N.S.C costs Rs. 20 and realises Rs. 25 after 10 years. Find the rate of interest involvedwhen it is added (a) yearly; (b) continuously.

(xvi) A bank issues ‘Re-investment certificates’ for a period of 2 years. If Rs. 5,000 are investedin these certificates, their maturity value is Rs. 6,500. Assuming that the interest is com-pounded every year, what is the rate of interest?

Answer: (i) Machine A: P.V. of a sequence of annual savings of Rs. 2,000 for 8 years @ 10% p.a. is given by:

VP

ii n= − + = − +− −1 1

2000

0 101 1 0 10 8a fo t a fo t

.. = − = × =20 000 1 0 4665 20 000 0 5335, . , .b g Rs. 10,670

∴ Net saving from Machine A = Rs. 10,670 − Rs. 10,000 (cost of machine) = Rs. 670

Machine B: P.V. of a sequence of annual savings of Rs. 1,800 for 6 years @ 10% p.a. is given by:

V = − + = −−1800

0 101 1 0 10 18000 1 0 564476

.. .a fo t k p = 18000 × 0.43553 = Rs. 7,839.54.

∴ Net savings from Machine B = Rs. (7,839.54 − 8,000) = −Rs. 160.46.

Hence, Machine A is preferable.


(ii) Let Rs. P be saved per year for 3 years to get Rs. 6,000.

Here, A = Rs. 6,000, R = 0.06, N = 3.

∴ A P e dN Pe

R

Pe eRN

RN

= ⋅ ⋅ =L

NM

O

QP ⇒ = −z

0

3

0

318 06 000

0 06,

.

⇒ 6 0000 06

11972 1,.

.= −P∴ P = × =6000 0 06

0 1972

.

.Rs. 1,825.56. ∵ e18 11972From Tableb g = .

Thus, Rs. 1,825.56 should be saved each year for 3 years.

(iii) Here, P = Rs. 1,200, n = 13, i = 0.04

Using the formula: VP

ii n= − + −1 1a fo t we have, V = − = − =−1200

0 041 1 04 30 000 1 0 6012

13

.. , .b g{ } l q Rs. 11,964

Let x = ∴ = − = − × = −−1 04 13 1 04 13 0 0170 0 22113. , log log . . .x b g

or, log or, xx = = ∴ =−1 7790 0 6012 1 04 0 601213. . , . . .

Hence the single sum equivalent to his pension is Rs. 11,964.

(iv) Out of 15 years, 10 years have elapsed. Hence the author is entitled to only 5 yearly instalments of Rs. 20,000.

Using the formula: VP

ii n= − + −1 1a fo t, where P i n= = =Rs. and we have,20 000 012 5, , . ,

V = − = −−20 000

0 121 112 166666 67 1 0 56745,

.. . .a f{ } k p = × =166666 67 0 4326. . .Rs. 72,100

Thus, the price at which the author sold his royalty is Rs. 72,100.

(v) (a) Here, A i N= = = = × =50010

2000 05 2 10 20, . ,

We know, P A i N= + =− −1 500 1 05 20a f a f. = × =500 0 3769. Rs. 188.45

∴ The present value is Rs. 188.45.

(b) Here, A = 500, R = 0.10, N = 10.

We know, A P e A eRN RN= ⋅ = ⋅ − or, P

or, P ee

= ⋅ = = =− ×500500 500

2 71820 10 10.

..Rs. 183 95

Thus, the present value (P) is Rs. 183.95.

(vi) As 10 paise/rupee/month is payable in advance, the interest rate p.m. will be 10

100 10

1

912

−= =

b g per month, N .

Now, Effective interest rate (E) = (1 + i)N − 1 = 11

91 3 541 1 2 541 254 1%

12

+FH

IK

− = − =. . , . ., .i e

Hence the effective rate of interest is 254.1%.(vii) (a) The annual compounding is given by;

A P i N PN= + = = =1 0 1 2 3000a f , . , , where i = = × =3000 11 3000 1 212. .a f Rs. 3,630.


(b) For semi-annual compounding, N = 2 × 2 = 4, i = =0 1

20 05

..

∴ A = = × =3000 1 05 3000 1 21554. .a f Rs. 3,646.52

(c) For monthly compounding, N i= × = = =2 12 240 1

120 00833,

.. .

∴ A = = × =3000 1 00833 3000 1 2202924

. .b g Rs. 3,660.87.

(d) For daily compounding, N = 2 × 365 = 730, i = 0.1/365 = .00027.

∴ A = = × =3000 1 00027 3000 1 21783730

. .b g Rs. 3,653.49.

(viii) P.V. of the bond =1000 1

114

1000 1

114

1000 1

114

1000

1142 3 3

×+

×+

×+

.

.

.

.

.

. .a f a f a f a f= 87.7193 + 76.9468 + 67.4972 + 674.9715 = 907.1348.

Thus purchase value of the bond is Rs. 907.1348.

(ix) 2P = P (1 + i)5 [∵ A = P (1 + i)N]

or, 2 1 1 1148698 15= + = + = +i i ib g b g or, 2 or, 1 5 .

or, i = 0.148698, ∴ Required rate of interest = 14.87%

(x) (a) The initial investment earned interest for 2 quarters (i.e., April to June and July to September).

Here, i N= = =5

41

1

42%, , and the compounded amount

= +FHG

IKJ

= + =900 1 11

4900 1 1 25% 900 1 0125

22 2

% . .b g b g = × =900 1 02516. Rs. 922.64.

∴ The additional amount = Rs. (1,200 − 922.64) = Rs. 277.36.

(b) Here, Time Deposit earned interest compounded monthly for 2 quarters, i.e., i N P= = =5

126 1200%, ,

∴ Maturity value = 1200 15

121200 1 0

66+F

HGIKJ

= +% .4167%a f

= + = × =1200 1 0 004167 1200 1 025266. .a f Rs. 1,230.31

(c) Total interest earned = Rs. (22.64 + 30.31) = Rs. 52.95

(xi) Interest for first quarter = 10 0001

40 05, .b g b g

FHGIKJ

= Rs. 125

∴ New principal at the end of 1st quarter = Rs. 10,125.

Interest for second quarter = 101251

40 05, .b g b g

FHGIKJ

= Rs. 126.56

∴ New principal at the end of 2nd quarter = Rs. 10,251.56

Interest for 3rd quarter = 10,251.56a f a f1

40 05 12814F

HIK

=. .


∴ New principal at the end of 3rd quarter = Rs. 10,379.70

Interest for 4th quarter = 10379 701

40 05. .b g b g

FHGIKJ

= Rs.129.75

∴ Interest accrued after 1 year = Rs. (125 + 126.56 + 128.14 + 129.75) = Rs. 509.45

For effective rate of interest, use I = PRt.

or, 509.45 = (10,000) R × 1 or, R = 0.0509

∴ Effective rate of interest is 5.09%

Alternative method:

E in= + − = = =1 1

5

41

1

44b g , % % where i and n ( )

441

1 1 % 1 1.0125 1 0.0509, i.e., 5.09%4

⎛ ⎞= + − = − =⎜ ⎟⎝ ⎠

(xii) Maturity value of the investment, i.e, A P iNN= +1b g ,

where P i N= = = = × =15 00012

121%, 5 12 60, ,

∴ AN = + = = ×15 000 1 1% 15 000 1 01 15 000 181669669960 60

, , . , .b g b g

= Rs. 27,250.45.

The present value P, of the amount AN due at the end of N interest periods @ 1% interest per period is given by,

P A iNN= ⋅ + −

1b g

Here, A i NN = = = = × =Rs. 27 250 456

41

1

25 4 20, . , %,

∴ P = +FHG

IKJ

−

27250 45 1 11

2

20

. % = + =− −27250 45 1 0 015 27250 45 1 015

20 20. . . .b g b g

= × =27250 45 0 742470418. . Rs. 20,232.65.

∴ Mr. Y should not sell the property for less than Rs. 20,232.65.

(xiii) (a) P.V. for the 10 year annuity (V)

= − + = = =−P

ii i n

n1 1 8000 5%, 20b g{ }, ,where P

∴ V = − + = −−8000

0 051 1 0 05 1 60 000 1 0 376889482

20

.. , , .b g{ } b gn s

= 1,60,000 × 0.623110518 = Rs. 99,697.68.

(b) We require the amount of single deposit which matures to Rs. 99,697.68 in 15 years at 10% compounded semi-annually,

A P i NNN

N= + = = × = = =1 99 697 68 15 2 3010

25%b g , , . , where A and i

∴ P A iNN= + −

1b g = + = ×−99697 68 1 0 05 99697 68 0 231377448

30. . . .b g

= Rs. 23,067.79.


(c) Required amount = Rs. 8000 × 20 = Rs. 1,60,000.

(xiv) (a) E (Effective rate) = (1 + i)n − 1

Here, i n= = =01

20 05 2

.. ,

∴ E i e= + − = − =1 0 05 1 11025 1 0 1025 10 25%2

. . . , . ., .b g

(b) Here, i n= = =0 1

40 025 4

.. ,

∴ E i e= + − = − =1 0 025 1 11038 1 01038 10 38%4

. . . , . ., .b g

(c) Here, i n= =0 1

1212

.,

∴ E i e= +FHG

IKJ

− = − =101

121 11047 1 01047 10 47%

12.

. . , . ., .

(d) The effective rate equivalent nominal rate 10% converted continuously is given by,

E e e i eR= − = − = − =1 1 11052 1 01052 10 52%0 1. . . , . ., .

(xv) (a) A P i A NNN

N= + = = =1 20 25 10b g , , , where P

or, 25 20 15

41

10 10= + = +i ib g b g or,

or, log log log5 4 10 1− = + ib g

or, loglog log . .

15 4

10

0 6990 0 6021

10+ = − = −

ib g

or, log .1 0 00969+ =ib g

or, 1 1 0226+ = =i Antilog 0.00969b g .

or, i i e= 0 0226 2 26%. , . ., .

(b) A P e eRN R R= ⋅ ⇒ = ⋅ =25 205

410 10 or, e

or, 10 5 4 0 6990 0 6021 0 0969R elog log log . . .= − = − =

or, Re

= =×

0 0969

10

0 0969

10 0 4343

.

log

.

. ∵ log .e = 0 4343

= 0 0223 2 23%. , . ., .i e

(xvi) If P be the principal, A be the amount, R% be the rate of interest and N be the number of years, then A PR N

= +FH

IK

1100

.

Here A = Rs. 6,500, P = Rs., 5,000, N = 2

∴ 1100

6500

50001

1001 3 11402

21 2+F

HGIKJ

= + = =R R or, . .b g

or, R

R100

01402 14 02%.= ∴ =. , .


EXERCISE 1(b)

1. A man borrows Rs. 40,000 at 4% compound interest and agrees to pay both the principal andinterest in 10 equal instalments at the end of each year. Find the amount of these instalments.

[B.Com. (Bangalore), 1991]

2. Find the amount of an immediate annuity of Rs. 100 p.a. left unpaid for 10 years allowing 5%p.a. compound interest. [B.Com. (C.U.)]

3. Shri Ajit bought a house paying Rs. 20,000 down payment and Rs. 4,000 at the end of eachyear 25 years. What amount he would have paid for house, if he had bought it cash down?Reckon interest at 5% p.a. C.I. [B.Com. (Bangalore), 1991]

[Hints: AP

ii

n= + − + −20 000 1 1, b g{ }

= + − −20 000

4 000

0 051 1 05

25,

,

..b g{ } ]

4. A machine costs Rs. 65,000 and its effective life is estimated to be 25 years. A sinking fund iscreated for replacing the machine at the end of its life time when its scrap realises a sum of Rs.2,500 only. Calculate what amount should be provided every year out of profits for the sinkingfund if it accumulates at 3½% p.a. compound. [ICWAI, Dec. 1992]

5. The cost of a machine is Rs. 1,00,000 and its effective life is 12 years. If the scrap realises onlyRs. 5,000, what amount should be retained out of profits at the end of each year to accumulateat C.I. 5% p.a.?[Given: log10 1.05 = 0.0212, log10 1.797 = 0.2544] [C.A. (Ent.), Nov. 1988; ICWAI, Dec. 1990]

6. A man retires at the age of 60 years and his employer gives him a pension of Rs. 1,200 a yearfor the rest of his life. Reckoning his expectations of life to be 13 years and that interest is at4% p.a., what single sum is equivalent to his pension? [Given: log 104 = 2.0170 and log 6012= 3.7790] [C.A. (Ent.), May 1991]

7. A man decides to deposit Rs. 500 at the end of each year in a bank which pays 8% p.a.compound interest. If the instalments are allowed to accumulate, what will be the total accu-mulation at the end of 7 years?

8. Find the amount of an annuity of Rs. 100 in 20 years, allowing compound interest at 4½%.[Given: log 1.045 = 0.191163 and log 24.117 = 1.3823260] [ICWAI]

9. A person desires to create an endowment fund to provide for a prize of Rs. 300 every year. Ifthe fund can be invested at 10% p.a. compound interest, find the amount of the endowment.

[ICWAI, June 1976]

10. What sum should be invested every year at 5% p.a. compound interest for 20 years, to replaceplant and machinery, which is expected to cost them 25% more over its present cost of Rs.60,000? [B.Com. (C.U.), 1970]

11. A machine costs a company Rs. 80,000 and its effective life is estimated to be 20 years. Asinking fund is created for replacing the machine at the end of its effective life time when itsscrap realises a sum of Rs. 5,000 only. Calculate to the nearest hundreds of rupees, the amount


which should be provided, every year, for the sinking fund if it accumulates at 9% p.a. com-pounded annually. [ICWAI, Dec. 1975]

12. A motor cycle is purchased on instalment basis such that Rs. 3,400 is to be paid on the signingof the contract and four yearly instalments of Rs. 2,400 each payable at the end of the first,second, third and fourth years. If interest is charged at 8% p.a., what would be the cash downprice? [Given: (1.08)4 = 1.36.] [B.Com. (Hons.) (C.U.), 1983]

13. A person invests Rs. 1,000 every year with a company which pays interest at 10% p.a. Heallows his deposits to accumulate with the company at compound rate. Find the amount stand-ing to his credit one year after he has made his yearly investment for the tenth time.

[ICWAI, Dec. 1975]

14. Calculate the present value of an annuity of Rs. 5,000 p.a. for 12 years, the interest being 4%p.a. compounded annually. [ICWAI, Dec. 1974]

15. A person purchases a house worth Rs. 70,000 on a hire purchase scheme. At the time ofgaining possession he has to pay 40% of the cost of the house and the rest amount is to be paidin 20 equal annual instalments. If compound interest is reckoned at 7½% p.a., what should bethe value of each instalment? [ICWAI, June 1980]

16. A person desires to endow a bed in a hospital the cost for which is Rs. 6,000 p.a. If the moneyis worth 10% p.a., how much should he provide in perpetuity? He further desires to providefor the building of a ward, at a cost of Rs. 1,00,000 and also to provide for periodical repairsthereto entailing 5% of the cost of the ward per year on the average. What should be the totalvalue of the endowment? [ICWAI, June 1977]

17. A person retires at the age of 58 and earns a pension of Rs. 6,000 a year. He wants to commuteone-fourth of his pension to ready money. If the expectation of life at this age be 12 years, findthe amount he will receive when money is worth 4% p.a. compound. (It is assumed that pen-sion for a year is due at the end of the year). [ICWAI, June 1981]

18. A loan of Rs. 8,000 is to be paid in 5 equal annual payments, interest being at 5% p.a. com-pound interest and first payment being made after a year. Find the amount of each of thepayments. [ICWAI, Jan. 1971]

19. On his 48th birthday a man decides to make a gift of Rs. 5,000 to a hospital. He decides to savethis amount by making equal annual payments upto and including his 60th birthday to a fundwhich gives 3½% compound interest, the first payment being made at once. Calculate theamount of each annual payment. [ICWAI, July 1972]

20. A person borrows a sum of Rs. 5,000 at 4% compound interest. If the principal and interest areto be repaid in 10 equal annual instalments, find the amount of each instalment; first paymentbeing made after 1 year. [ICWAI, Jan. 1969]

21. Find the sum of money received by a pensioner at 58 if he wants to commute his annualpension of Rs. 1,200 for a present payment when compound interest is reckoned at 4% p.a.and the expectation of his life is assessed as 10 years only. [ICWAI, July 1969]

22. A man purchases a house and takes a mortgage on it for Rs. 60,000 to be paid off in 12 yearsby equal annual payments. If the interest rate is 5% compounded annually, what amount willbe required to pay each year? [ICWAI, Jan. 1966]


23. A company sets aside for a reserve fund the sum of Rs. 20,000 annually to enable it to pay offa debenture issue of Rs. 2,39,000 at the end of 10 years. Assuming that the reserve accumu-lates at 4% p.a. compound, find the surplus after paying off the debenture stock.

[ICWAI, July 1970]

24. For endowing an annual scholarship of Rs. 10,000, Mr. Reddy wishes to make five equalannual contributions. The first award of the scholarship is to be made two years after the last ofhis five contributions. What would be the value of each contribution, assuming interest at 5%p.a. compounded annually?

[Assume that the first contribution is to be made now and the other four at an interval of oneyear thereafter.] [ICWAI, Dec. 1977]

ANSWERS

(1) Rs. 4,932; (2) Rs. 1,258; (3) Rs. 76,373; (4) Rs. 1,600; (5) Rs. 5,960; (6) Rs. 11,964;(7) Rs. 4,456.25; (8) Rs. 3,137.12; (9) Rs. 3,000; (10) Rs. 2,266; (11) Rs. 1,500;(12) Rs. 11,341.18; (13) Rs. 17,534; (14) Rs. 46,850; (15) Rs. 4,120; (16) Rs. 60,000,Rs. 1,10,000; (17) Rs. 14,055; (18) Rs. 1,847; (19) Rs. 343.14; (20) Rs. 617.50;(21) Rs. 9,717; (22) Rs. 6,767; (23) Rs. 500 (∵A = Rs. 2,39,500); (24) Rs. 34,440.

SUMMARY

The chapter begins with the discussion of simple and compound interests, time value of money,annuities, present values of annuity and perpetuity, etc and ends with the discussion of sinkingfund, amortization, present value of capital expenditure, free-hold and lease-hold estates, banker’sdiscount, etc. Each definition has been explained with adequate number of examples.

KEY-TERMS

(a) Some Important Formulae

1. A = P + S.I., where A = amount, P = principal, N = No. of years.

= P (1 + Ni), iR=

100; and R in % p.a. is rate of interest.

2. I = Compound interest = − = + − = + −A P P i P P iN N1 1 1a f n s( )

where, P = principal, N = No. of years, i = interest on Rs. 1 for 1 year.

3. Amount of an annuity = = + -AP

ii n( ) ,1 1 where P = Annuity, n = unpaid years, i = interest

on Rs. 1 for 1 year.

4. P.V. of an annuity = = − + −VP

ii n1 1( ) ,n s where P = Annuity to continue for n years.

5. P.V. of a perpetuity = =VP

i.


6. P.V. of a deferred annuity = =+ −+

RST

UVW

+VP

i

i

i

n

m n

1 1

1

a f

a f, where P = Annuity, Payment starts at the

end of m years and thereafter continues for n years.

7. P.V. of a deferred perpetuity = =+

L

NM

O

QPV

P

i i m

1

1a f, P to begin after m years.

8. If A is the amount of loan to be paid off at the end of n years and P is the accumulations of theannual sum, then

AP

ii n= + −1 1a f

9. If P1, P2, ..., Pn be the unequal cash inflows received at the end of year 1, 2, ..., n respectively,then the present value of these sums at interest rate ‘i’ is given by:

VP

i

P

i

P

in

n= ++

+ ++

1 221 1a f a f

……

10. Banker’s Gain = Banker’s Discount − True Discount.

11. (i) True Discount = − + −A i N1 1a fo t. (Compound Interest)

(ii) True Discount = A Ni

1+ Ni. (Simple Interest)

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