107475 1139046 Appendix a Mathematics of Finance

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J B GUPTA CLASSES 98184931932, [email protected] , www.jbguptaclasses.com Copyright: Dr JB Gupta APPENDIX A APPENDIX A APPENDIX A APPENDIX A MATHEMATICS OF FINANCE MATHEMATICS OF FINANCE MATHEMATICS OF FINANCE MATHEMATICS OF FINANCE Logarithms Logarithms Logarithms Logarithms Logarithms are of great use in calculations. They simplify typical calculations. With the help of Logarithms, we can make such calculations which other wise are difficult to make. Logarithms are of two types (i) simple Logarithms (mathematically called as log base to 10) (ii) natural log (mathematically called as natural log). In this note we shall be studying simple logarithms. The logarithm of a number consists of two parts – characteristic and mantissa. Characteristic is determined without any table. Mantissa is determined using log tables. Finding characteristic of a number 1 or greater than 1: Finding characteristic of a number 1 or greater than 1: Finding characteristic of a number 1 or greater than 1: Finding characteristic of a number 1 or greater than 1: In this case characteristic is equal to “number of digits before decimal” minus “one”. Number 7 56 567 5678 56432 5670.23 167.89 Characteristic 0 1 2 3 4 3 2 Finding Finding Finding Finding characteristic of a number less than one : characteristic of a number less than one : characteristic of a number less than one : characteristic of a number less than one : In this case characteristic is negative. Negative sign is written in the form of bar. For example -1 is written as 1 , -2 is written as 2 , -3 is written as 3 . Write the number (of which characteristic is to be determined) in proper decimal form. In this case characteristic is number of zeros before and just after decimal. Number .9 .08 .007 .0006 .00006 .00908 .002003 Number in proper Decimal form 0.9 0.08 0.007 0.0006 0.00006 0.00908 0.002003 Characteristic ī -2 -3 -4 -5 -3 -3 Mantissa is Mantissa is Mantissa is Mantissa is determined using log tables. Mantissa is always positive. For determining Mantissa decimal is ignored. Before finding Mantissa, the number (of which mantissa is to be determined) should be reduced to four digits by approximation. Number 233 2655 456.8 0.89 0.00902 Characteristic 2 3 2 ī -3

description

Notes

Transcript of 107475 1139046 Appendix a Mathematics of Finance

Page 1: 107475 1139046 Appendix a Mathematics of Finance

J B GUPTA CLASSES 98184931932, [email protected],

www.jbguptaclasses.com

Copyright: Dr JB Gupta

APPENDIX AAPPENDIX AAPPENDIX AAPPENDIX A

MATHEMATICS OF FINANCEMATHEMATICS OF FINANCEMATHEMATICS OF FINANCEMATHEMATICS OF FINANCE LogarithmsLogarithmsLogarithmsLogarithms

Logarithms are of great use in calculations. They simplify typical calculations.

With the help of Logarithms, we can make such calculations which other wise are

difficult to make. Logarithms are of two types (i) simple Logarithms

(mathematically called as log base to 10) (ii) natural log (mathematically called as

natural log). In this note we shall be studying simple logarithms.

The logarithm of a number consists of two parts – characteristic and mantissa.

Characteristic is determined without any table. Mantissa is determined using log

tables.

Finding characteristic of a number 1 or greater than 1: Finding characteristic of a number 1 or greater than 1: Finding characteristic of a number 1 or greater than 1: Finding characteristic of a number 1 or greater than 1: In this case characteristic

is equal to “number of digits before decimal” minus “one”.

Number 7 56 567 5678 56432 5670.23 167.89

Characteristic 0 1 2 3 4 3 2

Finding Finding Finding Finding characteristic of a number less than one : characteristic of a number less than one : characteristic of a number less than one : characteristic of a number less than one : In this case characteristic is

negative. Negative sign is written in the form of bar. For example -1 is written

as 1 , -2 is written as 2 , -3 is written as 3 . Write the number (of which

characteristic is to be determined) in proper decimal form. In this case

characteristic is number of zeros before and just after decimal.

Number .9 .08 .007 .0006 .00006 .00908 .002003

Number in proper

Decimal form

0.9 0.08 0.007 0.0006 0.00006 0.00908 0.002003

Characteristic ī -2

-3 -4 -5 -3 -3

Mantissa isMantissa isMantissa isMantissa is determined using log tables. Mantissa is always positive. For

determining Mantissa decimal is ignored. Before finding Mantissa, the number (of

which mantissa is to be determined) should be reduced to four digits by

approximation.

Number 233 2655 456.8 0.89 0.00902

Characteristic 2 3 2 ī -3

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Mantissa .3674 .4240 .6598 .9494 .9552

Logarithms 2.3674 3.4240 2.6598 -0.0506

-2.0448

Antilog Antilog Antilog Antilog is determined using Antilog tables. The table is consulted only for

Mantissa part. Place of decimal is “characteristic plus one”. This place is counted

from left hand side.

Number 233 2655 456.8 0.89 0.00902

Log 2.3674 3.4240 2.6598 -1+.9494

-3+.9552

Antilog 233.0 2655 456.8 .8900 .009020

Q. Q. Q. Q. No. 1:No. 1:No. 1:No. 1: Find log of (i) 2 (ii) 56 (iii) 567 (iv) 5678 (v) 56.78 (vi) 0.543 (vii)

55556.67

AnswerAnswerAnswerAnswer

(i) log 2 = 0.3010

(ii) log 56 = 1.7482

(iii) log 567 = 2.7536

(iv) log 5678 = 3.7542

(v) log 56.78 = 1.7542

(vi) log 0.543 = -1+.7348

(vii) log 55556.67 = 4.7448

Q. No. 2Q. No. 2Q. No. 2Q. No. 2: Find Antilog of above log values.

AnswerAnswerAnswerAnswer

(i) AL 0.3010 = 2.000

(ii) AL 1.7482 = 56.01

(iii) AL 2.7536 = 567.0

(iv) AL 3.7542 = 5678.00

(v) AL 1.7542 = 56.78

(vi) AL -1+.7348= .5430 (vii) AL 4.7448 = 55560.00

Finding the values of Finding the values of Finding the values of Finding the values of eeeex x x x / e/ e/ e/ e----xxxx

Remember loge = 0.4343

Q. No. 3Q. No. 3Q. No. 3Q. No. 3 : Find the values of e3 ; e1.40 ; e0.10 ; e0.005 ; e-3 ; e-1.60 ;

e-0.15 ; e-0.06 using the tables of these values.

AnswerAnswerAnswerAnswer

(i) e3 = 20.0855 (ii) e1.40 = 4.0552

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(iii) e0.10 = 1.1052

(iv) e0.005 = ?

e0 = 1

e0.01 = 1.0101

For 0.01↑ in e’s power, RHS ↑ by 0.01005

For 1 ↑ in e’s power, RHS ↑ by 0.01005/0.01 i.e. 1.005

For .005 ↑ in e’s power, RHS↑ by 1.005 x .005 i.e..005025

e(0+0.005) = 1 +0.005025

e0.005 = 1.005025

(v) e-3 = 0.0498 (vi) e-1.60 = 0.2019

(vii) e-0.15 = 0.8607

(viii) e-0.06 = 0.9418

Q. No. 4Q. No. 4Q. No. 4Q. No. 4 : Find the values of e3 ; e1.40 ; e0.10 ; e0.005 ; e-3 ; e-1.60 ; e-

0.15 ; e-0.06 using the log tables .

Answer Answer Answer Answer

(i) e3 = AL (log e3 ) = AL (3 X 0.4343) = AL (1.3029) = 20.08

(ii) e1.40 = AL (log e1.40 ) = AL (1.40 X 0.4343 ) = AL (0.6080) = 4.055

(iii) e0.10 = AL (log e0.10 ) = AL (0.10 X 0.4343 ) = AL (0.0434) = 1.105

(iv) e0.005 = AL (log e0.005 ) = AL (0.005x0.4343) =AL(0.0022) = 1.005

(v) e-3 = AL (log e-3) = AL ( -3x 0.4343) = AL(-1.3029) =

AL (-2 + 0.6971) = 0.04978

(vi) e-1.60 = AL (log e-1.60 ) = AL( -1.60X0.4343) = AL (-0.6949)

= AL( -1 + 0.3051) = 0.2018

(vii) e-0.15 = AL (log e-0.15 ) = AL(-0.15x0.4343) = AL(-0.0651)

= AL(-1 + 0.9349) = 0.8608

(viii) e-0.06 = AL(log e-0.06) = AL(-0.06X0.4343) = AL(-0.0261)

= AL(-1 + .9739) = 0.9417

DiscountingDiscountingDiscountingDiscounting

Finance managers generally take decisions on the basis of cash flows i.e. cash

coming in and cash going out because of the proposal under consideration. Cash

received or paid at different points of time has got different value because of

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existence of interest. To make these cash flows of different points of time

comparable, we find present value of future cash flows. (This process of finding

present value of future cash flows is known as discounting)

To find present value of future cash flows, we multiply future cash flows with

their relevant present value factors.

PVF 1 i.e. present value factor 1 i.e. present value of rupee one to be received

or paid after 1 period =

1

r1

1

+

Similarly, PVF 2 =

2

r1

1

+

PVF 3 =

3

r1

1

+ And so on…

Suppose, rate of interest is 10% per period:

PVF 1 i.e. present value of rupee one to be received or paid after one period

=

1

10.1

1

i.e. 0.909

PVF 2 i.e. present value of rupee one to be received or paid after two

periods =

2

10.1

1

i.e. 0.826

PVF 3 i.e. present value of rupee one to be received or paid after 3 periods

=

3

10.1

1

i.e. 0.751

Suppose for doing a job today, a professional will receive Rs.10,000 after one

year from today, Rs.20,000 after two years from today and Rs.30,000 after three

years from today Assuming that rate of interest is 20%, find the present value of

what he receives.

In this case rate of interest is annual, the ‘period’ is year. Hence,

present value factors are as follows: PVF 1 =

1

20.1

1

=0.833

PVF 2 =

2

20.1

1

=0.694

PVF 3 =

3

20.1

1

=0.579

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Hence, the present value of what he receives is equal to [(10,000x0.833)

+(20,000x0.694)+(30,000x0.579)] i.e. Rs.39,580.

If cash flow is equal amount year after year, to find present value we multiply

that equal amount of cash flow with annuity which is sum of present value

factors. For example, if in the case mentioned in the above paragraph, he was to

receive Rs.20,000 annually for three years (i.e. at the end of each year), the

present value is Rs. 20,000 x (0.833+0.694+ 0.579)= 42,120.

If interest is compounded half yearly, the rate of interest should be taken for half

year and the period should be taken in terms of number of half-year(s). For

example, if rate of interest is 20% p.a. compounded half yearly, find (i) present

value of rupee one to be received after two years (ii) present value of rupee one

to be received after four years and (iii) present value of rupee one to be received

after six years.

AnswerAnswerAnswerAnswer (i)

4

10.1

1

=0.683

(ii)

8

10.1

1

=0.467

(iii)

12

10.1

1

=0.319

Suppose, a person is to receive Rs20,000 after two years from today, Rs.20,000

after four years from today and Rs.20,000 after six years from today. Find the

present value of what he receives if rate of interest is 20% p.a. compounded half

yearly. The answer is Rs. 20,000 x (0.683+0.467+ 0.319) i.e. Rs 29, 380.

If interest is compounded quarterly, the rate of interest should be taken for a

quarter, and the period should be taken in terms of number of quarter(s).

Similarly, if interest is compounded monthly, the rate of interest should be taken

for a month, and the period should be taken in terms of number of month(s)

Q. No. 5:Q. No. 5:Q. No. 5:Q. No. 5: Rate of interest 20% p.a. A person is to receive Rs.1,00,000 after one

year from today. What is the present value of what he receives if interest is

compounded annually? What if half yearly? What if quarterly?

Answer Answer Answer Answer

If annually, it is 1,00,000 x

1

20.1

1

=83,300

If half yearly, it is 1,00,000 x

2

10.1

1

=82,600

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If quarterly, it is 1,00,000 x

4

05.1

1

=82,300

For “Options and Futures” it is assumed that interest is compounded

continuously.( Why this assumption? This question will be answered while

discussing the topic of Options and Futures)

The year is assumed to be divided into infinite small parts and it is also assumed

that interest compounds after each such small part of the year. (foot note 1) 1

In this case, the rate of interest for each such small period will be r/∞ (footnote

2)2 and the number of periods will be ∞. Hence, the present value factor will be equal to:

1

PVF = ————————

[(1) + (r / ∞)]∞

The mathematicians have provided its solution as equal to e-rt . Where r is annual interest rate; t=part /number of year(s).(For example if cash is

to be received or paid after six months, t = 0.50;if after 3 months, t = 0.25; if

after 1year, t= 1 ; if after 2years, t= 2)

Q. No. 6Q. No. 6Q. No. 6Q. No. 6:::: Find the present value of Rs.1,00,000 to be received after 1 year if rate

of interest is 20% p.a. continuously compounding.

AnswerAnswerAnswerAnswer

To find this value, we have to find the value of e-0.20x1

= Antilog(-0.20.log e ) =A.L.(-.20x.4343) =A.L.(-.08686)

=A.L. ( )91314.1 = 0.8187. (Footnote 3 )3 Hence, present value = Rs.1,00,000(.8187) i.e. Rs.81870.

Q. No. 7Q. No. 7Q. No. 7Q. No. 7: A person is to receive Rs.1,00,000 after six months.

Find the present value of what he receives in interest rate is 20% p.a. c.c.

1 In case of half yearly compounding, the year is divided into two parts; in case of

quarterly compounding the year is divided into four parts; in case of monthly

compounding the year is divided into 12 parts; in this case i.e. continuously

compounding it is assumed that the year is divided into infinite small parts

2 In case of half yearly compounding, r is taken as annual interest/2, in case of

quarterly compounding it is taken as annual interest/4, in case of monthly

compounding it is taken as annual interest rate /12; in this case i.e. continuously

compounding, it is taken as annual interest/∞

3 Alternatively, we can find this value in the table of values for e

-x .

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Answer Answer Answer Answer PV = 1,00,000.e-.20 x .5 = 90,480

CompoundingCompoundingCompoundingCompounding

Compounding is the process of obtaining the amount to which today’s rupee will

grow in future for the given rate of interest. We can understand compounding

with the help of Q.No.8.

Q. No. 8 : Q. No. 8 : Q. No. 8 : Q. No. 8 : A person invests Rs.1,00,000 for one year at interest rate of 20% p.a..

What amount he will receive after one year if interest is compounded

(a) annually (b) half yearly (c) quarterly and (d) continuously .(ignore tax)

AnswerAnswerAnswerAnswer

(a) = 1,00,000(1.20)1 =1,20,000

(b) = 1,00,000(1.10)2 =1,21,000

(c) = 1,00,000(1.05)4 =1,21,551

(d) = 1,00,000.e.20 x 1 =1,22,140

Q. No. 9 :Q. No. 9 :Q. No. 9 :Q. No. 9 : A finance company accepts deposits for any period of choice of the

depositor. Rate of interest is 10% p.a. compounded half yearly. What total amount

(principal and interest) a person will receive after 3 months if he has deposited

Rs 2,00,000 for three months? (ignore TDS)

Answer Answer Answer Answer

Amount to be received after 3 months = 2,00,000(1.05)(1/2) = 2,04,939

Alternative solution :Alternative solution :Alternative solution :Alternative solution :

Let interest rate per rupee for 3 months = x

1(1+x)(1+x) = 1.05

x =0.024695

Amount to be received = 2,00,000(1.024695) = 2,04,939

Q. No. 10:Q. No. 10:Q. No. 10:Q. No. 10: Under VARISTHA BIMA YOJANA, interest is 9% p.a. compounded or

payable monthly. An investor wants that he should get Rs.24,000 p.a. as interest

( payable at the end of each year ). What amount of investment is required? How

your answer change if he wants Rs.12,000 per half year , as interest , ( payable

at the end of each half year)? What should be the investment, if he wants Rs

6,000 per quarter, as interest, (payable at the end of each quarter)? Ignore tax.

Answer Answer Answer Answer

(a)(a)(a)(a) Interest requirement is annual:

Let amount of investment required be Rs X.

X(1+.0075)12 = X+24,000

X =2,55,864

(b)(b)(b)(b) Interest requirement is half-yearly:

Let amount of investment required be Rs. X.

X(1.0075)6 = X+12,000

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X = 2,61,713

(c)(c)(c)(c) Interest requirement is quarterly

Let the amount of investment required be X:

X(1.0075)3 = X + 6,000

X = 2,64,678

Q. No. 11:Q. No. 11:Q. No. 11:Q. No. 11: Find the total amount (principal + interest) to be received on maturity

in each of the following cases:

Amt. of Invest. Period Interest rate

(a) 1,000 1 year 10% p.a.

(b) 1,000 1 year 10% p.a. annually compounded

(c) 1,000 2 years 10% p.a.

(d) 1,000 2 years 10% p.a. annually compounded

(e) 1,000 6 months 21%p.a.

(f) 1,000 6 months 21% p.a. annually compounded

(g) 1,000 2 ½ years 10%p.a.

(h) 1,000 2 ½ years 10% p.a. annually compounded

(i) 1,000 3 months 10% p.a. annually compounded

Answer Answer Answer Answer

(a)(a)(a)(a) 1,100 (b)(b)(b)(b) 1,100 (c)(c)(c)(c) 1,000 +100 +110 =1,210

(d)(d)(d)(d) 1,000+100+110 =1,210

(e)(e)(e)(e) 1,000 +105 = 1,105

(f)(f)(f)(f) let interest rate per rupee for 6 months = x

100(1+x)(1+x) = 121

100(1+x)2 =121

x= 0.10

Hence, total amount on maturity = 1,000 +100 =1,100

Alternative solution ::::

Amount to be received after 6 months = = = = 1,000(1+.21)(1/2) = 1,00

(g)(g)(g)(g) 1,000 +100 +110 + 60.50 = 1,270.50

(h)(h)(h)(h) let interest rate per rupee for 6 months = x

100(1+x)(1+x) =110

100(1+x)2 = 110 x =0.048809

Amount to be received on maturity = 1,000 +100+110 + (.048809)(1210)

= 1,269.06

Alternative solution :

Amount to be received on maturity = 1,000(1.10)(2.50)

= 1,000[(1.10)(2)(1.10)(0.50)] = 1,269.06

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(i)(i)(i)(i) let interest rate per rupee for 3 months = x

100(1+x)(1+x)(1+x)(1+x) =110

100(1+x)4 =110

x = 0.024114

Hence total amount to be received on maturity = 1,000 +24.11 = 1,024.11

Alternative solution ::::

Amount to be received on maturity = 1,000(1.10)(0.25) = 1,024.11

Q. No. 12 Q. No. 12 Q. No. 12 Q. No. 12 A finance company accepts deposits for any period of choice of the

depositor. Rate of interest is 12% p.a. What total amount (principal and interest)

a person will receive after 13 months if he has deposited Rs 2,00,000 for 13

months?

Answer Answer Answer Answer

100 + 12 + 1.12 = 113.12

Q. No. 13Q. No. 13Q. No. 13Q. No. 13 An investor deposits Rs.5,000 to get Rs.5,400 at the end of 1 year.

Find the % return per annum if it is expressed as (a) Annual compounding (b)Half-

yearly compounding (c) Quarterly compounding (d) monthly compounding (v) cc.

AnsweAnsweAnsweAnswer

(i) Let annual compounding interest rate = r

5000(1+r) = 5400

1+r = 1.08

r = 0.08 = 8%.

Return is 8% p.a. annually compounding.

(ii) Let half-yearly compounding interest rate = r

5000(1+r)2 = 5400

1+r = 1.0392

r = 0.0392 = 3.92%.

Return is 7.84 % p.a. half-yearly compounding.

(iii) Let quarterly compounding interest rate = r

5000(1+r)4 = 5400

1+r = 1.0194265

r = 0.0194265 = 1.94265%.

Return is 7.7706 % p.a. quarterly compounding.

(iv) Let monthly compounding interest rate = r

5000(1+r)12 = 5400

(1+r)12 = 1.08

12log(1+r) = log 1.08

12log(1+r) = 0.0334

log(1+r) = 0.0028

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1+r = 1.007

r = 0.007 = 0.7%

Return is 8.40 % p.a. monthly compounding.

(v) Let cc rate = r

5000.er.1 = 5400

er.1 = 1.08

er = 1.08

r.loge = log 1.08

r( 0.4343) = 0.0334

r = 0.07690 = 7.69%. Return is 7.69% p.a. cc.

Q. No. 14Q. No. 14Q. No. 14Q. No. 14 You deposit $ 1,00,000 with a financial institution for 5 years. The

agreed rate of interest on the deposit is 12% p.a. c.c. The interest is to be paid

monthly. What amount of interest you will receive at the end of each month?

AnsweAnsweAnsweAnswer

Interest at the end of each month = 100000.e0.01 - 100000 = 1010

Q. No. 15:Q. No. 15:Q. No. 15:Q. No. 15: An investor invested Rs.100 for 2 months. Find the amount he will get

on maturity assuming (i) interest is 12 p.a. (ii) 12% p.a. monthly compounded (iii)

12 % p.a. annually compounded.

Answer Answer Answer Answer

(i)(i)(i)(i) 102

(ii)(ii)(ii)(ii) 100 + 1 +1.01 = 102.01

(iii)(iii)(iii)(iii) Let monthly compounded interest ( per rupee ) = x

1(1+x)12 = 1.12

(1+x)12 = 1.12

12.log(1+x) = log 1.12

log (1+x) = (log 1.12)/12 =(0.0492) / 12 = 0.0041

1+x = AL(.0041) = 1.009

x =0.009

Amt. to be received on maturity =100+0.90+(100.90)(0.009)=101.81

EQUALIZINGEQUALIZINGEQUALIZINGEQUALIZING

It is process of converting zero period cash flow to a number of equal amounts.

For this period, the zero period cash flow (i.e. present value of cash flow) is

divided by sum of relevant present value factors. We shall be understanding this

concept with he help of a few questions.

Q. No. 16:Q. No. 16:Q. No. 16:Q. No. 16:

(a) (a) (a) (a) Borrowed Rs. 1,00,000 at interest of 10% p.a. on 1.1.2004. Repayable in five

equal annual installments with interest. The first installment to be paid on

31.12.2004. Find the amount of each installment.

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Answer : Answer : Answer : Answer :

Date Amount of installment (Rs.)

31.12.2004 20,000 (Principal) +10,000 (interest) = 30,000

31.12.2005 20,000 (Principal) + 8,000 (interest) = 28,000

31.12.2006 20,000 (Principal) + 6,000 (interest) = 26,000

31.12.2007 20,000 (Principal) + 4,000 (interest) = 24,000

31.12.2008 20,000 (Principal) + 2,000 (interest) = 22,000

(b) (b) (b) (b) Borrowed Rs. 1,00,000 at interest of 10% p.a. on 1.1.2004. Repayable in five

equal annual installments including interest. The first installment to be paid on

31.12.2004. Find the amount of each installment. Also find the amount of interest

included in each installment.

Answer :Answer :Answer :Answer : 1,00,000

Amount of each installment = -------------- = 26,378

3.791

Amount due Payment towards

Principal

Payment towards

Interest

1.1.2004 1,00,000

31.12.2004 -16,378 16,378 10,000

1.1.2005 83,622

31.12.2005 -18,016 18,016 8,362

1.1.2006 65,606

31.12.2006 -19,817 19,817 6,561

1.1.2007 45,789

31.12.2008 -21,799 21,799 4,579

1.1.2008 23,990

31.12.2008 -23990 23,990 2,388 (Bal. fig.)

1.1.2008 nil

(c)(c)(c)(c) Borrowed Rs. 1,00,000 at interest of 10% p.a. on 1.1.2004. Repayable in five

equal annual payments including interest. The first payment to be made on

1.1.2004 itself. Find the amount of each payment. Also find the amount of

interest included in each payment.

Answer :Answer :Answer :Answer :

1,00,000

Amount of each payment = ----------- = 23,981

1 +3.17

Amount due Payment towards

Principal

Payment towards

Interest

1.1.2004 Bal. 1,00,000

1.1.2004 Payment -23,981 23,981 Nil

1.1.2004 Bal. 76,019

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1.1.2005 Payment -16,379 16,379 7,602

1.1.2005 Bal. 59,640

1.1.2006 Payment -18,017 18,017 5,964

1.1.2006 Bal. 41,623

1.1.2007 Payment -19,819 19,819 4,162

1.1.2007 Bal. 21,804

1.1.2008 Payment -21,804 21,804 2,177

1.1.2008 Balance nil

Q. No. 17 Q. No. 17 Q. No. 17 Q. No. 17 : X Ltd. acquires a machine on lease for a period of five years, cost of

the machine Rs.5,00,000, life 5 years, scrap value after 5 years nil. The lessor

expects a 10% (Pre-tax) return. What should be the lease rental if (i) lease rental

is equal annual amount payable in the beginning of each year (ii) lease rental is

equal annual amount payable at the end of each year (iii) lease rental is to be

paid in four equal amounts, first amount at the end of second year, next at the

end of third year, next at the end of fourth year and finally at the end of fifth

year (iv) Rs.1,00,000 to paid at the end of 3rd year and balance to be paid by way

of two equal installments one at the end of 4th year and other at the end of 5th

year.

Answer (i)Answer (i)Answer (i)Answer (i)

5,00,000

Annual lease rent = --------- = 1,19,904

1+3.17

Answer (ii)Answer (ii)Answer (ii)Answer (ii)

5,00,000

Annual lease rent = --------- = 1,31,891

3.791

Answer (iii)Answer (iii)Answer (iii)Answer (iii)

5,00,000

Annual lease rent = ---------------------------- = 1,73,551

0.826 + 0.751 + 0.683 + 0.621

Answer (iv) : Answer (iv) : Answer (iv) : Answer (iv) :

Net amount due at the end of 3rd year = 6,65,500-1,00,000 = 5,65,500

Amount of 2 equal annual installments :

5,65,500

= -------------- = 3,25,937

0.909 + 0.826

(Rs.3,25,937 to paid at the end of each of 4th year and 5th year of the lease)

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13

Alternative solution :

Net amount due at the end of 3rd year = 6,65,500-1,00,000 = 5,65,500

Present value of this amount = 5,65,500 x 0.751 = 4,24,691

Amount of 2 equal annual installments :

4,24,691

= -------------- = 3,25,683

0.683 + 0.621

(Rs.3,25,683 to paid at the end of each of 4th year and 5th year of the lease)

{Note Note Note Note : The difference in the two answers is due to calculation approximations.

The difference can be eliminated or reduced by increasing the number of decimal

places of present value factors}

Q. No. 18 : Q. No. 18 : Q. No. 18 : Q. No. 18 : Borrowed Rs.1,00,000. Repayable in six equal monthly installments

(including interest), first installment to be paid at the end of I month, second

installment at the end of II month and so on. Find the amount of each installment

assuming (i) interest is 12 p.a. (ii) 12% p.a. monthly compounded.

Answer (i) Answer (i) Answer (i) Answer (i) Amount of each installment :

1,00,000

= ------------------------------------ = 17,244.35

[1/(1.01)]+[1/(1.02)]+……+[1/(1.06)]

Answer (ii) Answer (ii) Answer (ii) Answer (ii) Amount of each installment :

1,00,000

= ------------------------------------ = 17,256.26

[1/(1.01)1]+[1/(1.01)2]+……+[1/(1.01)6]

Q.NoQ.NoQ.NoQ.No.19(a).19(a).19(a).19(a) A person borrowed Rs.1,00,000 on 1st January 2007. Interest rate

12% p.a. monthly compounded. He has to repay this amount in 60 monthly equal

installments including interest. Find the amount of each installment.

(b)(b)(b)(b) How your answer will change if the number of installments is 180 instead of

60?

Answer(a) Answer(a) Answer(a) Answer(a)

Annuity for 60 months at 12% p.a. monthly compounded =

[{1/(1.01)1}………..+{1/(1.01)20}]

+ [{1/(1.01)21}+ ………..+{1/(1.01)40}]

+ [{1/(1.01)41}+ ………..+{1/(1.01)60}]

=[{1/(1.01)1}+ ………..+{1/(1.01)20}]

+ {1/(1.01)20} [{1/(1.01)1}+ .………..+{1/(1.01)20}]

+ {1/(1.01)40} [{1/(1.01)1}+ ………..+{1/(1.01)20}]

= 18.046 + .820 [18.046] + (.820)2 [18.046] = 44.9778504

AMOUNT OF INSTALLMENT = 1,00,000/44.9778504 = 2223.32

Answer (b) Answer (b) Answer (b) Answer (b)

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14

Annuity for 180 months at 12% p.a. monthly compounded =

[{1/(1.01)1}+ ...........+{1/(1.01)60}] + +[{1/(1.01)61}+ ...........+{1/(1.01)120}]

+ [{1/(1.01)120}+ ...........+{1/(1.01)180}]

=[{1/(1.01)1}+ ...........+{1/(1.01)60}]

+ [1/(1.01)60]x[{1/(1.01)1}+ ...........+{1/(1.01)60}]

+ [1/(1.01)120]x [{1/(1.01)1}+ ...........+{1/(1.01)60}]

= 44.9778504 + (0.820)3 {44.9778504} + (0.820)6{44.9778504}

= 44.9778504 + 0.551368 {44.9778504} + 0.30400667142{44.9778504}

=83.4507644067

AMOUNT OF INSTALLMENT = 1,00,000 / 83.4507644067 = Rs.1198.31

Q. No. 20Q. No. 20Q. No. 20Q. No. 20: Mansukha borrows Rs. 1,00,000 from rural money lender. Interest rate

24% p.a. annually compounded. The loan is to be repaid in 24 monthly

installments including installments, the first installment being paid at the end of

1st month. Find the equated monthly installment (EMA.)

AnswerAnswerAnswerAnswer : As we have to pay monthly installment, we have to calculate the

monthly compounded interest rate.

Let monthly compounded interest rate for rupee one = x

(1+x)(1+x)(1+x)(1+x)(1+x)(1+x) (1+x)(1+x)(1+x) (1+x)(1+x)(1+x)

= 1.24

(1+x)12 = 1.24

log(1+x)12= log1.24

12.log(1+x) = 0.0934

log(1+x) = 0.0078 Taking antilog on both the sides,

1+x = = 1.018 x = 0.018 = 1.80%

1,00,000 1,00,000

EMI = ___________________________________________ =---------- = 5168

1 1 19.35

------ + …………………………+-----

(1.018)1 (1.018)24

Q. No. 21: Q. No. 21: Q. No. 21: Q. No. 21: Murali has been contributing Rs15,000 per year to each of the

following two schemes of a Mutual fund for getting post retirement annuity ; (i)

Debt fund (ii) Equity Fund. The current value of each of the two investments Rs.

1m. She plans to retire after 25 years from today and her life expectancy is 15

years after the retirement. The debt fund is expected to earn 3% p.a. while the

expected return on the equity fund is 6% p.a. What shall be her retirement

annuity? She is planning to increase her annual contribution to the debt fund so

that she may get a total annuity, from both the funds, of Rs. 8,00,000. Calculate

the increase in the annual contribution.

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Answer Answer Answer Answer Present value of funds accumulated after 25 years ( debt fund)

15000[(1.03)25+(1.03)24+……………+(1.03)1 ]

10,00,000+ ------------------------------------

[(1.03)25

1 1 1

10,00,000+ 15,000 [ 1 + ----- + -------+ …………+------]

(1.03)1 (1.03)2 (1.03)24

10,00,000 + 15000[ 1 + 16.9355] = 12,69,033

12,69,033

Annuity = -------------------------------

1 1

------ + …………………………+-----

(1.03)25 (1.03)39

12,69,033

Annuity = -------------------------------------------

1 1 1

--------- x[------ + ……………………+-----]

(1.03) 24 (1.03)1 (1.03)15

12,69,033

Annuity = ----------------------------- = 2,16,107

(0.4919)(1.9379)

Present value of funds accumulated after 25 years ( equity fund)Present value of funds accumulated after 25 years ( equity fund)Present value of funds accumulated after 25 years ( equity fund)Present value of funds accumulated after 25 years ( equity fund)

1 1 1

10,00,000+ 15,000 [ 1 + ----- + -------+ …………+------]

(1.06)1 (1.06)2 (1.06)24

10,00,000 + 15000[ 1 + 12.5504] = 11,88,256

11,88,256

Annuity = -------------------------------------------

1 1 1

--------- x[------ + ……………………+-----]

(1.06) 24 (1.06)1 (1.06)15

11,88,256

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Annuity = ----------------------------- = 495334

(0.2470)(9.7122)

Total annuity from existing fund 2,16,107 + 4.95,334 = 7,11,441

Target annuity 8,00,000

Short fall 88559

Let she invests Rs.x annually for 25 years in debt fund to get the annuity of

Rs.88,559.

88559 = x(17.9355) / 5.87225 = 28,995

Murali should invest Rs. 28,995 annually for 25 years, in debt fund, to get the

additional amount of annuity. If this action is taken, she will get the annuity of Rs.

8,00,000 for 15 years after her retirement.

Normal distributionNormal distributionNormal distributionNormal distribution

It is a statistical method of finding the probability. We use it when we are given

or we can calculate mean and standard deviation. Under this approach,

probability is equal to area under Normal curve.

The following three statements about Normal distribution are referred as

empirical rule :

(1) About 68% of the values described by a normal curve are within ± one

standard deviation of the mean.

(2) About 95% of the values described by a normal curve are within ±two

Standard deviations of the mean.

(3) About 99.7% of the values described by a normal curve are within three

Standard deviations of the mean.

Area under Normal curve is calculated with the help of z .

X - mean

z = ------------

SD

For this purpose, we consult the table “ Areas under Normal curve” ( also

referred as Areas under standard normal distribution)

Remember : Table gives us area from z=0 to a particular value of z.

QQQQ. No. 22. No. 22. No. 22. No. 22(a) Find the area for z = 1.26 Answer : 0.3962

(b) Find the area for z = -1.26 Answer : 0.3962

(c) Find the area to the right of z = 1.50

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Answer : 0.50 – 0.4332 = 0.0668

(d) Find the area to the left of z = 1.50

Answer : 0.50 + 0.4332= 0.9333

(e) Find the area to the left of z = -1.57

Answer : 0.50 -0.4418 = 0.0582

(f) Find the area between z = -1 to z = +1

Answer : 0.3413 + 0.3413 = 0.6826

Q. No. 23Q. No. 23Q. No. 23Q. No. 23 Mean marks of 100 students =60, SD =5, Find p of a randomly selected

student securing above 70.

Answer Answer Answer Answer

70 - 60

z = ----------- = 2

5

p = 0.50 -0.4772 = 0.0228

-3 -2 -1 0 1 2 3 Z

40 45 50 55 60 65 70 75 80

X

Q. No. 24Q. No. 24Q. No. 24Q. No. 24 Mean height of 50 soldiers = 170 cms., s.d. =10. Find the p of a

randomly selected soldier being less than 160.

AnswerAnswerAnswerAnswer

160 - 170

z = ----------- = -1

10

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p = 0.50 – 0.3413 = 0.1587

Q. No. 25Q. No. 25Q. No. 25Q. No. 25: A family uses a gas cylinder on an average for 40 days, s.d.10. Find

the p that the cylinder purchased today will be used for 30 to 50 days.

AnswerAnswerAnswerAnswer

For 30 days use :

30 - 40

z = ----------- = -1

10

For 50 days use :

50 - 40

z = ----------- = +1

10

p = 0.3413 + 0.3413 = 0.6826

(Extra practice questions are given at the end of the chapter)

Geometric meanGeometric meanGeometric meanGeometric mean

GM is the most suitable mean to find the rate of change over a period of time.

GM of a and b = (a x b)1/2

GM of a, b and c = (a x b x c)1/3

GM of a , b, c and d = (a x b x c x d)1/4

Finding the average rate of change over a period of time is being explained with

the help of following examples:

(a)(a)(a)(a) Find the average rate of change in price level over a period of 3 years if

prices increased by 10% in the I year, 20% in the II year and 30% in the III year.

(b)(b)(b)(b) A company paid the following equity dividend per share :

Year Dividend per share (Rs.)

20X1 2.00

20X2 2.50

20X3 3.00

20X4 3.30

20X5 3.60

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Find the rate of growth of dividend over 20X1 to 20X5.

Sum of Infinite Geometric Progression SeriesSum of Infinite Geometric Progression SeriesSum of Infinite Geometric Progression SeriesSum of Infinite Geometric Progression Series

Let the series be : a1 + a2 + a3 + a4 + a5+ ……… up to infinity

The series is referred as GP if (a2 /a1 ) = (a3 /a2 ) = (a4 /a3 ) and so on.

Sum of infinite GP = [(a) / (1-r)] Where a = first term

r = common ratio i.e. (a2 /a1 ) or (a3 /a2 ) or (a4 /a3 ).

ExampleExampleExampleExample : Find the sum of [( 1 ) / (1.10)1] +[( 1 ) / (1.10)2]+

[( 1 ) / (1.10)3] +[( 1 ) / (1.10)3] + ……… up to infinity.

[( 1 ) / (1.10)1]

Sum = ----------------- = 10

1- [(1) /(1.10)]

RULE OF 72

As per this rule, the time required to double the value of an investment is roughly

“72” / “interest rate in percent”.

(a) You invest Rs.1,00,000 at 12% p.a. annually compounded. Find the

approximate period in which the investment will double its value. ( Answer : 6

years)

(b) You invest Rs.1,00,000 at 10% p.a. half yearly compounded. Find the

approximate period in which the investment will double its value. ( Answer :

7.20 years)

(c) You invest Rs.1,00,000 at 8% p.a. quarterly compounded. Find the

approximate period in which the investment will double its value. ( Answer :

9 years)

(d) You invest Rs.1,00,000 at 6% p.a. monthly compounded. Find the approximate

period in which the investment will double its value. (Answer : 12 years

RULE OF 69

As per this rule, the time required to double the value of an investment is roughly

“69” / “interest rate in percent”, when the interest is continuously compounded.

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You invest Rs.1,00,000 at 12% p.a. continuously compounded. Find the

approximate period in which the investment will double its value. ( Answer: 5.75

years)

DIMINISHING BALANCE DEPRECIATION RATE

Where, Original cost (1-r)n = scrap value

r = Diminishing balance method depreciation rate, n = life of asset

Example: Original cost Rs. 1,00,000 . Life of the asset:10 years. Scrap value

after 10 years: Rs. 10,000. Find depreciation rate by (i) SLM (ii) WDV. State the

relationship between straight line rate and WDV rate.

SLM rate = 9 %

Calculation of WDV rate:

100000 (1-r)10 = 10,000

(1-r) 10 = 0.10

10.log(1-r) = - 1

Log(1-r) = -0.10

Log (1-r) = -1 +0.90

(1-r) = 0.7943

r = 0.2057 = 20.57 %

Relationship between straight line rate and WDV rate: WDV depreciation rate is

generally between 2 to 3 times of straight line rate. (We can say that WDV rate

is about 2.50 times of straight line; in other words we can say that straight line

rate is about 40 % of diminishing rate).

EXTRA PRACTICEEXTRA PRACTICEEXTRA PRACTICEEXTRA PRACTICE ( Must Do)( Must Do)( Must Do)( Must Do)

Q. No. Q. No. Q. No. Q. No. 26262626: If marks secured by the students in an exam are normally distributed

with a mean of 32 and a standard deviation of 4, what is the probability that a

student selected at random secures (a) greater than 35? (b) Greater than 40? (c)

Between 29 and 35?

AnswAnswAnswAnswer(a)er(a)er(a)er(a)

35 - 32

z = ----------- = 0.75

4

p = 0.50 – 0.2734 = 0.2266

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-3 -2 -1 0 1 2 3 Z

16 20 24 28 32 36 40 44 48

X

Answer(b)Answer(b)Answer(b)Answer(b) 40 -32

z = ----------- = 2

4

p = 0.50 – 0.4772 = 0.0228

-3 -2 -1 0 1 2 3 Z

16 20 24 28 32 36 40 44 48

X

Answer(c)Answer(c)Answer(c)Answer(c) 35 - 32

z = ----------- = 0.75

4

29 - 32

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z = ----------- = -0.75

4

p = 0.2734 + 0.2734 = 0.5468

-3 -2 -1 0 1 2 3 Z

16 20 24 28 32 36 40 44 48

X

Q. NoQ. NoQ. NoQ. No 27272727:::: A study of sleeping hours of 100 BA Final students reveals average of 6

hours with SD of 1. What is the probability that a randomly selected student sleeps for

more than 8 hours?

AnswerAnswerAnswerAnswer 8 - 6

z = ----------- = 2

1

p = 0.5000 – 0.4772 = 0.0228

-3 -2 -1 0 1 2 3 Z

2 3 4 5 6 7 8 9 10

X

Q. No. 28Q. No. 28Q. No. 28Q. No. 28 You know that the weight of 700 people chosen at random follows a normal

distribution with a mean of 80 Kg and a standard deviation of 10 Kg. What’s the

probability that the average weight a person selected at random exceeds 85 Kgs.

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AnswerAnswerAnswerAnswer 85 - 80

z = ----------- = 0.5

10

P = 0.3085

-3 -2 -1 0 1 2 3 Z

40 50 60 70 80 90 100 110 120

X