Taxmann's Fundamentals of Business Mathematics & Statistics (FBMS | Maths) | CRACKER

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© All rights reserved Price : ` 395 Fifth Edition : June 2026 Published by : Taxmann Publications (P.) Ltd. Sales & Marketing : 59/32, New Rohtak Road, New Delhi-110 005 India Phone : +91-11-45562222 Website : www.taxmann.com E-mail : sales@taxmann.com Regd. Office : 21/35, West Punjabi Bagh, New Delhi-110 026 India Printed at : Tan Prints (India) Pvt. Ltd. 44 Km. Mile Stone, National Highway, Rohtak Road Village Rohad, Distt. Jhajjar (Haryana) India E-mail : sales@tanprints.com Disclaimer Every effort has been made to avoid errors or omissions in this publication. In spite of this, errors may creep in. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the next edition. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, therefrom.

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Contents PAGE

Module-wise Marks Distribution

I-5

Module-wise Comparison with Study Material

I-7

MODULE 1 ARITHMETIC Chapter 1 RATIOS, VARIATIONS AND PROPORTIONS

3

Chapter 2

TIME VALUE OF MONEY AND ANNUITY - SIMPLE AND COMPOUND INTEREST

24

Chapter 3

ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION

37

Chapter 4

TIME AND DISTANCE

49

MODULE 2 ALGEBRA Chapter 1

SET THEORY, VENN DIAGRAM

55

Chapter 2

INDICES AND LOGARITHMS

64

Chapter 3

PERMUTATION AND COMBINATIONS

82

Chapter 4

QUADRATIC EQUATION

96

I-9

TAXMANN®

CONTENTS

I-10

PAGE

MODULE 3 CALCULUS - APPLICATION IN BUSINESS Chapter 1

CONCEPT OF CALCULUS AND ITS APPLICATION IN BUSINESS

109

Chapter 2

REVENUE AND COST FUNCTION

113

Chapter 3

OPTIMIZATION TECHNIQUES

117

MODULE 4 STATISTICAL REPRESENTATION OF DATA Chapter 1

TAXMANN®

INTRODUCTION

121

Chapter 2

DIAGRAMMATIC REPRESENTATION OF DATA

126

Chapter 3

FREQUENCY DISTRIBUTION

130

Chapter 4

GRAPHICAL REPRESENTATION OF FREQUENCY DISTRIBUTION – HISTOGRAM, FREQUENCY POLYGON CURVE, OGIVE, PIE-CHART 139

MODULE 5 MEASURES OF CENTRAL TENDENCY AND DISPERSION Chapter 1

MEAN, MEDIAN, MODE, MEAN DEVIATION

143

Chapter 2

RANGE, QUARTILES AND QUARTILE DEVIATION

171

Chapter 3

STANDARD DEVIATION

178

Chapter 4

CO-EFFICIENT OF VARIATION

189

Chapter 5

KARL PEARSON AND BOWLEY’S CO-EFFICIENT OF SKEWNESS

194

CONTENTS

I-11 PAGE

MODULE 6 CORRELATION AND REGRESSION Chapter 1

SCATTER DIAGRAM

203

Chapter 2

KARL PEARSON’S CO-EFFICIENT OF CORRELATION

206

Chapter 3

REGRESSION ANALYSIS

221

MODULE 7 PROBABILITY Chapter 1

CONCEPTS AND TERMINOLOGIES

243

PRIMARY APPLICATIONS OF PROBABILITY THEOREMS

257

MODULE 8 INDEX NUMBER AND TIME SERIES Chapter 1

USES OF INDEX NUMBERS, METHODS OF CONSTRUCTION OF INDEX NUMBER

269

Chapter 2

COMPONENTS OF TIME SERIES AND CALCULATION OF TREND BY MOVING AVERAGE METHOD

281

Solved Paper : December 2024 (Suggested Answers)

P.1

Solved Paper : June 2025 (Suggested Answers)

P.11

Solved Paper : December 2025 (Suggested Answers)

P.21

Solved Paper : June 2026 (Suggested Answers)

P.30

TAXMANN®

Chapter 2

2

8=3824B 0=3 ;>60A8C7<B

CHAPTER

QUICK REVISION OF THE CHAPTER IMPORTANT CONCEPTS Indices:

TAXMANN®

The power or exponent of any number or a variable is known as Index. Plural of Index is Indices, Indices provide a compact algebraic notation for repeated multiplication.

For solving expression with Indices, the law of exponents is applied. In the expression x5, the power 5 is known as Index and x is known as Base.

Basic rules of Indices: (1) am × an = am + n (2) am/an = am – n (3) 1/am = a–m (4) a0 = 1 (5) If in an equation bases on both sides are same, powers can be equated.

(6) If in an equation powers or indices on both sides are same, bases can be equated.

Logarithm: Definition: If ax= N and a > 0, N > 0 then x = loga N and read as logarithm of N to the base ‘a’ or in other words log N to the base ‘a’.

So, logarithm of a positive number N signifies that for a given base ‘a’ it is the power to which ‘a’ must be raised to produce the positive number ‘N’.

Common and Natural Logarithms: Common Logarithms are the logarithms for which the base is considered as 10.

For any calculation with logarithms, if base is taken as 10, the logarithms are considered to be Common Logarithms. In such scenarios, the base to a logarithm is not generally mentioned but is assumed to be 10.

When the base of Logarithm is e (where e means exponential and having value 2.718) then it is called Natural Logarithm. Its extensive use is seen in Calculus.

64

CH. 2 : INDICES AND LOGARITHMS

65

Laws followed for Logarithms: (1) Logarithm of the product of the numbers would be equal to the sum of their logarithms.

Loga(mn) = logam + logan (2) Logarithms of the quotient of two numbers would be equal to the difference of their logarithms.

Loga(m/n) = logam – logan (3) Logarithm of any number raised to a power would be equal to the product of the logarithm of the number and the power.

Loga mn = n Loga m (4) For any logarithm where base is same as the number, the value of logarithm would always be one (1). The value of base or number should not be equal to 1 and should be a positive number (> 0).

Loga a = 1 (5) For any logarithm where number is 1 and the base is a positive number, then value would always be zero.

Loga 1 = 0

(7) Change of Base for Logarithm: The logarithm of a number with a base can be changed with the introduction of the base and number splitted into two expressions of logarithms. The base and the number would be same with a new variable introduced both as base and number to the new logarithmic expressions.

Logam = Logan × Lognm

PAST EXAMINATION QUESTIONS THEORY QUESTIONS LOGARITHMS Q. 1. The logarithm of any number to the same base is:

Q. 2. The whole of the integral part of a logarithm is called: (a) Characteristic

(a) Unity

(b) Mantissa

(b) Zero

(c) Both

(c) Infinity

(d) None of these

(d) Non-existence Ans. (a) Unity

[March 2016]

Ans. (a) Characteristic

[June 2016]

TAXMANN®

(6) The base of a logarithm for a number would be equal to the reciprocal of the logarithm with the number replaced by base.

Logam = 1/Logma

66

MODULE 2: ALGEBRA

Q. 3. Logarithm of unity to any nonzero base is:

Note: log10 13.72 = 1.1374, Hence, Mantissa is 0.1374

(a) 1

Q. 5. The value of Log7 7 is:

(b) 10

(a) 0

(c) 0

(b) –1

(d) None of these [Dec. 2016] [Modified]

(c) 1/2 (d) 1

Ans. (c) 0

Ans. (d) 1

Q. 4. Given log10 1372 = 3.1374 The mantissa for log10 13.72 is:

Q. 6. The value of logarithm of 1 to the base 10 is:

(a) 1.374

(a) 2

(b) 0.1374

(b) 10

(c) 13.74

(c) 0

(d) None of these [Dec. 2016] [Modified]

(d) 1

TAXMANN®

Ans. (b) 0.1374

[June 2017]

[June 2018]

Ans. (c) 0

NUMERICAL QUESTIONS INDICES x+1

Q. 1. If 3 of x is:

x–1

+3

= 270 then the value

Q. 2. If x = 7 – 4 3 , then the value of

( )

x + 1 x is:

(a) 5

(a) 2 3

(b) 3

(b) 4 3

(c) 4 (d) 8

[June 2013]

(c) 4

Ans. (c) 4

(d) 6

Working Note:

Ans. (c) 4

3x + 1 + 3x – 1 = 270

Working Note:

3x 31 + 3x 3-1 = 270

x=7– 4 3

1

–1

3 (3 + 3 ) = 270 x

[June 2013]

3x (3 + 1/3) = 270 3x (10/3) = 270 3x = 270 (3/10) 3x = 81 3x = 34 Therefore, x = 4

x=4+3– 4 3

(

x = 2− 3

)

2

x =2− 3 Now,

(

x + 1/ x

)

67

CH. 2 : INDICES AND LOGARITHMS

(

)

(

= 2 − 3 + 1/ 2 − 3

(

)

(b) 27 abc

)

(c) –27 abc

(

2 = ⎡ 2 − 3 + 1⎤ / 2 − 3 ⎢⎣ ⎥⎦

(

= ⎡⎣7 − 4 3 + 1⎤⎦ / 2 − 3

)

)

(d) None of these

[June 2015]

Ans. (b) 27 abc Working Note: If (a + b + c) = 0 than, a3 + b3 + c3 = 3abc

( ) = 4 (2 − 3 ) / (2 − 3 ) = ⎡⎣8 − 4 3 ⎤⎦ / 2 − 3

Here, a = a1/3, b = b1/3, c = c1/3, Therefore, (a1/3)3 + (b1/3)3 + (c1/3)3 = 3 a1/3 b1/3 c1/3 a + b + c = 3 a1/3 b1/3 c1/3

=4

Doing cube on both sides we get, Q. 3. The value of to infinity is:

6 + 6 + 6 + ......

(a + b+ c)3 = (3 a1/3 b1/3 c1/3)3 (a + b+ c)3= 27abc Q. 5. If 3x = 9y, then (6y/x) – 1

(b) 4

(a) 1

(c) –2

(b) 2

(d) 3 [March 2015, Sept. 2015, March 2016]

(c) 1/2 (d) 0

Ans. (d) 3

Ans. (a) 1

Working Note:

Working Note:

[Sept. 2015]

3x = 9y Let x =

6 + 6 + 6 +......

x = 3y

Squaring Both sides we get,

Now, 6y/x – 1

x2 = 6 + x

= 6y/3y – 1

x2 – x – 6 = 0

=2–1

x2 – 3x + 2x – 6 = 0

=1

x (x – 3) + 2 (x – 3) = 0

Q. 6. If x + 4 = 4, 2x – 5y = 1 then x & y are:

(x + 2) (x – 3) = 0 x+2=0

(a) (1, 0)

x = –2 (Not possible. As Square root cannot be done for negative value)

(b) (0, –1/5)

x–3=0

(d) (1/5, 0)

x=3

Ans. (b) (0, –1/5)

Q. 4. If a1/3 + b1/3 + c1/3 = 0, then (a + b + c)³ =

Working Note:

(a) 3 abc

x=0

(c) (1, 1/5)

x+4=4

[Sept. 2015]

TAXMANN®

(a) 6

68

MODULE 2: ALGEBRA

2x – 5y = 1

y = –1/5

( 5 ) − 58 − 29 ( 5 ) x + 3x – 29x = 65 + 29 ( 5 ) − 58 − 29 ( 5 )

Q. 7. If x = 2 + 5 , then x3 + 3x2 –29x:

Q. 8. If 3x = 5y = (225)z, then Z:

x3 + 3x2 – 29x = 65 + 29

2 (0) – 5y = 1 –5y = 1

3

x3 + 3x2 – 29x = 7

(a) 7

(a) XY/X + Y

(b) 10

(b) XY/2(X + Y)

(c) 0

(c) 2(X + Y)

(d) None of these

[Dec. 2015]

(d) None of these

Ans. (a) 7

Ans. (b) XY/2(X + Y)

Working Note:

Working Note:

x = 2+ 5

Let, 3x = 5y = (225)z = k

Doing Cube on both sides we get,

3x = k

(

x3 = 2 + 5 TAXMANN®

2

3 = k1/x [Equation 1]

)

3

5y = k

( 5 ) + 3(2) ( 5 ) + ( 5 ) x = 8 + 12 ( 5 ) + 30 + 5 ( 5 ) x = 38 + 12 ( 5 ) + 5 ( 5 ) x = 38 + 17 ( 5 ) 2

x3 = 23 + 3(2)2

3

3

3

3

Adding 3x2 on both sides we get,

( 5 ) + 3 (2 + 5 ) x + 3x = 38 + 17 ( 5 ) + 3 ⎛ 2 + 2(2) ( 5 ) + ( 5 ) ⎞ ⎝ ⎠

2

x3 + 3x2 = 38 + 17 3

[March 2016]

2

2

2

5 = k1/y [Equation 2] 225z = k 152z = k 15 = k1/2z [Equation 3] Multiplying Equation 1 and 2, we get, (3) (5) = (k1/x) (k1/y) 15 = k1/x + 1/y [Inserting Value of 15 from Equation 3] k1/2z = k1/x + 1/y 1/2z = 1/x + 1/y 1/2z = (y + x)/xy xy/(x + y) = 2z z = xy/2(x + y)

( ) ( ( ) ) x + 3x = 38 + 17 ( 5 ) + 27 + 12 ( 5 ) x + 3x = 65 + 29 ( 5 )

Q. 9. 64x = 2 2 , then X:

Deducting 29x from both side we get,

Ans. (c) 1/4

x3 + 3x2 = 38 + 17 3

2

3

2

5 +3 4+4

x3 + 3x2 – 29x = 65 + 29

5 +5

( 5 ) − 29 (2 + 5 )

(a) 1/6 (b) 1/2 (c) 1/4 (d) None of these

[June 2016]

CH. 2 : INDICES AND LOGARITHMS

Working Note:

Working Note:

64x = 2 2

Let, ax= by = cz = k

(26)x =

ax = k

( 2)

( 2)

12 x

3

=

69

a = k1/x [Equation 1] by = k

( 2)

3

b = k1/y [Equation 2]

12x = 3

cz = k

x = 3/12

c = k1/z [Equation 3]

x = 1/4

Now, b2 = ac, Therefore,

Q. 10. The digit in the unit place of (2 (4)x)2 +1 (where x is a positive integer) is:

(k1/y)2 = (k1/x) (k1/z) [Putting values from Equation 1, 2 and 3]

(a) 1

2/y = 1/x + 1/z

(b) 5

2/y = (z + x)/xz

(c) 3

2xz = yz + xy

(d) None of these

[June 2016]

k2/y = k(1/x + 1/z)

Working Note:

(a) 3

Let x = 1, Therefore,

(b) 4

(2 (4)1)2 + 1

(c) –2

= 82 + 1

(d) –5

= 65

Ans. (a) 3

Let x = 2, Therefore,

Working Note:

(2 (4)2)2 + 1

2x – y = 5

= (2 (16))2 + 1

2 (4) – y = 5

= 322 + 1

8–y=5

= 1,025

y=3

Hence, for any value of x the digit in unit place shall be 5.

Q. 13. If 2x + (1/2x) = 5/2 the value of x is:

x

y

z

2

Q. 11. If a = b = c and b = ac then xy + yz = ? (a) x2

Ans. (c) 2xz

(b) r1 (c) 0 Ans. (b) 1

(c) 2xz (d) None of these

(a) –1

(d) 1

(b) –xz [June 2016]

[June 2016]

Working Note: Let 2x = a, Therefore, a + 1/a = 5/2

[Dec. 2016]

TAXMANN®

Ans. (b) 5

Q. 12. In the equation 2x – y = 5, if x = 4 then y = ?

70

MODULE 2: ALGEBRA

(a2 + 1)/a = 5/2

3a (2a – 3) – 2 (2a – 3) = 0

2a2 + 2 = 5a

(3a – 2) (2a – 3) = 0

2a2 – 5a + 2 = 0

3a – 2 = 0

2a2 – 4a – 1a + 2 = 0

3a = 2

2a (a – 2) – 1 (a – 2) = 0

a = 2/3 (Equation 1)

(2a – 1) (a – 2) = 0

or

2a – 1 = 0

2a – 3 = 0

2a = 1

2a = 3

a = ½ [Equation 1]

a = 3/2 (Equation 2)

a–2=0

Now, Putting the value of a in both the

a = 2 [Equation 2]

a = 2/3

Now Putting a = 2x in Equation 1 we get, 2 = 1/2 x

2x = 2–1

Squaring both sides, we get,

x = –1

t/(1 – t) = 4/9

Now Putting a = 2 in Equation 2 we get,

9t = 4 (1 – t)

2 =2

9t = 4 – 4t

x=1

13t = 4

x

TAXMANN®

t = 2/3 1−t

x

So, x = r1 Q. 14. Solve for t : t 1- t 13 + = 1- t t 6 (a) 4/13, 9/13

t = 4/13 or a = 3/2 t = 3/2 1−t

(b) 9/13, 9/13

Squaring both sides, we get,

(c) –4/13, 9/13

t/(1 – t) = 9/4

(d) –9/13, 4/13 [Dec. 2017] [Modified]

4t = 9 (1 – t)

Ans. (a) 4/13, 9/13 Working Note: t = a, Therefore, 1−t a + 1/a = 13/6 Let,

2

(a + 1)/a = 13/6 2

6a + 6 = 13a 2

6a – 13a + 6 = 0 2

6a – 9a – 4a + 6 = 0

4t = 9 – 9t 13t = 9 t = 9/13 Q. 15. If 3x – 1 – 3x – 3 = 8, then the value of (x2 – x + 4) is: (a) 10 (b) 8 (c) 12 (d) 15 Ans. (a) 10

[June 2018]

71

CH. 2 : INDICES AND LOGARITHMS

Working Note: 3

x–1

–3

x–3

=8

Q. 18. If p = 2 + 3 , then 1/p =

3x 3–1 – 3x 3–3 = 8

(a) 2 - 3

3x (3–1 – 3–3) = 8

(b)

3 +2

(c)

3−2

3 (1/3 – 1/27) = 8 x

3x (9 – 1)/27 = 8 3x 8/27 = 8 3x = 27 3

3 =3 x

x=3 x2 – x + 4

(d) None of these [June 2018] [Modified] Ans. (d) None of these Working Note:

(

)

1/p = 1/ 2 + 3 , Therefore, the correct answer is None of these.

=9–3+4

Q. 19. (–2)5 =

= 10

(a) 32

Q. 16. The value of (32)1/5 is:

(b) 1/32

(a) 1/2

(c) –32

(b) 2

(d) None of these [Dec. 2018] [Modified]

(c) 4 (d) None of these [June 2018] [Modified]

Ans. (c) –32 Working Note:

Ans. (b) 2

(–2)5 = –2 × –2 × –2 × –2 × –2 = –32

Working Note:

Q. 20. The value of (25/4)4 is:

(32)1/5

(a) 30

= (25)1/5

(b) 23

=2

(c) 31

Q. 17. The value of x when 2x = 3x is:

(d) 32

(a) 1

Ans. (d) 32

(b) 0

Working Note:

(c) 6

(25/4)4

(d) None of these [June 2018] [Modified]

= 25/4 × 4

Ans. (b) 0

= 32

Working Note:

Q. 21. If 2x – 1 + 2x + 1 = 80, then the value of x is:

Base of 2 and 3 with power x can only be same when x = 0, that will result into both 20 = 1 and 30 = 1.

[June 2019]

= 25

(a) 8 (b) 4

TAXMANN®

= (3)2 – 3 + 4

72

MODULE 2: ALGEBRA

(c) 5

Working Note:

(d) 10

[Dec. 2019]

Ans. (c) 5

3x = 1/35

Working Note:

3x = 3–5

2x – 1 + 2x + 1 = 80

x = –5

(2x) (2–1) + (2x) (21) = 80 2x [2–1 + 21] = 80

Q. 24. If 9 × 81x = 27(x – 3) then the value of x is:

2x [1/2 + 2] = 80

(a) 10

2x (5/2) = 80

(b) 11

2x = 80 × 2/5

(c) 9

2x = 32

(d) –11

2x = 25

Ans. (d) –11

x=5

Working Note:

⎛ 1⎞ Q. 22. Find the value of ⎜ ⎟ ⎝ 81⎠ TAXMANN®

3x = 1/243

-3 4

× (243)

-1 5

(a) 27

[Dec. 2022]

9 × 81x = 27(x – 3) 32 × 34x = 33(x – 3) 3(2 + 4x) = 3(3x – 9) 2 + 4x = 3x – 9

(b) 9

x = –11

(c) 81 (d) None of these [Dec. 2019] [Modified] Ans. (b) 9 Working Note: −3

−1 ⎛ 1⎞4 243 × ( )5 ⎜⎝ 82 ⎟⎠

= (81)3/4 × (243)–1/5 = (34)3/4 × (35)–1/5 = 33 – 3–1

Q. 25. If x = 5 + 2 6 then is equal to:

3

(b)

(c) 2 3 2

(d)

[Dec. 2022]

Ans. (c) 2 3

x=5+ 2 6

= 32 =9

3 2

x=3+2+2 x

Q. 23. If 3 = 1/243 then the value of x is:

( 3 + 2)

(a) 5

x=

(b) –5

Now,

(c) –1/5 (d) 1/5 Ans. (b) –5

[Dec. 2019] [Modified]

)

(a) 3 2

Working Note:

= 33 – 1

(

x + 1/ x

2

(

x + 1/ x

)

73

CH. 2 : INDICES AND LOGARITHMS

Working Note:

( 3 + 2) ) + 1/ ( 3 + 2 ) = ( 3 + 2) )( 3 + 2) ) + 1/ ( 3 + 2 ) = (5 + 2 6 + 1) / ( 3 + 2 ) (6 + 2 6 ) / ( 3 + 2 ) = 2 3 ( 3 + 2 3 2) / 3 + 2 = 2 3 ( 3 + 2) / ( 3 + 2)

Q. 28. If (300)8 =65.61 × 10x, then the value of x is:

= 2 3

Ans. (c) 18

Q. 26. 9 × 81x = 1/(27x – 3) find the value of x.

Working Note:

(a) 2

(3 × 100)8 = (6561/100) 10x

(b) –2

38 × 1016 = (38/102) 10x

(c) 1

1016 = 10x – 2

=

[June 2023]

2x + 4 = 3x 3x – 2x = 4 x=4

(a) 14 (b) 16 (c) 18 (d) 20

[June 2024]

(300)8 = 65.61 × 10x

16 = x – 2

Ans. (c) 1

x = 18

Working Note: 9 × 81x = 1/27(x – 3)

Q. 29. If 3x – y= 27 and 3x + y= 243, then x is equal to:

32 × 34x = 1/33(x – 3)

(a) 0

3(2 + 4x) = 1/3(3x – 9)

(b) 2

3(2 + 4x) × 3(3x – 9) = 1

(c) 4

3(2 + 4x + 3x – 9) = 1

(d) 6

3(7x – 7) = 1

Ans. (c) 4

3(7x – 7) = 30

Working Note:

7x – 7 = 0

3x – y= 27

7x = 7

3x – y= 33

x=1

x – y = 3 [Equation 1]

Q. 27. If 32x + 4 = 33x then the value of x is:

3x + y= 243 3x + y= 35

(a) 1

x + y = 5 [Equation 2]

(b) 2

Solving Equations 1 and 2, We get,

(c) 3

2x = 8

(d) 4 Ans. (d) 4

[Dec. 2023]

x=4

[June 2024]

TAXMANN®

(d) 0

32x + 4 = 33x

Fundamentals of Business Mathematics & Statistics (FBMS | Maths) | CRACKER AUTHOR : PUBLISHER : DATE OF PUBLICATION : EDITION : ISBN NO : No. of Pages : BINDING TYPE :

Tarun Agarwal Taxmann June 2026 5th Edition 9789375619741 336 Paperback

Rs. 395 DESCRIPTION Fundamentals of Business Mathematics & Statistics (FBMS | Maths) – CRACKER is an exam-oriented question bank built to help students crack Paper 3 of the CMA Foundation examination (ICMAI). Since the Foundation paper is entirely MCQ/OMR-based, it rewards conceptual recall and speed—so this title keeps theory to a minimum and puts the student straight into exam-style practice, question after question.

Every chapter opens with a quick revision of key concepts and formulae, then moves into a comprehensive MCQ bank. The questions are drawn from actual past exams, each tagged with its term and solved with a step-by-step ‘Working Note’ that teaches the method, not just the answer. In all, the book delivers 1,100+ MCQs (including original practice questions) and fully solved past-exam questions up to the June 2026 exam—making it one of the most current revision companions for this paper. Current for the December 2026 & June 2027 Attempts.

The Present Publication is the 5th Edition | June 2026, authored by CA. Tarun Agarwal, with the following noteworthy features: • [Fully Solved Past-Exam Questions till June 2026] Real past questions, each fully solved, with coverage up to the latest June 2026 paper—so practice reflects current trends, not dated material • [1,100+ MCQs Including Practice Questions] Past-exam questions plus original practice MCQs give the repetition needed to build accuracy and speed • [Descriptive Questions Converted into MCQs (Latest Pattern)] Earlier descriptive questions are recast as MCQs, preserving years of exam history in the current objective format • [Step-by-Step ‘Working Notes’ with Every Solution] Each answer is backed by a clear worked ‘Working Note’, turning the bank into a self-teaching tool • [Module-Wise Marks Distribution of Past Exams] A front-matter chart of topic-wise marks across June and December sittings (2023 to June 2026), with averages—a built-in map for prioritising study • [Module-Wise Comparison with the ICMAI Study Material] Every chapter is mapped to its official CMA module number for seamless cross-reference • [Tabular Quick-Revision Summary in Every Chapter] All 27 chapters open with a quick recap of key concepts, properties and formulae • [Four Fully Solved Recent Papers (Dec 2024 – June 2026)] Complete solved papers for December 2024, June 2025, December 2025 and June 2026, in the actual exam’s section format

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