250+ TOP MCQs on Groups – Closure and Associativity and Answers

Discrete Mathematics Multiple Choice Questions on “Groups – Closure and Associativity”.

1. Let (A7, ⊗7)=({1, 2, 3, 4, 5, 6}, ⊗7) is a group. It has two sub groups X and Y. X={1, 3, 6}, Y={2, 3, 5}. What is the order of union of subgroups?
a) 65
b) 5
c) 32
d) 18

Answer: b
Clarification: Given, (A7, ⊗7)=({1, 2, 3, 4, 5, 6}, ⊗7) and the union of two sub groups X and Y, X={1, 3, 6} Y={2, 3, 5} is X∪Y={1, 2, 3, 5, 6} i.e., 5. Here, the order of the union can not be divided by order of the group.

2. A relation (34 × 78) × 57 = 57 × (78 × 34) can have __________ property.
a) distributive
b) associative
c) commutative
d) closure

Answer: b
Clarification: For any three elements(numbers) a, b and c associative property describes a × ( b × c ) = ( a × b ) × c [for multiplication]. Hence associative property is true for multiplication and it is true for multiplication also.

3. B1: ({0, 1, 2….(n-1)}, xm) where xn stands for “multiplication-modulo-n” and B2: ({0, 1, 2….n}, xn) where xn stands for “multiplication-modulo-m” are the two statements. Both B1 and B2 are considered to be __________
a) groups
b) semigroups
c) subgroups
d) associative subgroup

Answer: b
Clarification: Here, B1 is the group and identity element is 0, means for all a∈B1, a+n.0=a. As a2 identity element does not exist. Here, 0 can not be the identity element. For example, for one of the member n of the set we have n+n.0=0, It will be n. So, B2 is not a group. Both B1 and B2 are semigroups as they satisfy closure and associativity property.

4. If group G has 65 elements and it has two subgroups namely K and L with order 14 and 30. What can be order of K intersection L?
a) 10
b) 42
c) 5
d) 35

Answer: c
Clarification: As it is an intersection so the order must divide both K and L. Here 3, 6, 30 does not divide 14. But 5 must be the order of the group as it divides the order of intersection of K and L as well as the order of the group.

5. Consider the binary operations on X, a*b = a+b+4, for a, b ∈ X. It satisfies the properties of _______
a) abelian group
b) semigroup
c) multiplicative group
d) isomorphic group

Answer: a
Clarification: Since * closed operation, a*b belongs to X. Hence, it is an abelian group.

6. Let * be the binary operation on the rational number given by a*b=a+b+ab. Which of the following property does not exist for the group?
a) closure property
b) identity property
c) symmetric property
d) associative property

Answer: b
Clarification: For identity e, a+e=e+a=e, a*e = a+e+ae = a => e=0 and e+a = e+a+ea = a => e=0. So e=0 will be identity, for e to be identity, a*e = a ⇒ a+e+ae = a ⇒ e+ae = 0 and e(1+a) = 0 which gives e=0 or a=-1. So, when a = -1, no identity element exist as e can be any value in that case.

7. Let G be a finite group with two sub groups M & N such that |M|=56 and |N|=123. Determine the value of |M⋂N|.
a) 1
b) 56
c) 14
d) 78

Answer: a
Clarification: We know that gcd(56, 123)=1. So, the value of |M⋂N|=1.

8. A group G, ({0}, +) under addition operation satisfies which of the following properties?
a) identity, multiplicity and inverse
b) closure, associativity, inverse and identity
c) multiplicity, associativity and closure
d) inverse and closure

Answer: b
Clarification: Closure for all a, b∈G, the result of the operation, a+b, is also in G. Since there is one element, hence a=b=0, and a+b=0+0=0∈G. Hence, closure property is satisfied. Associative for all a, b, c∈G, (a+b)+c=a+(b+c). For example, a=b=c=0. Hence (a+b)+c=a+(b+c)
⟹(0+0)+0=0+(0+0)⟹0=0. Hence, associativity property is satisfied. Suppose for an element e∈G such that, there exists an element a∈G and so the equation e+a=a+e=a holds. Such an element is unique, the identity element property is satisfied. For example, a=e=0. Hence e+a = a+e⟹0+0=0+0⟹0=a. Hence e=0 is the identity element. For each a∈G, there exists an element b∈G (denoted as a-1), such that a+b=b+a=e, where e is the identity element. The inverse element is 0 as the addition of 0 with 0 will be 0, which is also an identity element of the structure.

9. If (M, *) is a cyclic group of order 73, then number of generator of G is equal to ______
a) 89
b) 23
c) 72
d) 17

Answer: c
Clarification: We need to find the number of co-primes of 73 which are less than 73. As 73 itself is a prime, all the numbers less than that are co-prime to it and it makes a group of order 72 then it can be of {1, 3, 5, 7, 11….}.

10. The set of even natural numbers, {6, 8, 10, 12,..,} is closed under addition operation. Which of the following properties will it satisfy?
a) closure property
b) associative property
c) symmetric property
d) identity property

Answer: a
Clarification: The set of even natural numbers is closed by the addition as the sum of any two of them produces another even number. Hence, this closed set satisfies the closure property.

250+ TOP MCQs on Set Operations and Answers

Discrete Mathematics Interview Questions and Answers on “Set Operations”.

1. Let the set A is {1, 2, 3} and B is {2, 3, 4}. Then the number of elements in A U B is?
a) 4
b) 5
c) 6
d) 7

Answer: a
Clarification: AUB is {1, 2, 3, 4}.

2. Let the set A is {1, 2, 3} and B is { 2, 3, 4}. Then number of elements in A ∩ B is?
a) 1
b) 2
c) 3
d) 4

Answer: b
Clarification: A ∩ B is {2, 3}.

3. Let the set A is {1, 2, 3} and B is {2, 3, 4}. Then the set A – B is?
a) {1, -4}
b) {1, 2, 3}
c) {1}
d) {2, 3}

Answer: c
Clarification: In A – B the common elements get cancelled.

4. In which of the following sets A – B is equal to B – A?
a) A = {1, 2, 3}, B = {2, 3, 4}
b) A = {1, 2, 3}, B = {1, 2, 3, 4}
c) A = {1, 2, 3}, B = {2, 3, 1}
d) A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4, 5, 1}

Answer: c
Clarification: A- B= B-A = Empty set.

5. Let A be set of all prime numbers, B be the set of all even prime numbers, C be the set of all odd prime numbers, then which of the following is true?
a) A ≡ B U C
b) B is a singleton set.
c) A ≡ C U {2}
d) All of the mentioned

Answer: d
Clarification: 2 is the only even prime number.

6. If A has 4 elements B has 8 elements then the minimum and maximum number of elements in A U B are ____________
a) 4, 8
b) 8, 12
c) 4, 12
d) None of the mentioned

Answer: b
Clarification: Minimum would be when 4 elements are same as in 8, maximum would be when all are distinct.

7. If A is {{Φ}, {Φ, {Φ}}}, then the power set of A has how many element?
a) 2
b) 4
c) 6
d) 8

Answer: b
Clarification: The set A has got 2 elements so n(P(A))=4.

8. Two sets A and B contains a and b elements respectively. If power set of A contains 16 more elements than that of B, value of ‘b’ and ‘a’ are _______
a) 4, 5
b) 6, 7
c) 2, 3
d) None of the mentioned

Answer: a
Clarification: 32-16=16, hence a=5, b=4.

9. Let A be {1, 2, 3, 4}, U be set of all natural numbers, then U-A’(complement of A) is given by set.
a) {1, 2, 3, 4, 5, 6, ….}
b) {5, 6, 7, 8, 9, ……}
c) {1, 2, 3, 4}
d) All of the mentioned

Answer: c
Clarification: U – A’ ≡ A.

10. Which sets are not empty?
a) {x: x is a even prime greater than 3}
b) {x : x is a multiple of 2 and is odd}
c) {x: x is an even number and x+3 is even}
d) { x: x is a prime number less than 5 and is odd}

Answer: d
Clarification: Because the set is {3}.

250+ TOP MCQs on Types of Matrices and Answers

Discrete Mathematics Multiple Choice Questions on “Types of Matrices”.

1. If a matrix A = [A11 A12 ⋯ A1n A21 A2n ⋮ ⋮ An1 An2 ⋯ Ann], order(nxn) Aii = 1, Aij = 0 for i ≠ j. Then that matrix is known as ________
a) Identity matrix
b) Null matrix
c) Singular matrix
d) None of the mentioned

Answer: a
Clarification: In unit matrix all diagonal elements are 1 and all other 0.

2. A symmetric matrix is a one in which?
a) All diagonal elements are zero
b) All diagonal elements are 1
c) A = AT
d) A = -AT

Answer: c
Clarification: For symmetric matrices, matrix remains same even after transpose.

3. An anti-symmetric matrix is a one in which?
a) All diagonal elements are zero
b) All diagonal elements are 1
c) A = AT
d) A = -AT

Answer: d
Clarification: Foran anti-symmetric matrix, matrix changes it sign after transpose.

4. A matrix having one row and many columns is known as?
a) Row matrix
b) Column matrix
c) Diagonal matrix
d) None of the mentioned

Answer: a
Clarification: In row matrix there is only one row.

5. A matrix having many rows and one column is known as?
a) Row matrix
b) Column matrix
c) Diagonal matrix
d) None of the mentioned

Answer: b
Clarification: In column matrix there is only one column.

6. The trace of the matrix is defined as _______
a) Sum of all the elements of the matrix
b) Sum of all the elements of leading diagonal of matrix
c) Sum of all non-zero elements of matrix
d) None of the mentioned

Answer: b
Clarification: Trace is the sum of the elements of leading diagonal of matrix.

7. A square matrix A = [aij ]nxn, if aij = 0 for i > j then that matrix is known as _______
a) Upper triangular matrix
b) Lower triangular matrix
c) Unit matrix
d) Null matrix

Answer: a
Clarification: In upper triangular matrix A = [ aij ]nxn, if aij = 0 for i > j.

8. A square matrix A = [aij ]nxn, if aij = 0 for i < j then that matrix is known as _______
a) Upper triangular matrix
b) Lower triangular matrix
c) Unit matrix
d) Null matrix

Answer: b
Clarification: In lower triangular matrix A = [aij ]nxn, if aij = 0 for i < j.

9. Two matrix can be added if _______
a) rows of both the matrices are same
b) columns of both the matrices are same
c) both rows and columns of both the matrices are same
d) number of rows of first matrix should be equal to number of column of second

Answer: c
Clarification: Order of two matrices must be same.

10. For matrix A if AAT = I, I is identity matrix then A is?
a) Orthagonal matrix
b) Nilpotent matrix
c) Idempotent matrix
d) None of the mentioned

Answer: a
Clarification: For orthagonal matrices AAT = I = AT A.

250+ TOP MCQs on Number Theory – Base Conversion and Answers

Discrete Mathematics Multiple Choice Questions on “Number Theory – Base Conversion”.

1. Which of the number is not allowed in Binary representation of a number?
a) 0
b) 1
c) 2
d) None of the mentioned

Answer: c
Clarification: Binary numbers are formed with a combination of 0 & 1 only.

2. Which of the number is not allowed in Octal representation of a number?
a) 0
b) 4
c) 8
d) None of the mentioned

Answer: c
Clarification: Octal numbers are formed with a combination of 0 to 7 only.

3. Hexadecimal number equivalent of decimal 10 is?
a) 10
b) A
c) F
d) None of the mentioned

Answer: b
Clarification: In hexadecimal representation A is represented as decimal 10.

4. Decimal equivalent of binary number 1010 is?
a) 11
b) A
c) 10
d) None of the mentioned

Answer: c
Clarification: 0X1 + 1X2 + 0X4 + 1X8 = 10 in decimal.

5. Decimal 13 in base 8 can be represented as _________
a) 15
b) 12
c) 22
d) None of the mentioned

Answer: a
Clarification: 1X8 + 5X1 = 12, 15 is the octal representation of 13.

6. F in hexadecimal representation is equivalent to 9 in decimal.
a) True
b) False

Answer: b
Clarification: F in hexadecimal representation is equivalent to 15 in decimal.

7. Octal number may contain digits from 1 to 8.
a) True
b) False

Answer: b
Clarification: Octal number contains digits from 0 to 7, * is not allowed.

8. For some base r, the digits which are allowed in its representation are?
a) Digits from 1 to r
b) Digits from 0 to r-1
c) Digits from 1 to r-1
d) None of the mentioned

Answer: b
Clarification: A base r number may contain digits from 0 to r-1.

9. The binary number 100110 in octal is represented by _______________
a) 45
b) 10012
c) 46
d) 58

Answer: c
Clarification: Pairing 3 numbers from right hand side we get 110 as 6 and 100 as 4 in octal so the number is 46.

10. A number greater than 32 would require a minimum of how may bits in binary representation?
a) 5
b) 6
c) 4
d) 10

Answer: b
Clarification: Since through 5 bits we can only represent numbers till 31 since 25 = 32 we need greater than 5 bits, so minimum would be 6.

250+ TOP MCQs on Counting – Combinations and Answers

Discrete Mathematics Multiple Choice Questions on “Counting – Combinations”.

1. There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex heptagons of distinctly different areas can be drawn using these points as vertices?
a) 7! * 6
b) 7C5
c) 7!
d) same area

Answer: d
Clarification: Since all the points are equally spaced; hence the area of all the convex heptagons will be the same.

2. There are 2 twin sisters among a group of 15 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two sisters?
a) 15 *12! * 2!
b) 15! * 2!
c) 14C2
d) 16 * 15!

Answer: a
Clarification: We know that n objects can be arranged around a circle in (frac{(n−1)!}{2}). If we consider the two sisters and the person in between the brothers as a block, then there will 12 others and this block of three people to be arranged around a circle. The number of ways of arranging 13 objects around a circle is in 12! ways. Now the sisters can be arranged on either side of the person who is in between the sisters in 2! ways. The person who sits in between the two sisters can be any of the 15 in the group and can be selected in 15 ways. Therefore, the total number of ways 15 *12! * 2!.

3. The number of words of 4 consonants and 3 vowels can be made from 15 consonants and 5 vowels, if all the letters are different is ________
a) 3! * 12C5
b) 16C4 * 4C4
c) 15! * 4
d) 15C4 * 5C3 * 7!

Answer: d
Clarification: There are 4 consonants out of 15 can be selected in 15C4 ways and 3 vowels can be selected in 5C3 ways. Therefore, the total number of groups each containing 4 consonants and 3 vowels = 15C4 * 4C3. Each group contains 7 letters which can be arranged in 7! ways. Hence, required number of words = 15C4 * 5C3 * 7!.

4. How many ways are there to arrange 7 chocolate biscuits and 12 cheesecake biscuits into a row of 19 biscuits?
a) 52347
b) 50388
c) 87658
d) 24976

Answer: b
Clarification: Consider the situation as having 19 spots and filling them with 7 chocolate biscuits and 19 cheesecake biscuits. Then we just choose 7 spots for the chocolate biscuits and let the other 10 spots have cheesecake biscuits. The number of ways to do this job is 19C7 = 50388.

5. If a, b, c, d and e are five natural numbers, then find the number of ordered sets(a, b, c, d, e) possible such that a+b+c+d+e=75.
a) 65C5
b) 58C6
c) 72C7
d) 74C4

Answer: d
Clarification: Let assumes that there are 75 identical balls which are to be arranged in 5 different compartments (Since a, b, c, d, e are distinguishable). If the balls are arranged in the row. We have 74 gaps where we can place a ball in each gap since we need 5 compartments we need to place only 4 balls. We can do this in 74C4 ways.

6. There are 15 people in a committee. How many ways are there to group these 15 people into 3, 5, and 4?
a) 846
b) 2468
c) 658
d) 1317

Answer: d
Clarification: The number of ways to choose 3 people out of 9 is 15C3. Then, number of ways to choose 5 people out of (15-3) = 12 is 12C5. Finally, the number of ways to choose 4 people out of (12-4) = 8 is 8C4. Hence, by the rule of product, 15C3 + 12C5 + 8C4 = 1317.

7. There are six movie parts numbered from 1 to 6. Find the number of ways in which they be arranged so that part-1 and part-3 are never together.
a) 876
b) 480
c) 654
d) 237

Answer: b
Clarification: The total number of ways in which 6 part can be arranged = 6! = 720. The total number of ways in which part-1 and part-3 are always together: = 5!*2! = 240. Therefore, the total number of arrangements, in which they are not together is = 720 − 240 = 480.

8. How many ways are there to divide 4 Indian countries and 4 China countries into 4 groups of 2 each such that at least one group must have only Indian countries?
a) 6
b) 45
c) 12
d) 76

Answer: a
Clarification: The number of ways to divide 4+4=8 countries into 4 groups of 2 each is as follows: (10C2 * 10C2 * 10C2 * 10C2)/4! = 30. Since it is required that at least one group must have only Indian countries, we need to subtract 30 from the number of possible groupings where all 4 groups have 1 Indian country and 1 China country each. This is equivalent to the number of ways to match each of the 4 Indian countries with one China country: 4! = 24. Therefore, the answer is 30 – 24 = 6.

9. Find the number of factors of the product 58 * 75 * 23 which are perfect squares.
a) 47
b) 30
c) 65
d) 19

Answer: b
Clarification: Any factor of this number should be of the form 5a * 7b * 2c. For the factor to be a perfect square a, b, c has to be even. a can take values 0, 2, 4, 6, 8, b can take values 0, 2, 4 and c can take values 0, 2. Total number of perfect squares = 5 * 3 * 2 = 30.

10. From a group of 8 men and 6 women, five persons are to be selected to form a committee so that at least 3 women are there on the committee. In how many ways can it be done?
a) 686
b) 438
c) 732
d) 549

Answer: a
Clarification: We may have (2 men and 3 women) or (1 men and 4 woman) or (5 women only). The Required number of ways = (8C2 × 6C3) + (8C1 × 6C4) + (6C5) = 686.

250+ TOP MCQs on Discrete Probability – Logarithmic Series and Answers

Discrete Mathematics Multiple Choice Questions on “Discrete Probability – Logarithmic Series”.

1. Computation of the discrete logarithm is the basis of the cryptographic system _______
a) Symmetric cryptography
b) Asymmetric cryptography
c) Diffie-Hellman key exchange
d) Secret key cryptography

Answer: c
Clarification: A discrete logarithm modulo of an integer to the base is an integer such that ax ≡ b (mod g). The problem of computing the discrete logarithm is a well-known challenge in the field of cryptography and is the basis of the cryptographic system i.e., the Diffie-Hellman key exchange.

2. Solve the logarithmic function of ln((frac{1+5x}{1+3x})).
a) 2x – 8x2 + (frac{152x^3}{3}) – …
b) x2 + (frac{7x^2}{2} – frac{12x^3}{5}) + …
c) x – (frac{15x^2}{2} + frac{163x^3}{4}) – …
d) 1 – (frac{x^2}{2} + frac{x^4}{4}) – …

Answer: a
Clarification: To solve the logarithmic function ln((frac{1+5x}{1+3x})) = ln(1+5x) – ln(1+3x) = (5x – (frac{(5x)^2}{2} + frac{(5x)^3}{3}) – …) – (3x – (frac{(3x)^2}{2} + frac{(3x)^3}{3}) – …) = 2x – 8x2 + (frac{152x^3}{3}) – …

3. Determine the logarithmic function of ln(1+5x)-5.
a) 5x + (frac{25x^2}{2} + frac{125x^3}{3} + frac{625x^4}{4}) …
b) x – (frac{25x^2}{2} + frac{625x^3}{3} – frac{3125x^4}{4}) …
c) (frac{125x^2}{3} – 625x^3 + frac{3125x^4}{5}) …
d) -25x + (frac{125x^2}{2} – frac{625x^3}{3} + frac{3125x^4}{4}) …

Answer: d
Clarification: Apply the logarithmic law, that is logax = xlog(a). Now the function is ln(1+5x)-5 = -5log(1+5x). By taking the series = -5(5x – (frac{(5x)^2}{2} + frac{(5x)^3}{3} – frac{(5x)^4}{4}) + …) = -25x + (frac{125x^2}{2} – frac{625x^3}{3} + frac{3125x^4}{4}) …

4. Find the value of x: 3 x2 alogax = 348?
a) 7.1
b) 4.5
c) 6.2
d) 4.8

Answer: d
Clarification: Since, alogax = x. The given equation may be written as: 3x2 x = 348 ⇒ x = (116)1/3 = 4.8.

5. Solve for x: log2(x2-3x)=log2(5x-15).
a) 2, 5
b) 7
c) 23
d) 3, 5

Answer: d
Clarification: By using the property if logax = logay then x=y, gives 2x2-3x=10-6x. Now, to solve the equation x2-3x-5x+15=0 ⇒ x2-8x+15 ⇒ x=3, x=5
For x=3: log2(32-3*3) = log2(5*3-15) ⇒ true
For x=5: log2(52-3*5) = log2(5*5-15) ⇒ true
The solutions to the equation are : x=3 and x=5.

6. Solve for x the equation 2x + 3 = 5x + 2.
a) ln (24/8)
b) ln (25/8) / ln (2/5)
c) ln (32/5) / ln (2/3)
d) ln (3/25)

Answer: b
Clarification: Given that 2x + 3 = 5x + 2. By taking ln of both sides: ln (2x + 3) = ln (5x + 2)
⇒ (x + 3) ln 2 = (x + 2) ln 5
⇒ x ln 2 + 3 ln 2 = x ln 5 + 2 ln 5
⇒ x ln 2 – x ln 5 = 2 ln 5 – 3 ln 2
⇒ x = ( 2 ln 5 + 3 ln 2 ) / (ln 2 – ln 5) = ln (52 / 23) / ln (2/5) = ln (25/8) / ln (2/5).

7. Given: log4 z = B log2/3z, for all z > 0. Find the value of constant B.
a) 2/(3!*ln(2))
b) 1/ln(7)
c) (4*ln(9))
d) 1/(2*ln(3))

Answer: d
Clarification: By using change of base formula we can have ln (x) / ln(4) = B ln(x) / ln(2/3) ⇒
B = 1/(2*ln(3)).

8. Evaluate: 16x – 4x – 9 = 0.
a) ln [( 5 + (sqrt{21})) / 2] / ln 8
b) ln [( 2 + (sqrt{33})) / 2] / ln 5
c) ln [( 1 + (sqrt{37})) / 2] / ln 4
d) ln [( 1 – (sqrt{37})) / 2] / ln 3

Answer: c
Clarification: Given: 16x – 4x – 9 = 0. Since 16x = (4x)2, the equation may be written as: (4x)2 – 4x – 9 = 0. Let t = 3x and so t: t2 – t – 9 = 0 which gives t: t = (1 + (sqrt{37})) / 2 and (1 – (sqrt{37})) / 2
Since t = 4x, the acceptable solution is y = (1 + (sqrt{37})) / 2 ⇒ 4x = (1 + (sqrt{37}))/2. By using ln on both sides: ln 4x = ln [ (1 + (sqrt{37})) / 2] ⇒ x = ln [ ( 1 + (sqrt{37}))/2] / ln 3.

9. Transform 54y = n+1 into equivalent a logarithmic expression.
a) log12 (n+1)
b) log41 (n2)
c) log63 (n)
d) log54 (n+1)

Answer: d
Clarification: By using the equivalent expression: ay = x ⇔ y = loga (x) to write 3x = m as a logarithm: y = log54 (n+1).

10. If loga((frac{1}{8}) = -frac{3}{4}), than what is x?
a) 287
b) 469
c) 512
d) 623

Answer: c
Clarification: By using exponential form: a-5/9 = 2/8. Now, raise both sides of the above equation to the power -9/5: (x-5/9)-9/5 = (1/32)-9/5. By simplifying we get, a = 329/5 = 29 = 512.