Category: Discrete Mathematics Objective Questions

Discrete Mathematics Multiple Choice Questions on “Boolean Algebra – Prime Implicants and Essentials”.

1. What is the maximum number of prime implicants with 34-variable minimized expression?
a) 34
b) 764
c) 2³³
d) 2³¹

Answer: c
Clarification: For n-variable K Map, we have = 2^n-1 prime implicants. In this case, n=34 and the maximum number of prime implicants will be 2^34-1 = 2³³.

2. How many cells are there for an 8-variable K-Map?
a) 421
b) 1048
c) 256
d) 375

Answer: c
Clarification: Any Boolean expression or a function comprising of 8 variables can be solved using an 8-variable K-Map. So, an 8-variable K-Map must contain 2⁸ = 256.

3. Determine the number of essential prime implicants of the function f(a, b, c, d) = Σm(1, 3, 4, 8, 10, 13) + d(2, 5, 7, 12), where m denote the minterm and d denotes the don’t care condition.
a) 2³
b) 3
c) 643
d) 128

Answer: b
Clarification: A prime implicant that cannot be replaced by any other implicant for getting the output is called the essential prime implicants. Here, we have 3 essential prime implicants by using the K-map representation.

4. How many number of prime implicants are there in the expression F(x, y, z) = y’z’ + xy + x’z.
a) 7
b) 19
c) 3
d) 53

Answer: c
Clarification: An implicant of a function is a product term which is included in the function.
Hence, for the given function, y’z’, xy and x’z all are prime implicants.

5. f(x, y, z) = xy’+yz’+xyz, what are essential prime implicants of this switching function?
a) 8
b) 0
c) 4
d) 3

Answer: b
Clarification: There are no essential prime implicants for this switching function. We can get this solution by using K-Map.

6. How many essential prime implicants are there in the K-Map of the function F = Σ(0, 1, 2, 4, 7, 11, 12, 13, 15)?
a) 4
b) 1
c) 3
d) 7

Answer: b
Clarification: By, solving the minimization expression using K-Map, there is only 1 essential prime implicant exist as it is not covered by any other input variable.

7. Determine the number of prime implicants of the following function F?
F(a, b, c, d) = Σm(1, 3, 7, 9, 10, 11, 13, 15)
a) 621
b) 187
c) 3⁵
d) 5

Answer: d
Clarification: There are 5 prime implicants for the function (a+b+d’)(a+c’+d’)
(a’+c+d)(a’+b+d’)(a’+b’+c’+d). Hence, the required answer is 5.

8. For an 18-variable k-map determine the number of prime implicants?
a) 2¹⁸
b) 35
c) 253
d) 721

Answer: a
Clarification: The maximum number of implicants for the n-variable k-map is 2ⁿ. Hence, the required answer is 2¹⁸.

9. How many false essential prime implicants for the given Boolean functions f(A, B, C) = ∑m(2, 5, 6)?
a) 1024
b) 2
c) 16
d) 435

Answer: b
Clarification: There are two essential prime implicants such as (B+C) and (B+C’) for the given function. Hence, the required answer is 2.

10. How many minimal forms are there in the function F(A, B, C) = ∑(1, 3, 2, 5, 6, 7) if it is having cyclic prime implicants k-map?
a) 216
b) 2
c) 14
d) 82

Answer: b
Clarification: In cyclic prime implicant, min terms will be (1, 3, 5, 7, 9, 11). Hence, either we can have [(1,3), (7,11), (5,9)] or [(1,5), (11,9), (3.7)]. So, there can be 2 minimal forms.

Discrete Mathematics Assessment Questions and Answers on “Types of Proofs”.

1. Let the statement be “If n is not an odd integer then square of n is not odd.”, then if P(n) is “n is an not an odd integer” and Q(n) is “(square of n) is not odd.” For direct proof we should prove _________
a) ∀nP ((n) → Q(n))
b) ∃ nP ((n) → Q(n))
c) ∀n~(P ((n)) → Q(n))
d) ∀nP ((n) → ~(Q(n)))

Answer: a
Clarification: Definition of direct proof.

2. Which of the following can only be used in disproving the statements?
a) Direct proof
b) Contrapositive proofs
c) Counter Example
d) Mathematical Induction

Answer: c
Clarification: Counter examples cannot be used to prove results.

3. Let the statement be “If n is not an odd integer then sum of n with some not odd number will not be odd.”, then if P(n) is “n is an not an odd integer” and Q(n) is “sum of n with some not odd number will not be odd.” A proof by contraposition will be ________
a) ∀nP ((n) → Q(n))
b) ∃ nP ((n) → Q(n))
c) ∀n~(P ((n)) → Q(n))
d) ∀n(~Q ((n)) → ~(P(n)))

Answer: d
Clarification: Definition of proof by contraposition.

4. When to proof P→Q true, we proof P false, that type of proof is known as ___________
a) Direct proof
b) Contrapositive proofs
c) Vacuous proof
d) Mathematical Induction

Answer: c
Clarification: Definition of vacuous proof.

5. In proving √5 as irrational, we begin with assumption √5 is rational in which type of proof?
a) Direct proof
b) Proof by Contradiction
c) Vacuous proof
d) Mathematical Induction

Answer: b
Clarification: Definition of proof by contradiction.

6. A proof covering all the possible cases, such type of proofs are known as ___________
a) Direct proof
b) Proof by Contradiction
c) Vacuous proof
d) Exhaustive proof

Answer: d
Clarification: Definition of exhaustive proof.

7. Which of the arguments is not valid in proving sum of two odd number is not odd.
a) 3 + 3 = 6, hence true for all
b) 2n +1 + 2m +1 = 2(n+m+1) hence true for all
c) All of the mentioned
d) None of the mentioned

Answer: a
Clarification: Some examples are not valid in proving results.

8. A proof broken into distinct cases, where these cases cover all prospects, such proofs are known as ___________
a) Direct proof
b) Contrapositive proofs
c) Vacuous proof
d) Proof by cases

Answer: c
Clarification: Definition of proof by cases.

9. A proof that p → q is true based on the fact that q is true, such proofs are known as ___________
a) Direct proof
b) Contrapositive proofs
c) Trivial proof
d) Proof by cases

Answer: c
Clarification: Definition of trivial proof.

10. A theorem used to prove other theorems is known as _______________
a) Lemma
b) Corollary
c) Conjecture
d) None of the mentioned

Answer: a
Clarification: Definition of lemma.

Discrete Mathematics Multiple Choice Questions & Answers on “Arithmetic and Geometric Mean”.

1. Let A₁, A₂, be two AM’s and G₁, G₂ be two GM’s between a and b,then (A₁ + A₂) / G₁G₂ is equal to _______
a) (a+b) / 2ab
b) 2ab/(a+b)
c) (a+b)/(ab)
d) None of the mentioned

Answer: c
Clarification: A₁ + A₂ = a + b, G₁G₂ = ab.

2. The series a,(a+b)/2, b is in _______
a) AP
b) GP
c) HP
d) None of the mentioned

Answer: a
Clarification: (a+b)/2 is AM between a, b. Hence series is in AP.

3. The series a, (ab)^1/2, b is in _______
a) AP
b) GP
c) HP
d) None of the mentioned

Answer: b
Clarification: (ab)^1/2 is GM between a, b. Hence series is in GP.

4. If A and G be the A.M and G.M between two positive number then the numbers are A + (A² – G²)^1/2, A – (A² – G²)^1/2.
a) True
b) False

Answer: a
Clarification: The equation having its roots as given equation is
x² – 2Ax + G² = 0 which implies
x = A + (A² – G²)^1/2, A – (A² – G²)^1/2.

5. If one geometric mean G and two airthmetic mean A₁, A₂ are inserted between two numbers, then (2A₁ – A₂) (2A₂ – A₁) is equal to _______
a) 2G
b) G
c) G²
d) None of the mentioned

Answer: c
Clarification: Let a and b be two numbers then, G = (ab)^1/2, A₁ = (2a+b)/3, A₂ = (a+2b)/3, (2A₁ – A₂) = a, (2A₂ – A₁) = b, (2A₁ – A₂)(2A₂ – A₁) = G².

6. State whether the given statement is true or false.
AM ≤ GM.
a) True
b) False

Answer: b
Clarification: Airthmetic Mean is always greater or equal to the geometric mean.

7. If between two numbers which are root of given equation. x² – 18x + 16 = 0, a GM is inserted then the value of that GM is?
a) 4
b) 5
c) 6
d) 16

Answer: a
Clarification: x² – 2Ax + G² = 0, here G² = 16 and therefore G = 4.

8. If a₁, a₂, a₃ are in airthemetic as well as geometric progression then which of the following is/are correct?
a) 2a₂ = a₁ + a₃
b) a₂ = (a₁a₃)^1/2
c) a₂ – a₁ = a₃ -a₂
d) All of the mentioned are correct

Answer: d
Clarification: a₂ is AM, GM between a₁, a₃, also the series is in AP so common difference should be same.

9. If a₁, a₂, a₃ are in GP then 1/a₁, 1/a₂, 1/a₃ are in ___________
a) AP
b) GP
c) HP
d) None of the mentioned

Answer: b
Clarification: Let the terms be ar, a, a/r then reciprocals are 1/(ar), 1/a, r/a. Still the terms are in GP.

10. If a₁, a₂, a₃…….. are in AP then if a₇ = 15, then the value of common difference that would make a₂ a₇ a₁₂ greatest is?
a) 2
b) 0
c) 4
d) 9

Answer: b
Clarification: Let d be common difference of the AP. Then,
a₂ a₇ a₁₂ = (15 – 5d)(15)(15 + 5d) = 375(9 – d²)
For maximum value d=0.

Discrete Mathematics Multiple Choice Questions on “Number Theory – Prime Numbers”.

1. The number of factors of prime numbers are ___________
a) 2
b) 3
c) Depends on the prime number
d) None of the mentioned

Answer: a
Clarification: A prime number is only divisible by 1 and itself.

2. What is the number ‘ 1’?
a) Prime number
b) Composite number
c) Neither Prime nor Composite
d) None of the mentioned

Answer: c
Clarification: 1 is neither prime number nor composite.

3. All prime numbers are odd.
a) True
b) False

Answer: b
Clarification: 2 is even as well as prime.

4. 3 is the smallest prime number possible.
a) True
b) False

Answer: b
Clarification: 2 is also a prime number.

5. How many prime numbers are there between 1 to 20?
a) 5
b) 6
c) 7
d) None of the mentioned

Answer: d
Clarification: The prime numbers between 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19.

6. There are finite number of prime numbers.
a) True
b) False

Answer: b
Clarification: There are infinite numbers of primes.

7. Sum of two different prime number is a ____________
a) Prime number
b) Composite number
c) Either Prime or Composite
d) None of the mentioned

Answer: c
Clarification: Eg:- 2 + 3 = 5 a prime, 3 + 7 = 10 a composite.

8. Difference of two distinct prime numbers is?
a) Odd and prime
b) Even and composite
c) None of the mentioned
d) All of the mentioned

Answer: c
Clarification: 3 – 2 = 1 is neither prime nor composite.

9. If a, b, c, d are distinct prime numbers with an as smallest prime then a * b * c * d is a ___________
a) Odd number
b) Even number
c) Prime number
d) None of the mentioned

Answer: b
Clarification: Since a is 2, 2 * b * c * d = Even number.

10. If a, b are two distinct prime number than a highest common factor of a, b is ___________
a) 2
b) 0
c) 1
d) ab

Answer: c
Clarification: HCF of two prime numbers is 1.

Discrete Mathematics Multiple Choice Questions on “Fundamental Principle of Counting”.

1. How many even 4 digit whole numbers are there?
a) 1358
b) 7250
c) 4500
d) 3600

Answer: c
Clarification: The thousands digit cannot be zero, so there are 9 choices. There are 10 possibilities for the hundreds digit and 10 possibilities for the tens digit. The units digit can be 0, 2, 4, 6 or 8, so there are 5 choices. By the basic counting principle, the number of even five digit whole numbers is 9 × 10 × 10 × 5 = 45,00.

2. In a multiple-choice question paper of 15 questions, the answers can be A, B, C or D. The number of different ways of answering the question paper are ________
a) 65536 x 4⁷
b) 194536 x 4⁵
c) 23650 x 4⁹
d) 11287435

Answer: a
Clarification: There are 4¹⁵ = 65536 x 4⁷ different ways of answering the exam paper of 15 MCQs.

3. How many words with seven letters are there that start with a vowel and end with an A? Note that they don’t have to be real words and letters can be repeated.
a) 45087902
b) 64387659
c) 12765800
d) 59406880

Answer: d
Clarification: The first letter must be a vowel, so there are 5 choices. The second letter can be any one of 26, the third letter can be any one of 26, the fourth letter can be any one of 26 and fifth and sixth letters can be any of 26 choices. The last letter must be an A, so there is only 1 choice. By the basic counting principle, the number of ‘words’ is 5 × 26 × 26 × 26 × 26 × 26 × 1 = 59406880.

4. Neela has twelve different skirts, ten different tops, eight different pairs of shoes, three different necklaces and five different bracelets. In how many ways can Neela dress up?
a) 50057
b) 14400
c) 34870
d) 56732

Answer: b
Clarification: By the basic counting principle, the number of different ways = 12 × 10 × 8 × 3 × 5 = 14400. Note that shoes come in pairs. So she must choose one pair of shoes from ten pairs, not one shoe from twenty.

5. How many five-digit numbers can be made from the digits 1 to 7 if repetition is allowed?
a) 16807
b) 54629
c) 23467
d) 32354

Answer: a
Clarification: 7⁵ = 16807 ways of making the numbers consisting of five digits if repetition is allowed.

6. For her English literature course, Ruchika has to choose one novel to study from a list of ten, one poem from a list of fifteen and one short story from a list of seven. How many different choices does Rachel have?
a) 34900
b) 26500
c) 12000
d) 10500

Answer: d
Clarification: By the Basic Counting Principle, the number of different choices is 10 × 15 × 7 = 10500.

7. There are two different Geography books, five different Natural Sciences books, three different History books and four different Mathematics books on a shelf. In how many different ways can they be arranged if all the books of the same subjects stand together?
a) 353450
b) 638364
c) 829440
d) 768700

Answer: c
Clarification: There are four groups of books which can be arranged in 4! different ways. Among those books, two are Geography books, five are Natural Sciences books, three are History books and four are Mathematics books. Therefore, there are 4! × 2! × 5! × 3! × 4! = 829440 ways to arrange the books.

8. The code for a safe is of the form PPPQQQQ where P is any number from 0 to 9 and Q represents the letters of the alphabet. How many codes are possible for each of the following cases? Note that the digits and letters of the alphabet can be repeated.
a) 874261140
b) 537856330
c) 549872700
d) 456976000

Answer: d
Clarification: 10³ × 2⁶⁴ = 456976000 possible codes are formed for the safe with the alphanumeric digits.

9. Amit must choose a seven-digit PIN number and each digit can be chosen from 0 to 9. How many different possible PIN numbers can Amit choose?
a) 10000000
b) 9900000
c) 67285000
d) 39654900

Answer: a
Clarification: By the basic counting principle, the total number of PIN numbers Amit can choose is 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10,000000.

10. A head boy, two deputy head boys, a head girl and 3 deputy head girls must be chosen out of a student council consisting of 14 girls and 16 boys. In how many ways can they are chosen?
a) 98072
b) 27384
c) 36428
d) 44389

Answer: b
Clarification: There are 16 × 15 × 14 + 14 × 13 × 12 × 11 = 27384 ways to choose from a student council.

Discrete Mathematics Multiple Choice Questions on “Discrete Probability – Mean and Variance of Random Variables”.

1. Two t-shirts are drawn at random in succession without replacement from a drawer containing 5 red t-shirts and 8 white t-shirts. Find the probabilities of all the possible outcomes.
a) 1
b) 13
c) 40
d) 346

Answer: a
Clarification: Let X denote the number of red t-shirts in the outcome. Here, x₁ = 2, x₂ = 1, x₃ = 1, x₄ = 1, x₅ = 0. Probability of first t-shirt being red = (frac{5}{13}).
Probability of second t-shirt being red = (frac{4}{12}).
So: P(x₁) = (frac{5}{13} × frac{4}{12} = frac{20}{146}). Likewise, for the probability of red first followed by black is (frac{8}{12}) (as there are 8 red t-shirts still in the drawer and 12 t-shirts all together).
So, P(x₂) = (frac{5}{13} *frac{8}{12} = frac{40}{146}). Similarly for white then red: P(x₃) = (frac{8}{13} × frac{4}{12} = frac{32}{146}). Finally, for 2 black balls: P(x₄) = (frac{8}{13} × frac{7}{12} = frac{56}{146}). So, (frac{20}{146} + frac{40}{146} + frac{32}{146} + frac{40}{146} = 1). Hence, all the t-shirts have been found.

2. A jar of pickle is picked at random using a filling process in which an automatic machine is filling pickle jars with 2.5 kg of pickle in each jar. Due to few faults in the automatic process, the weight of a jar could vary from jar to jar in the range 1.7 kg to 2.9 kg excluding the latter. Let X denote the weight of a jar of pickle selected. Find the range of X.
a) 3.7 ≤ X < 3.9
b) 1.6 ≤ X < 3.2
c) 1.7 ≤ X < 2.9
d) 1 ≤ X < 5

Answer: c
Clarification: Possible outcomes should be 1.7 ≤ X < 2.9. That is the probable range of X for the answer.

3. A probability density function f(x) for the continuous random variable X is denoted as _______
a) ∫ f(x)dx = ∞, -1<=x<=1
b) ∫ f(x)dx = 1, -∞<=x<=∞
c) ∫ f(x)dx = 0, -∞<=x<=∞
d) ∫ f(x+2)dx = .5, -∞<=x<=∞

Answer: b
Clarification: A probability density function f(x) for the continuous random variable X is denoted as ∫ f(x)dx = 1, -∞<=x<=∞. The area under the curve between any two ordinates x = a and x = b is a probability that X lies between a and b. So, ∫f(x)dx = P(a≤X≤b).

4. Let X is denoted as the number of heads in three tosses of a coin. Determine the mean and variance for the random variable X.
a) 4.8
b) 6
c) 3.2
d) 1.5

Answer: d
Clarification: Let H represents a head and T be a tail. X denotes the number of heads in three tosses of a coin. X can take the value 0, 1, 2, 3. P(X = 0) = (frac{1}{8}), P(X = 1) = (frac{3}{8}), P(X = 2) = (frac{3}{8}), P(X = 3) = (frac{1}{8}). The probability distribution of X is E(X) = Σ_ix_ip_i = 1 × (frac{3}{8} + 2 × frac{3}{8} + 3 × frac{1}{8}) = 1.5. E(X₂) = (12 × frac{3}{8} + 22 × frac{3}{8} + 32 × frac{1}{8}) = 3. So, Variance of X = V(X) = E(X²) – [E(X)]² = 3 – 1.5 = 1.5.

5. A football player makes 75% of his 5-point shots and 25% his 7-point shots. Determine the expected value for a 7-point shot of the player.
a) 4.59
b) 12.35
c) 5.25
d) 42.8

Answer: c
Clarification: Multiply the outcome by its probability, so the expected value becomes 0.75 * 7 points = 5.25.

6. In a card game Reena wins 3 Rs. if she draws a king or a spade and 7 Rs. if a heart or a queen from an pack of 52 playing cards. If she pays a certain amount of money each time she will lose the game. What will be the amount so that the game will come out a fair game?
a) 15
b) 6
c) 23
d) 2

Answer: d
Clarification: We know that E(X) = ∑{x_i * P(xi)} = 3 * (frac{2}{13} + 7 * frac{2}{13} − x * frac{10}{13} = frac{20}{13} − frac{10x}{13}). Suppose the expected value should be 0 Rs. for the game to be fair. So (frac{20}{13} − frac{10x}{13}) = 0 ⇒ x=2. So she should pay Rs.2 for it to be a fair game.

7. A Random Variable X can take only two values, 4 and 5 such that P(4) = 0.32 and P(5) = 0.47. Determine the Variance of X.
a) 8.21
b) 12
c) 3.7
d) 4.8

Answer: c
Clarification: Expected Value: μ = E(X) = ∑x * P(x) = 4 × 0.32 + 5 × 0.47 = 3.63. Next find ∑x² * P(x): ∑x² * P(x) = 16 × 0.32 + 25 × 0.47 = 16.87. Therefore, Var(X) = ∑x²P(x) − μ² = 16.87 − 13.17 = 3.7.

8. A 6-sided die is biased. Now, the numbers one to four are equally likely to happen, but five and six is thrice as likely to land face up as each of the other numbers. If X is the number shown on the uppermost face, determine the expected value of X when 6 is shown on the uppermost face.
a) (frac{13}{4})
b) (frac{3}{5})
c) (frac{2}{7})
d) (frac{21}{87})

Answer: a
Clarification: Let P(1) = P(2) = P(3) = P(4) = p; P(5) = P(6) = 2p. We know that the sum of all probabilities must be 1 ⇒ p + p + p + p + 2p + 2p = 1
⇒ 8p = 1 ⇒ p = (frac{1}{8})
Expected Value:
μ = E(X) = ∑x * P(x) = (1 * frac{1}{8} + 2 * frac{1}{8} + 3 * frac{1}{8} + 4 * frac{1}{8} + 5 * frac{2}{8} + 6 * frac{2}{8} = frac{13}{4}).

9. A fair cubical die is thrown twice and their scores summed up. If the sum of the scores of upper side faces by throwing two times a die is an event. Find the Expected Value of that event.
a) 48
b) 76
c) 7
d) 132

Answer: c
Clarification: Sample space = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.Suppose: P(2) = (frac{1}{36}), P(3) = (frac{2}{36}), P(4) = (frac{3}{36}), P(5) = (frac{4}{36}), P(6) = (frac{5}{36}), P(7) = (frac{6}{36}), P(8) = (frac{5}{36}), P(9) = (frac{4}{36}), P(10) = (frac{3}{36}), P(11) = (frac{2}{36}) and P(12) = (frac{1}{36}). Now, Expected Value:
μ = E(A) = ∑x * P(x) = (2 * frac{1}{36} + 3 * frac{2}{36} + 4 * frac{3}{36} + 5 * frac{4}{36} + 6 * frac{5}{36} )
(+ 7 * frac{6}{36} + 8 * frac{5}{36} + 9 * frac{4}{36} + 10 * frac{3}{36} + 11 * frac{2}{36} + 12 * frac{1}{36} = frac{252}{36}) = 7.

10. A random variable X can take only two values, 2 and 4 i.e., P(2) = 0.45 and P(4) = 0.97. What is the Expected value of X?
a) 3.8
b) 2.9
c) 4.78
d) 5.32

Answer: c
Clarification: We know that E(X) = ∑ x*P(x) = 2 × 0.45 + 4 × 0.97 = 4.78, where x={2,4}.