MCQ Questions

Answer:(c) P (A | B) = P(A ∩ B)/P(B)
Explanation:
(a) P(A ∩ B) = P(A) P(B) is false since this is true if and only if A and B are independent events.
(b) P (A ∪ B) = P(A) + P(B) is false since P(A ∩ B) is zero if and only if A and B are mutually exclusive.
(c) P (A | B) = P(A ∩ B)/P(B) is true.
(d) P (A ∪ B) < P(A) + P(B) is false.
Since P(A ∪ B) ≤ P(A) + P(B)

Answer: (c) \(\frac { 1 }{ 2 }\)
Explanation:

The bag contains 10 white balls and 15 black balls. Required probability

Answer:(d) \(\frac { 11 }{ 36 }\)
Explanation:
P(atleast one of dice will have 6 facing = 1 – P(none of dice have 6 facing up) = 1 – [\(\frac { 5 }{ 6 }\) × \(\frac { 5 }{ 6 }\)] = 1 – \(\frac { 25 }{ 36 }\) = \(\frac { 11 }{ 36 }\)

Answer:(c) \(\frac { 4 }{ 52 }\) x \(\frac { 3 }{ 51 }\)
Explanation:
The probability that the bottom card of a randomly shuffled deck is ace = \(\frac { 4 }{ 52 }\) because there are 4 aces out of total 52 cards. From the remaining 3 aces out of 51 cards the probability that the top card is also an ace = \(\frac { 3 }{ 51 }\). So required probability = \(\frac { 4 }{ 52 }\) × \(\frac { 3 }{ 51 }\).

Answer:(d) 0.4
Explanation:
P(Rain today) = 0.5
P(Rain tomorrow) = 0.6
P(Rain today ∪ Rain tomorrow) = 0.7
P(Rain today ∩ Rain tomorrow) = ?
P(rain today ∩ Rain tomorrow) = P(Rain today) + P(rain tomorrow) – P(Rain today ∩ Rain tomorrow)
So; 0.7 = 0.5 + 0.6 – P(Rain today ∩ Rain tomorrow) P(Rain today ∩ Rain tomorrow) = 0.5+ 0.6-0.7 = 0.4
So, the probability that it will rain today and tomorrow is 0.4.

Answer:(b) \(\frac { 3 }{ 8 }\)
Explanation:
Probability of getting an odd number in rolling of a die = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\). Now using binomial distribution P(Exactly one odd number among three outcomes)
= ³C₁( \(\frac { 1 }{ 2 }\))¹ ( \(\frac { 1 }{ 2 }\))² = 3 × ( \(\frac { 1 }{ 2 }\))³ = ( \(\frac { 3 }{ 8 }\))

Answer:(c) There is a sample point at which X has a value greater than or equal to 5.
Explanation:
If all the points have X < 5, then the expectation of a random variable X is surely less than 5. So according to this, there should be at least a sample point at which X ≥ 5.

Answer:(c) Events E₁ and E₂ are not independent
Explanation:
P(E₁)= \(\frac { 1 }{ 2 }\) , P(E₂) = \(\frac { 1 }{ 3 }\) and P(E₁ ∩ E₂) = \(\frac { 1 }{ 5 }\)

(a) P(E₁ or E₂)
= P(E₁) + P(E₂) – P(E₁ ∩ E₂)
= \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 3 }\) – \(\frac { 1 }{ 5 }\) = \(\frac { 19 }{ 30 }\)
However given in option (a) is \(\frac { 2 }{ 3 }\).
So option (a) is not true.

(b) For independent events P(E₁ ∩ E₂) = P(E₁) P(E₂)
Here, P(E₁ ∩ E₂) = \(\frac { 2 }{ 5 }\)
P(E₁) P(E₂) = \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 3 }\) = \(\frac { 1 }{ 6 }\)
So; (P E₁ ∩ E₂) ≠ P(E₁) P(E₂)
So events E₁ and E₂ are not independent. Option (b) is not true.

(c) Since E₁ and E₂ are not independent So option (c) is true.

(d) P(E₁/E₂) =\(\frac{\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{2}\right)}\) = \(\frac { 3 }{ 5 }\) So option (d) P(E₁/E₂) = \(\frac { 4 }{ 5 }\) is false.

Answer:(d) 1
Explanation:
Constraints are
(i) P(E₁) = P(E₂) = x
(ii) P(E₁ ∪ E₂) = 1
(iii) E₁ and E₂ are independent so
P(E₁ ∩ E₂) = P(E₁) P(E₂) = x × x = x²

NOW,
P(E₁ ∪ E₂) = P(E₁)+P(E₂) – P(E₁ ∩ E₂)
1 = x + x – x²
1 = 2x – x²
x² – 2x + 1 = 0
(x – 1)² = 0
x = 1
So; P(E₁) = P(E₂) = x = 1

Answer:(b) \(\frac{1}{7^{6}}\)
Explanation:
Sample space = 7⁷
All accidents on the same day = 7 ways (all on Monday, all on Tuesday…)
So, required probability = \(\frac{7}{7^{7}}\) = \(\frac{7}{7^{6}}\)

Answer:(c) \(\frac { 7 }{ 8 }\)
Explanation:
p(atleast one head and one tail)
= 1 – p(no head or no tail)
= 1 – [p(no head) + p(no tail) – p(no head and no tail)]
= 1 – [p (all tails) + p(all heads) – 0]
= 1 – [\(\left[\frac{1}{2^{4}}+\frac{1}{2^{4}}-0\right]\) = 1 – \(\frac { 2 }{ 16 }\) = 1 – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\).

Or

Alternate Method:
We need atleast one head (≥1H) and atleast one tail (≥ 1 T). First we satisfy ≥ 1H as follows.
1 H, 3T
2 H, 2 T
3H, 1 T
and 4 H, 0 T
now to satisfy the second condition of ≥ 1 T, we have to remove 4 H, 0 T.
So, the favorable cases are only 1 H, 2 H, and 3 H. The probability of this by binomial distribution formula is
= \(\frac{{ }^{4} C_{1}}{2^{4}}\) + \(\frac{{ }^{4} C_{2}}{2^{4}}\) + \(\frac{{ }^{4} C_{3}}{2^{4}}\)= \(\frac { 7 }{ 8 }\)

Answer:(d) 1 , \(\frac { 1 }{ 2 }\)
Explanation:
Given, P(A) = 1
P(B) = \(\frac { 1 }{ 2 }\)
Since both events are independent
P(A|B) = P(A) = 1
P(B|A) = P(B) = \(\frac { 1 }{ 2 }\)

Answer:(c) \(\int_{t}^{0}\) f₁ (X) f₂ (t – X)dX
Explanation:
Let the time taken for first and second modules be represented by x and y and total time = t.
∴ t = x + y is a random variable. Now the joint density function
g(t) = \(\int_{0}^{t}\)f(x,y)dx
= \(\int_{0}^{t}\)f (x, t – x)dx
= \(\int_{0}^{t}\)f₁(x) f₂(t – x)dx
which is also called as convolution of f₁ and f₂, abbreviated as f₁ * f₂.Correct answer is therefore, choice (c).

Answer:(a) \(\frac { 3 }{ 8 }\)
Explanation:
Given, P(H) = \(\frac { 1 }{ 2 }\)
P(T) = \(\frac { 1 }{ 2 }\)
Apply Bernoulli’s formula for binomial distribution,
P(X = 2) = \({ }^{4} \mathrm{C}_{2}\)(1/2)²(1 – \(\frac { 1 }{ 2 }\))^{4 – 2}
= \({ }^{4} \mathrm{C}_{2}\) ( \(\frac { 1 }{ 2 }\) )² (1/2)²
= \(\frac{{ }^{4} \mathrm{C}_{2}}{2^{4}}\) = \(\frac { 6 }{ 16 }\) = \(\frac { 3 }{ 8 }\)

Answer:(b) \(\frac { 6 }{ 23 }\)
Explanation:

Total number of children = 50 × 3 + 30 × 2 + 20 × 1 = 230
Favorable cases = The number of children who come from 2 children families = 30 × 2 = 60
So the probability that a randomly picked child belongs to a 2 children families = \(\frac { 60 }{ 230 }\) = \(\frac { 6 }{ 23 }\)

Answer:(c) (αβ/(α + β))
Explanation:
f(x) = λ.e^-λx, x ≥ 0
X and Y are two independent exponentially distributed random variables. Let λ₁ and λ₂ parameters of X and Y respectively.
P(X ≥ x)= e^-λx, x > 0
P(Y ≥ x)=e^-λ₂x, x > 0
Given Z = min (X, Y)
P(Z ≥ x) = P(X ≥ x, Y ≥ x)
= P(X ≥ x) P(Y ≥ x)
= e^-λ₁x . e^-λ₂x = e^{-(λ₁ + λ₂₎x}
Since mean of exponential distribution = 1/Parameter
So, α = 1/λ₁ ⇒ λ₁ = 1/α
β = 1/λ₂ ⇒ λ₂ = 1/β
∴ Z is random variable with parameter
Mean of Z = IMG

Answer:(d) 9375
Explanation:
Let the marks obtained per question be a random variable X. Its probability distribution table is given below:

Expected marks per question = E(x)=ΣXP(X)
= 1 × \(\frac { 1 }{ 4 }\) + (-o.25) × \(\frac { 3 }{ 4 }\)
= \(\frac { 1 }{ 4 }\) – \(\frac { 3 }{ 16 }\) = \(\frac { 1 }{ 16 }\) marks
Total marks expected for 150 questions = \(\frac { 1 }{ 16 }\) × 150 = \(\frac { 75 }{ 8 }\) marks per student
Total expected marks of 1000 students = \(\frac { 75 }{ 8 }\) ×1000 =9375 marks
So correct answer is (d).

Answer:(a) ⁿC_d/2ⁿ
Explanation:
If the hamming distance between two n bit strings is d, we are asking that d out of n trials be successful (success here means that the bits are different). So this is a binomial distribution with n trials and d successes and probability of success

P = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(Since out of the 4 possibilities {(0,0), (0,1), (1, 0), (1,1)} only two of them (0,1) and (1,0) are success)
So p(X = d) = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{d}}(1 / 2)^{\mathrm{d}}(1 / 2)^{\mathrm{n}-\mathrm{d}}\) = \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{d}}}{2^{\mathrm{n}}}\)
Correct choice is therefore (a).

Answer:
(d) \(\frac { 5 }{ 3 }\)
Explanation:

Length of position vector of point = p = \(\sqrt{x^{2}+y^{2}}\)
p² = x² + y²
E(p²) = E(x² + y²) = E(x²) + E(y²)
Now x and y are uniformly distributes 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2 Probability density function of x = \(\frac { 1 }{ 1 – 0 }\) = 1
The probability density function of

Answer:(c)\(\int_{b}^{a}\) f(X)dX
Explanation:
If f(x) is the continuous probability density function of a random variable X then, p(a < x ≤ b) = p(a ≤ x ≤ b) = \(\int_{b}^{a}\)f (x)dx

Answer:
(b) \(\frac { 1 }{ 6 }\)
Explanation:
The given condition corresponds to sampling with replacement and with order. No 2 marbles have the same color i.e. Drawn 3 different marble. So total number of ways for picking 3 different marbles = 3! = 6. Probability of getting blue, green, red in order
= \(\frac { 10 }{ 60 }\) × \(\frac { 20 }{ 60 }\) × \(\frac { 30 }{ 60 }\) × 6
[Since 6 ways to get the marbles] = \(\frac { 1 }{ 6 }\)

Answer:(a) 3
Explanation:
E(X) = S(X_i × P_i)
Where X_i = number of tosses until we get successive HEAD or TAIL.
P_i = Probability that we get in X_i tosses. Total possibilities by tossing a coin two times i.e. HT, HH, TH, TT. Out of which favorable cases are HH, TT.
P(X = 2) = \(\frac { 2 }{ 4 }\)
Similarly for X = 3 only THH and HTT are favorable out of a total of 8 outcomes.
So, P(X = 3) = \(\frac { 2 }{ 8 }\)
Similarly, we can see that in every case we will have only 2 favorable cases and 2ⁿ sample space.
So for n^th toss probability = P(X = n) = 2/2ⁿ So the probability distribution table for X (the number of Tosses) is given below:

E(X) = 2 × \(\frac { 2 }{ 4 }\) + 3 × \(\frac { 2 }{ 8 }\) + 4 × \(\frac { 2 }{ 16 }\) + …… It is combination of AP and GP form. So multiplying E(X) by \(\frac { 1 }{ 2 }\) and subtracting from E(X)
E(X) = 2 × \(\frac { 1 }{ 2 }\) + 3 × \(\frac { 1 }{ 4 }\) + 4 × \(\frac { 1 }{ 8 }\) + …..
\(\frac { 1 }{ 2 }\) × E(X) = 2 × \(\frac { 1 }{ 4 }\) + 3 × \(\frac { 1 }{ 8 }\) + 4 × \(\frac { 1 }{ 16 }\) + …..
After solving both equations we get,
\(\frac { 1 }{ 2 }\) × E(X) = 1 + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 16 }\) + ……
\(\frac { 1 }{ 2 }\) × E(X) = 1 + \(\frac{1 / 4}{(1-1 / 2)}\) ⇒ E(X) = \(\frac { 6 }{ 4 }\) × 2 = 3

Answer:(c) 0.8
Explanation:
Let RA : Rain in the afternoon T > 25: Temperature more than 25°C
Let the desired probability = P(RA | T ≤ 25) = x
The tree diagram for this problem is given below:

Given P(RA) = 0.6 by rule of total probability P(RA)
= \(\frac { 1 }{ 2 }\) × 0.4 + \(\frac { 1 }{ 2 }\) × x = 0.6 ⇒ x = 0.8

Answer:(a) 1/p
Explanation:
The number of attempts to first succeed follows a geometric distribution. It is well known that the expected value in geometric distribution E(X) = 1/p. Where p is the probability of success in any one attempt.

Alternate Method:

Let X be the number of attempts to first success. Let p be the probability of success in any one attempt. Now the probability distribution table of X is given below:

E(X) = Σ xp(x) = 1 × p + 2 × (p – 1) p + 3 × (p – 1)² p + ……
This is the arithmetico-geometric series which can be solved as E(X) = 1/p

Answer:(c) \(\frac { 7 }{ 16 }\)
Explanation:
The tree diagram for the problem is given below:

Answer:(d) None of these
Explanation:
The number of permutations with ‘2’ in the first position = 19!
The number of permutations with ‘2’ in the second position = 10 × 18!
(Fill the first space with any of the 10 add numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
A number of permutations with ‘2’ in 3^rd position = 10 × 9 × 17!
(Fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with the remaining 17 numbers) and so on until ‘2’ is in 11^th place. After that, it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations that satisfy the given condition is 19! + 10 × 18! + 10 × 9 × 17! + 10 × 9 × 8 × 16! + … + 10! × 9!

Now the probability of this happening is given by \(\frac{19 !+10 \times 18 !+10 \times 9 \times 17 ! \ldots+10 ! \times 9 !}{20 !}\)
which is clearly not choices (a), (b) or (c)
∴ Answer is (d) none of these.

Answer:(b) 1.02 × 10⁶ × e^-20
Explanation:
The requests follow poisson distribution with α = 20 requests/hr
The observation period = △T = 45 minutes = \(\frac { 45 }{ 60 }\) = \(\frac { 3 }{ 4 }\)hr
So the parameter for the poisson distribution = λ = α △ T= 20 × \(\frac { 3 }{ 4 }\) = 15
The required probability P(X = 1) + P(X = 3) + P(X = 5)
= \(\frac{e^{-15} 15^{1}}{1 !}\) + \(\frac{e^{-15} 15^{3}}{3 !}\) + \(\frac{e^{-15} 15^{5}}{5 !}\) = 1.02 × 10⁶ × e^-20

Answer:(b) \(\frac{10}{12}\)
Explanation:
We know that:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(\overline{\mathrm{P}(\mathrm{A})}\) = 1 – P(A) and \(\overline{\mathrm{P}(\mathrm{B})}\) = 1 – P(B)
So, P(A) = ( 1 – \(\overline{\mathrm{P}(\mathrm{A})}\)) and P(B) = (1 – \(\overline{\mathrm{P}(\mathrm{B})}\))
P(A ∪ B) = (1 – \(\overline{\mathrm{P}(\mathrm{A})}\)) + (1 – \(\overline{\mathrm{P}(\mathrm{B})}\)) – P(A ∩ B)
= (1 – 1/3) + (1 – 1/3) – (1/2)
= (2/3) + (2/3) – (1/2)
= (4/3) – (1/2) = (5/6) = (10/12)

Answer:
(c) 1 – \(\frac{365.364 \ldots(365-r+1)}{365^{r}}\) – ^rC₂ . 365 . \(\frac{364.363 \ldots(364-(r-2)+1)}{364^{r-2}}\)
Explanation:
P(at least three people have the same birthday) = 1 – P (all have different b’days) – P(exactly two people have same b’day)
Now, P(all have different b’days) = \(\frac{365.364 \ldots(365-r+1)}{365^{r}}\)
P(exactly two people have same b’day) = \({ }^{r} C_{2} \cdot 365 \cdot \frac{364.363 \ldots(364-(r-2)+1)}{364^{r-2}}\)
[^rC₂ ways to choose who those two people with same b’day are, 365 ways to choose what the b’day is]
Now,
P (at least three people have the same birthday)
= 1 – \(\frac{365.364 \ldots(365-r+1)}{365^{r}}\)
= \({ }^{r} C_{2} \cdot 365 \cdot \frac{364.363 \ldots(364-(r-2)+1)}{364^{r-2}}\)
Which is option (c).

Answer:(c) 0.4
Explanation:
Let C denote computes science study and M denotes maths study. The tree diagram for the problem can be represented as shown below:

Now by the rule of total probability we total up the desired branches (✓) and get the answer as shown below:
p(C on Monday and C on Wednesday)
= p(C on Monday, C on Tuesday and C on Wednesday) + p(C on Monday, M on Tuesday, and C on Wednesday)
= 1 × 0.4 × 0.4 + 1 × 0.6 × 0.4 = 0.16 + 0.24 = 0.40

Answer:(a) 3
Explanation:
Given µ_X = 1, σ_x² = 4 => σ_x = 2 and µ_Y = -1.σ_Y is unknow
Given, p(X ≤ -1) = p(Y ≥ 2)
Converting into standard normal variates

Now since we know that in a standard normal distribution,
p(z ≤ -1) = p(z ≥ 1) … (ii)
Comparing (i) and (ii) we can say that
\(\frac{3}{\sigma_{\mathrm{y}}}\) = 1 ⇒σ_y = 3

Answer:(b) 0.468
Explanation:
It is given that, p(odd) = 0.9p(even)
Now since, Σp(x) = 1
∴ p(odd) + p (even) = 1
⇒ 0.9 p (even) + p (even) = 1
⇒ p(even) = \(\frac { 1 }{ 1.9 }\) = 0.5263
Now, it is given that p (any even face) is same i.e. p(2) = p(4) = p(6)
Now since,
p(even) = p(2) or p(4) or p(6)
= p(2) + p(4) + p(6)
∴ p(2) = p(4) = p(6)
= \(\frac { 1 }{ 3 }\) p (even) = \(\frac { 1 }{ 3 }\) (0.5263) = 0.1754
It is given that
p(even | face > 3) = 0.75
⇒ \(\frac{p(\text { even } \cap \text { face }>3)}{p(\text { face }>3)}\) = 0.75
⇒ \(\frac{\mathrm{p}(\text { face }=4,6)}{\mathrm{p}(\text { face }>3)}\) = 0.75
⇒ p (face>3) = \(\frac{p(\text { face }=4,6)}{0.75}\)
= \(\frac{p(4)+p(6)}{0.75}\)
= \(\frac{0.1754+0.1754}{0.75}\)
= 0.4677 ≃ 0.468

Answer:(a) pq+(1 – p) (1 – q)
Explanation:

The tree diagram of probabilities is shown above. From above tree, by rule of total probability, p(declared faulty) = pq + (1 – p) (1 – q)

Answer:(a) \(\frac{1}{625}\)
Explanation:
p(multiple of 10⁹⁶ | divisor of 10⁹⁹) = \(\frac{\mathrm{p}\left(\text { multiple of } 10^{96} \& \text { divisor of } 10^{99}\right)}{\mathrm{p}\left(\text { divisor of } 10^{99}\right)}\) = \(\frac{\mathrm{n}\left(\text { multiple of } 10^{96} \& \text { divisor of } 10^{99}\right)}{\mathrm{n}\left(\text { divisor of } 10^{99}\right)}\)
Since, 10 = 2 . 5
10⁹⁹ = 2⁹⁹. 5⁹⁹
Any divisor of 10⁹⁹ is of the form 2^a . 5^b where 0 ≤ a ≤ 99 and 0 ≤ b ≤ 99. The number of such divisors is given by (99+1) × (99 + 1) = 100 × 100.
So, no. of divisors of 10″ = 100 × 100.
Any number which is a multiple of 10⁹⁶ as well as divisor of 10⁹⁹ is of the form 2^a. 5^b where 96 ≤ a ≤ 99 and 96 ≤ b ≤ 99. The number of such combinations of 4 values of a and 4 values of b is 4 × 4 combinations, each of which will be a multiple of 10⁹⁶ as well as a divisor of 10⁹⁹.
∴ p(multiple of 10⁹⁶ | divisor of 10⁹⁹)
= \(\frac { 4 × 4 }{ 100 × 100 }\) = \(\frac { 1 }{ 625 }\)

Answer:(a) \(\frac{1}{3}\)
Explanation:
Let A be the event of head in one coin. B be the event of head in second coin. The required probability is
P(A ∩ B | A ∪ B) = \(\frac { P{(A ∩ B) ∩ (A ∪ B)} }{ P(A ∪ B }\) = \(\frac { P(A ∩ B }{ P (A ∪ B)}\)
P( A ∩ B) = p(both coin heads) = p(H , H) = \(\frac { 1 }{ 4 }\)
P(A ∪ B) = p(at least one head)
p(HH, HT, TH) = \(\frac { 3 }{ 4 }\)
So required probability = \(\frac { 1/4 }{ 3/4 }\) = \(\frac { 1 }{ 3 }\)

Answer:
(c) R ≥ 0
Explanation:
V(x) = E(x²) – [E(x)]² = R
where V(x) is the variance of x,
Since variance is σ²_x and hence never negative, R ≥ 0.

Answer:
(d) σ_y = aσ_x + b
Explanation:
Standard deviation is affected by scale but not by the shift of origin.
So, y_i = ax_i + b
⇒ σ_y = aσ_x
if a could be negative then σ_y = | a | σ_x is more correct since standard deviation cannot be negative)
Clearly, σ_y = aσ_x + b is false
So (d) is incorrect.

Answer:
(a) \(\frac { 1 }{ 5 }\)
Explanation:
The five cards are {1, 2, 3, 4, 5}
Sample space = 5 × 4 ordered pairs.
[Since there is a I^st card and II^nd card we have to take ordered pairs]
p(I^st card = II^nd card + 1)
= P{(2, 1), (3, 2), (4, 3), (5, 4)} = \(\frac { 4 }{ 5 × 4 }\) = \(\frac { 1 }{ 5 }\)

Answer:
(c) 0.5 and 1
Explanation:
The p.d.t of the random variable is

The cumulative distribution function F(x) is the probability up to x as given below:

So the correct option is (c).

Answer:(b) \(\frac { 5 }{ 12 }\)
Explanation:
If first throw is 1, 2 or 3 then sample space is only 18 possible ordered pairs. Out of this only (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5) and (3, 6) i.e. 9 out of 18 ordered pairs gives a Sum ≥ 6.
If first throw is 4, 5 or 6 then second throw is not made and therefore the only way Sum ≥ 6 is if the throw was 6. Which is one out of 3 possible. So the tree diagram becomes as follows:

From above diagram
p(sum ≥ 6) = \(\frac { 1 }{ 2 }\) × \(\frac { 9 }{ 18 }\) + \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 3 }\) = \(\frac { 15 }{ 36 }\) = \(\frac { 5 }{ 2 }\)

Answer:
(c) 17/(2e³)
Explanation:
Poisson formula for (P = x) given as \(\frac{\mathrm{e}^{-\lambda} \lambda^{\mathrm{x}}}{\mathrm{x} !}\)
α = 3 cars/minute
△T = 1 minute
So λ = α △ T = 3 × 1 = 3
Probability of observing fewer than 3 cars = P(X < 3) = P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{\mathrm{e}^{-3} 3^{0}}{0 !}\) + \(\frac{\mathrm{e}^{-3} 3^{1}}{1 !}\) + \(\frac{e^{-3} 3^{2}}{2 !}\) = \(\frac{17}{2 \mathrm{e}^{3}}\)
Option (c) is correct.

Answer:
Explanation:
Suppose you break a stick of unit length at a point chosen uniformly at random, then let the length of the shorter stick = x
x has uniform distribution in the interval [0,1/2] i.e. a = 0 and p – 1/2
In uniform distribution, E(x) = (β + α)/2
So the expected length of the shorter stick = E(x) = (1/2 + 0)/2 = 1/4 = 0.25

Explanation:
Sample space = 6⁴= 1296
6,6,6,4 ⇒ 4 ways (\(\frac { 4 ! }{ 3 !}\) = 4)
6,6,5,5 ⇒ 6 ways (\(\frac { 4 ! }{ 2!2!}\) = 6)
probability of sum to be 22 = \(\frac { 6 + 4 }{ 1296 }\) = \(\frac { x }{ 1296 }\)
⇒ x = 10</de

Explanation:
The tree diagram for the problem is shown below.

Required probability = \(\frac{{ }^{4} \mathrm{C}_{3} \cdot{ }^{6} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{4}}\) + \(\frac{{ }^{4} \mathrm{C}_{3} \cdot{ }^{6} \mathrm{C}_{0}}{{ }^{10} \mathrm{C}_{4}}\) = \(\frac { 24 }{ 210}\)
+ \(\frac { 1 }{ 210 }\) = \(\frac { 25 }{ 210 }\)
p = 0.1190
⇒ 100 p = 11.90

Explanation:
1 ≤ x ≤ 100
P(x is not divisible by 2, 3 or 5) = 1 – P(x is divisible by 2, 3 or 5)

Explanation:
It is given that A and B are mutually exclusive also it is given that A ∪ B = S which means that A and B are collectively exhaustive. Now if two events A and B are both mutually exclusive and collectively exhaustive, then P(A) + P(B) = 1 ⇒P(B) = 1-P(A)
Now we wish to maximize P(A) P(B) = P(A)(1-P(A))
Let, P(A) = x
Now P(A) (1 – P(A)) = x(1 – x) = x – x²
Say y = x- x²
\(\frac { dy }{ dx }\) = 1 – 2x = 0 ⇒ x = \(\frac { 1 }{ 2 }\)
= \(\frac { d^2y }{ dx^2 }\) = -2 < 0; (\(\frac { d^2y }{ dx^2 }\))_x = \(\frac { 1 }{ 2 }\) = -2 < 0
y has maximum at x = \(\frac { 1 }{ 2 }\)
y_max = \(\frac { 1 }{ 2 }\) – (\(\frac { 1 }{ 2 }\))² = 0.25

Explanation:
The p.d.t. for X¹, X², and X³ as given in the problem is shown below:

Given, Y = X₁X₂ ⊕ X₃
The required probability – p(Y = 0 | X₃ = 0)
= \(\frac{p\left(Y=0 \cap X_{3}=0\right)}{p\left(X_{3}=0\right)}\) …..(1)
Now, p{X₃ = 0) = \(\frac { 1 }{ 2 }\) (from p.d.t. of X₃) …(2)
p(Y = 0 ∩ X₃ = 0)
= p (X₁X₂ ⊕ X₃ = 0 ∩ X₃ = 0)
= P((X₁X₂ ⊕ 0 ) = 0 ∩ X₃ = 0)
= p(X₁X₂ = 0 ∩ X₃ = 0)
= p(X₁ = 0, X₂ = 0, X₃ = 0) + p(X₁ = 0, X₂ = 1, X₃ = 0) + p(X₁ = 1, X₂ = 0, X₃ = 0) + P(X₁ = 1, X₂ = 0, X₃ = 0)
= \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 8 }\)
So, p(Y = 0 ∩ X₃ = 0) = \(\frac { 3 }{ 8 }\) ….(3)
Now substituting (2) and (3) in (1), we get
The required probability = p(Y = 0|X₃ = 0) = \(\frac { 3/8 }{ 1/2 }\) = \(\frac { 3 }{ 4 }\) = 0.75

Explanation:

Explanation:

The tree diagram for the problem is given above. The desired output is Y. Now by rule of total probability P(output = Y) = 0.5 × 0.5 + 0.5 × 0.5 × 0.5 × 0.5 +… Infinite geometric series with α = 0.5 × 0.5 and r = 0.5 × 0.5 So p (output = Y)
= \(\frac { 0.5 × 0.5 }{ 1 -0.5 × 0.5 }\) = \(\frac { 0.25 }{ 0.75 }\) = \(\frac { 1 }{ 3 }\) = 0.33 (upto 2 decimal places)

Explanation:

P(losts > 100 hr) = 0.5 × 0.7 + 0.5 × 0.4 = 0.35 + 0.2 = 0.55

Explanation:
The Gaussian random variable is the same as the normal random variable.
So, the distribution of X is N(0, σ²)
For X, Median = Mean = Mode = 0 For Y = Max (X, 0)
Now half the data are below 0 and half the data above 0 for X. If we apply Y = Max(X, 0) now, all the negative values will become 0 and all the positive values will remain positive. So, the Median being positional average will not be affected and will remain at 0, since now half will be at 0 and a half will be positive. So, Median (Y) = 0.

Answer:(a) \(\frac { 4 }{ 5 }\)
Explanation:
Given that,
p(P) = \(\frac { 1 }{ 4 }\) …..(1)
p(P | Q) = \(\frac { 1 }{ 2 }\) …..(2)
p(Q | p) = \(\frac { 1 }{ 3 }\) …..(3)
p(\(\bar{P}\) | \(\bar{Q}\) = ?
First solve for p(Q) and p (P ∩ Q) from equation (2) and (3) as follows:
From equation (2)
p(P | Q) = \(\frac { p(P ∩ Q) }{ P(Q) }\) = \(\frac { 1 }{ 2 }\) …(4)
From equation (3)
p(Q | P) = \(\frac { p(P ∩ Q) }{ P(P) }\) = \(\frac { 1 }{ 3 }\)
⇒ p(P ∩ Q) = \(\frac { 1 }{ 3 }\) × p(P) = \(\frac { 1 }{ 3 }\) × \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 12 }\)
Now substitute in equation (4) and get
P(Q) = \(\frac { p(P ∩ Q) }{ 1/2 }\) = \(\frac { 1/12 }{ 1/2 }\) = \(\frac { 2 }{ 12 }\) = \(\frac { 1 }{ 6 }\)
So now we have p(P) = \(\frac { 1 }{ 4 }\)
p(Q) = \(\frac { 1 }{ 6 }\)
and p(P ∩ Q) = \(\frac { 1 }{ 12 }\) we need to find
p(\(\bar{P}\) | \(\bar{Q}\) = \(\frac{p(\bar{P} \cap \bar{Q})}{p(\bar{Q})}\)
= \(\frac{1-(P \cup Q)}{1-p(Q)}\) = 1 – \(\frac{[p(P)+p(Q)-p(P \cap Q)]}{1-p(Q)}\)
= \(\frac{1-\left[\frac{1}{4}+\frac{1}{6}-\frac{1}{12}\right]}{1-\frac{1}{6}}\) = \(\frac{\frac{2}{3}}{\frac{5}{6}}\) = \(\frac { 4 }{ 5 }\)
So, p(\(\bar{P}\) | \(\bar{Q}\) = \(\frac { 4 }{ 5 }\)

Answer:(b) Nβ
Explanation:
g_y(z) = ((1 – β) + βz)^N
If g_y(z) is expanded, we would get a bionomial distribution with n = N and p = β So the E(Y) = np = Nβ

Explanation:
Given, Poisson distribution λ = 5 We know that in Poisson distribution E (X) = V(X) = λ
So here E(X) = V(X) = 5 Now, we need E[(X + 2)²]
= E(X² +4X + 4)
= E(X²) + 4E(X) + 4
To find E(X²) we write, V(X) = E(X²) – (E(X))²
5 = E(X²) – 5² So, E(X²) = 5² + 5 = 30
Required value = 30+ 4 × 5 + 4 = 54

Explanation:
P(one of them wins in 3^rd trial)
= P(I^st trial is Tie) × P(II^nd trial is Tie) × P(one of them wins 3^rd trial)
P(Tie in any trial) = P(P = 1 and Q = 1) + P(P = 2 and Q = 2) + …. + P(P = 6 and Q = 6)
= \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) +\(\frac { 1 }{ 36 }\) = \(\frac { 1 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
P(one of them wins) = 1 = p(Tie) = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
So required probability = \(\frac { 1 }{ 6 }\) × \(\frac { 1 }{ 6 }\) × \(\frac { 5 }{ 6 }\) = \(\frac { 5 }{ 216 }\) = 0.023
(rounded to 3 decimal places)

Explanation:
The condition probability table given is

P(H_G) = 0.2, P(M_G) = 0.5, P(L_G) = 0.3 Drawing the tree diagram for HD we get,

From diagram, P(H_G ∩ H_D) = 0.2 × 0.4
P(H_D) = 0.2 × 0.4+ 0.5 × 0.1 + 0.3 × 0.01 = 0.133 Required probability,
P(H_G | H_D) = \(\frac { 0.2 × 0.4 }{ 0.133 }\) = 0.60 (rounding upto 2 decimal place)

Explanation:

p(2 elements out of 13 elements chosen randomly and independently) = (Number of MSB_‘0’ × Number of MSB_‘0’ + Number of MSB_‘1’ × Number of MSB_‘1’) / (Total × Total) = \(\frac{7 \times 7+6 \times 6}{13 \times 13}\) = \(\frac{49+36}{169}\) = \(\frac{85}{169}\) = 0.503

Explanation:
polynomial 3x² + 6xY + 3Y + 6 has only real roots
b² – 4ax ≥ 0
(6Y)² – 4(3) (3Y + 6) ≥ 0
Y² – Y + 2 ≥ 0
Y ∈ (-∞, – 1] ∩ [2, ∞)
⇒ Y ∈ [2 , 6)
Since y is uniformly distributed in (1, 6) probability distributed function f(Y) = \(\frac { 1 }{ 5 }\) 1 p(2 ≤ y < 6) = \(\int_{2}^{6}\)f(Y)dy = \(\frac { 1 }{ 5 }\) \([Y]_{2}^{6}\) = \(\frac { 4 }{ 5 }\) = 0.8

Explanation:

Fixed non-Zero
a_i = 0 or 1
x_i = 0 or 1
\(\sum_{i=1}^{n} a_{i} x_{i}\) value lies between 0 to n.
Total number cases = 2ⁿ

Answer:
(a) Both S₁ and S₂ are false.
Explanation:
S₂ : Cov(x,y) = E{|x – \(\bar{x}\)||y – \(\bar{y}\)|} is false
Case-I: If x > \(\bar{x}\) and y > \(\bar{y}\) then above is true.
Case-II: If x < \(\bar{x}\) and y < \(\bar{y}\) then above is true. Case-III: If x > \(\bar{x}\) but y < \(\bar{y}\) then above is false.
Case-IV: If x < \(\bar{x}\) but y > \(\bar{y}\) then above is false.
∵ Given expression is not always true. So we can conclude that it is false.
S₁: It is obviously false,
∵ True statement is
[E{(x – \(\bar{x}\))(y – \(\bar{y}\))}]² < var(x) . Var(y)
So both S₁ and S₂ are false.

Explanation:
Let, t = {lifetime of component} and μ = 2
Expected lifetime = \(\frac { 1 }{ μ }\)

Question 62.
A sender (S) transmits a signal, which can be one of the two kinds: H and L with probabilities 0.1 and 0.9 respectively, to a receiver (R). In the graph below, the weight of edge (u, v) is the probability of receiving v when u is transmitted, where u,v ∈ {H, L}. For example, the probability that the received signal is L given the transmitted signal was H, is 0.7.

If the received signal is H, the probability that the transmitted signal was H (rounded to 2 decimal places) is .
Answer:
Explanation:

Explanation:
n = 1000, p = 0.4, q = 0.6
It is a binomially distributed random variable.
So, S.D. = \(\sqrt{n p q}\)
= \(\sqrt{1000 \times 0.4 \times 0.6}\) = \(\sqrt{240}\) = 15.49

Answer:
(d) First QuesA and then QuesB. Expected marks 16.
Explanation:
X → Random variable which represents total marks record.
P(X) → Probability of getting those marks.

Now, if QuestionA is attempted first and it is correct.
Case-I:
E(x) = Σ(x) . P(x)
= 0.4 × 10 + 0.4 × 30 = 4 + 12 = 16
Case-II:
If QuestionB is attempted first and is correct.
E(x) – Σ(x) . P(x)
= 0.1 (20) + 0.4 (30)
= 2 + 12 = 14

So case-I is giving maximum expected marks. Hence option (d) is correct.

Answer:
Explanation:
There are 10 favorable ways to calculate the probability of red ball in 4^th trial.
(RFR)R = R or (BRR)R ≈ 1 way or (RRR)R ≈ 3 ways or (BBR)R ≈ 3 ways

Explanation:
A relation is said to be equivalence relation is

• Reflexive
• Symmetric, and
• Transitive

The union of two reflexive relations and two symmetric relations are reflexive and symmetric respectively. However, the union of two transitive relations need not be transitive. Therefore, the union of two equivalence relations need not be an equivalence relation.
Example:
Let R₁ and R₂ on set A = {1, 2, 3}
R₁ = {(1, 1), (2, 2,), (3, 3) (1, 2), (2, 1)} is an equivalence relation
R₂ = {(1, 1),(2, 2), (3, 3), (2, 3), (3, 2)} is an equivalence relation
R₁ ∪ R₂ = {(1, 1), (2, 2), (3, 3,), (1, 2), (2, 1), (2, 3), (3,2)} is not an equivalence relation, because (1, 2) & (2, 3) needs (1, 3) element to be in transitive relation.

Explanation: (a) Set A contains n elements. Every subset of A × A is a binary relation on set A.
∴ Number of binary relations on a set A with n elements 2^n²
(b) Number of one-to-one functions from a set A with m-elements to a set B with n – elements are nP_m So the number of one-to-one functions from a set A with n-elements to itself is nP_n = n!.

Explanation: The complement of an element x is x’ iff LUB of x and x’ is 1 (greatest element) and GLB of x and x’ is 0 (least element).
∴ The complement of the element ‘a’ in the lattice = {b, c, d, e}.

Explanation:
The relation has (1, 2) (2, 3) then add (1, 3) to the relation. Since relation have (2, 3) (3,4) then add (2, 4) to relation.
So the resultant relation is = {(1, 2), (2, 3), (3, 4), (5, 4), (1, 3), (2, 4)} Now the resultant relation have (1,2) (2, 4) then add (1, 4) to the relation
∴ {(1,2), (2,3), (3,4), (5,4), (1,3), (2,4), (1,4)}
So the transitive closure of the relation is = {(1,2), (2,3), (3,4), (5,4), (1,3), (2,4), (1,4)}

Answer: (c) At least one of sets S; is infinite
Explanation:
S = S₁ ∪ S₂ ∪ S₃ ∪ … ∪ S_n.
For S to be an infinite set, at least one of sets S_i must be infinite, since if all S_i were finite, then S will also be finite.

Answer: (b) 2^n²
Explanation:
Number of elements in A × A = n²
∴ A number of elements in the power set of A × A = 2^n².

Answer: (c) (goh)² = g² o h² for every g, h ∈ G.
Explanation:
(goh)² = (goh)o(goh)
Since group is abelian so it is commutative as well as associative.
= go (hog)h
= go (goh) oh
= (gog) o(hh) = g² o h²
So (goh)² = g² o h² for every g, h∈ G

Answer: (d) None of the above
Explanation:
A relation that is symmetric and transitive, need not be reflexive relation.

• R = { }; on the set A = {a, b}. The relation R is symmetric and transitive but not reflexive.
• R = {(a, a), (b, b)}; on the set A = {a, b}. The relation R is symmetric, transitive and also reflexive.

∴ A relation is transitive and symmetric relation but need not be reflexive relation.

Answer: (c) 8
Explanation:
If a set has n elements then its powers set has 2ⁿ elements.
Given set S = {(Φ), 1, (2, 3)}
Number of elements in S = 3
∴ The number of elements in powerset (S)
= 2³ = 8.

Answer: (d) <A, *> is a group but not an abelian group
Explanation:

• Closure Property: Multiplication of two non-singular matrices is also a non-singular matrix. Matrix multiplication over non-singular matrices follows closure properties.
• Associative Property: Multi-plication over any set of matrices is associative. (AB)C = A(BC) Where A, B, and C are non-singular matrices
• Identity Element: Identity matrix I am the identity element for matrix multiplication over matrices and I is non-singular.
• Inverse Element: For every non-singular matrix its inverse exists. So, for non¬singular matrices, inverse elements exist.
• Commutative: Matrix multiplication is not commutative
  AB ≠ BA
  Where A, B are non-singular matrices. Matrices multiplication is not commutative. So <A, *> is a group but not an abelian group

Answer: (a) A ∪ B
Explanation:
(A – B) ∪ (B – A) ∪ (A ∩ B)
Representing the above set using the Venn diagram as follows.

∴ By Venn diagram,
(A – B) ∪ (B – A) ∪ (A ∩ B) = A ∪ B

Answer: (b) 4
Explanation:
x = {2, 3, 6, 12, 24}
The Hasse diagram of (x, ≤) is

Therefore, the number of edges in the Hasse diagram = 4

Answer: (a) |X| = 1, |Y| =97
Explanation:
X and Y are set. The cardinalities of X and Y are |X| and |Y| respectively.
The number of functions from X to Y = (|Y|)^|X| Given that number of functions from X to Y = 97
∴ 97 = (|Y|)^|X|
So above implies that |X| = 1 and |Y| = 97

Answer: (c) The set of rational numbers form an abelian group under multiplication
Explanation:
0 is also a rational number and for 0, the inverse doesn’t exist under multiplication. Therefore, the set of rational numbers doesn’t form an abelian group under multiplication.

Answer: (c) f^-1(x,y) = \(\left(\frac{x+y}{2}, \frac{x-y}{2}\right)\)
Explanation:
f(x, y) = ((x + y), (x — y))
So let z₁ = x+ y ……(i)
z₂=x—y …(ii)
f(x, y) = (z₁, z₂)
So f^-1(z₁, z₂) = (x, y)
Adding (i) and (ii)
z₁ + z₂ = 2x
X = \(\frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\)
Subtracting (i) and (ii)
z₁ – z₂ = 2y
y = \(\mathrm{y}=\frac{\mathrm{z}_{1}-\mathrm{z}_{2}}{2}\)
So the inverse function of F is given by
f^-1(z₁ , z₂) = (x,y) = \(\left(\frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}, \frac{\mathrm{z}_{1}-\mathrm{z}_{2}}{2}\right)\)
Since z₁ and z₂ are just dummy variables so replacing z1 and z₂ by x and y respectively
f^-1(x,y) = \(\left(\frac{x+y}{2}, \frac{x-y}{2}\right)\)

Alternate Method:

Given f(x, y) = ((x + y), (x — y))
Take some random ordered pair say (2, 3) f(2, 3) ((2 + 3), (2 — 3)) = (5, —1)
Now the correct inverse must map (5. —1) back to(2,3).
Trying the options one by one we find that only option (e) maps (5, -1) to (2, 3).

Answer: (b) R is symmetric and not transitive
Explanation:
(i) Reflexive
A ∩ A = A ≠ Φ
So; (A, A) doesn’t belong to relation R.
∴ Relation R is not reflexive.

(ii) Symmetric
If A ∩ B = Φ then B ∩ A Φ is also true
∴ relation R is a symmetric relation.

(iii) Transitive
If A ∩ B = Φ and B ∩ C = Φ. it need not be true that A ∩ C = Φ.
For example:
A = {1, 2}, B = {3, 4}, C = {1, 5, 6}
A ∩ B = Φ and B ∩ C = Φ but
A ∩ C = {1} ≠ Ø
∴ Relation R is not transitive relation.

Answer: (c) The set of all strings over a finite alphabet Σ forms a group under concatenation.
Explanation:
In option (c). the set of all strings over a finite alphabet Σ doesn’t form a group under concatenation because the inverse of a string doesn’t exist with respect to concatenation.

Answer: (a) 15
Explanation:
Corresponding to every partition of the set 1, 2, 3, 4, there exists a unique equivalence relation. So we count every type of unordered partition of the set of 4 elements into one block, two-block, three-block, and four block partitions. as shown below:

Total 1+7+6+1=15
So the number of equivalence relations on the set {1, 2, 3. 4} is 15.

Answer: (b) n²
Explanation:
A is a finite set with n elements. The largest equivalence relation of A is the cross. product A × A and the number of elements in the cross product A × A is n².

Answer: (c) assertion (ii) is true but assertion (i) is not true
Explanation:
A relation is said to be an equivalence relation if the relation is
(i) Reflexive
(ii) Symmetric
(iii) Transitive
Reflexive and symmetric properties are both closed under ∪ & ∩.
Transitive property is closed under n but not u. So equivalence relations are closed under ∩ hut not ∪.
Therefore R₁ ∩ R₂ is an equivalence relation but R₁ ∪ R₂ is not necessarily an equivalence relation.

Answer: (c) n^m
Explanation:
The number of functions from an m element set to an n clement set is n^m.

Answer: (b) Neither reflexive nor irreflexive but transitive
Explanation:
The relation R doesn’t contain (4, 4), so R is not reflexive relation. Since relation R contains (1, 1), (2, 2), and (3, 3). Therefore, relation R is also not irreflexive.
That R is transitive, can be checked by systematically checking for all (a. b) and (b, c) in R, whether (a, e) also exists in R. So option (b) is correct.

Answer: Given partition of A is
π₁ = {(a, b, c), (d)}
(a) The ordered pairs of the equivalence relations induced by π₁ is
R = (a, b, c) × (a, b, c) ∪ (d) × (d)
R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c), (d, d)}
(b) The digraph of the above equivalence relation is as follows:

Answer:
Explanation:
Given a ≠ b ⇒ a * b ≠ b * a The contrapositive version of this is a*b = b*a ⇒ a = b
(a) Now since (A, *) is given to be a semigroup, * is associative
i. e. a * (a * a) = (a * a) * a
Hence a = a * a
(b) (a * b * a) * a = a * b * (a * a) = a * b * a
⇒ a*(a*b*a) = (a*a)*b*a = a * b * a
as a * a = a [proved in part a]
So (a * b * a) * a = a * (a * b * a)
Hence a * b * a = a
(c) (a * b * c) * (a * c) = a * b * (c * a * c) = a * b * c
(a * c) * (a * b * c) = (a * c * a) * b * c = a * b * c
So (a *b * c) * (a * c) = (a * c) * (a * b * c)
Hence a*b*c = a*c

Answer: (c) 2ⁿ²
Explanation:
The maximum number of elements in a binary relation on a set A with n elements = Number of elements in A × A = n²
Each element has two choices, either to appear on a binary relation or doesn’t appear on a binary relation.
∴ The number of binary relations = 2^n².

Explanation:
(a) According to Mr. X if xRy is in relation then yRx is also in relation because of symmetricity. Now because R is transitive, xRy and yRx implies xRx.
The flaw in Mr. X claim is that if xRy is present then only it implies yRx and hence xRx. However, if xRy is not present then according to symmetricity yRx is also not present. So xRx need not be present in relation. So it need not be reflexive.
For Example, Empty relation Φ is both symmetric and transitive. However, Φ is not reflexive relation.
(b) Examples of relations that are symmetric and transitive but not reflexive.
(i) Empty relation Φ is symmetric and transitive but not reflexive.
(ii) Relation R = {(a, b), (b, a), (a, a), (b, b)} over A = {a, b, c} is both symmetric and transitive but not reflexive.

Answer: (c) R is an equivalence relation having 2 equivalence classes
Explanation:
A relation R is defined as xRy iff (x + y) is even over a set of integers.
(x + y) is even iff
(i) both x and y are even
(ii) both x and y are odd
Therefore, relation R is equivalence relation because relation is
(i) Reflexive
x + x = 2x = even
So (x, x) belongs to R. So relation is reflexive.
(ii) Symmetric
If x + y = even then y + x is also even So relation is symmetric.
(iii) Transitive
If x + y = even and y + z = even
Then x + y + y + z = even + even
⇒ x + z + 2y = even
⇒ x + z = even – 2y
⇒ x + z = even
∴ Relation R is transitive.
So relation R is an equivalence relation which divides the set of integer into two equivalence classes: One is of all even and other is of odd integer.

Equivalence classes of R are
[0] = { ….. -6, -4, -2, 0, 2, 4, 6, …}
[1] = { ….. -7, -5, -3, -1, 1, 3, 5, 7, …}

Answer: (b) P(S) ∩ P(P(S)) = {Ø}
Explanation:
Φ always present in any powerset of a set and Φ is the only common element between P(S) and P(P(S))
∴ P(S) ∩ P(P(S)) = {Φ}

Answer: (b) R₁ and R₃ are equivalence relations, R₂ and R₄ are not
Explanation:
(I) Relation R₁(a, b) iff (a + b) is even over the set of integers.
(i) a + a = 2a which is even So (a, a) belongs to R₁
∴ R₁ is reflexive relation

(ii) If (a + b) is even, then (b + a) is also even
∴ R₁ is symmetric relation

(iii) If (a + b) and (b + c) are even then a + c = (a + b) + (b + c) – 2b = even + even — even = even
∴ R₁ is transitive relation.
Since R₁ is reflexive, symmetric and transitive so R₁ is an equivalence relation.

(II) RR₂(a, b) iff (a +b) is odd over set of integers,
(i) a + a = 2a which is not odd So (a, a) doesn’t belong to R₂
∴ R₂ is not reflexive relation Since R₂ is not reflexive, it is not an equivalence relation.

(III) R₃(a, b) iff a . b > 0 over set of non-zero relational numbers.
(i) a. a. > 0 for every non-zero rational number.
∴ R₃ is reflexive relation.
(ii) If a.b > 0 then b.a > 0
∴ R₃ is symmetric relation.
(iii) a.b > 0 and b.c > 0 ⇒ All a,b,c are positive or all a,b,c are negative.
So a.c > 0
∴ R₃ is transitive relation.
So R₃ is an equivalence relation.

(IV) R₄ (a, b) iff |a – b| ≤ 2 over the set of natural number
(i) |a – a| ≤ 2
0 ≤ 2
∴ R₄ is reflexive relation.

(ii) If |a — b| ≤ 2 then also |b – a| ≤ 2 ∴ R₄ is symmetric relation
(iii) If |a — b| ≤ 2 and |b — c| ≤ 2 then it is not necessary that |a — c| ≤ 2
Ex. |3 – 5| ≤ 2 and |5 — 7| ≤ 2 but |3 —7| ≤ 2
∴ R₄ is not transitive.
Since R₄ is reflexive and symmetric not transitive, so R₄ is not an equivalence relation.

Answer: (c) Both S₁ and S₂ are correct
Explanation:
S₁: Let A = set of integers
B = set of odd integers
C = set of even integers
A ∩ (B ∩ C) = Φ and Φ is finite set.
Therefore, S₁ is true.
S₂: Let two irrational number x and y are respectively \((1+\sqrt{2})\) and \((1-\sqrt{2})\).
So x + y= 1 + \(\sqrt{2}\) +1 – \(\sqrt{2}\)
= 2 which is rational number Therefore, S₂ is true. Since both S₁ and S₂ are true, option (c) is true.

Answer: (a) Only S₁ is correct
Explanation:
Given a function f: x → y and subsets E and F of A then we have f(E ∪ F) = f(E) ∪ f(F) and f(E ∩ F) ⊆ f(E) ∩ f(F)
Therefore S₁ is correct and S₂ is false.

Answer: (d) Transitive and symmetric
Explanation:
The empty relation on any set is always transitive and symmetric but not reflexive.

Answer: (a) does not form a group
Explanation:
Σ = {0, 1}
Σ* = {0,1}*
= {ε, 0, 1, 01, 10, 11, 000, …}
So (Σ*, •) is an algebraic system, where • (concatenation) is a binary operation.
So (Σ*, •) is a group if and only if the following conditions are satisfied.
1. • (Concatenation) is a closed operation.
2. • is an associative operation.
3. There is an identity.
4. Every element of Σ* has an inverse

Condition 1: * is a closed operation because for any ω₁,∈ Σ* and ω₂ ∈ Σ*, ω₁ . ω₂∈ Σ*
Condition 2: For any string x, y, z ∈ Σ*, x . (y . z) = (x . y) . z
So it is associative for example let
x = 01, y = 11, z = 00 then L.H.S. = x . (y.z)
= 01(1100) = 01(1100) = 011100 R.H.S. = (x.y).z
= (0111)00 = (0111)00 = 011100
Condition 3: The Identity is e or empty string because for any string ω ∈ Σ*,
εω = ωε = ω
Now, since ε ∈ Σ*, identity exists.
Condition 4: There is no inverse exist for Σ* because any string ω ∈ Σ*, there is no string ω^-1 such that ω . ω^-1 = ε = ω^-1 ω.
So Σ* with the concatenation operator for strings doesn’t form a group but it does form a monoid.

Answer: (d) P(x) = False for all x ∈ S such that a ≤ x and b ≤ x
Explanation:
If a ≤ x, since p(x) ⇒ p(y) whenever x ≤ y ∴ p(a) ⇒ p(x)
Now since p(a) = True, p(x) = cannot be false. ∴ (d) cannot be true.

Answer: (c) 2
Explanation:
The possible solution pairs are (a, a), (a, b), (a, e), (b, a), (b, b), (b, c), (c, a,), (c, b) and (e, c). Substitute them one by one in both equations and see which of them satisfies both the equations.
The given equations are:
(a × x)+(a × y) = C ……(i)
(b × x)+(c × y) = C ……(ii)
Substitute first (x. y) = (a, a)
LHS of equation (i) becomes (a × a) + (a × a) = a +a = b
Now RHS of equation (i) = e
Therefore LHS ≠ RHS. This means that (a, a) is not a solution pair. Similarly, try each of the remaining seven possible solution pairs. It will be found that only two pairs (b, c) and (c, b) will satisfy both equations (i) & (ii) simultaneously. Therefore choice (c) is correct.

Answer: (b) {(x, y) | y ≥ x and x,y ∈ {0, 1, 2, …..}}
Explanation:
S = {(x, y) | y = x + 1, and x, y ∈ (0, 1, 2, …}} = {(0, 1), (1, 2), (2, 3), (3, 4), …}
Now let T₁ be the reflexive closure of S.
T₁ = {(0, 0), (1, 1), (2, 2), (3, 3)…} ∪ {(0, 1), (1, 2), (2, 3), (3, 4)…}
= {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3), (3, 4) …}

Let T₂ be the transitive closure of S.
(0, 1), (1, 2) ∈ S ⇒ (0, 2) ∈ T₂
(0, 2), (2, 3) ∈ S ⇒ (0, 3) ∈ T₂
(0, 3), (3, 4) ∈ S ⇒ (0, 4) ∈ T₂
and so on…..
Also (1, 2), (2, 3) ∈ S ⇒ (1, 3) ∈ T₂
(1, 3), (3, 4) ∈ S ⇒ (1, 4) ∈ T₂
(1, 4), (4, 5) ∈ S ⇒ (1, 5) ∈ T₂
and so on….
∴ T₂ = {(0, 1), (0, 2), (0, 3), …, (1, 2), (1, 3), (1, 4), …}

Now the reflexive, transitive closure of S will be T₃ = T₁ ∪ T₂ = {(0, 0),(0, 1), (0, 2),…, (1,1), (1, 2), (1, 3),…,(2, 2) (2, 3), (2, 4),…}.
Option (b) is correct.

Answer: (c) power(2, (n² + n)/2)
Explanation:
In a symmetric matrix, the lower triangle must be the mirror image of the upper triangle using the diagonal as a mirror. Diagonal elements may be anything. Therefore, when we are counting symmetric matrices we count how many ways are there to fill the upper triangle and diagonal elements. Since the first row has n elements, second (n -1) elements, third row (n – 2) elements, and so on up to the last row, one element.
Total number of elements in diagonal + upper triangle
= n + (n – 1) + (n – 2) + … + 1 = n(n + 1)/2
Now, each one of these elements can be either 0 or 1. So total number of ways we can fill these elements is
\(2 \frac{n(n+1)}{2}\) = power (2, (n² + n)/2)
Since there is no choice for lower triangle elements the answer is power (2, (n2 + n)/2) which is choice (c).

Answer: (c) 25
Explanation:

According to inclusion – exclusion formula:
| PL ∪ DS ∪ CO | = | PL | + | DS | + | CO | – | PL ∩ DS|- |PL – CO|-|DS ∩ CO| + |PL ∩ DS ∩ CO| PL = students who have taken programming. DS = Students who have taken Data structures. CO = Students who have taken Computer Organization.
So, the number of students who have taken atleast 1 of the 3 courses is given by:
= 125 + 85 + 65- 50- 35- 30 + 15 = 175
Therefore, the number of students who have not taken any of the 3 courses is = Total students – students taking at least 1 course = 200 – 175 = 25

Answer: (c) R₁R₂={1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
Explanation:
R₁ = {(An element x in A, An element y in B) if x + y divisible by 3}
R₁ = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)}
R₂ = {(An element x in B, An element y in C) if x + y is even but not divisible by 3}
R₂ = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)}
So, R₁ o R₂ = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}.

Answer: (d) c e a b
Explanation:
Step 1:
By looking at the row for e, we see that it is a copy of the column headers. So e must be the identity element. Since right identity and left identity elements must both be the same. The column corresponding to e must be a copy of the row headers. We can now say that the operation table is:

Step 2:
From the table above we see that a*c = e
∴ c*a must also be = e (if a is the inverse of c, then c is the inverse of a)
Now the operation table looks like

Step 3:
The blank in the second column must he c (since in a Cayley table, every row and every column is a unique permutation of the row and column headers).
Now the operation table looks like

Step 4:
Now the blanks in the third row can be filled as a, e or e, a. Let us try each one in turn. If we fill a, e in the third row the operation table will look like

Now the blank in the fourth row and third column must be filled by e. However, this is not possible since e is already entered in the fourth row and second column. Therefore filling a, e in third-row blanks is wrong. So let us try filling the third-row blanks with e, a Now the operation table looks like

Now the blanks in the fourth row have to be filled with a, b The final operation table looks like

which is consistent with all the rules of a Cayley Table. The last row of this table is c, e, a, b Therefore the correct answer is (d).

Answer: (A) {1}
Explanation:
The hasse diagram of the given post is:

In a complete lattice L, every nonempty subset of L has both LUB and GLB. Now it is necessary to add {1} since GLB of {1,2} and {1, 3, 5} is {1}. The hasse diagram now becomes:

Now the above hasse diagram represents a complete lattice since every nonempty subset has both LUB and GLB. Therefore adding {1} is not only necessary but it is also sufficient to make the given lattice a complete lattice. Therefore the correct choice is (a).

Answer: (a) X = Y
Explanation:
X = (A-B) – C
= (A ∩ B’) – C
= (A ∩ B’) ∩ C’
= AB’C’
Y= (A-C)-(B-C)
= (A ∩ C) – (B ∩ C’)
= (AC’) – (BC’)
= (AC’) ∩ (BC’)’
= (AC’) ∩ (B’ + C)
= (AC’) . (B’ + C)
= AC’B’ + AC’C
= AC’B’ (Since C’C = 0)
= AB’C’ (commutative property)
∴ X =Y

Answer:(b) a lattice but not a distributive lattice
Explanation:

The poset [{a, b, c, d, e}, ≤ ] is a lattice (since every pair of elements has LUB and GLB) but it is not a distributive lattice. Because distributive lattice satisfies the following conditions. For any x, y, z,
x ∧ (y v z) — (x ∧ y) v (x ∧ z)
x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)
Where ∧ and ∨ are meet and join operations but for given poset [{a, b, c, d, e} ≤]
b ∧ (c ∨ d) = b ∧ a = b
(b ∧ c) ∨ (b ∧ d) = e ∨ e = e
So it is not distributive. (Also, element b has 2 complements c and d, which is not possible in a distributive lattice, since, in a distributive lattice, complement if it exists, is always unique).

Answer: (c) 4 and 13
Explanation:
The set S = {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15.
The identity element for this group is e = 1 since,
∀x ∈ S, 1 . x mod 15 = x
Now let the inverse of 4 be 4^-1.
Now (4 . 4^-1) mod 15 = e = 1
Since (4 . 4) mod 15=1
∴ 4^-1 = 4 (This inverse is unique)
Similarly let the inverse of 7 be 7^-1
(7 . 7^-1) mod 15=1
putting each element of set as 7^-1 by trial and error we get
(7 . 13) mod 15 = 91 mod 15=1
∴ 7^-1 = 13
So, 4^-1 and 7^-1 are respectively 4 and 13. The correct choice is (c).

Answer: (c) R ∩ S is an equivalence relation
Explanation:
R ∩ S is an equivalence relation as can be seen from proof given below.
Let ∀x ∈ A (x, x) ∈ R and (x, x) ∈ S (since R and S are reflexive)
∴ (x, x) ∈ R n S also ∴ R ∩ S is reflexive.
Now, (x, y) ∈ R ∩ S
⇒ (x, y) ∈ R and (x, y) ∈ S
⇒ (y, x) ∈ R and (y, x) ∈ S
(Since R and S are symmetric)
⇒ (y,x) ∈ R ∩ S
∴ (x, y) ∈ R ∩ S
⇒ (y, x) ∈ R ∩ S
R ∩ S is therefore symmetric Now consider
(x, y) and (y, z) ∈ R ∩ S ⇒ (x, y) and (y, z) ∈ R and (x, y) and (y, z) ∈ S ⇒ (x, z) ∈ R and (x, z) ∈ S
(Since R and S are transitive)
⇒ (x, z) ∈ R ∩ S
∴ R ∩ S is transitive also. Since R ∩ S is reflexive, symmetric, and transitive.
∴ R ∩ S is an equivalence relation.
Note: A similar argument cannot be made from R ∪ S.

Answer: (b) f should be onto but g need not be onto
Explanation:
Consider the arrow diagram shown below:

h(a) = f . g(a) = α
h(b) = f . g(b) = β
Here f is onto but g is not onto, yet h is onto. As can be seen from the diagram if f is not onto, h cannot be onto.
∴ f should be onto, but g need not be onto.
∴ The answer is (b).

Answer: (a) a group
Explanation:
(i) The set H is closed since the multiplication of upper triangular matrices will result only in the upper triangular matrix.
(ii) Matrix multiplication is associative, i.e.
A * (B * C) = (A * B) * C.
(iii) Identity element is

and this belongs to H as I is an upper triangular as well as a lower triangular matrix.

(iv) If A ∈ H, then | A | = abc. Since it is given that abc ≠ 0, this means that |A | ≠ 0 i.e. every matrix belonging to H is non-singular and has a unique inverse.
∴ the set H along with matrix multiplication is a group.

Answer: (c) h is onto ⇒ g is onto
Explanation:
Given, h = g(f(x)) = g.f
Consider the following arrow diagram:

From the above diagram it is clear that g is not onto ⇒ h = g.f is also not onto, since the co-domain of g is the same as the co-domain of g.f. The contrapositive version of the above implication is h is onto ⇒ g is onto which also has to be true since direct = contrapositive. So option (c) is true.

Answer: (b) n+1
Explanation:
The way C is defined in the question contains only comparable subsets of A. i.e. the set C is the set of all comparable subsets of set A. Such a set is called a chain. Consider an example with set A = {1, 2} Subsets are Φ, {1}, {2} and {1, 2} Maximum cardinality of collection of distinct subsets is {Φ, {1} and {1, 2}} i.e. 3. So with n elements, the maximum cardinality is n + 1.

Answer: (d) None of these
Explanation:
I. f(x) = x + y – 3
x + a – 3 = x = a + x – 3
so a = 3
Now 3 is unique, and 3 ∈ N So I has identity.
II: f(x) = max(x, y)
max(x, a) = x = max(a, x)
In N, the only value of a which will satisfy the above equation is a = 1.
Since 3 is unique, and 3 ∈ N So II has an identity.

III. f(x) = x^y
x^a = x = a^x
Now x^a = x ⇒ a = 1, but x= a^x has no solution for a in the set N.
So III has no identity.
So only I and II have an identity.

Answer: (d) 2^xyz
Explanation:
Given | X | = x, | Y | = y and | Z | = z W = X × Y
So | W| = xy
| E | = 2|W| =2^xy
So the number of functions for Z to E = | E | ^|z|
= (2^xy)^z = 2xyz

Answer: (c) 3 does not have an inverse
Explanation:
Let A = {1, 2, 3, 5, 7, 8, 9}
Construct the table for any x, y ∈ A such that x * y = (x.y) mod 10

We know that 0 ∉ A. So it is not closed. Therefore, (a) is true.
The identity element = 1
∴ (2 . 2^-1) mod 10 = 1
From the table, we see that 2^-1does not exist. Since, (3 . 7) mod 10 = 1
∴ 7 is the inverse of 3 and 7 ∈ A.
(c) is false.
(d) is true since 8 does not have an inverse.

Answer: (a) Neither a Partial Order nor an Equivalence Relation
Explanation:
(x, y) R (u, v) iff x < u and y > v
(x, x) (x, x) since x / x and x / x
So R is not reflexive,
∴ R is neither a partial order, nor an equivalence relation.

Answer: (c) X = Y
Explanation:
Consider the following Venn diagram for X = (E ∩ F) – (F ∩ G)

Y = (E -(E ∩ G)) – (F – F)

So, X = Y
or alternatively the solution can be obtained from boolean algebra as follows:
X = (E ∩ F) – (F ∩ G)
= EF – FG
= EF ∩ (FG)’
= EF . (F’ + G’)
= EFF’ + EFG’
= EFG’
Similarly, Y = (E -(E ∩ G)) – (E – F)
= (E – EG) – (E . F’)
= E . (EG)’ – EF’
= E . (E’ + G’) – EF’
= EG’ – EF’
= EG’ . (EF’)’
= EG’ . (E’ + F)
= EE’ G’ + EFG’
= EFG’
Therefore X = Y

Answer: (b) 3n
Explanation:
The problem can be solved by considering the cases m = 4 and m = 5 etc.
Let, m = 4
S = {1, 2, 3, 4}
n = number of 3 element subsets
= ⁴C₃ = ⁴C₁ = 4
n = 4
The 4 subsets are {1, 2, 3}, {1, 2, 4}, {1, 3, 4} and {2, 3, 4}
f(1) = number of subsets having 1 as an element = 3
f(2) = number of subsets having 2 as an element = 3
f(3) = 3 and f(4) = 3
∴ \(\sum_{i=1}^{4} f(i)\) = 3 + 3 + 3 + 3=12

Both choice (a) and choice (b) are matching the answer since
3m = 3n = 12
Now let us try m = 5
S = {1, 2, 3, 4, 5}
n = number of 3 element subsets = ⁵C₃ = 10
∴ n = 10
The 10 subsets are {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}
f(1) = f(2) = f(3) = f(4) = f(5) = 6
\(\sum_{i=1}^{5} f(i)\) = 6 + 6 + 6 + 6 + 6 = 30

Clearly 3m = 3 × 5 = 15{is not matching Σf(i)} But
3n = 3 × 10 = 30 {is matching Σf(i) = 30}
∴ 3 n is the only correct answer.
The correct choice is (b).
The problem can also be solved in a more general way as follows:

\(\sum_{i=1}^{m} f(i)\) = f(1) + f(2) + ……+ f(m)
Since, f(1) = f(2) =…….= f(m) = ^(m-1)C₂
= \(\frac{\mathrm{m} \times(\mathrm{m}-1) \times(\mathrm{m}-2)}{2}\)
= \(\frac{3 \times \mathrm{m} \times(\mathrm{m}-1) \times(\mathrm{m}-2)}{1 \times 2 \times 3}\)
= 3 × ^mC₃
Since n = Number of there elements subsets of a set of m elements = ^mC₃
Therefore, \(\sum_{i=1}^{m} f(i)\) = 3 × ^mC₃ = 3n

Answer: (b) II only
Explanation:
P △ Q = PQ’ + P’Q
where △ is symmetric difference between P and Q.
I. LHS = P △ Q ∩ R = P△(QR) = P(QR)’ + P'(QR) = P(Q’ + R’) + P’QR
RHS = (P △ Q) ∩ (P △ R) = (PQ’ + P’Q) (PR’ + P’R) = PQ’R’ + P’QR
LHS ≠ RHS So statement I is false.

II. LHS = P ∩ Q △ R = P(QR’ + Q’R)
= PQR’ + PQ’R
RHS = (P ∩ Q) △ (P ∩ R) = PQ △ PR = PQ(PR)’ + (PQ)’ PR = PQ(P’ + R’) + (P’+Q’)PR = PQR’ + PQ’R
LHS = RHS
So statement II is true.

Answer: (b) 33
Explanation:
We know that, number of elements divisible by number ‘x’ in set of integer (from 1 to X) = \(\frac{X}{x}\)
So, according to inclusion exclusion formula:
n(2 or 3 or 5) = n(2) + n(3) + n(5) – n(2,3) – n(2,5) – n(3,5) – n (2,3,5))
No’s divisible by 2 in X = \(\left[\frac{123}{2}\right\rfloor\) = 61
No’s divisible by 3 in X = \(\left[\frac{123}{3}\right\rfloor\) = 41
No’s divisible by 5 in X = \(\left[\frac{123}{5}\right\rfloor\) = 24
No’s divisible by 2 and 3 = \(\left.\mid \frac{123}{2 \times 3}\right\rfloor\) = \(\left[\frac{123}{6}\right\rfloor\) = 20
No’s divisible by 3 and 5 = \(\left.\mid \frac{123}{3 \times 5}\right\rfloor\) = \(\left[\frac{123}{15}\right\rfloor\) = 8
No’s divisible by 2 and 5 = \(\)
No’s divisible by 2 and 3 and 5
The problem can be solved by considering the cases m = 4 and m = 5 etc.
= \(\left[\frac{123}{2 \times 3 \times 5}\right\rfloor\) = \(\left[\frac{123}{30}\right\rfloor\) = 4
So, n(2 or 3 or 5) = 61 + 41 +24 — 20 — 8-12 + 4 = 90
n(2 or 3 or 5) = Total number – n(2 or 3 or 5) = 123-90 = 33

Answer: (b) n² and n
Explanation:
Let S be a set of n elements say {1, 2, 3,…, n}. Now the smallest equivalence relation on S must contain all the reflexive elements {(1, 1), (2, 2), (3, 3),…, (n, n)} and its cardinality is therefore n. The largest equivalence relation on S is S × S, which has a cardinality of n × n = n².
∴ The largest and smallest equivalence relations on S have cardinalities of n² and n respectively. The correct choice is (b).

Answer: (c)
Explanation:
A partition P₁ is called a refinement of the partition P₂ if every set in P₁, is a subset of one of the sets in P₂.
π₄ is a refinement of π₂,π_3, and π₁
π₂ and n₃ are refinements of π₁
π₂ and π₃ are not comparable since neither is a refinement of the other.
So the poset diagram for (S’, <) would

Which is choice (c).

Answer: (d) (iii) and (iv)
Explanation:
In this problem
a % b means the mod function (i.e. residue when a is divided by b).
a/b means integer division (i.e. quotient when a is divided by b)
(i) (101,22):
101% 10 ≤ 22% 10 ⇒ 1 ≤ 2 which is true. (101/10, 22/10) = (10, 2) ∈ P need to check IS (10, 2) ∈ P?
10% 10 ≤ 2% 10 ⇒ 0 ≤ 2 which is True.
Then (10/10, 2/10) = (1, 0) fails since 1 < 0.
∴ (101, 22) ∉ P.

(ii) (22, 101)
22% 10 ≤ 101% 10 ⇒ 2 ≤ 1 is False.
∴ (22, 101)∉ P

(iii) (145, 265)
145% 10 ≤ 265% 10 ⇒ 5 ≤ 5 is true and (145/10, 265/10) = (14, 26) e P has to be checked.
Now consider (14, 26).
14% 10 ≤ 26% 10 ⇒ 4 ≤ 6 is true and (14/10, 26/10) = (1, 2) ∈ P has to be checked. Now consider (1, 2)
1% 10 ≤ 2% 10 ⇒ 1 ≤ 2 is true and (1/10, 2/10) = (0, 0) ∈ P which is given to be true. Therefore (145, 265) ∈ P.

(iv) (0, 153):
0% 10 ≤ 153% 10 ⇒ 0 ≤ 3 is true.
Then (0/10, 153/10) = (0, 15) should be in P. (0, 15):
0% 10 ≤ 15% 10 ⇒ 0 ≤ 5 is true.
Then (0/10, 15/10) = (0, 1) should be in P. (0, 1):
0% 10 ≤ 1% 10 ⇒ 0 ≤ 1 is true.
Then (0/10, 1/10) = (0, 0) should be in P. It is given that (0, 0) ∈ P Therefore (0, 153) ∈ P.
So (iii) and (iv) are contained in P.

Answer: (d) U
Explanation:
The given set theory expression can be converted into equivalent boolean algebra expression as follows:
(p∩q∩r) ∪ (p^c ∩ q ∩ r) ∪ q^c ∪ r^c
= p.q.r + p’.q.r + q’ + r’
= qr (p + p’) + q’ + r’
= qr + q’ + r’
= (q + q’) . (r + q’) + r’
= r + q’ + r’
= r + r’ + q’
= 1 + q’
= 1 = U

Answer: (a) (i) and (iv) only
Explanation:
In lattice, every pair in the Hasse diagram must have Least Upper Bound (LUB) and Greatest Lower Bound (GLB). In figure

For element d, e LUB (d, e) = {b, c} but LUB must be unique i.e. only 1 element present in LUB. Same for GLB (b, c) = {d, e} Hence not lattice.

GLB (b, d) = {c, e} but GLB must be unique i.e. only 1 element present in GLB. Hence not lattice. So, (ii) and (iii) are not lattice

Answer: (a) Commutativity
Explanation:
Group properties are closure, associativity existence of identity, and existence of inverse for every element. Commutativity is not required for a mathematical structure to become a group.

Answer: (d) R is neither symmetric nor antisymmetric
Explanation:
Given, R {(x, y), (x, z), (z, x), (z, y)) on set {x, y, z}, here (x, y) ∈ R and (y, x) ∉ R.
∴ R is not symmetric
Also(x, z)ER and(z, x) ∈ R.
∴ R is not an autism metric.
R is neither symmetric nor anti-symmetric.

Answer: (c) c, d are generators
Explanation:
If an element is a generator, all elements must be obtained as powers of that element.
Try a, b, c, d one by one to see which are the generators.
a = a
a² = a.a = a
a³ = a.a² = a.a = a and so on.
∴ a is not the generator,
b = b
b² = b.b = a
b³ = b.b² = b.a = b
b⁴ = b.b³ = b.b = a and so on

∴ b is not the generator c = c
c² c.c = b
c³ = c.c² = c.b = d
c⁴ = c.c³ = c.d = a
Since all of a, b, c, d have been generated as powers of c

∴ c is a generator of this group.
Similarly,
d = d
d² = d.d = b
d³ = d.d² d.b = c
d⁴ = d.d³ = d.c a
∴ d is the other generator.

Answer: (c) 2²⁰
Explanation:
A number of reflexive relations on a set with n elements = 2^{n² – n}.
Here n = 5. So, answer is 2^{5² – 5} = 2²⁰.

Answer: (a) a group
Explanation:
The structure ({n, n^th roots of unity}, ×) always forms a group. When n = 3 we get the set ({1, w, w²}, ×) which must also form a group.
or
The fact that ({1, w, w²}, × ) forms a group can also be seen by the fact that it satisfies the four group properties.

From the above operation table, we can see that the given operation is closed, on the given set.
2. Associative: “*” is an associative operation.
3. Identity: From the operation table, we can see that the identity element, is “1”.
4. Inverse: From the operation table, we can see that the inverse of 1 is 1, the inverse of w is w² and the inverse of w² is w.
So, ({1, w, w²}, *) is a group.
To be a ring, integral domain, or field, we need two binary operations to be specified, whereas here we have only one operation given. So, choice (a) is correct.

Answer: (c) 2ⁿ– 2
Explanation:
Let n = 2. There are only 2 onto functions as shown below:

For, n = 2
Option (a) 2ⁿ = 2² = 4
Option (b) 2ⁿ – 1 = 2² – 1 = 3
Option (c) 2ⁿ — 2 = 2² — 2 = 2
Option (d) 2(2ⁿ — 2) = 2(2² – 2) = 4
So only option (c) gives correct answer.
Alternate method:
The number of onto functions from a set A with m elements to set B with n elements where n < m is given by
n^{m – n}C₁(n- 1)^m+ ⁿC₂(n —2)^m … + ⁿC_{n- 1} 1^m
Here, m = n = 2
So number of onto functions = 2ⁿ—²C₁ 1^m = 2ⁿ – 2
which is choice (c).

Answer: (a) Commutative but not associative
Explanation:
x ⊕ y = x² + y²
y ⊕ x = y² + x²
As ‘+’ sign in commutative so x² + y² is equal to y² + x² so x ⊕ y is commutative.
Now check associativity
x ⊕ (y ⊕ z) = x ⊕ (y² + z²)
= x² + (y² + z²)²
= x² + y⁴ + z⁴ + 2y²z²
(x ⊕ y) ⊕ z = (x² + y²) ⊕ z
= (x² + y²)² + z²
= x⁴ + y⁴ + 2x²y² + z²
x ⊕ (y ⊕ z) t- (x ⊕ y) ⊕ z
So not associative Option (a) is correct.

Explanation:
f : {0, 1}⁴ → {0,1} ⇒ S is the set of all functions from a 16 element set to a 2 element set.
| S | =2¹⁶
N = Number of functions from S to 2 element set {0, 1} = 2²¹⁶ ]
N = 22
∴ log log N = log log 22’6 = log2²¹⁶ = 16

Answer: (a) Both S₁ and S₂ are true
Explanation:
Given the following details:
S = {1,2,3,4,. ..2014}
U < V if the minimum element in the symmetric difference of the two sets is in U.
S₁: There is a subset of S that is larger than every other subset.
S₂: There is a subset of S that is smaller than every other subset.
S₁ is true since Φ (which is a subset of S) is larger than every other subset.
i.e. ∀u v < Φ is true since the minimum element in v △ Φ = v, is in v.
(Note: v △ Φ = uΦ + v’Φ = v)
S₂ is also true since S (which is a subset of S) is smaller than every other subset, i.e. ∀u S < v is true since the minimum element in S △ v = v’, is in <S.
(Note: S △ v = Sv’ + S’v = lv’ + Ov =v’)
∴ Both S1 and S2 are correct.

Answer: (d) For any subsets S and Tof Y, f^-1(S ∩ T) = f ^-1(S) ∩ f^-1(T)
Explanation:
Example:

Let A = {1, 2}, B = {3, 4}
(a) |f (A ∪ B)| = |f(A) |+ |f(B)|
⇒ 3 = 2 + 2 is false

(b) f (A ∩ B) = f (A) ∩ f (B)
⇒ Φ = {a} is false

(c) |f(A ∩ B)| =min{|f(A)|, |f(B)|}
⇒ 0 = min (2, 2) = 2 is false

(d) Let S = {a, b}, T- {a, c}
f^-1 (S ∩ T)=f^-1(S)∩ f^-1(T)
⇒ {1, 3} = {1, 2, 3} n {1, 3, 4}
⇒ {1, 3} = {1, 3} is true

Explanation:
Order of subgroup divides the order of group (Lagrange’s theorem). So the order of L has to be 1, 3, 5, or 15. As it is given that the subgroup has at least 4 elements and it is not equal to the given group, therefore the order of the subgroup can’t be 1, 3, or 15. Hence it is 5.

Answer: (b) Only Q and R are true
Explanation:
Since it is given that ∀i f(f(i)) = i
It means f . f = I i.e. f = f’^-1 That is f is a symmetric function.
Statement P means that every symmetric function is an identity function which is false since there are many symmetric functions other than the identity function.

Example: {(0,1), (1,0), (2, 3), (3,2),…(2012,2013),(2013, 2012), (2014, 2014)} is a symmetric function but not the identity function.

Statement Q means that in every such symmetric function at least one element is mapped to itself and this is true since there is an odd number of elements in the set {0, 1, 2, … 2014}.

Statement R means that every symmetric function is onto which is true since it is impossible to make an into function symmetric. See the below diagram for an into function shown with only 3 elements:

In the above function note that (2, 1) exists in the function but not (1,2) and so it is not symmetric.

Explanation:
(i) e is identity element
(ii) x*x = e, so x = x^-1.
(iii) y*y = e, so y = y^-1.
(iv) (x*y)*(x*y) = e, so x*y = (x*y)^-1 and (y*x)*(y*x) = e, so y*x = (y*x)^-1.
Now (x*y)*(y*x) = x*(y*y)*x = x*e*x = x*x = e
So, (x*y)^-1. = (y*x)
From (i) and (ii) we get x*y* = y*x
There are only 4 distinct elements possible in this group
1. e
2. x
3. y
4. xy
All other combinations are equal to one of these four as can be seen below:
yx = xy (already proved)
xxx = xe = x
xyy = xe = x
xxy = ey = y
xyx = xxy = y
and so on…….
So the group is G = {e, x, y, x*y
⇒ |G| = 4

Answer: (a) \(\frac{h(x)}{g(x)}\)
Explanation:
g(x) = 1 – x, h(x) = \(\frac{x}{x-1}\)
\(\frac{g(h(x))}{h(g(x))}\) = \(\frac{g\left(\frac{x}{x-1}\right)}{h(1-x)}\) = \(\frac{1-\frac{x}{x-1}}{\frac{1-x}{1-x-1}}\) = \(\frac{x-1-x}{\frac{x-1}{\frac{1-x}{-x}}}\) = \(\frac{\frac{-1}{x-1}}{\frac{1-x}{-x}}\) = \(\frac{x}{(x-1)(1-x)}\) = \(\frac{-x}{(1-x)^{2}}\)
\(\frac{h(x)}{g(x)}\) = \(\frac{x}{\frac{x-1}{1-x}}\) = \(\frac{x}{(1-x)(x-1)}\) = \(\frac{-x}{(1-x)^{2}}\)
∴ \(\frac{g(h(x))}{h(g(x))}\) = \(\frac{h(x)}{g(x)}\)

Answer: (c) 1, 2, and 3 only
Explanation:
A = {5, {6}, {7}}
1. Φ ∈ 2^A is true. Any power set contains an empty set.
2. Φ ⊆ 2^A is true. Empty set is subset of any set.
3. {5, {6}}e 2^A is true, power set of A contain {5, {6}} as a 2 element subset of A.
4. {5, {6}} ⊆ 2^A is false. Power set of A not contain 5 and {6} as elements.
So {5, {6}} can not be subset of 2^A.

Answer: (d) \(\frac { 1 }{ 5 }\) < p_r < 1
Explanation:
L has 5 elements
L³ has all ordered triplets of elements of L
⇒ L³ contain 5 × 5 × 5 = 5³ = 125 elements.

If q, r, s are chosen, then only it will violate the distributive property. Number of ways to choose q, r, s in any triplet order = 3! = 3 × 2 × 1 = 6.
∴ p(satisfying distributive property)
= 1 – p(violate the distributive property)
= 1 – \(\frac{6}{5^{3}}\) = 0.952 which is between \(\frac{1}{5}\) and 1.

Answer: (d) R is symmetric but not reflexive and not transitive
Explanation:
aRb iff a and b are distinct a and b have a common divisor other than 1.
(i) R is not reflexive since a and b are distinct i.e, (a, a) ∉ R
(ii) R is symmetric If a and b are distinct and have a common divisor other than 1, then b and a also are distinct and have a common divisor other than 1.
(iii) R is not transitive
If (a, b) ∈ R and (b, c) ∈ R then (a, c) need not be in R.
Example: (2, 6) ∈ R and (6, 2) ∈ R, but (2, 2) ∉ R
∴ R is symmetric but not reflexive and not transitive.

Explanation:
Let X = {0, 1,2,…..10}
|X| = 1
|P(X)| = 2¹¹ = 2048

Explanation:

The number of onto functions from A → B where | A | = m and | B | = n is given by the formula
n^{m – n}C₁(n – 1)^{m + n}C₂(n – 2)^m +…
+ (-1)^{n – 1 n}C_{n – 1}1^m

Here, m = 4 and n = 3
Substituting in above formula we get, 3⁴ – ³C₁(3 – 1)⁴ + ³C₂ (3 – 2)⁴ = 81 – 48 + 3 = 36

Explanation:

Total possible functions = 20² = 400
Number of one-to-one functions = 20P₂ = 20 × 19
∴ Required probability = \(\frac{20 \times 19}{20 \times 20}\) = 0.95

Answer: (b) Only S₂ is True
Explanation:
X # Y = X’ + Y’
This is the NAND operation. NAND is known to be commutative but not associative. So, only S₂ is true.

Answer: (d) ∀X ∈U ∀ Y ∈U(X\Y =Y’\X’)
Explanation:
S = {1, 2, 3, 4, 5, 6}
U is the power set of S ⇒ U = P(S)
U ={{ } {1}, {2}, …{1, 2}, {1, 3}, …{1, 2, 3}, … {1, 2, 3, 4}, …{1, 2, 3, 4, 5} …{1, 2, 3, 4, 5, 6}}
| U |= 2⁶ = 64 elements.
T ∈ U ⇒ T’ ∈ U
R ∈U, T\R=T – R

(a)∀X ∈ U(| X | = | X’ |) means that every subset of S has same size as its complement. Clearly this is False.
(For example, {1} ∈ U, and complement of {1} = {2, 3, 4, 5, 6})
(b)∃X ∈ U∃Y ∈ U(|X|=5,|Y|=5 and X ∩ Y = Φ) means that there are two 5 element subsets of S which have nothing in common. This is clearly False.
Since any two 5 element subsets will have atleast 4 elements in common.
(c) ∀X ∈ U ∀ Y ∈ U(|X| = 2,|y|=3 and X\Y = Φ) means that every 2 element subset X and every 3 element subset Y will have X – Y = Φ i.e., X ⊆ Y.
This is clearly False as can be seen from the example X= {1,2}, Y = {3,4,5} where X’^Y.
(d) ∀X ∈ U ∀T ∈ U(X \Y = Y’\X’) means that for any two subsets X and Y, X\ Y = Y \ X’ i.e. X – Y = Y’ – X’.
This is clearly True since by boolean algebra LHS = X – Y = XY’.
RHS =Y’ -X’ = Y’Xand therefore LHS = RHS.

Answer: (c) Not reflexive but symmetric
Explanation:
R = {(p, q), (r, s) I p – s = q – r}
(i) Check reflexive property
∀(p, q) ∈ Z⁺ × Z⁺ ((p, q), (p, q)) ∈ R
is true iff p – q = q – p which is false. So relation R is not reflexive.
(ii) Check symmetry property
If ((p, q), (r, .s’))∈ R then ((r, s), (p, q)) ∈ R
((p, q), (r, s)) ∈ R ⇒ p – s = q – r
((r, s), (p, q)) ∈ R ⇒ r – q = s – p
If p – s = q – r is true, then r — q = s – p is also true by rearranging the equation.
∴ R is symmetric.

Explanation:
f(n) = f\(\left(\frac{n}{2}\right)\) if n is even
f(n) = f{n + 5) if n is odd
f : N⁺ → N⁺
Now f(2) = f\(\left(\frac{2}{2}\right)\) = f(1)
f(3) = f(3 + 5) = f(8) = f\(\left(\frac{8}{2}\right)\) = f(4) = f\(\left(\frac{4}{2}\right)\) = f(2) = (1)
So, f(1) = f(2) = f(3) = f(4) = f(8)
Now let us find f(5) = f(5 +5) = f(10) = f\(\left(\frac{10}{2}\right)\)
= f(5) so f(5) = f(10)
Now let us find f(9)
f(9) = f(9 +5) = f(14) = f\(\left(\frac{14}{2}\right)\) = f(7)
= f(7 + 5) = f(12) = f\(\left(\frac{12}{2}\right)\) = f(6)
= f\(\left(\frac{6}{2}\right)\) = f(3)
So f(9) = f(7) = f(6) = f(3) = f(1) = f(2) = f(4) = f(8)
For n > 10, the function will be equal to one of f(1), f(2)….f(10)
So the maximum no. of distinct values / takes is only 2.
First is f(1) = f(2) = f(3) = f(4) = f(8) = f(9) = f(7) = f(6)
Second is f(5) = f(10)
All other n values will give only one of these two values.

Answer: (b) P is true and Q are false
Explanation:
(a, b) R (c, d) if a ≤ c or b ≤ d
P : R is reflexive
Q : R is transitive
Since, (a, b) R (a, b) ⇒ a ≤ a or b ≤ b ⇒ t or t
Which is always True, R is reflexive. Now let us check transitive property
Let(a,b) R (c, d) ⇒ a ≤ c or b ≤ d
and(c, d) R {e, f) ⇒ c ≤ e or d ≤ f
Now let us take a situation
a ≤ c (True) or b ≤ d ( false)
and c ≤ e (False) or d ≤ f (True)
Now we can get neither a ≤ e nor b ≤ f So, (a, b) R (c, d) and (c, d) R (e, f) ≠ (a, b) R (e, f). So clearly R is not transitive.
So Pis true and Q is false. Choice (b) is correct.

Answer: (b) Only II
Explanation:
This is actually a graph theory Question. Let the components be represented by vertices and if component x reacts with component y then let there be an undirected edge between x and y.Now, 9 of the 23 vertices will have 3 degrees each.
Note: U \ S means U — S, which is the remaining 23 – 9 = 14 vertices.
By Handshaking theorem,
9 × 3 + degree of 14 vertices = 2e = Even. i.e., 27 + degree of 14 vertices = 2e = Even. So, the total degree of 14 vertices must be odd.
Statement I means that the degree of each of the 14 vertices is odd, which means that the total degree of 14 vertices will be even (since even number of odd numbers is always even). But we need the total degree of 14 vertices to be odd so, statement I is false.
Statement III means that everyone of the 14 vertices has even degree, which guarantees that the total degree of 14 vertices will be even. So, statement III is false. Since statement II is the negation of statement III, it must be true.

Explanation:
Since the given hasse diagram is already a lattice, three is no need to add any ordered pair. So the answer is 0.

Explanation:
Let A → -5- by 3
B → ÷ by 5
C →÷ by 7
n(A ∪ B ∪ C) = n(A)+n(B) + n(C) – n(A ∩ B) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)
= \(\left[\frac{500}{3}\right\rfloor\) + \(\left[\frac{500}{5}\right\rfloor\) + \(\left[\frac{500}{7}\right\rfloor\) – \(\left[\frac{500}{15}\right\rfloor\) – \(\left[\frac{500}{21}\right\rfloor\) – \(\left[\frac{500}{35}\right\rfloor\) + \(\left[\frac{500}{105}\right\rfloor\) = 166 + 100 + 71 – 33 – 23 – 14 + 4 = 271

Explanation:
Given | G | =84
By Lagrang’s theorem any subgroup size is a divisior of 84.
But a proper subgroup cannot have same size as group.
So largest divisor of 84, other than 84 is 42. So, largest proper subgroup can have in size of 42.

Answer: (d) P, Q, and S only

Explanation:
P: Set of rational number → countable
Q: Set of functions from {0, 1} to N → N

0 can be assigned in N ways
1 can be assigned in N ways
There are N × N functions, the cross product of countable set in countable.
R: Set of functions from N to {0,1}

Each of these boxes can be assigned to 0 or 1 so each such function is a binary number with an infinite number of bits.
Example: 0000 is the binary number corresponding to 0 is assigned to all boxes and so on. Since each such binary number represents a subset of N (the set of natural numbers) by characteristic function method, therefore, the set of such function is the same as the power set of N which is uncountable due to Cantor’s theorem, which says that power set of a countably infinite set is always uncountably infinite.
S: Set of finite subsets of N → countably infinite since we are counting only finite subsets. So P, Q, and S are countable.

Answer: (b) R₁ only
Explanation:
R₁: ∀a, b ∈ G,a R₁ b if and only if ∃g∈ G such that a = g^-1bg
Reflexive: a = g^-1ag can be satisfied by putting g = e, identity “e” always exists in a group.
So reflexive
Symmetric: aRb ⇒ a = g^-1ag for some g
⇒ b = gag^-1 = (g^-1)^-1 ag^-1 g^-1 always exists for every g ∈ G.
So symmetric
Transitive: aRb and bRc ⇒ a = g₁^-1bg₁ and b = g₂^-1 cg2 for some g₁g₂ ∈ G.
Now a=g₁^-1g₂^-1cg₂g₁(g₂g₁)^-1 cg₂g₁
g₁ ∈ G and g₂ ∈ G ⇒ g₂g₁ ∈ G since group is closed so aRb and aRb ⇒ aRc hence transitive
Clearly R₁ is equivalence relation.
R₂ is not equivalenced it need not even be reflexive, since aR₂ a => a = a^-1 ∀a which not be true in a group.
R₁ is equivalence relation is the correct answer.

Explanation:
A = {1, 2, 3}
n = |A| = 3
Number of relations on A = 2^n² = 2^3² = 2⁹
Number of reflexive relations on A
= 2^{n² – n}  = 2^{3² – 3} = 2⁶

Explanation:
Size of group = O(G) = 35
Lagrange’s theorem says if H is a subgroup of G
then O(H) | 0(G).
So 0(H) can be 1, 5, 7 or 35.
So largest subgroup size other than G itself is 7.

Answer: (a) If a relation S is reflexive and circular, then S is an equivalence relation.
Explanation:
Let S be reflexive and circular,
Let us checking symmetry:
Symmetry:
Let xSy
Now since S is reflexive ySy true.
So xSy and ySy is true
Now by circular property we get, ySx
So xSy => ySx
So S is symmetric.
Transitive:
Let xSy and ySz
Now by circular property we get zSx and by symmetry property proved above, we get
zSx => xSz
So xSy and ySz => xSz
So S is transitive.
So S is reflexive, symmetric and transitive and hence an equivalence relation.
So option (a) is true.
Option (b): Let S be circular and symmetric. Let S be defined on set {1, 2, 3}
Now empty relation is circular and symmetric but not reflexive. So S need not be an equivalence relation.
So option (b) is false.
Option (c): Let S be transitive and circular. Let S be defined on the set {1, 2, 3}
Now empty relation again satisfies transitive and circular but is not reflexive. So S need not be an equivalence relation.
So option (c) is false.
Option (d): Reflexive and symmetric need not be transitive for example on {1, 2, 3}.
S = {(L 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
is reflexive and symmetric. But it is not transitive because (1, 2) and (2, 3) belong to S but (1, 3) does not.
So option (d) is false.

Answer: (c) G may not be cyclic, but H is always cyclic.
Explanation:
|G| = 6
H is subgroup, so by Lagrange’s theorem
| H | = 1, 2, 3 or 6 (Divisor’s of 6) Now it is given that 1 < | H | <6 or | H | = 2 or 3
Since 2 and 3 are both prime and since every group of prime order is cyclic, H is surely cyclic. But order of | G | =6 which is not prime. So G may or may not be cyclic. So G may not be cyclic, but H is always cyclic. Option (c) is correct.

Answer:
(a) There exists a surjection from S₁ to S₂.
(d) There exists a bijection from S₁ to S₂.
Explanation:
| S₁ | = 3ⁿ
Since each of the n2 entries in n × n matrix can be fills in 3 ways.
|S₂| = 3ⁿ2
Since | {0, 1, 2} | =3 and | {0, 1, 2, n2 — 1} | = n²)
Now the theorem says A bijection f_{A → B} exists iff |A| = |B|.
Here |S₁| = |S₂|
So, there has to be a bijection from S₁ to S₂. So, option (d) is correct. If bijection exists surely surjection also exists. So, option (a) is also correct. Option (b) and (c) are incorrect.

Explanation:
The Venn diagram for this is

Now every element x in S has only 3 options. It can be x ∈ A or x ∈ B – A or x ∈ S — B. So the number of ways to choose A and B such that A ⊆ B ⊆ S is 3¹⁰.

Answer: (a) 10
Explanation:
L is the set of 10-dimensional orthogonal vectors.
So cardinality of L ≤ 10.
i.e., Maximum cardinality of L = 10.
So, option (a) is correct.

Answer: (a) ((P ⇒ Q) ∧ (Q ⇒ R)) ⇒ (P ⇒ R)
Explanation:
Option (a) is well known valid formula (tautology) because it is a rule of inference called a hypothetical syllogism. By applying Boolean algebra and simplifying, we can show that (b), (c), and (d) are invalid.
For example,
Choice (b) ≡(P ⇒ Q) ⇒ (¬P ⇒ ¬Q)
≡(P ⇒ Q) ⇒ (P ⇒Q’)
≡(P’ + Q) ⇒ (P + Q’)
≡(P’ + Q)’ + (P+Q’)
≡PQ’ + P+Q’
≡P+ Q’
≠1
So, invalid.

Answer: (a) (∀x) P(x)∨ (∀x) Q(x) → (∀x) {P(x) ∨ Q(x)}
Explanation:
According to distributive properties
∀x (P(x) ∧ Q(x)) ↔ ∀xP(x) ∧ ∀xQ(x)
(∀xP(x) ∨∀xQ(x)) → ∀x(P(x) ∨ Q(x))
∃x (P(x) ∨ Q(x)) ↔ ∃xP(x)∨∃xQ(x)
∃x(P(x) ∧Q(X)) → ∃xP(x) ∧∃xQ(X)
So option (a) is valid.

Answer: (b) (a ∧ b) → (b ∨ c)
Explanation:
(a) (a ∨ b) → (b ∧ c)
≡ (a + b)’ + bc
≡ a’b’+ bc
Therefore, ((a ∨ b) → (b ∧ C)) is contingency and not tautology.
(b) (a ∧ b) → (b ∨ c)
≡ ab → b + c
≡ (ab)’ + b + c
≡a’ + b’ + b + c
≡a’ + 1 + c
≡1
So ((a ∧b) → (b ∨c)) is tautology.
(c) (a ∨ b) → (b → c)
≡ (a + b) → (b’ + c)
≡(a +b)’ +b’ + c
≡a’b’ + b’ + c
≡b’ + c
So ((a ∨b) → (b → c)) is contingency but not a tautology.
(d) (a → b) → (b → c)
≡(a’ + b) → (b’ + c )
≡(a’ + b)’ + b’ + c
≡ ab’ + b’ + c
≡b’ + c
Therefore, ((a → b) → (b → c)) is contingency but not tautology.

Answer: (b) ⇔ (p ∧ q)
Explanation:
The proposition p ∧ (~p ∨ q)
≡p(p’ + q) ≡pq
≡p ∧ q

Answer: TRUE
Explanation:

From the truth table, (p ↔ q) → (p → ~q) is not a tautology, hence it is true that p ↔ q doesn’t imply p →¬q.

Answer: (d) Cannot be determined
Explanation:

Now since ¬p → q is given true, We reduce the truth table as follows.

In the reduced truth table we need to find the true value of ¬p ∨(p → q) ≡p’ +(p → q) ≡p’ + p’ +q ≡p’ + q
The truth value of p’ + q in the reduced truth table is given below.

Since in the reduced truth table also, the given expression is sometimes true and sometimes false, therefore the truth value of proposition ¬p ∨(p → q) can not be determined.

Answer: (d) ((x ∨ y)↔(~ x → ~ y))
Explanation:
(d) ((x ∨y) ↔ (~x →~y))
= (x ∨ y) ↔ (~(~x ) ∨~y)
= (x ∨ y) ↔ (~x ∨~y)
= ((x ∨ y) ∧ (~x ∨~y)) ∨(~(x ∨y) ∧~(x ∨~y))
= ((x ∨(y ∧ ~y)) ∨(( ~x ∧ ~y) ∧ (~x ∧ y))
= (x ∨ F) ∨ ((~x ∧ ~y) ∧(~x ∧ y))
= x ∨ (~x ∧(y ∧~y)) = x ∨(~x ∧F)
= x ∨ F = x

Answer: (a) True
Explanation:
. a ↔ (b ∨¬b)
a ↔ True
So a is true, i.e. a = 1
. b ↔ c holds. So b = c
Now the given expression is
(a ∧ b) → ((a ∧ c) ∨d) ≡(a . b) → ((a . c) + d)
putting a=1 in above expression We get
1 . b →((1 . c) +d)
≡ b → c + d
≡ b’ + c + d
Now putting b=c in above expression We get
≡c’ + c + d ≡1 + d ≡1
So the expression is always true.

Answer: (a) I stay if you go
Explanation:
Let p: I stay
q: you go
I stay only if you go
p → q
The converse of p → q is q → p
Now convert the answers one-by-one into boolean form. Only option (a) i.e. “I stay if you go” converts to q → p.

Answer: (a) F₁ is satisfiable, F₂ is valid
Explanation:
F₁: P → ~P ≡ p → p’ ≡ p’ + p’ ≡p’
So F₁ is contingency. Hence, F₁ is satisfiable but not valid.
F₂: (P → ~P) ∨ (~P → P)
≡ (p → p’) + (p’ → p)
≡ (p’+p’) + (p + p)
≡ p’ + p ≡ 1
So F₂ is tautology and therefore valid.

Answer: (a) (X ∧¬ Z) → Y
Explanation:
If X then Y unless Z is represented by
X →Y unless Z ≡ (X → Y) + Z
≡ X’ + Y + Z
Now convert the answers one-by-one into boolean form only choice (a) converts to X’+ Y+ Z as can be seen below:
(X∧~Z) →Y ≡ XZ’ → Y
≡ (XZ’)’ + y
= X’+ Y+Z

Answer: (d) (∀x) [α ⇒ β ] ⇒ ((∀x) [α] ⇒ (∀x) [β])
Explanation:
(∀x) [α ⇒ β] ⇒ ((∀x) [α] ⇒ ∀(x) [[β]) is a logical equivalence and therefore, a valid first order formula.

Answer: (d) Both I₁ and I₂ satisfy α
Explanation:
Q_xy ≡ “y divides y” is always true
∴Q_xy ⇔ ¬ Q_xy is the same as Q_xy ⇔False
Now α becomes
(∀x)[P(x)⇔(∀y)(Q_xy⇔false)]
⇒ (∀x) [¬ P(x)]
Now consider I₁: P(x) = “x is a prime number”. α becomes
(∀x x is a prime number if and only if ∀y (y does not divide x)) ⇒ (∀x) x is not prime. which means that x is a prime number if and only if no number divides x implies that no number is prime. Since x always divides x, the above sentence is true.

Now consider I₂: P(x) = “x is a composite number”. Now α becomes

(∀x x is a composite number if and only if ∀y (y does not divide x)) ⇒ (∀x) x is not composite. This means that x is a composite number if and only if no number divides x implies that no number is composite.
Since x always divides x, the above sentence is true.
∴ Both I₁ and I₂ satisfy α.

Answer: (b) (P ∨ Q) ⇒ (P ∨ R) ∧ (Q ∨¬ R) is logically valid
Explanation:
Derive clause P ∨ Q from clauses P ∨ R, Q ∨¬R means that (P ∨ R) ∧ (Q ∨ ¬R) ⇒ P ∨ Q
.’. (a) is true
Since x ⇒ y does not imply that y ⇒ x
.’. P ∨ Q ⇒ (P ∨ R) ∧ (Q ∨¬R)
.’. may or may not be true.
Hence (b) is false.

Answer: (d) (∃x) (boy(x) ∧ (∀y) (girl(y) → taller (x, y)))
Explanation:
The statement is “some boys in the class are taller than all the girls”. so the notation for the given statement is (∃x) (boy(x) ∧(∀y) (girl(y) → taller (x,y)))

Answer: (c) ¬(∀x)(∃y)[(a(x,y) ∧ b(x,y)) → c(x,y)]
Explanation:
Choice (c) is
¬(∀x)(∃y)[(a(x,y) ∧ b(x,y)) → c(x,y)]
= ¬(∀x)(∃y)[a ∧ b → c]
= ¬(∀x)(∃y)[(ab)’ + c]
= ∃x ∀y[(ab)’ + c]’
= ∃x ∀y [abc’] = ∃x ∀y [a ∧ b ∧ ¬C] which is same as the given expression. (∃x)(∀y)[(a(x,y) ∧ b(x,y)) ∧ ¬c(x,y)]

Answer: (c) P and S only
Explanation:
p: [(¬p ∨q) ∧(r → s) ∧(p ∨r)] →q
≡p(p’ +r)(q + r’) →q
≡pr(q +r’) →q
≡prq →q
≡(prq)’ + q
≡p’ + r’ + q’ + q
≡p’ + r’ + 1
≡1
Therefore S is valid.
Q and R can be similarly simplified in boolean algebra to show that is both not equivalent to 1.

Answer: (a) Satisfiable but not valid
Explanation:
(p → (Q ∨R)) → ((P ∧Q) → R)
≡(P → Q + R) → (PQ → R)
≡[P’ + Q + R] → [(PQ)’ + R]
≡[P’ + Q + R] → [P’ + Q’ + R]
≡(P’ + Q + R)’ + P’ + Q’ + R
≡P Q’ R’ + P’ + Q’ + R
≡Q’ + Q’ P R’ + P’ + R
≡Q’ + P’ + R (by absorption law)
which is a contingency (i.e. satisfiable but not valid).

Answer: (b) X → Y
Explanation:
X: (P ∨ Q) → R
Y: (P → R) ∨(Q → R)
X: P + Q → R ≡(P + Q)’ + R ≡P’ Q’ + R
Y: (P’ + R) + (Q’ +R) ≡P’ + Q’ + R
Clearly X ≠ Y
Consider X → Y
≡(P’Q’ +R) →(P’ + Q’ + R)
≡(P’Q’+R)’ + P’ + Q’ + R
≡(P’Q’)’ . R’ + P’ + Q’ + R
≡(P + Q) . R’ + P’ + Q’ + R
≡ PR’ + QR’ + P’ + Q’ + R
≡ (PR’ + R) + (QR’ + Q’) + P’
≡(P + R) (R’ + R) + (Q + Q’) × (R’ +Q’) + P’
≡(P + R) + (R’ + Q’) + P’
≡P + P’ + R + R’ + Q’
≡1 + 1 + Q’
≡ 1
∴X → Y is a tautology.

Answer: (b)∀(x) [teacher (x) → ∃ (y) [student (y) ∧ likes (y, x)]]
Explanation:
Every teacher is liked by some student: then the logical expression is ∀(x) [teacher(x) → ∃(y) [student (y) ∧ likes (y, x)]] Where likes (y, x) means y likes x, such that y represent the student and x represents the teacher.

Answer: (b) (∀x(P(x) ⇒ Q(x))) ⇒ ((∀xP(x)) ⇒ (∀xQ(x)))
Explanation
Consider choice (b)
(∀x(P(x) ⇒ Q(x))) ⇒ ((∀xP(x)) ⇒ (∀xQ(x))) Let the LHS of this implication be true This means that
P₁ → Q₁
p₂ → Q₂

.
.
.
P_n → Q_n
Now we need to check if the RHS is also true. The RHS is ((∀xP(x)) ⇒ (∀xQ(x)))
To check this let us take the LHS of this as true i.e. take∀xP(x) to be true. This means that (P₁ P₂,…P_n ) is taken to be true. Now P₁ along with P₁ → Q₁ will imply that is true. Similarly
P₂ along with P₂ → Q₂, will imply that Q₂ is true. And so on…
Therefore (Q₁, Q₂,……Q_n) all true.
i.e. ∀x Q(x) is true. Therefore statement (b) is a valid predicate statement.

Answer: (b) Satisfiable and so is its negation
Explanation:
Since a relation may or may not be symmetric, the given predicate is satisfiable but not valid. So (a) is clearly false.
Whenever a predicate is satisfiable its negation also is satisfiable. So option (b) is the correct answer.

Answer: (d) ∀x [(tiger (x) ∨ lion (x)) → {(hungry (x) ∨ threatened (x)) → attacks (x)}]
Explanation:
The given statement should be read as “If an animal is a tiger or a lion, then (if the animal is hungry or threatened, then it
will attack). Therefore the correct translation is ∀x [(tiger (x) ∨ lion (x)) → {(hungry (x) ∨ threatened (x)) → attacks (x)}] which is choice (d).

Answer: (d) Both P₁ and P₂ are not tautologies
Explanation:
P₁: ((A ∧ B) → c) ≡ ((A → c) ∧ (B → c))
LHS :
(A ∧ B) → C
≡ AB → C
≡ (A B)’ + C
≡ A’ + B’ + C
RHS:
(A → C) ∧ (B → C)
≡ (A’+ C) (B’ + C)
≡ A’B’ + C
Clearly, LHS ≠ RHS
P₁ is not a tautology
P₂: ((A ∨ B → C)) ≡ ((A → C) ∨(B → C))
LHS ≡ (A + B → C)
≡ (A + B)’+C
≡ A’B’ + C
RHS = (A → C) ∨ (B → C)
≡ (A’+ C) + (B’ + C)
≡ A’+ B’ + C
Clearly, LHS ≠ RHS ⇒ P₂ is also not a tautology. Therefore, both P₁ and P₂ are not tautologies. The correct choice is (d).

Answer: (d) ~ (~ A ⊙ B)
Explanation:
By using min terms we can define
A⊙B = AB + AB’ + A’B’
= A + A’B’
= (A + A’) . (A + B’) = A + B’
(a) ~ A⊙B = A’⊙ B = A’ + B’
(b) ~ (A ⊙ ~B) = (A ⊙ B’)’ =(A + (B’)’)’
= (A + B)’ = A’B’
(c) ~ (~A ⊙ ~B) = (A’ ⊙ B’)’ = (A’ + (B’)’)’
= (A’ + B)’ = AB’
(d) ~ (~A ⊙ B) = (A’ ⊙ B)’ = (A’ + B’)’
= A . B = A ∧ B
∴ Only, choice (d) = A ∧ B
Note: This problem can also be done by constructing a truth table for each choice and comparing it with the truth table for A ∧ B.

Answer: (d) ∀x (Graph(x) ⇒ ¬ Connected (x))
Answer:
Explanation:
The statement “Not every graph is connected” is the same as “There exists some graph which is not connected” which is the same as
∃x {graph (x) ∧ ¬ connected (x)}
Which is choice (b)
By boolean algebra, we can see that options (a) and (c) are the same as (b). Only option (d) is not the same as (b).
In fact option (d) means that “all graphs are not connected”.
Alternate solution:
We can translate the given statement “NOT (every graph is connected)” as ¬(∀x graph (x) → connected (x))
≡∃x¬(graph (x) →connected (x))
≡ ∃x¬(¬graph (x) ∨connected (x))
≡ ∃x (graph (x) ∧ ¬ connected (x))
By boolean algebra, we can see that options (a) and (c) are the same as (b). Only option (d) is not the same as (b).
In fact option (d) means that “all graphs are not connected”.

Answer: (a) For any formula, there is a truth assignment for which at least half the clauses evaluate to true.
Explanation:
In conjunction normal form, for any particular assignment of truth values, all except one clause, will always evaluate to true. So, the proportion of clauses which evaluate to true to the total

number of clauses is equal to \(\frac{2^{n}-1}{2^{n}}\)
Now putting n = 1, 2, . . . . ., We get \(\frac{1}{2}\) , \(\frac{3}{4}\) , \(\frac{7}{8}\) ….
All of these proportions are ≥ \(\frac{1}{2}\) and so choice (a) at least half of the clauses evaluate to true, is the correct answer.

Answer: (a) ∀x(P(x) ⇒ Q(x)) ⇒ ((∀xP(x)) ⇒ (∀xQ(x)))
Explanation:
Option (a) is a standard one-way distributive property of predicates.

Answer: (a) ∀x (fsa(x) ⇒ ∃y (pda(y) ∧ equivalent (x, y)))
Explanation:
“For x which is a fsa, there exists a y which is a pda and which is equivalent to x.” ∀x (fsa(x) ⇒ ∃y (pda(y) ∧ equivalent (x, y))) is the logical representation.

Answer: (c) [(∃x, α(x)) → β] → [∀x,α(x) → β]
Explanation:
Option (c) is [(∃x, α(x)) → β] → [∀x,α(x) → β] Let us check the validity of this predicate. Let the LHS of this predicate be true.
This means that some α → β.
Let a₅ → β
Now we will check if the RHS is true. The RHS is [∀x,α(x) → β] to check this implication let us take ∀x,α(x)to be true.
This means that all the a are true. It means that a5 is also true.
But α₅ → β. Therefore β is true.
So the RHS [∀x,α(x) → β] is true.
Whenever the LHS [(∃x, α(x)) → β] is true. So option (c) is valid.

Answer: (d) [∀x,α ∧ (∃y, β ∧ (∃U, ∀U, ¬Y))]
Explanation:
The given predicate is
[∀x, α → (∃y,β → (∀u, ∃u, y))]
The negation is of this predicate is
¬[∀x, α → (∃y,β → ∀u, ∃u, y)]
¬[∀x, α →(∀y, ¬β ∨∀u, ∃u, y)]
¬[∃x, ¬α ∨(∀y, ¬β ∨∀u, ∃u, y)]
[∀x, α ∧ (∃y, β ∧(∃y, β ∧ (∃u, ∀u, ¬y))]
which is an option (d).

Answer: (b) Only 1, 2 and 3
Explanation:
1. P ∨ ~Q ≡ P + Q’
2. ~(~P ∧ Q) ≡ (P’Q)’ ≡P+Q’
3. (P ∧ Q) ∨ (P ∧ ~Q) ∨ (~P ∧ ~Q)
≡ PQ + PQ’ + P’Q’
≡ P(Q + Q’) + P’Q’
≡ P + P’Q’
≡ (P + P’) (P + Q’)
≡ P + Q’
4. (P ∧ Q) ∨ (P ∧ ~Q) ∨ (~P ∧ Q)
≡ PQ + PQ’ + P’Q
≡ P(Q + Q’) + P’Q
≡ P + P’Q
≡ (P + P’) (P + Q) = P + Q
Clearly (i), (ii) and (iii) are equivalent. Correct choice is (b).

Answer: (d) ∀x ((G(x) ∨ S(x)) → P(x))
Explanation:
The correct translation of “Gold and silver ornaments are precious” is choice (d)
∀x((G(x) ∨ S(x)) → P(x))
which is read as “if an ornament is a gold or silver, then it is precious”. Now since a given ornament cannot be both gold and silver at the same time.
Choice (b) ∀x ((G(x) ∧ S(X)) → P(x)) is incorrect.

Answer: (b) P □ ¬ Q
Explanation:
The given table can be converted into a boolean function by adding minterms corresponding to true rows.

Since there is only one false in the above truth table, we can represent the function P □ Q more efficiently, in conjunctive normal form. Translates P □ Q = P + Q’ (the max-term corresponding to the third row, where the function is false).
Now, we can easily translate the choices into boolean algebra as follows:
Choice (a) ¬Q □ ¬P ≡ Q’ □ P’ ≡ Q’ + P
Choice (b) P □ ¬Q ≡ P □ Q’ ≡ P + Q
Choice (c) ¬P □ Q ≡ P’ □ Q ≡ P’ + Q’
Choice (d) ¬P □ ¬Q ≡ P’ □ Q’ ≡ P’ + Q
As we can clearly see only choice (b) P □ ¬ Q is equivalent to P + Q.

Answer: (b) I and IV
Explanation:
I ¬∀xP(x) ≡ ∃x¬P(x)
and IV ∃x ¬P(x)
Clearly, choices I and IV are equivalent.
II ¬∃xP(x) ≡ ∀x ¬P(x)
and III ¬∃x[¬P(x)] ≡ ∀xP(x)
Clearly, II and III are not equivalent to each other or to I and IV.

Answer: (b) No one can fool everyone all the time
Explanation:
∀x ∃y ∃t ¬ F(x, y, t)
≡ ¬{∃x ∀y ∀t F(x, y, t)}
≡ it is not true that (someone can fool all people at all times)
≡ no one can fool everyone all the time

Answer: (a) P(x) being true means that x is a prime number
Explanation:
If P(x) is true, then
x ≠ 1 and also
x is broken into two factors, only if, one of the factors is x itself and the other factor is 1, which is exactly the definition of a prime number.
So P(x) is true means x is a prime number.

Answer: (b) I₁ is correct but I₂ is not a correct inference
Explanation:
Let p: It rains
q: cricket match will not be played.
I₁: p ⇒ q
∴ \(\underline{\sim q}{\sim p}\)
Clearly I₁ is correct since it is in the form of modus Tollens (rule of contrapositive)

I₂: p ⇒ q
∴ \(\underline{\sim q}{\sim p}\)
which corresponds [p ⇒ q ∧~p] ⇒ ~ q
≡ [(p’ + q) p’] ⇒ q’
≡ [p’ + qp’] ⇒ q’
≡p’ ⇒ q’
≡(p’)’ + q’ ≡p + q’
which is not a tautology. So I₂ is incorrect inference.

Answer: (c) ∃x (real (x) ∧ rational (x))
Explanation:
Some real numbers are rational ≡∃x [real (x) ∧rational (x)]

Answer: (d) ¬∃x (F(x) ∧ P(x))
Explanation:
None of my friends are perfect i.e., all of my friends are not perfect
∀x((F(x) → ¬ P(x))
∀x(¬ F(x) ∨¬ P(x))
¬ ∃x (F(x) ∧ P(x))
Alternatively
∃x (F(x) ∧ P(x)) gives
there exist some of my friends who are perfect, ¬∃x (F(x) ∧ P(x))
there does not exist any friend who is perfect i.e., none of my friends are perfect. So (d) is the correct option.

Answer: (a) ∀x(∃z(¬β) → ∀y(α))
Explanation:
Let, ∀y(α) = p, ∀z(β) = Q
then, ∃y(¬α) = ¬p and ∃z (¬β) = ¬Q
Given, ¬∃x(∀y(α) ∧∀z(β))
= ¬∃x(P ∧ Q) = ¬∀x (¬P ∨¬Q)
= ∀x (P →¬Q)
(a) ∀x(∃z(¬β) → ∀y(α)) = ∀x (¬Q → P)
(b) ∀x(∀z(β) → ∃y(¬α)) = ∀x (Q → ¬ P)
= ∀x (P →¬Q)
(c) ∀x (∀x(α) → ∃z (¬β)) = ∀x (P → ¬ Q)
(d) ∀x (∃y(¬α) ∨ ∃z (¬β)) = ∀x (¬ P ∨ ¬ Q)
= ∀x (P → ¬ Q)
∴ Only (a) is not logocally equivalent to ∀x(P → ¬ Q).

Answer: (d) ∃x: glitters(x) ∧¬gold(x)
Explanation:
(a) ∀x glitters (x) ⇒¬gold (x) All glitters are not gold
(b) ∀x gold(x) ⇒ glitters(x) All golds are glitters
(c) ∃x gold(x) ∧ ¬ glitters(x) There exist gold which is not glitter i.e. not all golds are glitters.
(d) ∃x glitters(x) ∧ ¬ gold(x) Not all that glitters is gold i.e., there exist some which glitters and which is not gold.

Answer: (b) (~ (p ↔ q) ∧ r) ∨ (p ∧ q ∧ ~ r)
Explanation:
Option (b) is (~(p ↔ q) ∧ r) ∨ (p ∧ q ∧ ~r)
≡((p ⊕ q)r) + pqr’ ≡(pq’ + p’q)r + pqr’
≡pq’r + p’qr + pqr’ ≡pqr’ + pq’r + pq’r
This is exactly the mm-term form of a logical formula which is true when exactly two variables are true (only p, q true or only p, r true or the only q, r true).

Answer: (b) (a ↔ c) → (~b → (a ∧ c))
Explanation:
(a ↔ c) → (~ b → (a ∧c))
≡ (a ↔ c )’ + (b’ → ac)
≡(a ⊕ c)’ + (b’ → ac)
≡ac’ + a’c + b + ac
≡a(c’ + c) + a’c + b
≡a + a’c + b ≡a + c + b
which is not a tautology.

Answer: (d) L, M, and N are TRUE.
Explanation:
g: mobile is good
C: mobile is cheap
P: Good mobile phones are not cheap
g → c’ ≅ (g’ + c’)
Q: Cheap mobile phones are not good
c → g’ ≅ (c’ + g’)
:. Both P and Q are equivalent.
L: P → Q
M: Q → P
N: P ≡ Q
Since both P and Q are equivalent, all three of L. M, N are true. So the correct option is (d).

Answer: (d) ∃d (Rainy (d)∧ ~ Cold (d))
Explanation:
Not (all rainy days are cold):
~ (∀d (Rainy (d) → Cold (d)))
≡~ (∀d (~ Rainy (d) ∨ Cold (d)))
≡ ∃d (Rainy (d)∧ ~ Cold (d))
Alternate Method:
Not all rainy days are cold is same as some rainy days are not cold which is the same as ∃d (Rainy (d)∧ ~ Cold (d))

Answer: (c) (¬p ∧ q) ∨ (p ∧ ¬q)
Explanation:
Here, option (a) and (b) can be reduced to (p → q) ∧ (q → p) and hence = p ↔ q
Option (d) is p’q’ + pq ≡ p ↔ q.
Option (c) is p’q + pq’ = p ⊕ q which is not equivalent to p ↔ q.

Answer: (a) Both commutative and associative
Explanation:
The given truth table corresponds to p’q + pq’ = P ⊕ q. ⊕ is known to be both commutative and associative.

Answer: (c) If a person is kind, he is not known to be corrupt
Explanation:
C: Person is corrupt
K: Person is kind.
E : Person is elected
S₁ : C → \(\bar{E}\)
S₂: K → E
S₂ ↔ \(\bar{E}\) → \(\bar{K}\)
So from S₁ and S₂:(C → \(\bar{E}\)) ∧ (\(\bar{E}\) → \(\bar{K}\)) ≅ C → \(\bar{K}\) We can conclude C → \(\bar{K}\) which is same as K → \(\bar{C}\), which is same as option (c).

Answer: (c) [∀x ∃y (P(x, y) → R(x, y))] ↔ [∀x ∃y (¬P(x, y) ∨ R(x, y))]
Explanation:
Since P → R ≡¬P ∨R, and both the sides are the same (∀x∃y) Option (e) is clearly a tautology.

Answer: (a) The result is head
Explanation:
Type 1 always tells truth.
Type 2 always tells lies.
The statement is “toss is head if and only if I am telling the truth”.
There are two cases possible.
Case-1: Let the person be type 1.
Case-2: Let the person be type 2.
In each case, we shall prove that result is head.
Case-1: Let the person be type 1. Type 1 always tells truth.
So the statement “toss is head if and only if I am telling the truth” is true.
So toss head ⇔ telling truth. Since type 1 is telling truth so toss head is also true. So in case, 1 the result is that the toss is head.
Case-2: Let the person be type 2. Type 2 always tells lies.
So the statement “toss is head if and only if I am telling the truth” is false.
So toss head ⇔ telling truth is false. So toss head ⊕ telling truth. So toss head and telling truth to have opposite truth values. Now, since type 2 telling truth is false, so toss head has to have the opposite truth value which is true.
So toss head is true. So in case 2 also, the result is head.
So in both cases, we have proved that the result is head.
So option (a) is correct.

Explanation: We wish to make
¬ ((p ⇒ q) ∧ (¬r ∨ ¬s)) = 1
⇒ (p ⇒ q) ∧ (¬r ∨ ¬s) = 0
⇒ (p ⇒ q) = 0
or ¬r ∨ ¬s = 0

Now (1) is satisfies only when p 1 and q = 0.
Equation (2) ¬r ∨ ¬s = 0, iff r ∧ s = 1
i.e. r 1 and s= 1
i.e. x is a perfect square and x is a prime number. Which is not possible so condition (2) cannot be satisfied by any x. So condition (1) must be satisfies which is p = 1 and q = 0 i.e. x e {8, 9, 10, 11, 12,) and x is not a composite. Now the only value of x which satisfies this is x = 11. So correct answer is x = 11.

Solution:
p∧(p ⇒ q) = p(p’ + q) =pq
Take (i) false
pq ⇒ false ≡ pq ⇒ 0
≡ (pq)’ + 0
≡ p’ + q’ + 0
≡ not valid

Take (ii)
pq ⇒ q = (pq)’ + q
≡ (pq)’ + q
≡ p’ + 1 = 1
≡ valid

Take (iii)
pq ⇒ true ≡ pq ⇒ 1
≡ (pq)’ + 1 ≡ 1
≡ valid

Take (iv)
pq ⇒ p + q ≡ (pq)’ + p + q
≡ P’ + q’ + P + q
≡ 1 + 1
≡ 1
≡ valid

Take (v)
pq ⇒ q’ + p ≡ (pq)’ + q’ + p
≡ p’ + q’ + q’ + p
≡ 1
≡ valid

So the number of expressions that are logically implied by p ∧ (p ⇒ q) is 4.

Answer: (d) ∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀xq(x))

Explanation:
∀x is only one way distribution over “∨” so let us check (d)
(d) ∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀xq(x))
Let LHS be true, so we have,
P₁ ∨ q₁ (true)
p₂ ∨ q₂ (true)
p₃ ∨ q₃ (true)
.
.
.

Now take p₁ is true, q₁ is false and p₂ is false, q₂ is true.
Now LHS is true, but RHS, ∀x p(x) is false (since p₂ is false) and ∀x q(x) is also false (since q₁ is false)
So, LHS ≠ RHS.

Answer: (b) I and IV only
Explanation:
I. ∀x ∃y R(x,y) →∃y (∃x R(x,y)) is true, since ∃y (∃x R(x,y))
II. ∀x ∃y R(x,y) →∃y (∀x R(x,y)) is false Since ∃y when it is outside is stronger than when it is inside.
III. ∀x ∃y R(x,y) → ∀y ∃x R(x,y) is false Since R(x, y) may not be symmetric in x and y.
IV. ∀x ∃y R(x,y) →¬ (∃x ∀y ¬R(x,y)) is true since ¬ (∃x ∀y ¬R(x,y)) ≡∀x ∃y R (x,y)
So, IV will reduce to ∀x ∃y R(x,y) → ∀x ∃y R(x,y) which is trivially true.
So the correct answer is I and IV only which is an option (b).

Answer: (d) II and III only
Explanation:
Statement p’ ⇒ q’ ≡p + q’
I. p → q ≡p’ + q
II. q → p ≡ q’ + p
III. (¬q ∨ p) ≡ q’ + p.
IV. (¬p ∨ q) ≡p’ + q
So, clearly, the given statement is the same as II and III only.

Question 57.
Let p, q and r be propositions and the expression (p → q)→r be a contradiction. Then, the expression (r→p)→q is
(a) A tautology
(b) A contradiction
(c) Always TRUE when p is FALSE
(d) Always TRUE when q is TRUE

Question 58.
Let p, q, r denote the statements “It is raining”, “It is cold”, and “It is pleasant”, respectively. Then the statement “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold” is represented by
(a) (¬ p ∧ r) ∧ (¬r → (p ∧ q))
(b) (¬ p ∧ r) ∧ ((p ∧ q) → ¬r)
(c) (¬ p ∧ r) ∨ ((p ∧ q) → ¬r)
(d) (¬ p ∧ r) ∨ (r → (P ∧ q))

Question 59.
Consider the first order predicate formula φ:∀x[∀z z|x ⇒∃w (w >x) ∧(∀z z | w ⇒ (( w = z) ∨ (z = 1)))]
Here ‘a | b’ denotes that ‘a divides b’, where a and b are integers. Consider the following sets:
S₁:{1, 2, 3, …, 100}
S₂: Set of all positive integers
S₃: Set of all integers
Which of the above sets satisfy φ?
(a) S₁ and S₃
(b) S₂ and S₃
(c) S₁ S₂ and S₃
(d) S₁ and S₂

Question 60.
Which one of the following predicate formulae is NOT logically valid?
Note that W is a predicate formula without any free occurrence of x.
(a) ∃x(p(x) ∧ W) ≡ ∃x p (x) ∧ W
(b) ∀x(p(x) → W) ≡ ∀xp(x) → W
(c) ∃x(p(x) → W) ≡ ∀xp(x) → W
(d) ∀x(p(x) ∨ W) ≡ ∀x p(x) ∨ W

Question 61.
Let p and q be two propositions. Consider the following two formulae in propositional logic.
S₁: (¬ p ∧ (p ∨ q)) → p
S₂: q → ( ¬ p ∧ (p ∨ q))
Which one of the following choices is correct?
(a) Neither S₁ nor S₂ is a tautology.
(b) Both S₁ and S₂ are tautologies.
(c) S₁ is a tautology but S₂ is not a tautology.
(d) S₁ is not a tautology S₂ is a tautology.

Question 62.
Choose the correct choice(s) regarding the following propositional logic assertion S:
S : ((p ∧ Q) →R) → ( (P ∧Q) →(Q → R))
(a) S is a tautology.
(b) S is neither a tautology nor a contradiction.
(c) The antecedent of S is logically equivalent to the consequent of S.
(d) S is a contradiction.