EngineeringMathematics
Question 1 
If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)^{2}] equals _________.
54  
55  
56  
57 
Mean = Variance
E(X) = E(X^{2})  (E(X))^{2} = 5
E(X^{2}) = 5 + (E(X))^{2} = 5 + 25 = 30
So, E[(X + 2)^{2}] = E[X^{2} + 4 + 4X]
= E(X^{2}) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Question 2 
If the ordinary generating function of a sequence is , then a_{3}  a_{0} is equal to ________.
15  
16  
17  
18 
Question 3 
1  
Limit does not exist  
53/12  
108/7 
Question 4 
R: ∀a,b ∈ G, aRb if and only if ∃g ∈ G such that a = gbg
R: ∀a,b ∈ G, aRb if and only if a = b1
Which of the above is/are equivalence relation/relations?
R_{2} only  
R_{1} and R_{2}  
Neither R_{1} and R_{2}  
R_{1} only 
Consider Statement R_{1}:
Reflexive:
aR_{1}a
⇒ a = g^{1}ag
Left multiply both sides by g
⇒ ga = gg^{1}ag
Right multiply both sides by g^{1}
⇒ gag^{1} = gg^{1}agg^{1}
⇒ gag^{1} = a [∴ The relation is reflexive]
Symmetric:
If aR_{1}b, then ∃g ∈ G such that gag^{1} = b then a = g^{1}bg, which is Correct.
⇒ So, given relation is symmetric.
Transitive:
The given relation is Transitive.
So, the given relation R_{1} is equivalence.
R_{2}:
The given relation is not reflexive.
So, which is not equivalence relation.
Such that a ≠ a^{1}.
So, only R_{1} is true.
Question 5 
I. X is invertible.
II. Determinant of X is nonzero.
Which one of the following is TRUE?
I implies II; II does not imply I.  
II implies I; I does not imply II.  
I and II are equivalent statements.  
I does not imply II; II does not imply I. 
That means we can also say that determinant of X is nonzero.
Question 6 
Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to
n!  
1  
(n1)!  
The total number of hamiltonian cycles in a complete graph are
(n1)!/2, where n is number of vertices.
Question 7 
Which of the above statements is/are TRUE?
Only II  
Only I  
Neither I nor II  
Both I and II 
and given A = {(x, X), x∈X and X⊆U}
Possible sets for U = {Φ, {1}, {2}, {1, 2}}
if x=1 then no. of possible sets = 2
x=2 then no. of possible sets = 2
⇒ No. of possible sets for A = (no. of sets at x=1) + (no. of sets at x=2) = 2 + 2 = 4
Consider statement (i) & (ii) and put n=2
Statement (i) is true
Statement (i) and (ii) both are true.
Answer: (C)
Video Explanation
Question 8 
Suppose Y is distributed uniformly in the open interval (1,6). The probability that the polynomial 3x^{2} + 6xY + 3Y + 6 has only real roots is (rounded off to 1 decimal place) _____.
0.3  
0.9  
0.1  
0.8 
3x^{2} + 6xY + 3Y + 6
= 3x^{2} + (6Y)x + (3Y + 6)
which is in the form: ax^{2} + bx + c
For real roots: b^{2}  4ac ≥ 0
⇒ (6Y)^{2}  4(3)(3Y + 6) ≥ 0
⇒ 36Y^{2}  36Y  72 ≥ 0
⇒ Y^{2}  Y  2 ≥ 0
⇒ (Y+1)(Y2) ≥ 0
Y = 1 (or) 2
The given interval is (1,6).
So, we need to consider the range (2,6).
The probability = (1/(61)) * (62) = 1/5 * 4 = 0.8
Question 9 
Consider the first order predicate formula φ:
 ∀x[(∀z zx ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z zw ⇒ ((w = z) ∨ (z = 1)))]
Here 'ab' 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?
S1 and S3  
S1, S2 and S3  
S2 and S3  
S1 and S2 
One of the case:
If 7 is a number which is prime (either divided by 7 or 1 only). then there exists some number like 3 which is larger than 7 also satisfy the property (either divided by 3 or 1 only).
So, S3 is correct
It's true for all integers too.
Question 10 
Consider the following matrix:
The absolute value of the product of Eigen values of R is ______.
12  
17  
10  
8 
Question 11 
The largest eigenvalue of A is ________
3  
4  
5  
6 
→ Correction in Explanation:
⇒ (1  λ)(2  λ)  2 = 0
⇒ λ^{2}  3λ=0
λ = 0, 3
So maximum is 3.
Question 12 
Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equiprobable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is __________.
0.021  
0.022  
0.023  
0.024 
⇾ A person wins who gets lower number compared to other person.
⇾ There could be “tie”, if they get same number.
Favorable cases = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Probability (tie) = 6/36 (when two dice are thrown, sample space = 6 × 6 = 36)
= 1/6
“Find the probability that one of them wins in the third attempt"
⇾ Which means, first & second time it should be tie and third time it should not be tie
⇾ P (tie) * P (tie) * P (not tie)
⇒ 1/6* 1/6 * (1  1/6)
⇒ (5/36×6)
= 0.138/6
= 0.023
Question 13 
The chromatic number of the following graph is _______.
1  
2  
3  
4 
Question 14 
Let G be a finite group on 84 elements. The size of a largest possible proper subgroup of G is _________.
41  
42  
43  
44 
For any group ‘G’ with order ‘n’, every subgroup ‘H’ has order ‘k’ such that ‘n’ is divisible by ‘k’.
Solution:
Given order n = 84
Then the order of subgroups = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}
As per the proper subgroup definition, it should be “42”.
Question 15 
Which one of the following is a closed form expression for the generating function of the sequence {a_{n}}, where a_{n} = 2n+3 for all n = 0, 1, 2, …?
3/(1x)^{2}  
3x/(1x)^{2}  
2x/(1x)^{2}  
3x/(1x)^{2} 
Question 16 
0.289  
0.298  
0.28  
0.29 
Question 17 
Assume that multiplying a matrix G_{1} of dimension p×q with another matrix G_{2} of dimension q×r requires pqr scalar multiplications. Computing the product of n matrices G_{1}G_{2}G_{3}…G_{n} can be done by parenthesizing in different ways. Define G_{i}G_{i+1} as an explicitly computed pair for a given parenthesization if they are directly multiplied. For example, in the matrix multiplication chain G_{1}G_{2}G_{3}G_{4}G_{5}G_{6} using parenthesization(G_{1}(G_{2}G_{3}))(G_{4}(G_{5}G_{6})), G_{2}G_{3} and G_{5}G_{6} are the only explicitly computed pairs.
Consider a matrix multiplication chain F_{1}F_{2}F_{3}F_{4}F_{5}, where matrices F_{1}, F_{2}, F_{3}, F_{4} and F_{5} are of dimensions 2×25, 25×3, 3×16, 16×1 and 1×1000, respectively. In the parenthesization of F_{1}F_{2}F_{3}F_{4}F_{5} that minimizes the total number of scalar multiplications, the explicitly computed pairs is/ are
F_{1}F_{2} and F_{3}F_{4} only
 
F_{2}F_{3} only  
F_{3}F_{4} only  
F_{1}F_{2} and F_{4}F_{5} only

→ Optimal Parenthesization is:
((F_{1}(F_{2}(F_{3} F_{4})))F_{5})
→ But according to the problem statement we are only considering F_{3}, F_{4} explicitly computed pairs.
Question 18 
Consider the firstorder logic sentence
where ψ(s,t,u,v,w,x,y) is a quantifierfree firstorder logic formula using only predicate symbols, and possibly equality, but no function symbols. Suppose φ has a model with a universe containing 7 elements.
Which one of the following statements is necessarily true?
There exists at least one model of φ with universe of size less than or equal to 3.  
There exists no model of φ with universe of size less than or equal to 3.
 
There exists no model of φ with universe of size greater than 7.  
Every model of φ has a universe of size equal to 7. 
"∃" there exists quantifier decides whether a sentence belong to the model or not.
i.e., ~∃ will make it not belong to the model. (1) We have ‘7’ elements in the universe, So max. size of universe in a model = ‘7’
(2) There are three '∃' quantifiers, which makes that a model have atleast “3” elements. So, min. size of universe in model = ‘7’.
(A) is False because: (2)
(B) is true
(C) is false because of (1)
(D) is false, because these all models with size {3 to 7} not only ‘7’.
Question 19 
Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (H), medium (M) and low (L). Let P(H_{G}) denote the probability that Guwahati has high temperature. Similarly, P(M_{G}) and P(L_{G}) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(H_{D}), P(M_{D}) and P(L_{D}) for Delhi.
The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.
Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (H_{G}) then the probability of Delhi also having a high temperature (H_{D}) is 0.40; i.e., P(H_{D} ∣ H_{G}) = 0.40. Similarly, the next two entries are P(M_{D} ∣ H_{G}) = 0.48 and P(L_{D} ∣ H_{G}) = 0.12. Similarly for the other rows.
If it is known that P(H_{G}) = 0.2, P(M_{G}) = 0.5, and P(L_{G}) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______ .
0.60  
0.61  
0.62  
0.63 
The first entry denotes that if Guwahati has high temperature (H_{G} ) then the probability that Delhi also having a high temperature (H_{D} ) is 0.40.
P (H_{D} / H_{G} ) = 0.40
We need to find out the probability that Guwahati has high temperature.
Given that Delhi has high temperature (P(H_{G} / H_{D} )).
P (H_{D} / H_{G} ) = P(H_{G} ∩ H_{D} ) / P(H_{D} )
= 0.2×0.4 / 0.2×0.4+0.5×0.1+0.3×0.01
= 0.60
Question 20 
P: Set of Rational numbers (positive and negative)
Q: Set of functions from {0, 1} to N
R: Set of functions from N to {0, 1}
S: Set of finite subsets of N
Which of the above sets are countable?
Q and S only  
P and S only  
P and R only  
P, Q and S only 
Set of functions from {0,1} to N is countable as it has one to one correspondence to N.
Set of functions from N to {0,1} is uncountable, as it has one to one correspondence to set of real numbers between (0 and 1).
Set of finite subsets of N is countable.
Question 21 
Consider the following statements.
(I)P does not have an inverse
(II)P has a repeated eigenvalue
(III)P cannot be diagonalized
Which one of the following options is correct?
Only I and III are necessarily true  
Only II is necessarily true  
Only I and II are necessarily true  
Only II and III are necessarily true 
Though the multiple of a vector represents same vector, and each eigen vector has distinct eigen value, we can conclude that ‘p’ has repeated eigen value.
If the unique eigen value corresponds to an eigen vector e, but the repeated eigen value corresponds to an entire plane, then the matrix can be diagonalized, using ‘e’ together with any two vectors that lie in plane.
But, if all eigen values are repeated, then the matrix cannot be diagonalized unless it is already diagonal.
So (III) holds correct.
A diagonal matrix can have inverse, So (I) is false.
Then (II) and (III) are necessarily True.
Question 22 
Let G be a graph with 100! vertices, with each vertex labeled by a distinct permutation of the numbers 1, 2, …, 100. There is an edge between vertices u and v if and only if the label of u can be obtained by swapping two adjacent numbers in the label of v. Let y denote the degree of a vertex in G, and z denote the number of connected components in G.
Then, y + 10z = ___________.
109  
110  
111  
112 
There exists edge between two vertices iff label of ‘u’ is obtained by swapping two adjacent numbers in label of ‘v’.
Example:
12 & 21, 23 & 34
The sets of the swapping numbers be (1, 2) (2, 3) (3, 4) … (99).
The no. of such sets are 99 i.e., no. of edges = 99.
As this is regular, each vertex has ‘99’ edges correspond to it.
So degree of each vertex = 99 = y.
As the vertices are connected together, the number of components formed = 1 = z
y + 102 = 99 + 10(1) = 109
Question 23 
I. p ⇒ q
II. q ⇒ p
III. (¬q) ∨ p
IV. (¬p) ∨ q
I only  
I and IV only  
II only  
II and III only 
Construct Truth tables:
~p ⇒ ~q
II, III are equivalent to (~p) ⇒ (~q)
Method 2:
(I) p⇒q ≡ ~p∨q
(II) q⇒p ≡ ~q∨p
(III) (~q) ∨ p ≡ ~q∨p
(IV) (~p) ∨ p ≡ ~p∨q
Also, from question:
(~p) ⇒ (~q)
≡ p∨~q
So, (II) & (III) are equivalent to the statement given in question.
Question 24 
I. ∃y(∃xR(x,y))
II. ∃y(∀xR(x,y))
III. ∀y(∃xR(x,y))
IV. ¬∃x(∀y¬R(x,y))
IV only  
I and IV only  
II only  
II and III only 
F: ∀x(∃yR(x,y)) (given)
: For all girls there exist a boyfriend
(x for girls and y for boys)
I: ∃y(∃xR(x,y))
: There exist some boys who have girlfriends.
(Subset of statement F, so True)
II: ∃y(∀xR(x,y))
: There exists some boys for which all the girls are girlfriend. (False)
III: ∀y(∃xR(x,y))
: For all boys exists a girlfriend. (False)
IV: ~∃x(∀y~R(x,y))
= ∀x(~∀y~R(x,y))
= ∀x(∃yR(x,y)) (∵ ~∀y=∃y, ~∃x=∀x)
(True)
Question 25 
Let c^{1}, c^{n} be scalars not all zero. Such that the following expression holds:
where a_{i} is column vectors in R^{n}. Consider the set of linear equations.
Ax = B.where A = [a_{1}.......a_{n}] and
Then, Set of equations has
a unique solution at x = J_{n} where J_{n} denotes a ndimensional vector of all 1  
no solution  
infinitely many solutions  
finitely many solutions 
AX = B
As given that
and c_{1}&c_{n} ≠ 0
means c_{0}a_{0} + c_{1}a_{1} + ...c_{n}a_{n} = 0, represents that a_{0}, a_{1}... a_{n} are linearly dependent.
So rank of 'A' = 0, (so if ‘B’ rank is = 0 infinite solution, ‘B’ rank>0 no solution) ⇾(1)
Another condition given here is, 'Σa_{i} = b',
so for c_{1}c_{2}...c_{n} = {1,1,...1} set, it is having value 'b',
so there exists a solution if AX = b →(2)
From (1)&(2), we can conclude that AX = B has infinitely many solutions.
Question 26 
Let T be a binary search tree with 15 nodes. The minimum and maximum possible heights of T are:
Note: The height of a tree with a single node is 0.4 and 15 respectively  
3 and 14 respectively  
4 and 14 respectively  
3 and 15 respectively 
The height of a tree with single node is 0.
Minimum possible height is when it is a complete binary tree.
Maximum possible height is when it is a skewed tree left/right.
So the minimum and maximum possible heights of T are: 3 and 14 respectively.
Question 27 
Let X be a Gaussian random variable with mean 0 and variance σ^{2}. Let Y = max(X, 0) where max(a,b) is the maximum of a and b. The median of Y is __________.
0  
1  
2  
3 
Median is a point, where the probability of getting less than median is 1/2 and probability of getting greater than median is 1/2.
From the given details, we can simply conclude that, median is 0. (0 lies exactly between positive and negative values)
Question 28 
is 0  
is 1  
is 1  
does not exist 
If "x=1" is substituted we get 0/0 form, so apply LHospital rule
Substitute x=1
⇒ (7(1)^{6}10(1)^{4})/(3(1)^{2}6(1)) = (710)/(36) = (3)/(3) = 1
Question 29 
Let p, q and r be prepositions and the expression (p → q) → r be a contradiction. Then, the expression (r → p) → q is
a tautology  
a contradiction  
always TRUE when p is FALSE  
always TRUE when q is TRUE 
So r = F and (p→q) = T.
We have to evaluate the expression
(r→p)→q
Since r = F, (r→p) = T (As F→p, is always true)
The final expression is T→q and this is true when q is true, hence option D.
Question 30 
Let u and v be two vectors in R^{2} whose Euclidean norms satisfy u=2v. What is the value of α such that w = u + αv bisects the angle between u and v?
2  
1/2  
1  
1/2 
Let u, v be vectors in R^{2}, increasing at a point, with an angle θ.
A vector bisecting the angle should split θ into θ/2, θ/2
Means ‘w’ should have the same angle with u, v and it should be half of the angle between u and v.
Assume that the angle between u, v be 2θ (thus angle between u,w = θ and v,w = θ)
Cosθ = (u∙w)/(∥u∥ ∥w∥) ⇾(1)
Cosθ = (v∙w)/(∥v∥ ∥w∥) ⇾(2)
(1)/(2) ⇒ 1/1 = ((u∙w)/(∥u∥ ∥w∥))/((v∙w)/(∥v∥ ∥w∥)) ⇒ 1 = ((u∙w)/(∥u∥))/((v∙w)/(∥v∥))
⇒ (u∙w)/(v∙w) = (∥u∥)/(∥v∥) which is given that ∥u∥ = 2 ∥v∥
⇒ (u∙w)/(v∙w) = (2∥v∥)/(∥v∥) = 2 ⇾(3)
Given ∥u∥ = 2∥v∥
u∙v = ∥u∥ ∥v∥Cosθ
=2∙∥v∥^{2} Cosθ
w = u+αv
(u∙w)/(v∙w) = 2
(u∙(u+αv))/(v∙(u+αv)) = 2
(u∙u+α∙u∙v)/(u∙v+α∙v∙v) = 2a∙a = ∥a∥^{2}
4∥v∥^{2}+α∙2∙∥v∥^{2} Cosθ = 2(2∥v∥^{2} Cosθ+α∙∥v∥^{2})
4+2αCosθ = 2(2Cosθ+α)
4+2αCosθ = 4Cosθ+2α ⇒ Cosθ(uv)+2α4 = 0
42α = Cosθ(42α)
(42α)(Cosθ1) = 0
42α = 0
Question 31 
Consider the following statements
(i) One eigenvalue must be in [5, 5].
(ii) The eigenvalue with the largest magnitude must be strictly greater than 5.
Which of the above statements about eigenvalues of A is/are necessarily CORRECT?
Both (I) and (II)  
(I) only  
(II) only  
Neither (I) nor (II) 
be a real valued, rank = 2 matrix.
a^{2}+b^{2}+c^{2}+d^{2} = 50
Square values are of order 0, 1, 4, 9, 16, 25, 36, …
So consider (0, 0, 5, 5) then Sum of this square = 0+0+25+25=50
To get rank 2, the 2×2 matrix can be
The eigen values are,
AλI = 0 (The characteristic equation)
λ(λ)25 = 0
λ^{2}25 = 0
So, the eigen values are within [5, 5], Statement I is correct.
The eigen values with largest magnitude must be strictly greater than 5: False.
So, only Statement I is correct.
Question 32 
The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is _________.
271  
272  
273  
274 
Let A = number divisible by 3
B = numbers divisible by 5
C = number divisible by 7
We need to find “The number of integers between 1 and 500 that are divisible by 3 or 5 or 7" i.e., A∪B∪C
We know,
A∪B∪C = A+B+CA∩BA∩CB∩C+A∩B
A = number of integers divisible by 3
[500/3 = 166.6 ≈ 166 = 166]
B = 100
[500/5 = 100]
C = 71
[500/7 = 71.42]
A∩B = number of integers divisible by both 3 and 5 we need to compute with LCM (15)
i.e.,⌊500/15⌋ ≈ 33
A∩B = 33
A∩C = 500/LCM(3,7) 500/21 = 23.8 ≈ 28
B∩C = 500/LCM(5,3) = 500/35 = 14.48 ≈ 14
A∩B∩C = 500/LCM(3,5,7) = 500/163 = 4.76 ≈ 4
A∪B∪C = A+B+CA∩BA∩CB∩C+A∩B∩C
= 166+100+71332814+4
= 271
Question 33 
If f(x) = Rsin(πx/2) + S, f'(1/2) = √2 and , then the constants R and S are, respectively.
Question 34 
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
(¬p ∧ r) ∧ (¬r → (p ∧ q))  
(¬p ∧ r) ∧ ((p ∧ q) → ¬r)  
(¬p ∧ r) ∨ ((p ∧ q) → ¬r)  
(¬p ∧ r) ∨ (r → (p ∧ q)) 
q: It is cold
r: It is pleasant
“If it is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold.”
We can divide the statement into two parts with “Conjunction”.
i.e., ¬r→(p∧q) ⇾(2)
From (1) & (2), the given statement can be represented as
Question 35 
Consider the set X = {a,b,c,d e} under the partial ordering
 R = {(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}.
The Hasse diagram of the partial order (X,R) is shown below.
The minimum number of ordered pairs that need to be added to R to make (X,R) a lattice is _________.
0  
1  
2  
3 
As per the definition of lattice, each pair should have GLB, LUB.
The given ‘R’ has GLB, LUB for each and every pair.
So, no need to add extra pair.
Thus no. of required pair such that Hasse diagram to become lattice is “0”.
Question 36 
Then the rank of P+Q is _________.
2  
3  
4  
5 
R_{2}→R_{2}+R_{1}
The number of nonzero rows of P + Q = 2,
So Rank = 2
Note: “Rank” is the number of independent vectors.
Method1:
Each vector is a row in matrix.
Echelon form of a matrix has no. of zeroes increasing each rows.
The total nonzero rows left give the rank.
Method2:
Find determinant of matrix, for 3×5, if determinant is ‘0’, the max rank can be 2.
If determinant of any n×n is nonzero, then rows proceed with (n1)×(n1).
Question 37 
G is an undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is ___________.
16  
17  
18  
19 
Degree of each vertex ≥ 3
v = 2E
The relation between max and min degree of graph are
m ≤ 2E / v ≤ M
Given minimum degree = 3
So, 3 ≤2 E / v
3v ≤ 2E
3(n) ≤ 2(25)
n ≤ 50/3
n ≤ 16.6
(n = 16)
Question 38 
P and Q are considering to apply for job. The probability that p applies for job is 1/4. The probability that P applies for job given that Q applies for the job 1/2 and The probability that Q applies for job given that P applies for the job 1/3.The probability that P does not apply for job given that Q does not apply for the job
Probability that ‘P’ applies for the job given that Q applies for the job = P(p/q) = 1/2 ⇾ (2)
Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job = P(p/q) = 1/3 ⇾ (3)
Bayes Theorem:
(P(A/B) = (P(B/A)∙P(A))/P(B) ; P(A/B) = P(A∩B)/P(B))
⇒ P(p/q) = (P(q/p)∙P(p))/p(q)
⇒ 1/2 = (1/3×1/4)/p(q)
p(q) = 1/12×2 = 1/(6) (P(q) = 1/6) ⇾ (4)
From Bayes,
P(p/q) = (P(p∩q))/(P(q))
1/2 = P(p∩q)/(1⁄6)
(p(p∩q) = 1/12)
We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job = P(p'/q')
From Bayes theorem,
P(p'/q') = (P(p'∩q'))/P(q') ⇾ (5)
We know,
p(A∩B) = P(A) + P(B)  P(A∪B)
also (P(A'∩B') = 1  P(A∪B))
P(p'∩q') = 1  P(p∪q)
= 1  (P(p) + P(q)  P(p∩q))
= 1  (P(p) + P(q)  P(p) ∙ P(q))
= 1  (1/4 + 1/6  1/12)
= 1  (10/24  2/24)
= 1  (8/24)
= 2/3
(P(p'∩q') = 2/3) ⇾ (6)
Substitute in (5),
P(p'⁄q') = (2⁄3)/(1P(q)) = (2⁄3)/(11/6) = (2⁄3)/(5⁄6) = 4/5
(P(p'/q') = 4/5)
Question 39 
For any discrete random variable X, with probability mass function P(X=j)=p_{j}, p_{j}≥0, j∈{0, ..., N} and , define the polynomial function . For a certain discrete random variable Y, there exists a scalar β∈[0,1] such that g_{y}(Z)=(1β+βz)^{N}. The expectation of Y is
Nβ(1  β)  
Nβ  
N(1  β)  
Not expressible in terms of N and β alone 
Given g_{y} (z) = (1  β + βz)^{N} ⇾ it is a binomial distribution like (x+y)^{n}
Expectation (i.e., mean) of a binomial distribution will be np.
The polynomial function ,
given
Mean of Binomial distribution of b(x_{j},n,p)=
The probability Mass function,
Given:
Question 40 
If the characteristic polynomial of a 3 × 3 matrix M over R (the set of real numbers) is λ^{3}  4λ^{2} + aλ + 30, a ∈ ℝ, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.
5  
6  
7  
8 
λ^{3}  4λ^{2} + aλ + 30 = 0 ⇾ (1)
One eigen value is ‘2’, so substitute it
2^{3}  4(2)^{2} + a(2) + 30 = 0
8  16 + 2a + 30 = 0
2a = 22
a = 11
Substitute in (1),
λ^{3}  4λ^{2}  11 + 30 = 0
So, (1) can be written as
(λ  2)(λ^{2}  2λ  15) = 0
(λ  2)(λ^{2}  5λ + 3λ  15) = 0
(λ  2)(λ  3)(λ  5) = 0
λ = 2, 3, 5
Max λ=5
Question 41 
p: x ∈ {8,9,10,11,12}
q: x is a composite number
r: x is a perfect square<
s: x is a prime number
The integer x≥2 which satisﬁes ¬((p ⇒ q) ∧ (¬r ∨ ¬s)) is _________.
11  
12  
13  
14 
~((p→q) ∧ (~r ∨ ~S))
⇒ first simplify the given statement by converging them to ∧, ∨
⇒ [~(p→q) ∨ (~(~r ∨ ~s)]
Demorgan’s law:
⇒ [~(~p ∨ q) ∨ (r ∧ s)]
∵ p→q ≡ ~p ∨ q
⇒ [(p ∧ ~q) ∨ (r ∧ s)]
p ∧ ~q is {8,9,10,11,12} ∧ {not a composite number} i.e. {11}
r ∧ s is {perfect square} ∧ {prime} i.e. no answer
So, the one and only answer is 11.
Question 42 
Let a_{n} be the number of nbit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for a_{n}?
a_{n} = a_{(n1)} + 2a_{(n2)}  
a_{n} = a_{(n1)} + a_{(n2)}  
a_{n} = 2a_{(n1)} + a_{(n2)}  
a_{n} = 2a_{(n1)} + 2a_{(n2)} 
If n=1, we have {0,1}
# Occurrences = 2
If n=2, we have {00,01,10}
# Occurrences = 3
If n=3, we have {000,001,010,100,101}
# Occurrences = 5
It is evident that a_{3} = a_{1} + a_{2}
Similarly, a_{n} = a_{n1} + a_{n2}
Question 43 
4  
3  
2  
1 
Question 44 
A probability density function on the interval [a,1] is given by 1/x^{2} and outside this interval the value of the function is zero. The value of a is _________.
0.7  
0.6  
0.5  
0.8 
or
where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x^{2} , a≤x≤1
The area under curve,
 1 + 1/a = 1
1/a = 2
a = 0.5
Question 45 
Two eigenvalues of a 3 × 3 real matrix P are (2 + √1) and 3. The determinant of P is __________.
18  
15  
17  
16 
So, For the given 3×3 matrix there would be 3 eigen values.
Given eigen values are : 2+i and 3.
So the third eigen value should be 2i.
As per the theorems, the determinant of the matrix is the product of the eigen values.
So the determinant is (2+i)*(2i)*3 = 15.
Question 46 
The coefﬁcient of x^{12} in (x^{3} + x^{4} + x^{5} + x^{6} + ...)^{3} is _________.
10  
11  
12  
13 
⇒ [x^{3}(1 + x + x^{2} + x^{3} + ...)]^{3}
= x^{9}(1 + x + x^{2} + x^{3} + ...)^{3}
First Reduction:
As x^{9} is out of the series, we need to find the coefficient of x^{3} in (1 + x + x^{2} + ⋯)^{3}
Here, m=3, k=3, the coefficient
= ^{5}C_{3} = 5!/2!3! = 10
Question 47 
Consider the recurrence relation a_{1} = 8, a_{n} = 6n^{2} + 2n + a_{n1}. Let a_{99} = K × 10^{4}. The value of K is ___________.
198  
199  
200  
201 
Replace a_{(n1)}
⇒ a_{n} = 6n^{2} + 2n + 6(n1)^{2} + 2(n1) + 6(n2)^{2} + 2(n2) + ⋯ a_{1}
Given that a_{1} = 8, replace it
⇒ a_{n} = 6n^{2} + 2n + 6(n1)^{2} + 2(n1) + 6(n2)^{2} + 2(n2) + ⋯8
= 6n^{2} + 2n + 6(n1)^{2} + 2(n1) + 6(n2)^{2} + 2(n2) + ⋯ + 6(1)^{2} + 2(1)
= 6(n^{2} + (n1)^{2} + (n2)^{2} + ⋯ + 2^{2} + 1^{2}) + 2(n + (n1) + ⋯1)
Sum of n^{2} = (n(n+1)(2n+1))/6
Sum of n = (n(n+1))/2
= 6 × (n(n+1)(2n+1))/6 + 2×(n(n+1))/2
= n(n+1)[1+2n+1]
= n(n+1)[2n+2]
= 2n(n+1)^{2}
Given a_{99} = k×10^{4}
a_{99} = 2(99)(100)^{2} = 198 × 10^{4}
∴k = 198
Question 48 
f(n) = f(n/2) if n is even
f(n) = f(n+5) if n is odd
Let R = {i∃j: f(j)=i} be the set of distinct values that f takes. The maximum possible size of R is __________.
2  
3  
4  
5 
f(n)= f(n+5) if n is odd
We can observe that
and f(5) = f(10) = f(15) = f(20)
Observe that f(11) = f(8)
f(12) = f(6) = f(3)
f(13) = f(9) = f(14) = f(7) = f(12) = f(6) = f(3)
f(14) = f(9) = f(12) = f(6) = f(3)
f(16) = f(8) = f(4) = f(2) = f(1) [repeating]
So, we can conclude that
‘R’ can have size only ‘two’ [one: multiple of 5’s, other: other than 5 multiples]
Question 49 
Step1. Flip a fair coin twice.
Step2. If the outcomes are (TAILS, HEADS) then output Y and stop.
Step3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop.
Step4. If the outcomes are (TAILS, TAILS), then go to Step 1.
The probability that the output of the experiment is Y is (up to two decimal places) ________.
0.33  
0.34  
0.35  
0.36 
Stop conditions:
If outcome = TH then Stop [output 4]  (1)
else
outcome = HH/ HT then Stop [output N]  (2)
We get ‘y’ when we have (1) i.e., ‘TH’ is output.
(1) can be preceded by ‘TT’ also, as ‘TT’ will reset (1) again
Probability of getting y = TH + (TT)(TH) + (TT)(TT)(TH) + …
= 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + ...
= (1/4)/(11/4)
= 1/3
= 0.33
Question 50 
(i) false
(ii) Q
(iii) true
(iv) P ∨ Q
(v) ¬Q ∨ P
The number of expressions given above that are logically implied by P ∧ (P ⇒ Q) is _________.
4  
5  
6  
7 
(P ∧ (P → Q))→ expression is a tautology. So we have to find
How many tautological formulas are there for the given inputs.
(P ∧ (P → Q)) → True is always tautology
(P ∧ (P → Q)) → False is not a tautology
(P ∧ (P → Q)) → Q is a tautology
(P ∧ (P → Q)) → ¬Q ∨ P is a tautology
(P ∧ (P → Q)) → P ∨ Q is a tautology
So there are 4 expressions logically implied by (P ∧ (P → Q))
Question 51 
Let f(x) be a polynomial and g(x) = f'(x) be its derivative. If the degree of (f(x) + f(x)) is 10, then the degree of (g(x)  g(x)) is __________.
9  
10  
11  
12 
It is given that f(x) + f(x) degree is 10.
It means f(x) is a polynomial of degree 10.
Then obviously the degree of g(x) which is f’(x) will be 9.
Question 52 
The minimum number of colours that is sufﬁcient to vertexcolour any planar graph is ________.
4  
5  
6  
7 
Here it is asked about the sufficient number of colors, so with the worst case of 4 colors we can color any planar graph.
Question 53 
I. If m < n, then all such systems have a solution
II. If m > n, then none of these systems has a solution
III. If m = n, then there exists a system which has a solution
Which one of the following is CORRECT?
I, II and III are true  
Only II and III are true  
Only III is true  
None of them is true 
If R(A) ≠ R(AB)
then there will be no solution.
ii) False, because if R(A) = R(AB),
then there will be solution possible.
iii) True, if R(A) = R(AB),
then there exists a solution.
Question 54 
Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is _________.
0.55  
0.56  
0.57  
0.58 
The bulbs of Type 1, Type 2 are same in number.
So, the probability to choose a type is 1/2.
The probability to choose quadrant ‘A’ in diagram is
P(last more than 100 hours/ type1) = 1/2 × 0.7
P(last more than 100 hours/ type2) = 1/2 × 0.4
Total probability = 1/2 × 0.7 + 1/2 × 0.4 = 0.55
Question 55 
Suppose that the eigenvalues of matrix A are 1, 2, 4. The determinant of (A^{1})^{T} is _________.
0.125  
0.126  
0.127  
0.128 
Given that eigen values are 1, 2, 4.
So, its determinant is 1*2*4 = 8
The determinant of (A^{1})^{T} = 1/ A^{T} = 1/A = 1/8 = 0.125
Question 56 
P:R is reﬂexive
Q:R is transitive
Which one of the following statements is TRUE?
Both P and Q are true.  
P is true and Q is false.  
P is false and Q is true.  
Both P and Q are false. 
a≤c ∨ b≤d
Let a≤a ∨ b≤b is true for all a,b ∈ N
So there exists (a,a) ∀ a∈N.
It is Reflexive relation.
Consider an example
c = (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f)
This does not hold for any (a>e) or (b>f)
eg:
(2,2)R(1,2) as 2≤2
(1,2)R(1,1) as 1≤1
but (2,2) R (1,1) is False
So, Not transitive.
Question 57 
Which one of the following wellformed formulae in predicate calculus is NOT valid?
(∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬p(x) ∨ ∀x q(x))  
(∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))  
∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))  
∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x)) 
But in option (D), we can generate T → F.
Hence, not valid.
Question 58 
I. Each compound in U\S reacts with an odd number of compounds.
U\S reacts with an odd number of compounds.
U\S reacts with an even number of compounds.
ALWAYS TRUE?
Only I  
Only II  
Only III  
None 
U = 23
∃S ∋ (S⊂U)
Each component in ‘S’ reacts with exactly ‘3’ compounds of U,
If a component ‘a’ reacts with ‘b’, then it is obvious that ‘b’ also reacts with ‘a’.
It’s a kind of symmetric relation.>br> If we connect the react able compounds, it will be an undirected graph.
The sum of degree of vertices = 9 × 3 = 27
But, in the graph of ‘23’ vertices the sum of degree of vertices should be even because
(d_{i} = degree of vertex i.e., = no. of edges)
But ‘27’ is not an even number.
To make it an even, one odd number should be added.
So, there exists atleast one compound in U/S reacts with an odd number of compounds.
Question 59 
The value of the expression 13^{99}(mod 17), in the range 0 to 16, is ________.
4  
5  
6  
7 
a^{(p1)} ≡ 1 mod p (p is prime)
From given question,
p = 17
a^{(171)} ≡ 1 mod 17
a^{16} ≡ 1 mod 17
13^{16} ≡ 1 mod 17
Given:
13^{99} mod 17
13^{3} mod 17
2197 mod 17
4
Question 60 
In the LU decomposition of the matrix ,if the diagonal elements of U are both 1, then the lower diagonal entry l_{22} of L is
5  
6  
7  
8 
l_{11} = 2 (1)
l_{11}u_{12} = 2
u_{12} = 2/2
u_{12} = 1  (2)
l_{21} = 4  (3)
l_{21}u_{12}+l_{22} = 9
l_{22} = 9  l_{21}u_{12} = 9  4 × 1 = 5
Question 61 
If g(x) = 1  x and h(x) , then is:
h(x)/g(x)  
1/x  
g(x)/h(x)  
x/(1x)^{2} 
Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1h(x)=1x/x1=1/x1
h(g(x))=g(x)/g(x)1=1x/x
⇒ g(h(x))/h(g(x))=x/(x1)(1x)=(x/x1)/1x=h(x)/g(x)
Question 62 
Suppose L = {p, q, r, s, t} is a lattice represented by the following Hasse diagram:
For any x, y ∈ L, not necessarily distinct, x ∨ y and x ∧ y are join and meet of x, y respectively. Let L^{3} = {(x,y,z): x, y, z ∈ L} be the set of all ordered triplets of the elements of L. Let p_{r} be the probability that an element (x,y,z) ∈ L^{3} chosen equiprobably satisfies x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). Then
pr = 0  
pr = 1  
0 < pr ≤ 1/5  
1/5 < pr < 1 
Let A be the event that an element (x,y,z)∈ L^{3} satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s) = t∧t = t q∨(r∧s) ≠ (q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 1256 = 119
∴ required probability is 119/125
⇒ 1/5
Question 63 
Consider the following 2 × 2 matrix A where two elements are unknown and are marked by a and b. The eigenvalues of this matrix are –1 and 7. What are the values of a and b?
a=6, b=4  
a=4, b=6  
a=3, b=5  
a=5, b=3 
By properties,
⇒ 6=1+a and 7=a4b
⇒ a=5 ⇒ 7=54b
⇒ b=3
Question 64 
Let G = (V, E) be a simple undirected graph, and s be a particular vertex in it called the source. For x ∈ V, let d(x)denote the shortest distance in G from s to x. A breadth first search (BFS) is performed starting at s. Let T be the resultant BFS tree. If (u, v) is an edge of G that is not in T, then which one of the following CANNOT be the value of d(u)  d(v) ?
1  
0  
1  
2 
Then the difference between the d(u) and d(v) is not more than '1'.
In the option 'D' the difference is given as '2' it is not possible in the undirected graph.
Question 66 
Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three, then the number of edges in G is _______________.
24  
25  
26  
27 
V + R = E + 2 (1) where V, E, R are respectively number of vertices, edges and faces (regions)
Given V = 10 (2) and number of edges on each face is three
∴3R = 2E ⇒ R = 2/3E (3)
Substituting (2), (3) in (1), we get
10 + 2/3E = E + 2 ⇒ E/3 = 8 ⇒ E = 24
Question 67 
Consider the operations f(X, Y, Z) = X'YZ + XY' + Y'Z' and g(X′, Y, Z) = X′YZ + X′YZ′ + XY Which one of the following is correct?
Both {f} and {g} are functionally complete  
Only {f} is functionally complete  
Only {g} is functionally complete
 
Neither {f} nor {g} is functionally complete 
f(X,X,X) = X'XX'+XX'+X'X'
= 0+0+X'
= X'
Similarly, f(Y,Y,Y) = Y' and f(X,Z,Z) = Z'
f(Y',Y',Z') = (Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
= YY'Z'+Y'Y+YZ
= 0+0+YZ
= YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z) = X'YZ+X'YZ'+XY
g(X,X,X) = X'XX+X'XZ'+XX
= 0+0+X
= X
Similarly, g(Y,Y,Y) = Y and g(Z,Z,Z) = Z
NOT is not derived. Hence, g is not functionally complete.
Question 68 
0.99  
1.00  
2.00  
3.00 
= 21/1(2)+32/2(3)+43/3(4)+…+10099/99(100)
= 1/11/2+1/21/3+1/3…+1/981/99+1/991/100
= 11/100
= 99/100
= 0.99
Question 69 
Let R be the relation on the set of positive integers such that aRb if and only if a and b are distinct and have a common divisor other than 1. Which one of the following statements about R is true?
R is symmetric and reflexive but not transitive  
R is reflexive but not symmetric and not transitive  
R is transitive but not reflexive and not symmetric  
R is symmetric but not reflexive and not transitive 
In aRb, 'a' and 'b' are distinct. So it can never be reflexive.
Symmetric:
In aRb, if 'a' and 'b' have common divisor other than 1, then bRa, i.e., 'b' and 'a' also will have common divisor other than 1. So, yes symmetric.
Transitive:
Take (3, 6) and (6, 2) elements of R. For transitivity (3, 2) must be the element of R, but 3 and 2 don't have a common divisor. So not transitive.
Question 70 
Consider the following statements:

S1: If a candidate is known to be corrupt, then he will not be elected.
S2: If a candidate is kind, he will be elected.
Which one of the following statements follows from S1 and S2 as per sound inference rules of logic?
If a person is known to corrupt, he is kind  
If a person is not known to be corrupt, he is not kind
 
If a person is kind, he is not known to be corrupt
 
If a person is not kind, he is not known to be corrupt 
q: candidate will be elected
r: candidate is kind
then S1 = p→~q
= q→~p (conrapositive rule)
and S2: r→q ⇒ r→~p (transitive rule)
i.e., If a person is kind, he is not known to be corrupt. ∴ Option is C
Question 71 
The larger of the two eigenvalues of the matrix is _________.
6  
7  
8  
9 
⇒ λ^{2}  5λ  6 = 0 ⇒ (λ  6)(λ + 1) = 0 ⇒ λ = 6, 1
∴ Larger eigen value is 6.
Question 72 
The cardinality of the power set of {0, 1, 2, … 10} is _________.
2046  
2047  
2048  
2049 
Question 73 
The number of divisors of 2100 is ______.
36  
37  
38  
39 
= 2^{2}+3×5^{2}×7 (i.e., product of primes)
Then the number of divisions of 2100 is
(2+1)∙(1+1)∙(2+1)∙(1+1) i.e., (3)(2)(3)(2) i.e., 36.
Question 74 
In a connected graph, a bridge is an edge whose removal disconnects a graph. Which one of the following statements is true?
A tree has no bridges  
A bridge cannot be part of a simple cycle  
Every edge of a clique with size 3 is a bridge (A clique is any complete sub graph of a graph)  
A graph with bridges cannot have a cycle 
∴ (A) is false
Since, every edge in a complete graph kn(n≥3) is not a bridge ⇒
(C) is false
Let us consider the following graph G:
This graph has a bridge i.e., edge ‘e’ and a cycle of length ‘3’
∴ (D) is false
Since, in a cycle every edge is not a bridge
∴ (B) is true
Question 75 
Consider six memory partitions of sizes 200 KB, 400 KB, 600 KB, 500 KB, 300 KB and 250 KB, where KB refers to kilobyte. These partitions need to be allotted to four processes of sizes 357 KB, 210KB, 468 KB and 491 KB in that order. If the best fit algorithm is used, which partitions are NOT allotted to any process?
200KBand 300 KB  
200KBand 250 KB  
250KBand 300 KB  
300KBand 400 KB 
Since Best fit algorithm is used. So, process of size,
357KB will occupy 400KB
210KB will occupy 250KB
468KB will occupy 500KB
491KB will occupy 600KB
So, partitions 200KB and 300KB are NOT alloted to any process.
Question 76 
The number of onto function (surjective function) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is ______.
36  
37  
38  
39 
m = 4, n = 3 ⇒ number of onto function is
Question 77 
Perform the following operations on the matrix
 (i) add the third row to the second row
(ii) Subtract the third column from the first column
The determinant of the resultant matrix is ________.
0  
1  
2  
3 
Method 2: Determinant is unaltered by the operations (i) and (ii)
∴ Determinant of the resultant matrix = Determinant of the given matrix
(Since C_{1}, C_{3} are proportional i.e., C_{3} = 15C_{1})
Question 78 
Which one of the following well formed formulae is a tautology?
∀x ∃y R(x,y) ↔ ∃y ∀x R(x,y)
 
(∀x [∃y R(x,y) → S(x,y)]) → ∀x∃y S(x,y)
 
[∀x ∃y (P(x,y) → R(x,y)] ↔ [∀x ∃y (¬ P(x,y)∨R(x,y)]  
∀x ∀y P(x,y) → ∀x ∀y P(y,x) 
[∀x ∃y (P(x,y) → R(x,y)] ↔ [∀x ∃y (¬ P(x,y)∨R(x,y)] is a tautology.
Question 79 
A graph is selfcomplementary if it is isomorphic to its complement for all selfcomplementary graphs on n vertices, n is
A multiple of 4  
Even  
Odd  
Congruent to 0 mod 4, or, 1 mod 4

Question 80 
The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates x_{a} and x_{b} for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if f(x_{b}) is very small and then x_{b} is the solution. The procedure is given below. Observe that there is an expression which is missing and is marked by? Which is the suitable expression that is to be put in place of? So that it follows all steps of the secant method?
Secant
Initialize: x_{a}, x_{b}, ε, N // ε = convergence indicator f_{b} = f(x_{b}) i = 0 while (i < N and f_{b} > ε) do i = i + 1 // update counter x_{t} = ? // missing expression for // intermediate value x_{a} = x_{b} // reset x_{a} x_{b} = x_{t} // reset x_{b} f_{b} = f(x_{b}) // function value at new x_{b} end while if f_{b} > ε then // loop is terminated with i = N write “Nonconvergence” else write “return x_{b}” end if
x_{b} – (f_{b}–f(x_{a}))f_{b} /(x_{b}–x_{a})  
x_{a} – (f_{a}–f(x_{a}))f_{a} /(x_{b}–x_{a})  
x_{b} – (x_{b}–x_{a})f_{b} /(f_{b}–f(x_{a}))  
x_{a} – (x_{b}–x_{a}) f_{a} /(f_{b}–f(x_{a})) 
The first two iterations of the secant method. The red curve shows the function f and the blue lines are the secants. For this particular case, the secant method will not converge.
Question 81 
Let f(x) = x ^{(1/3)} and A denote the area of the region bounded bu f(x) and the Xaxis, when x varies from 1 to 1. Which of the following statements is/are TRUE?

I) f is continuous in [1,1]
II) f is not bounded in [1,1]
III) A is nonzero and finite
II only  
III only  
II and III only  
I, II and III 
∴ f is not bounced in [1, 1] and hence f is not continuous in [1, 1].
∴ Statement II & III are true.
Question 82 
Let X and Y denote the sets containing 2 and 20 distinct objects respectively and F denote the set of all possible functions defined from X to Y. Let f be randomly chosen from F. The probability of f being onetoone is ______.
0.95  
0.96  
0.97  
0.98 
Number of functions from X to Y is 20^{2} i.e., 400 and number of oneone functions from X to Y is ^{20}P_{2} i.e., 20×19 = 380
∴ Probability of a function f being oneone is 380/400 i.e., 0.95
Question 83 
In a room there are only two types of people, namely Type 1 and Type 2. Type 1 people always tell the truth and Type 2 people always lie. You give a fair coin to a person in that room, without knowing which type he is from and tell him to toss it and hide the result from you till you ask for it. Upon asking, the person replies the following:
“The result of the toss is head if and only if I am telling the truth.”
Which of the following options is correct?
The result is head  
The result is tail
 
If the person is of Type 2, then the result is tail  
If the person is of Type 1, then the result is tail 
Case 1:
The person who speaks truth. This definitely implies that result of toss is Head.
Case 2:
The person who lies. In this the reality will be the negation of the statement.
The negation of (x⇔y) is exactly one of x or y holds. "The result of the toss is head if and only if I am telling the truth". So here two possibilities are there,
→ It is head and lie spoken.
→ It is not head and truth spoken.
Clearly, the second one cannot speaks the truth. So finally it is head.
Hence, option (A).
Question 84 
Suppose U is the power set of the set S = {1, 2, 3, 4, 5, 6}. For any T ∈ U, let T denote the number of element in T and T' denote the complement of T. For any T, R ∈ U, let TR be the set of all elements in T which are not in R. Which one of the following is true?
∀X ∈ U (X = X')  
∃X ∈ U ∃Y ∈ U (X = 5, Y = 5 and X ∩ Y = ∅)  
∀X ∈ U ∀Y ∈ U (X = 2, Y = 3 and X \ Y = ∅)  
∀X ∈ U ∀Y ∈ U (X \ Y = Y' \ X') 
(A) False. Consider X = {1,2}. Therefore, X' = {3,4,5,6}, X = 2 and X' = 4.
(B) False. Because for any two possible subsets of S with 5 elements should have atleast 4 elements in common. Hence X∩Y cannot be null.
(C) False. Consider X = {1,4}, Y= {1,2,3} then X\Y = {4} which is not null.
(D) True. Take any possible cases.
Question 85 
The number of 4 digit numbers having their digits in nondecreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is _____.
15  
16  
17  
18 
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
Hence, total 15 4digit no. are possible.
Question 86 
In the given matrix, one of the eigenvalues is 1. the eigenvectors corresponding to the eigenvalue 1 are
⎡ 1 1 2 ⎤ ⎢ 0 1 0 ⎥ ⎣ 1 2 1 ⎦
{α(4,2,1)  α≠0, α∈R}  
{α(4,2,1)  α≠0, α∈R}  
{α(2,0,1)  α≠0, α∈R}  
{α(2,0,1)  α≠0, α∈R} 
AX = λX ⇒ (A  I)X = 0
⇒ y+2z = 0 and x+2y = 0
⇒ y = 2z and x/(2) = y
∴ x/(2) = y = 2z ⇒ x/(4) = y/2 = z/1 = α(say)
∴ Eigen vectors are {α(4,2,1  α≠0, α∈R}
Question 87 
Consider a machine with a byte addressable main memory of 2^{20} bytes, block size of 16 bytes and a direct mapped cache having 2^{12} cache lines. Let the addresses of two consecutive bytes in main memory be (E201F)_{16} and (E2020)_{16}. What are the tag and cache line address (in hex) for main memory address (E201F)_{16}?
E, 201  
F, 201  
E, E20  
2, 01F 
No. of cache lines in cache is 2^{12} bytes which needs 12 bits. So next lower 12 bits are line indexing bits.
And the remaining top 4 bits are tag bits (out of 20). So answer is (A).
Question 88 
If for nonzero x, where a≠b then is
af(x) + bf(1/x) = 1/x  25  (1)
Put x = 1/x,
af(1/x) + bf(x) = x  25  (2)
Multiply equation (1) with 'a' and Multiply equation (2) with 'b', then
abf(1/x) + a^{2} = a/x  25a  (3)
abf(1/x) + b^{2} = bk  25b  (4)
Subtract (3)  (4), we get
(a^{2}  b^{2}) f(x) = a/x 25a  bx + 25b
f(x) = 1/(a^{2}  b^{2}) (a/x  25a  bx +25b)
Now from equation,
Hence option (A) is the answer.
Question 89 
The velocity v (in kilometer/minute) of a motorbike which starts from rest, is given at fixed intervals of time t(in minutes) as follows:
t 2 4 6 8 10 12 14 16 18 20 v 10 18 25 29 32 20 11 5 2 0
The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson’s 1/3rd rule is _________.
309.33  
309.34  
309.35  
309.36 
∵v = velocity
= 2/3[(0+0)+4(10+25+32+11+2)+2(18+29+20+5)]
= 309.33 km
(Here length of each of the subinterval is h = 2)
Question 90 
If the following system has nontrivial solution,
 px + qy + rz = 0
qx + ry + pz = 0
rx + py + qz = 0
then which one of the following options is True?
pq+r = 0 or p = q = r  
p+qr = 0 or p = q = r  
p+q+r = 0 or p = q = r  
pq+r = 0 or p = q = r 
Question 91 
Let R be a relation on the set of ordered pairs of positive integers such that ((p,q),(r,s)) ∈ R if and only if p  s = q  r. Which one of the following is true about R?
Both reflexive and symmetric  
Reflexive but not symmetric  
Not reflexive but symmetric  
Neither reflexive nor symmetric 
∴(p,q) R (p,q)
⇒ R is not reflexive.
Let (p,q) R (r,s) then ps = qr
⇒ rq = sp
⇒ (r,s) R (p,q)
⇒ R is symmetric.
Question 92 
Let G = (V,E) be a directed graph where V is the set of vertices and E the set of edges. Then which one of the following graphs has the same strongly connected components as G?
G_{1}=(V,E_{1}) where E_{1}={(u,v)(u,v)∉E}  
G_{2}=(V,E_{2} )where E_{2}={(u,v)│(u,v)∈E}  
G_{3}=(V,E_{3}) where E_{3}={(u,v)there is a path of length≤2 from u to v in E}  
G_{4}=(V_{4},E) where V_{4} is the set of vertices in G which are not isolated 
→ It strongly connected.
(A) G_{1}=(V,E_{1}) where E_{1}={(u,v)(u,v)∉E}
If (u, v) does not belong to the edge set ‘E’, then it indicates there are no edges. So, it is not connected.
(B) G_{2}=(V,E_{2} )where E_{2}={(u,v)│(u,v)∈E}
Given that ‘G’ is directed graph, i.e., it has path from each vertex to every other vertex.
Though direction is changed from (u, v) to (v, u), it is still connected component same as ‘G’.
(C) G_{3}=(V,E_{3}) where E_{3}={(u,v)there is a path of length≤2 from u to v in E}
This can also be true.
eg:
Both from each vertex to other vertex is also exists. So it is also strongly connected graph.
(D) G_{4}=(V_{4},E) where V_{4} is the set of vertices in G which are not isolated.
If ‘G’ has same ‘x’ no. of isolated vertices, one strongly connected component
then no. of SCC = x + 1
G_{4} contain only ‘1’ component, which is not same as G.
Question 93 
The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4by4 symmetric positive definite matrix is ________.
0  
1  
2  
3 
The finite dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix.
Question 94 
Let the function
where and f(θ) denote the derivative of f with respect to θ. Which of the following is/are TRUE?

(I) There exists such that such that f(θ)=0.
(II) There exists such that such that f(θ)≠0.
I only  
II only  
Both I and II  
Neither I nor II 
Rolle’s theorem states that for any continuous, differentiable function that has two equal values at two distinct points, the function must have a point on the function where the first derivative is zero. The technical way to state this is: if f is continuous and differentiable on a closed interval [a,b] and if f(a) = f(b), then f has a minimum of one value c in the open interval [a, b] so that f'(c) = 0.
We can observe that, sin, cos are continuous, but, Tan is not continuous at π/2. As the mentioned interval does not contain π/2, we can conclude that it is continuous.
As per Rolls theorem both statement 1 and statement 2 are true.
Question 95 
The function f(x) = x sinx satisfies the following equation: f''(x) + f(x) + tcosx = 0. The value of t is __________.
2  
3  
4  
5 
f ’(x) = x(Sinx)’ + Sin(x)(x’)
= xCosx + Sinx ①
f ’’(x) = x (Cosx)’ + Cos (x)’+ Cos x
= x Sinx + 2Cosx ②
Given: f ’’(x) + f(x) + t Cosx = 0
Replace ① & ②,
xSinx + 2Cosx + xSinx + tCosx = 0
2Cosx + tCosx = 0
t = 2
Question 96 
A function f(x) is continuous in the interval [0,2]. It is known that f(0) = f(2) = 1 and f(1) = 1. Which one of the following statements must be true?
There exists a y in the interval (0,1) such that f(y) = f(y+1)  
For every y in the interval (0,1), f(y) = f(2y)  
The maximum value of the function in the interval (0,2) is 1  
There exists a y in the interval (0, 1) such that f(y) = f(2y) 
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1]
g(0) = 2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0,1]. Therefore at some point g would be 0 in (0,1).
g=0 ⇒ f(y) = f(y+1) for some y in (0,1).
Apply similar logic to option D, Let g(y) = f(y) + f(2  y)
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1] (sum of two continuous functions is continuous)
g(0) = 2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0, 1]. Therefore at some point g would be 0 in (0, 1).
There exists y in the interval (0, 1) such that:
g=0 ⇒ f(y) = f(2 – y)
Both A, D are answers.
Question 97 
Four fair sixsided dice are rolled. The probability that the sum of the results being 22 is X/1296. The value of X is ___________.
10  
11  
12  
13 
To get ‘22’ as Sum of four outcomes
x_{1} + x_{2} + x_{3} + x_{4} = 22
The maximum Sum = 6+6+6+6 = 24 which is near to 22
So, keeping three 6’s, 6+6+6+x = 22
x = 4 combination① = 6 6 6 4
Keeping two 6’s, 6+6+x_{1}+x_{2} = 22
x_{1}+x_{2} = 10 possible x’s (5, 5) only
combination② = 6 6 5 5
No. of permutation with 6664 = 4!/ 3! = 4
“ “ “ 6655 = 4!/ 2!2! = 6
Total = 4+6 = 10 ways out of 6×6×6×6 = 1296
Pnb (22) = 10/1296 ⇒ x = 10
Question 98 
A pennant is a sequence of numbers, each number being 1 or 2. An npennant is a sequence of numbers with sum equal to n. For example, (1,1,2) is a 4pennant. The set of all possible 1pennants is {(1)}, the set of all possible 2pennants is {(2), (1,1)}and the set of all 3pennants is {(2,1), (1,1,1), (1,2)}. Note that the pennant (1,2) is not the same as the pennant (2,1). The number of 10pennants is ________.
89  
90  
91  
92 
Single two: 211111111 ⇒ 9!/8!1! = 9 pennants
Two twos: 22111111 ⇒ 8!/6!2! = 28
Three twos: 2221111 ⇒ 7!/3!4! = 35
Four twos: 222211 ⇒ 6!/4!2! = 15
Five twos: 22222 ⇒ 1
Total = 89 pennants.
Question 99 
Let S denote the set of all functions f:{0,1}^{4 }→ {0,1}. Denote by N the number of functions from S to the set {0,1}. The value of log_{2}log_{2}N is ______.
16  
17  
18  
19 
{0,1}^{4} = {0,1}×{0,1}×{0,1}×{0,1} = 16
S = 2^{16}
N = 2^{S}
loglogN=loglog2^{S} = log S = log2^{16} = 16
Question 100 
Consider an undirected graph where selfloops are not allowed. The vertex set of G is {(i,j): 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a,b) and (c,d) if a  c ≤ 1 and b  d ≤ 1. The number of edges in this graph is __________.
506  
507  
508  
509 
If we observe the graph, it looks like a 12 by 12 grid. Each corner vertex has a degree of 3 and we have 4 corner vertices. 40 external vertices of degree 5 and remaining 100 vertices of degree 8.
From Handshaking theorem, sum of the degrees of the vertices is equal to the 2*number of edges in the graph.
⇒ (4*3) + (40*5) + (100*8) = 2*E
⇒ 1012 = 2*E
⇒ E = 506
Question 101 
An ordered tuple (d_{1}, d_{2}, ..., d_{n}) with d_{1} ≥ d_{2} ≥ ... d_{n} is called graphic if there exists a simple undirected graph with n vertices having degrees d_{1}, d_{2}, ..., d_{n} respectively. Which of the following 6tuples is NOT graphic?
(1, 1, 1, 1, 1, 1)  
(2, 2, 2, 2, 2, 2)  
(3, 3, 3, 1, 0, 0)  
(3, 2, 1, 1, 1, 0) 
A) (1, 1, 1, 1, 1, 1)
Yes, it is a graph.
We will see that option (C) is not graphic.
Question 102 
Which one of the following propositional logic formulas is TRUE when exactly two of p, q, and r are TRUE?
((p↔q)∧r)∨(p∧q∧∼r)  
(∼(p↔q )∧r)∨(p∧q∧∼r)  
((p→q)∧r)∨(p∧q∧∼r)  
(∼(p↔q)∧r)∧(p∧q∧∼r) 
Method2: directly check with one of {TTF, TFT, FTT} options.
As there are two T’s in each option, replace them and check with the third value.
Eg: Place p=q= T
(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔T)∧r)∨(T∧T∧∼r)
=(∼(T)∧r)∨(T∧∼r)
=(F∧r)∨(T∧∼r)
=(F)∨(∼r)
=∼r
This is true for r=F.
Similarly with p=r=T and q=F.
q=r=T and p=F
Option B is the answer.
Question 103 
The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p =_____________.
11.90  
11.91  
11.92  
11.93 
Out of which ‘4’ are chosen at random without replaced and created.
If at least three of them are working then system is deemed functional
i.e., there should be only ‘one’ nonworking system in set of ‘4’.
It is possible with combination
W – W – W – N,
W – W – N – W,
W – N – W – W,
N – W – W – W.
For W – W – W – N, the probability = (choosing working out of 10) × (choosing working out of 9) × (choosing working out of 8) × (choosing nonworking out of 7)
= 4/10×3/9×2/8×1/7
where 4/10 ⇒ 4 working out of 10
3/9 ⇒ 3 working are remaining out of ‘9’ as ‘1’ is already taken
For ‘4’ Sum combinations
Total probability = 4×[4/10×3/9×2/8×1/7] = 600/5040
We need 100p ⇒ 100×600/5040 = 11.90
Question 104 
Each of the nine words in the sentence "The quick brown fox jumps over the lazy dog" is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)
3.9  
4.0  
4.1  
4.2 
There are ‘9’ words in this sentence.
No. of characters in each word
The (3)
quick (5)
brown (5)
fox (3)
jumps (5)
over (4)
the (3)
lazy (4)
dog (3)
Each word has equal probability.
So expected length = 3×1/9+5×1/9+5×1/9+3×1/9+5×1/9+ 4×1/9+3×1/9+4×1/9+3×1/9
= 35/9
= 3.9
Question 105 
The maximum number of edges in a bipartite graph on 12 vertices is ______.
36  
37  
38  
39 
Total no. of edges = 6×6 = 36
Question 106 
If the matrix A is such that
then the determinant of A is equal to
0  
1  
2  
3 
Question 107 
A nonzero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?
f(0)f(4) < 0  
f(0)f(4) > 0  
f(0) + f(4) > 0  
f(0) + f(4) < 0 
Polynomial will be
f(x) = (x1)(x2)(x3)
f(0) = 1 × 2 × 3 = 6
f(4) = 3 × 2 × 1 = 6
f(0) ∙ f(4) =  36
f(0) + f(4) = 6  6 = 0
Option (A) is correct.
Question 108 
In the NewtonRaphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2 … is obtained for the function
0.75x^{3} – 2x^{2} – 2x + 4 = 0
Consider the statements
(I) x_{3} = 0. (II) The method converges to a solution in a finite number of iterations.
Which of the following is TRUE?
Only I  
Only II  
Both I and II  
Neither I nor II 
Question 109 
The product of the nonzero eigenvalues of the matrix
1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 1
is ______.
6  
7  
8  
9 
AX = λX
x_{1} + x_{5} = λx_{1}  (1)
x_{1} + x_{5} = λx_{5}  (2)
(1) + (2) ⇒ 2(x_{1} + x_{5}) = λ(x_{1} + x_{5}) ⇒ λ_{1} = 2
x_{2} + x_{3} + x_{4} = λ∙x_{2}  (4)
x_{2} + x_{3} + x_{4} = λ∙x_{3}  (5)
x_{2} + x_{3} + x_{4} = λ∙x_{4}  (6)
(4)+(5)+(6) = 3(x_{2} + x_{3} + x_{4}) = λ(x_{2} + x_{3} + x_{4} ) ⇒ λ_{2} = 3
Product = λ_{1} × λ_{2} = 2×3 = 6
Question 110 
The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______.
0.259 to 0.261  
0.260 to 0.262  
0.261 to 0.263  
0.262 to 0.264 
n(A)=50, n(B)=33, n(C)=20
n(A∩B)=16, n(B∩C)=6, n(A∩C)=10
n(A∩B∩C)=3
P(A∪B∪C) = P(A)+P(B)+P(C)P(A∩B)P(B∩C) P(A∩C)+P(A∩B∩C) = 74/100
∴ Required probability is P(A∩B∩C) = 1P(A∪B∪C) = 0.26
Question 111 
The number of distinct positive integral factors of 2014 is _________.
0.26  
0.27  
8  
0.29 
= 2' × 19' × 53'
Now number of distinct integral factors of 2014 will be,
(1+1)×(1+1)×(1+1) = 2×2×2 = 8
Question 112 
Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and V of S we say U < V if the minimum element in the symmetric difference of the two sets is in U. Consider the following two statements:
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.
Which one of the following is CORRECT?
Both S1 and S2 are true  
S1 is true and S2 is false  
S2 is true and S1 is false  
Neither S1 nor S2 is true 
U⊂S, V⊂S
Let U = {1, 2, 3}
V = {2, 3, 4}
Symmetric difference:
(U – V) ∪ (V – U) = {1} ∪ {4} = {1, 4}
The minimum element in the symmetric difference is 1 and 1∈U.
S1: Let S = Universal set = {1, 2, … 2014}
This universal set is larger than every other subset.
S2: Null set is smaller than every other set.
Let U = { }, V = {1}
Symmetric difference = ({ } – {1}) ∪ ({1} – { }) = { } ∪ {1} = {1}
So, U < V because { } ∈ U.
Question 113 
A cycle on n vertices is isomorphic to its complement. The value of n is _____.
5  
6  
7  
8 
In a cycle of n vertices, each vertex is connected to other two vertices. So each vertex degree is 2.
When we complement it, each vertex will be connected to remaining n3 vertices ( one is self and two other vertices in actual graph).
As per given question,
n3 =2
n=5
Cycle of 5 vertices is
Complement of the above graph1 is
Graph1 and Graph2 are complement each other.
So, the value of n is 5.
Question 114 
Which one of the following Boolean expressions is NOT a tautology?
((a ⟶ b) ∧ (b ⟶ c)) ⟶ (a ⟶ c)  
(a ⟷ c) ⟶ (∽ b ⟶ (a ∧ c))  
(a ∧ b ∧ c) ⟶ (c ∨ a)  
a ⟶ (b ⟶ a) 
((a → b) ∧ (b → c)) → (a → c)
If (a → b) is false with a = T, b = F,
then (F ∧ (b → c)) → (a → c)
F → (a → c)
which is True for any (a → c)
This is tautology.
B:
(a ⟷ c) ⟶ (∽b ⟶ (a ∧ c))
For (a ⟷ c) be True and
∽b → (a ∧ c) should be False
Let a = c = F
(F → F) → (∽b (F ∩ F))
T → (∽b → F)
This is False for b = F
So, this is not True.
C:
(a ∧ b ∧ c) ⟶ (c ∨ a)
(c ∨ a) is False only for a = c = F
if (c ∨ a) is False
(F ∧ b ∧ F) → F
F → F which is Tautology
True always.
D:
a ⟶ (b ⟶ a)
a ⟶ (~b ∨ a)
(~a ∨ a) ∨ ~b = T ∨ ~b = T which is tautology
Question 115 
Consider the following statements:
P: Good mobile phones are not cheap Q: Cheap mobile phones are not good L: P implies Q M: Q implies P N: P is equivalent to Q
Which one of the following about L, M, and N is CORRECT?
Only L is TRUE.  
Only M is TRUE.  
Only N is TRUE.  
L, M and N are TRUE. 
So, given statement can be sub divided such that we can utilize the negation of this atomic statements.
Suppose, X is Good mobile and Y is cheap then
P: (Good(x) → ~cheap(x)) → (~good(x) ∨ ~cheap(x))
Q: cheap(x) → ¬good(x) ⟺ ((¬cheap(x) ∨ good(x)) ⟺ ¬good(x) ∨ ¬cheap(x))
All these are contra positive.
All L, M, N are true.
Question 116 
Let X and Y be finite sets and f: X→Y be a function. Which one of the following statements is TRUE?
For any subsets A and B of X, f(A ∪ B) = f(A)+f(B)  
For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B)  
For any subsets A and B of X, f(A ∩ B) = min{ f(A),f(B)}  
For any subsets S and T of Y, f^{ 1} (S ∩ T) = f^{ 1} (S) ∩ f^{ 1} (T) 
We need to consider subsets of 'x', which are A & B (A, B can have common elements are exclusive).
Similarly S, T are subsets of 'y'.
To be a function, each element should be mapped with only one element.
(a) f(A∪B) = f(A)+f(B)
{a,b,c}∪{c,d,e} = {a,b,c} + {c,d,e}
{a,b,c,d,e} = 3+3
5 = 6 FALSE
(d) To get inverse, the function should be oneone & onto.
The above diagram fulfills it. So we can proceed with inverse.
f^{1} (S∩T ) = f^{1} (S)∩f^{1} (T)
f^{1} (c) = f^{1} ({a,b,c})∩f^{1} ({c,d,e})
2 = {1,2,3}∩{2,4,5}
2 = 2 TRUE
Question 117 
Let G be a group with 15 elements. Let L be a subgroup of G. It is known that L ≠ G and that the size of L is at least 4. The size of L is __________.
5  
6  
7  
8 
So, 15 is divided by {1, 3, 5, 15}.
As minimum is 4 and total is 15, we eliminate 1,3,15.
Answer is 5.
Question 118 
Which one of the following statements is TRUE about every n × n matrix with only real eigenvalues?
If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.  
If the trace of the matrix is positive, all its eigenvalues are positive.  
If the determinant of the matrix is positive, all its eigenvalues are positive.  
If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. 
• The product of the n eigenvalues of A is the same as the determinant of A. •
A: Yes, for sum to be negative there should be atleast one negative number.
B: There can be one small negative number and remaining positive, where sum is positive.
C: Product of two negative numbers is positive. So, there no need of all positive eigen values.
D: There is no need for all eigen values to be positive, as product of two negative numbers is positive.
Question 119 
If V_{1} and V_{2} are 4dimensional subspace of a 6dimensional vector space V, then the smallest possible dimension of V_{1}∩V_{2} is ______.
2  
3  
4  
5 
For eg: a two dimensional vector space have x, y axis. For dimensional vector space, it have x, y, z axis.
In the same manner, 6 dimensional vector space has x, y, z, p, q, r (assume).
Any subspace of it, with 4 dimensional subspace consists any 4 of the above. Then their intersection will be atmost 2.
[{x,y,z,p} ∩ {r,q,p,z}] = #2
V_{1} ∩ V_{2} = V_{1} + V_{2}  V_{1} ∪ V_{2} = 4 + 4 + (6) = 2
Question 120 
If , then the value of k is equal to ________.
4  
5  
6  
7 
We have xSinx,
We can observe that it is positive from 0 to π and negative in π to 2π.
To get positive value from π to 2π we put ‘‘ sign in the (π, 2π)
Question 121 
With respect to the numerical evaluation of the definite integral , where a and b are given, which of the following statements is/are TRUE?
I) The value of obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral.
II) The value of obtained using the Simpson’s rule is always equal to the exact value of the definite integral.
I only  
II only  
Both I and II  
Neither I nor II 
Question 122 
The value of the integral given below is
2π  
π  
π  
2π 
Question 123 
Let S be a sample space and two mutually exclusive events A and B be such that A∪B = S. If P(∙) denotes the probability of the event, the maximum value of P(A)P(B) is __________.
0.25  
0.26  
0.27  
0.28 
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
Question 124 
Consider the set of all functions f: {0,1, … ,2014} → {0,1, … ,2014} such that f(f(i)) = i, for all 0 ≤ i ≤ 2014. Consider the following statements:
 P. For each such function it must be the case that for every i, f(i) = i.
Q. For each such function it must be the case that for some i, f(i) = i.
R. Each such function must be onto.
Which one of the following is CORRECT?
P, Q and R are true  
Only Q and R are true  
Only P and Q are true  
Only R is true 
So f(i)should be resulting only {0, 1, …2014}
So, every element in range has a result value to domain. This is onto. (Option R is correct)
We have ‘2015’ elements in domain.
So atleast one element can have f(i) = i,
so option ‘Q’ is also True.
∴ Q, R are correct.
Question 125 
There are two elements x, y in a group (G,∗) such that every element in the group can be written as a product of some number of x's and y's in some order. It is known that
x * x = y * y = x * y * x = y * x * y * x = e
where e is the identity element. The maximum number of elements in such a group is ________.
4  
5  
6  
7 
a*a^{1} = e
1. x*x = e So x^{1} is x ⇒ x is element of Group
2. y*y = e So y^{1} = y ⇒ y is element of Group
4. (y*x)*(y*x) = x*y*y*x = x*x*e = e So (y*x)^{1} = (y*x)
In ③, ④
x*y, y*x has same inverse, there should be unique inverse for each element.
x*y = y*x (even with cumulative law, we can conclude)
So {x, y, e, x*y} are element of Group.
Question 126 
If G is a forest with n vertices and k connected components, how many edges does G have?
⌊n/k⌋  
⌈n/k⌉  
n–k  
nk+1 
Option 1, 2 will give answer 1. (i.e. one edge among them),
Option 3: nk = 0 edges.
Option 4: nk+1 = 1 edge, which is false.
Question 127 
Let δ denote the minimum degree of a vertex in a graph. For all planar graphs on n vertices with δ ≥ 3, which one of the following is TRUE?
In any planar embedding, the number of faces is at least n/2 + 2  
In any planar embedding, the number of faces is less than n/2 + 2  
There is a planar embedding in which the number of faces is less than n/2 + 2  
There is a planar embedding in which the number of faces is at most n/(δ+1) 
v – e + f = 2 →①
Point ① degree of each vertex is minimum ‘3’.
3×n ≥ 2e
e ≤ 3n/2
From ① :
n3n/2+f = 2 ⇒
Question 128 
The CORRECT formula for the sentence, “not all rainy days are cold” is
∀d (Rainy(d) ∧∼Cold(d))  
∀d (∼Rainy(d) → Cold(d))  
∃d (∼Rainy(d) → Cold(d))  
∃d (Rainy(d) ∧∼Cold(d)) 
= ∼[∀rainy days are cold]
= ∼[∀ days (rainy days ⇒ cold days]
= ∃ days[∼(cold days ∨ ∼rainy days)]
= ∃ days[rainy days ∧ ∼cold days]
Question 129 
A binary operation ⊕ on a set of integers is defined as x ⊕ y = x^{2 }+ y^{2}. Which one of the following statements is TRUE about ⊕?
Commutative but not associative  
Both commutative and associative  
Associative but not commutative  
Neither commutative nor associative 
A binary relation on a set S is called cumulative if a*b = b*a ∀ x,y∈S.
Associative property:
A binary relation on set is called associative if (a*b)*c = a*(b*c) ∀ x,y∈S.
Given x⊕y = x^{2} + y^{2} (1)
Replace x, y in (1)
y⊕x = y^{2} + x^{2} which is same as (1), so this is cumulative
(x⊕y)⊕z = (x^{2} + y^{2}) ⊕ z
= (x^{2} + y^{2}) + z^{2}
= x^{2} + y^{2} + z^{2} + 2x^{2}y^{2} (2)
x⊕(y ⊕ z) = x ⊕ (y^{2} + z^{2})
= x^{2} + (y^{2} + z^{2})^{2}
= x^{2} + y^{2} + z^{2} + 2y^{2z2  (3) (2) & (3) are not same so this is not associative. }
Question 130 
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
8/(2e^{3})  
9/(2e^{3})  
17/(2e^{3})  
26/(2e^{3}) 
P(x : λ) = (e^{λ} λ^{x})/x! for x = 0,1,2….
‘λ’ is the average number (mean)
Given that mean = λ = 3
The probability of observing fewer than three cars is
P(zero car) + P(one car) + P(two cars)
= (e^{3} 3^{0})/0!+(e^{3} 3^{1})/1!+(e^{3} 3^{2})/2!
= e^{3}+e^{3}∙3+(e^{3})∙9)/2
= (17e^{3<})/2
= 17/(2e^{3} )
Question 131 
Which one of the following does NOT equal to
Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C_{2} → C_{1} + C_{2}
Question 132 
Which one of the following functions is continuous at x = 3?
At x = 3, f(x) = 2
LHL (3), f(3^{}) = (x+3 )/3 = (3+3)/3 = 2
RHL (3), f(3^{+}) = x1 = 31 = 2
∴ f(x) is continuous.
Option B:
At x = 3, f(x) = 4
For x ≠ 3, f(x) = f(3^{+}) = f(3^{}) = 8^{x} = 83 = 5
This is not continuous.
Option C:
At x ≤ 3, f(x) = x+3 = 3+3 = 6
At RHL(3), f(3^{+}) = 4
This is not continuous.
Option D:
f(x) at x = 3 is not defined.
There is a break at x = 3, so this is not continuous.
Question 133 
Function f is known at the following points:
The value of computed using the trapezoidal rule is
8.983  
9.003  
9.017  
9.045 
Question 134 
What is the logical translation of the following statement?
"None of my friends are perfect."
∃x(F(x)∧¬P(x))  
∃x(¬F(x)∧P(x))  
∃x(¬F(x)∧¬P(x))  
¬∃x(F(x)∧P(x)) 
P(x) = x is perfect
The meaning of ∃x(P(x)∧F(x)) is atleast one person who is my friend and perfect.
The negation of ∃x(P(x)∧F(x)) is “This is not the case that atlease one person who is my friend and perfect”.
So ~∃x(P(x)∧F(x)) is none of my friends are perfect.
Question 135 
The following code segment is executed on a processor which allows only register operands in its instructions. Each instruction can have atmost two source operands and one destination operand. Assume that all variables are dead after this code segment.
c = a + b; d = c * a; e = c + a; x = c * c; if (x > a) { y = a * a; } else { d = d * d; e = e * e; }
Suppose the instruction set architecture of the processor has only two registers. The only allowed compiler optimization is code motion, which moves statements from one place to another while preserving correctness. What is the minimum number of spills to memory in the compiled code?
0  
1  
2  
3 
In the above code total number of spills to memory is 1.
Question 136 
Consider the following logical inferences.
 I1: If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.
I2: If it rains then the cricket match will not be played.
It did not rain.
Inference: The cricket match was played.
Which of the following is TRUE?
Both I_{1} and I_{2} are correct inferences  
I_{1} is correct but I_{2} is not a correct inference  
I_{1} is not correct but I_{2} is a correct inference  
Both I_{1} and I_{2} are not correct inferences 
The cricket match was played.
Let p = it rains
q = playing cricket/ match played
If (it rains) then (the match will not be played)
p ⇒ (∼q)
Inference: There was no rain. (i.e., p = F)
So for any F ⇒ (∼q) is true.
So this inference is valid.
I_{2}: If it rains then the cricket match will not be played.
It did not rain.
p ⇒ (∼q)
Inference: The cricket match was played.
q = T
p ⇒ (∼q)
p ⇒ (∼T)
p ⇒ F
This is false for p = T, so this is not true.
Question 137 
Consider the function f(x) = sin(x) in the interval x ∈ [π/4, 7π/4]. The number and location(s) of the local minima of this function are
One, at π/2  
One, at 3π/2  
Two, at π/2 and 3π/2  
Two, at π/4 and 3π/2 
f’(x) = cos x
[just consider the given interval (π/4, 7π/4)]
f'(x) = 0 at π/2, 3π/2
To get local minima f ’’(x) > 0, f ’’(x) =  sin x
f ’’(x) at π/2, 3π/2
f ’’(x) = 1< 0 local maxima
f ’’ (3π/2) = 1 > 0 this is local minima
In the interval [π/4, π/2] the f(x) is increasing, so f(x) at π/4 is also a local minima.
So there are two local minima for f(x) at π/4, 3π/2.
Question 138 
Let A be the 2×2 matrix with elements a_{11} = a_{12} = a_{21} = +1 and a_{22} = 1. Then the eigenvalues of the matrix A^{19} are
1024 and 1024  
1024√2 and 1024√2  
4√2 and 4√2  
512√2 and 512√2 
The 2×2 matrix =
Cayley Hamilton theorem:
If matrix A has ‘λ’ as eigen value, A^{n} has eigen value as λ^{n}.
Eigen value of
AλI = 0
(1λ)(1+λ)1 = 0
(1λ^{2} )1 = 0
1 = 1λ^{2}
λ^{2} = 2
λ = ±√2
A^{19} has (√2)^{19} = 2^{9}×√2 (or) (√2)^{19} = 512√2
= 512√2
Question 139 
What is the correct translation of the following statement into mathematical logic?
“Some real numbers are rational”
∃x (real(x) ∨ rational(x))  
∀x (real(x) → rational(x))  
∃x (real(x) ∧ rational(x))  
∃x (rational(x) → real(x)) 
∃x (real(x) ∧ rational(x))
(A) ∃x(real(x) ∨ rational(x))
means There exists some number, which are either real or rational.
(B) ∀x (real(x)→rational(x))
If a number is real then it is rational.
(D) ∃x (rational(x)→real(x))
There exists a number such that if it is rational then it is real.
Question 140 
Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to
3  
4  
5  
6 
ve+f = 2
Given 10 vertices & 15 edges
1015+f = 2
f = 2+1510
f = 7
There will be an unbounded face always. So, number of faces = 6.
Question 141 
Which of the following graphs is isomorphic to
(A) 3 cycle graph not in original one.
(B) Correct 5 cycles & max degree is 4.
(C) Original graph doesn’t have a degree of 3.
(D) 4 cycles not in original one.
Question 142 
The bisection method is applied to compute a zero of the function f(x) = x^{4}  x^{3}  x^{2}  4 in the interval [1,9]. The method converges to a solution after ________ iterations.
1  
3  
5  
7 
Question 143 
Suppose a fair sixsided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
10/21  
5/12  
2/3  
1/6 
The value on second time can be {1, 2, 3, 4, 5, 6}
So the Sum can be
We have Sample space = 36
The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}
So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①
If we get ‘6’ in the first time itself, then we do not go for rolling die again.
So, its probability = 1/6
Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12
Question 144 
How many onto (or surjective) functions are there from an nelement (n ≥ 2) set to a 2element set?
2^{n}  
2^{n}1  
2^{n}2  
2(2^{n}– 2) 
Onto function is possible if m ≥ n. So, no. of onto functions possible is,
n^{m}  ^{n}C_{1} (n1)^{m} + ^{n}C_{2} (n2)^{m} + .......
Here in Question,
m = n, n = 2
So, the final answer will be,
= 2^{n}  ^{2}C_{1} (21)^{n} + ^{2}C_{2} (22)^{n}
= 2^{n}  2 × 1 + 0
= 2^{n}  2
Question 145 
Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled, then the number of distinct cycles of length 4 in G is equal to
15  
30  
45  
360 
It is asked to find the distinct cycle of length 4. As it is complete graph, if we chose any two vertices, there will be an edge.
So, to get a cycle of length 4 (means selecting the 4 edges which can form a cycle) we can select any four vertices.
The number of such selection of 4 vertices from 6 vertices is ^{6}C_{4} => 15.
From each set of 4 vertices, suppose a set {a, b, c, d} we can have cycles like
abcd
abdc
acbd
acdb
adbc
adcb (Total 6, which is equal to number of cyclic permutations (n1)! )
As they are labelled you can observe, abcd and adcb are same, in different directions.
So, we get only three combinations from the above 6.
So, total number of distinct cycles of length 4 will be 15*3 = 45.
If it is asked about just number of cycles then 15*6 = 90
Question 146 
If the difference between the expectation of the square of a random variable (E[X^{2}]) and the square of the expectation of the random variable (E[X])^{2} is denoted by R, then
R = 0  
R < 0  
R ≥ 0  
R > 0 
So the answer will be R≥0.
Question 147 
Consider the matrix as given below.
Which one of the following provides the CORRECT values of eigenvalues of the matrix?
1, 4, 3  
3, 7, 3  
7, 3, 2  
1, 2, 3 
Question 148 
Which one of the following options is CORRECT given three positive integers x,y and z, and a predicate
P(x) = ¬(x=1)∧∀y(∃z(x=y*z) ⇒ (y=x)∨(y=1))P(x) being true means that x is a prime number  
P(x) being true means that x is a number other than 1  
P(x) is always true irrespective of the value of x
 
P(x) being true means that x has exactly two factors other than 1 and x 
This is the definition of prime numbers.
Question 149 
Given i=√1, what will be the evaluation of the definite integral
0  
2  
i  
i 
Question 150 
Consider a finite sequence of random values X = [x_{1}, x_{2}, …, x_{n}]. Let μx be the mean and σx be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as y_{i} = a * x_{i} + b, where a and b are positive constants. Let μ_{y} be the mean and σ_{y} be the standard deviation of this sequence. Which one of the following statements is INCORRECT?
Index position of mode of X in X is the same as the index position of mode of Y in Y.  
Index position of median of X in X is the same as the index position of median of Y in Y.  
μ_{y} = aμ_{x} + b  
σ_{y} = aσ_{x} + b 
(σ_{y})^{2} is variance so,
y_{i} = a * x_{i} + b
(σ_{y})^{2} = a^{2 }(σ_{x})^{2}
⇒ σ_{y} = a σ_{x}
Hence option (D) is incorrect.
Question 151 
A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second.
1/5  
4/25  
1/4  
2/5 
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
^{5}C_{1} × ^{4}C_{1} = 20
So probability = 4/20 = 1/5
Question 152 
Let G = (V,E) be a graph. Define ξ(G) = Σd id x d, where id is the number of vertices of degree d in G. If S and T are two different trees with ξ(S) = ξ(T),then
S = 2T  
S = T  1  
S = T  
S = T + 1 
i_{d}= no. of vertices of degree ‘d’ in ‘G’
Eg:
No. of vertices with degree ‘2’ = 3
ξ(G') = 3 × 2 = '6' i.e., sum of degrees
By Handshaking Theorem,
The sum of degrees would be equal to twice the no. of edges
V = 2E
It is given that ξ(G) = ξ(S) then
Sum of degrees of vertices in G is equal to sum of degrees of vertices in S
i.e., 2*(no. of edges in G) = 2*no. of edges in S no. of edges in G = no. of edges in S
Eg:
ξ(G) = (2 × 2) + (2 × 3) = 4 + 6 = 10
ξ(S) = 2 × 5 = 10
You can observe that, though no. of vertices are different, but still no. of edges are same.
Question 153 
NewtonRaphson method is used to compute a root of the equation x^{2}  13 = 0 with 3.5 as the initial value. The approximation after one iteration is
3.575  
3.676  
3.667  
3.607 
Question 154 
What is the possible number of reflexive relations on a set of 5 elements?
2^{10}  
2^{15}  
2^{20}  
2^{25} 
Definition of Reflexive relation:
A relation ‘R’ is reflexive if it contains xRx ∀ x∈A
A relation with all diagonal elements, it can contain any combination of nondiagonal elements.
Eg:
A={1, 2, 3}
So for a relation to be reflexive, it should contain all diagonal elements. In addition to them, we can have possible combination of (n^{2}n)nondiagonal elements (i.e., 2^{n2n})
Ex:
{(1,1)(2,2)(3,3)}  ‘0’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)}  ‘1’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)(1,3)} “
___________ “
___________ “
{(1,1)(2,2)(3,3)(1,2)(1,3)(2,1)(2,3)(3,1)(3,2)} (n^{2}n) diagonal elements
____________________
Total: 2^{n2n}
For the given question n = 5.
The number of reflexive relations = 2^{(255)} = 2^{20}
Question 155 
Consider the set S = {1, ω, ω^{2}}, where ω and ω^{2} are cube roots of unity. If * denotes the multiplication operation, the structure (S,*) forms
A group  
A ring  
An integral domain  
A field 
1) Closure
2) Associativity
3) Have Identity element
4) Invertible
Over ‘*’ operation the S = {1, ω, ω^{2}} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w^{2}' and inverse of 'w^{2}' is 'w'.
Question 156 
What is the value of
0  
e^{2}  
e^{1/2}  
1 
Question 157 
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?
pq + (1  p)(1  q)  
(1  q)p  
(1  p)q  
pq 
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of testing process giving incorrect result × Probability that computer is not faulty
= p × q + (1  p) (1  q)
Question 158 
What is the probability that divisor of 10^{99} is a multiple of 10^{96}?
1/625  
4/625  
12/625  
16/625 
We can write 10^{99} as 10^{96}×10^{3}
So, (10^{99})/(10^{96}) to be a whole number, [10^{96}×10^{3}/10^{96}] ➝ (1)
We can observe that every divisor of 10^{3} is a multiple of 10^{96}
So number of divisor of 10^{3} to be found first
⇒ 10^{3} = (5×2)^{3} = 2^{3}×5^{3}
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 10^{99} are 10^{99} = 2^{99}×5^{99} = 100×100 = 10000
Probability that divisor of 10^{99} is a multiple of 10^{96} is
⇒ 16/10,000
Question 159 
The degree sequence of a simple graph is the sequence of the degrees of the nodes in the graph in decreasing order. Which of the following sequences can not be the degree sequence of any graph?
 (I) 7, 6, 5, 4, 4, 3, 2, 1
(II) 6, 6, 6, 6, 3, 3, 2, 2
(III) 7, 6, 6, 4, 4, 3, 2, 2
(IV) 8, 7, 7, 6, 4, 2, 1, 1
I and II  
III and IV  
IV only  
II and IV 
⇾ Arrange the degree of vertices in descending order
eg. d_{1}, d_{2}, d_{3}... d_{n}
⇾ Discard d_{1}, subtrack ‘1’ from the next 'd_{1}' degrees
eg:
⇒ 1 1 0 1
⇾ We should not get any negative value if its negative, this is not valid sequence
⇾ Repeat it till we get ‘0’ sequence
I. 7, 6, 5, 4, 4, 3, 2, 1
➡️5, 4, 3, 3, 2, 1, 0
➡️3, 2, 2, 1, 0, 0
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
II. 6, 6, 6, 6, 3, 3, 2, 2
➡️5, 5, 5, 2, 2, 1, 2
put them in descending order
➡️5, 5, 5, 2, 2, 2, 1
➡️4, 4, 1, 1, 1, 1
➡️3, 0, 0, 0, 1 (descending order)
➡️3, 1, 0, 0, 0
➡️0, 1, 1, 0
[This is not valid]
III. 7, 6, 6, 4, 4, 3, 2, 2
➡️5, 5, 3, 3, 2, 1, 1
➡️4, 2, 2, 1, 0, 1
➡️4, 2, 2, 1, 1, 0 (descending order)
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
IV. 8, 7, 7, 6, 4, 2, 1, 1
There is a degree ‘8’, but there are only ‘8’ vertices.
A vertex cannot have edge to itself in a simple graph. This is not valid sequence.
Question 160 
Consider the following matrix . If the eigenvalues of A are 4 and 8, then
x=4, y=10  
x=5, y=8  
x=3, y=9  
x=4, y=10 
Trace = {Sum of diagonal elements of matrix}
Here given that eigen values are 4, 8
Sum = 8 + 4 = 12
Trace = 2 + y
⇒ 2 + y = 12
y = 10
Determinant = 2y  3x
Product of eigen values = 8 × 4 = 32
2y  3x = 32
(y = 10)
20  3x = 32
12 = 3x
x = 4
∴ x = 4, y = 10
Question 161 
Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. Which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x,y,t))?
Everyone can fool some person at some time  
No one can fool everyone all the time  
Everyone cannot fool some person all the time
 
No one can fool some person at some time 
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
Question 162 
Which one of the following in NOT necessarily a property of a Group?
Commutativity
 
Associativity  
Existence of inverse for every element
 
Existence of identity 
So, commutativity is not required.
Question 163 
What is the chromatic number of an nvertex simple connected graph which does not contain any odd length cycle? Assume n ≥ 2.
2  
3  
n1  
n 
Eg: Consider a square, which has 4 edges. It can be represented as bipartite ,with chromatic number 2.
Question 164 
Which one of the following is TRUE for any simple connected undirected graph with more than 2 vertices?
No two vertices have the same degree.  
At least two vertices have the same degree.  
At least three vertices have the same degree.  
All vertices have the same degree.

If all vertices have different degrees, then the degree sequence will be {1,2,3,....n1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n1) vertices). Degree sequence has only (n1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) Consider a triangle, all vertices has same degree, so it is false
C) Consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) Consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
Question 165 
Consider the binary relation R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?
R is symmetric but NOT antisymmetric
 
R is NOT symmetric but antisymmetric  
R is both symmetric and antisymmetric  
R is neither symmetric nor antisymmetric 
Antisymmetric Relation: A relation R on a set A is called antisymmetric if (a,b)€ R and (b,a) € R then a = b is called antisymmetric.
In the given relation R, for (x,y) there is no (y,x). So, this is not Symmetric. (x,z) is in R also (z,x) is in R, but as per antisymmetric relation, x should be equal to z, where this fails.
So, R is neither Symmetric nor Antisymmetric.
Question 166 
An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the
If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3?
0.453  
0.468  
0.485  
0.492 
P(e) = Probability of getting even no. face.
It is given that,
P(0) = 0.9 P(e)  (I)
Also we know that,
P(0) + P(e) = 1  (II)
Solving equation (I) and (II) we get,
P(e) = 0.52
Also even no. can be 2 or 4 or 6.
And given in question that P(2) = P(4) = P(6).
So, 3 × P(2) = 0.52
P(2) = 0.175
So, P(2) = P(4) = P(6) = 0.175
Also in question it is given that,
P(e/>3) = 0.75
P(even no. greater than 3)/ P(no. greater than 3) = 0.75
P(4,6)/P(>3) = 0.75
(0.175×2)/P(>3) = 0.75
P(>3) = 0.35/0.75 = 0.467
Question 167 
For the composition table of a cyclic group shown below
Which one of the following choices is correct?
a, b are generators  
b, c are generators  
c, d are generators  
d, a are generators

We can observe that, a is an identity element. ( a *x = x ). An identity element cannot be a generator, as it cannot produce any other element ( always a*a*... = a).
Also, b*b =a, so it also cannot produce all other elements ( always b*b*... =a , where a is identify element).
c,d are able to produce other elements like { c*c =b, c*(c*c) = c*b= d, c*(c*(c*c))) = c*(c*b)= c*d=a. }. Similar with d.
Question 168 
Which one of the following is the most appropriate logical formula to represent the statement?
“Gold and silver ornaments are precious”.
The following notations are used:
G(x): x is a gold ornament S(x): x is a silver ornament P(x): x is precious
∀x(P(x) → (G(x) ∧ S(x)))  
∀x((G(x) ∧ S(x)) → P(x))  
∃x((G(x) ∧ S(x)) → P(x)  
∀x((G(x) ∨ S(x)) → P(x))

(A) for all ornaments, if it is precious then they should be gold and silver.
But, given statement does not says that, “ only gold and silver are precious “ . So this is wrong.
(B) For all ornaments, which contains gold and silver are precious.
Which is only the shaded region in the venn diagrams. But, it misses p,r regions. So, this is wrong option.
C) Some ornaments, which are gold and silver are precious. It is false, because all gold or silver ornaments are precious.
D) For all ornaments, Any ornament which is gold or silver is precious. Which is true.
Question 169 
The binary operation □ is defined as follows
Which one of the following is equivalent to P ∨ Q ?
¬Q□¬P  
P□¬Q  
¬P□Q  
¬P□¬Q 
P∨Q = P□️Q
So, option B is correct.
Question 170 
0  
1  
ln 2  
1/2 ln 2 
Question 171 
Consider the following wellformed formulae:
 I. ¬∀x(P(x))
II. ¬∃(P(x))
III. ¬∃(¬P(x))
IV. ∃x(¬(P(x))
Which of the above are equivalent?
I and III  
I and IV  
II and III  
II and IV 
II ) ¬∃x(P(x))= ∀x(~P(x))
III) ¬∃x(¬P(x)) = ∀x(P(x))
Question 173 
If P, Q, R are subsets of the universal set U, then (P∩Q∩R) ∪ (P^{c}∩Q∩R) ∪ Q^{c }∪ R^{c} is
Q^{c} ∪ R^{c}
 
P ∪ Q^{c} ∪ R^{c}  
P^{c} ∪ Q^{c} ∪ R^{c}  
U 
(P∩Q∩R)∪(P^{c}∩Q∩R)∪Q^{c}∪R^{c}
It can be written as the p.q.r + p'.q.r +q' + r'
=> (p+p').q.r + q' + r'
=> q.r + (q'+r')
=> q.r + q' + r' = 1 i.e., U
Question 174 
The following system of equations

x_{1} + x_{2} + 2x_{3} = 1
x_{1} + 2x_{2} + 3x_{3} = 2
x_{1} + 4x_{2} + ax_{3} = 4
has a unique solution. The only possible value(s) for a is/are
0  
either 0 or 1  
one of 0, 1 or 1  
any real number 
When a5 = 0, then rank(A) = rank[AB]<3,
So infinite number of solutions.
But, it is given that the given system has unique solution i.e., rank(A) = rank[AB] = 3 will be retain only if a5 ≠ 0.
Question 175 
The minimum number of equal length subintervals needed to approximate to an accuracy of atleast using the trapezoidal rule is
1000e  
1000  
100e  
100 
Question 176 
The NewtonRaphson iteration can be used to compute the
square of R
 
reciprocal of R  
square root of R  
logarithm of R 
Question 177 
Which of the following statements is true for every planar graph on n vertices?
The graph is connected  
The graph is Eulerian
 
The graph has a vertexcover of size at most 3n/4  
The graph has an independent set of size at least n/3

(A) Consider the following disconnected graph which is planar.
So false.
(B) A graph is Eulerian if all vertices have even degree but a planar graph can have vertices with odd degree.
So false.
(D) Consider K_{4} graph. It has independent set size 1 which is less than 4/3.
So false.
Hence, option (C) is correct.
Question 178 
Let and , where k is a positive integer. Then
P = Q  k  
P = Q + k  
P = Q  
P = Q + 2 k 
P = 1+3+5+7+...+(2k1)
= (21)+(41)+(61)+(81)+...+(2k1)
= (2+4+6+8+...+2k)+(1+1+1+k times)
= Q(1+1+...+k times)
= Qk
Question 179 
A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve 3x^{4}  16x^{3} + 24x^{2} + 37 is
0  
1  
2  
3 
f’(x) = 12x^{3} + 48x^{2} + 48x = 0
12x(x^{2}  4x + 4) = 0
x=0; (x2)^{2} = 0
x=2
f’’(x) = 36x^{2}  96x + 48
f ”(0) = 48
f ”(2) = 36(4)  96(2) + 48
= 144  192 + 48
= 0
At x=2, we can’t apply the second derivative test.
f’(1) = 12; f’(3) = 36, on either side of 2 there is no sign change then this is neither minimum or maximum.
Finally, we have only one Extremum i.e., x=0.
Question 180 
Aishwarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
0.24  
0.36  
0.4  
0.6 
(i) She study Mathematics on Tuesday and computer science on wednesday.
⇒ 0.6×0.4
⇒ 0.24
(ii) She study computer science on Tuesday and computer science on wednesday.
⇒ 0.4×0.4
⇒ 0.16
→ The probability that she study computer science on wednesday is
0.24+0.16 = 0.40
Question 181 
How many of the following matrices have an eigenvalue 1?
one  
two  
three  
four 
Answer: We have only one matrix with eigen value 1.
Question 182 
Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean 1 and variance unknown. If P(X ≤ 1) = P(Y ≥ 2), the standard deviation of Y is
3  
2  
√2  
1 
We can compare their values using standard normal distributions.
The above equation satisfies when σ_{y} will be equal to 3.
Question 183 
Let fsa and pda be two predicates such that fsa(x) means x is a finite state automaton, and pda(y) means that y is a pushdown automaton. Let equivalent be another predicate such that equivalent (a, b) means a and b are equivalent. Which of the following first order logic statements represents the following:
Each finite state automaton has an equivalent pushdown automaton
(∀x fsa(x)) ⇒ (∃y pda(y) ∧ equivalent(x,y))  
∼∀y(∃x fsa(x) ⇒ pda(y) ∧ equivalent(x,y))  
∀x ∃y(fsa(x) ∧ pda(y) ∧ equivalent(x,y))  
∀x ∃y(fsa(y)∧ pda(x) ∧ equivalent(x,y)) 
Option A:
If everything is a FSA. Then there exists an equivalent PDA for everything.
Option B:
Not for the case Y, if there exists a FSA then it can have equivalent PDA.
Option C:
Everything is a PDA and consists equivalent PDA.
Option D:
Everything is a PDA and has exist an equivalent FSA. In option A we are getting the equivalent of a and b.
So answer is option A.
Question 184 
P and Q are two propositions. Which of the following logical expressions are equivalent?
 I. P∨∼Q
II. ∼(∼P∧Q)
III. (P∧Q)∨(P∧∼Q)∨(∼P∧∼Q)
IV. (P∧Q)∨(P∧∼Q)∨(∼P∧Q)
Only I and II  
Only I, II and III  
Only I, II and IV  
All of I, II, III and IV 
II. ∼(∼P∧Q)⇒(P∨∼Q)≡I (✔️)
III. (P×Q)∨(P×∼Q)∨(∼P×∼Q)
P∧(Q∨∼Q)∨(∼P∧∼Q)
P∨(∼P×∼Q)
(P∨∼P)×(P∨∼Q)
(P∨∼Q)≡I=II (✔️)
IV. (P×Q)∨(P∧∼Q)∨(∼P×Q)
P×(Q∨∼Q)∨(∼P∧Q)
P∨(∼P×Q)
(P∨∼P)×(P∨Q)
(P∨Q)≠I (❌)
So I≡II≡III (✔️)
Question 185 
Consider the following two statements about the function f(x)=x
P. f(x) is continuous for all real values of x Q. f(x) is differentiable for all real values of x
Which of the following is TRUE?
P is true and Q is false.  
P is false and Q is true.  
Both P and Q are true.  
Both P and Q are false. 
→ f(x) is continuous for all real values of x
For every value of x, there is corresponding value of f(x).
For x is positive, f(x) is also positive
x is negative, f(x) is positive.
So, f(x) is continuous for all real values of x.
→ f(x) is not differentiable for all real values of x. For x<0, derivative is negative
x>0, derivative is positive.
Here, left derivative and right derivatives are not equal.
Question 186 
Let S be a set of n elements. The number of ordered pairs in the largest and the smallest equivalence relations on S are:
n and n  
n^{2} and n  
n^{2} and 0  
n and 1 
→ Reflexive
→ Symmetric
→ Transitive
Let a set S be,
S = {1, 2, 3}
Now, the smallest relation which is equivalence relation is,
S×S = {(1,1), (2,2), (3,3)}
= 3
= n (for set of n elements)
And, the largest relation which is equivalence relation is,
S×S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
= 9
= 3^{2}
= n^{2} (for set of n elements)
Question 187 
Let G be the nonplanar graph with the minimum possible number of edges. Then G has
9 edges and 5 vertices  
9 edges and 6 vertices  
10 edges and 5 vertices  
10 edges and 6 vertices 
if n ≥ 3 then e ≤ 3n6 (for planarity)
where n = no. of vertices
e = no. of edges
Now lets check the options.
A) e=9, n=5
9 ≤ 3(5)  6
9 ≤ 15  6
9 ≤ 9
Yes, it is planar.
B) e=9, n=6
9 ≤ 3(6)  6
9 ≤ 18  6 9 ≤ 12 Yes, it is planar.
iii) e=10, n=5
10 ≤ 3(5)  6
10 ≤ 15  6
10 ≤ 9 No, it is not planar.
So, option C is nonplanar graph.
iv) e=10, n=6
10 ≤ 3(6)  6
10 ≤ 18  6
10 ≤ 12
Yes, it is planar.
Question 188 
How many different nonisomorphic Abelian groups of order 4 are there?
2  
3  
4  
5 
4 = 2^{2}
So, prime no. is 2 and power of 2 is 2. So exponent value 2 is considered now.
Now the no. of ways we can divide 2 into sets will be the answer.
So division can be done as,
{1,1}, {0,2}
in two ways. Hence, answer is 2.
Question 189 
Let Graph(x) be a predicate which denotes that x is a graph. Let Connected(x) be a predicate which denotes that x is connected. Which of the following first order logic sentences DOES NOT represent the statement: “Not every graph is connected”?
¬∀x (Graph (x) ⇒ Connected (x))  
¬∃x (Graph (x) ∧ ¬Connected (x))  
¬∀x (¬Graph (x) ∨ Connected (x))  
∀x (Graph (x) ⇒ ¬Connected (x))

Given expression is
¬∀x(¬Graph(x) ∨ Connected(x)
which can be rewritten as,
¬∀x(Graph(x) ⇒ Connected(x)
which is equivalent to option (A)
(∵ ¬p∨q ≡ p→q)
So, option (A) and (C) cannot be the answer.
Coming to option (B), the given expression is,
∃x (Graph (x) ∧ ¬Connected (x))
"There exist some graph which is not connected", which is equivalent in saying that "Not every graph is connected".
Coming to option (D),
For all x graph is not connected, which is not correct.
Hence, option (D) is the answer.
Question 190 
Which of the following graphs has an Eulerian circuit?
Any kregular graph where k is an even number.  
A complete graph on 90 vertices.  
The complement of a cycle on 25 vertices.
 
None of the above. 
→ all vertices in the graph have an "even degree".
→ And the graph must be corrected.
Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.
Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24  2 = 22.
There is a theorem which says "G be a graph with n vertices and if every vertex has a degree of atleast n1/2 then G is connected."
So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.
Question 191 
Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1, 2, 3,….., 20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?
1/2  
1/10  
9!/20!  
None of these 
→ Total no. of possible even number = 10
→ Here we are not considering odd number.
→ The probability that 2 appears at an earlier position than any other even number is =1/10
Question 192 
Let A be a 4 x 4 matrix with eigenvalues 5, 2, 1, 4. Which of the following is an eigenvalue of
[A I] [I A]
where I is the 4 x 4 identity matrix?
5  
7  
2  
1 
(AλI)^{2}I = 0 [a^{2}b^{2} = (a+b)(ab)]
(AλI+I)(AλII) = 0
(A(λI)I)(A(λ+I)I = 0
Let us assume:
λ1=k & λ +1=k
λ =k+1 λ =k1
⇓ ⇓
for k=5; λ=4 λ =6
k=2; λ=1 λ =3
k=1; λ=2 λ = 0
k=4; λ=5 λ = 3
So, λ = 4,1,2,5,6,3,0,3
Check with the option
Option C = 2
Question 193 
Consider the set S = {a,b,c,d}. Consider the following 4 partitions π_{1}, π_{2}, π_{3}, π_{4} on Let p be the partial order on the set of partitions S' = {π_{1}, π_{2}, π_{3}, π_{4}} defined as follows: π_{i} p π_{j} if and only if π_{i} refines π_{j}. The poset diagram for (S', p) is:
And, neither π_{2} refines π_{3}, nor π_{3} refines π_{2}.
Here, only π_{1} refined by every set, so it has to be at the top.
Finally, option C satisfies all the property.
Question 194 
Consider the set of (column) vectors defined by X = {x ∈ R^{3} x_{1}+x_{2}+x_{3} = 0, where x^{T} = [x_{1},x_{2},x_{3}]^{T}}. Which of the following is TRUE?
{[1,1,0]^{T}, [1,0,1]^{T}} is a basis for the subspace X.  
{[1,1,0]^{T}, [1,0,1]^{T}} is a linearly independent set, but it does not span X and therefore is not a basis of X.  
X is not a subspace of R^{3}  
None of the above

Question 195 
Consider the series obtained from the NewtonRaphson method. The series converges to
1.5  
√2  
1.6  
1.4 
x_{n+1} = x_{n}/2+9/8x_{n} ⟶ (I); x_{0} = 0.5
Equation based on NewtonRaphson is
x_{n+1} = x_{n}f(x_{n})/f'(x_{n}) ⟶ (II)
Equate I and II
x_{n}f(x_{n})/f'(x_{n}) = x_{n}/2+9/8x_{n}
x_{n}f(x_{n})/f'(x_{n}) = x_{n}x_{n}/2+9/8x_{n}
x_{n}f(x_{n})/f'(x_{n}) = x_{n}(4x^{n2}9)/8x_{n}
So, f(x) = 4x^{n2}9
4x^{2}9 = 0
4x^{2} = 9
x^{2} = 9/4
x = ±3/2
x=±1.5
Question 196 
Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i+1,j) or (i,j+1).
How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)?
2^{20}  
2^{10}  
None of the above 
So now we have 10 u's and 10 r's, i.e.,
uuuuuuuuuurrrrrrrrrr
So, finally the no. of arrangements of above sequences is,
Question 197 
Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i+1,j) or (i,j+1).
Suppose that the robot is not allowed to traverse the line segment from (4,4) to (5,4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)?
2^{9}  
2^{19}  
So, no. of paths possible if line segment from (4,4) to (5,4) is taken is,
= paths possible from (0,0) to (4,4) * paths possible from (5,4) to (10,10)
= {uuuurrrr} * {uuuuuurrrrr}
Hence, the final answer is
Question 198 
Consider the polynomial p(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3}, where a_{i} ≠ 0, ∀_{i}. The minimum number of multiplications needed to evaluate p on an input x is:
3  
4  
6  
9 
p(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} where a_{i}≠0
This can be written as
p(x) = a_{0} +x( a_{1} + a_{2}x + a_{3}x^{2})=a_{0}+(a_{1}+(a_{2}+a_{3}x)x)x
Total no. of multiplications required is 3
i.e., a_{3}x = K.....(i)
(a_{2}+K)x = M..... (ii)
(a_{1}+M)x=N...... (iii)
Question 199 
Let X, Y, Z be sets of sizes x, y and z respectively. Let W = X×Y and E be the set of all subsets of W. The number of functions from Z to E is:
Z^{2}^{xy}  
Z×2^{xy}  
Z^{2}^{x+y}  
2^{xyz} 
A set ‘P’ consists of m elements and ‘Q’ consists of n elements then total number of function from P to Q is m^{n}.
⇒ E be the no. of subsets of W = 2^{w} = 2^{xxy} = 2^{xy}
No. of function from Z to E is = (2xy)^{z} = (2xy)^{z} = 2^{xyz}
Question 200 
The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four plausible reasons. Which one of them is false?
It is not closed  
2 does not have an inverse  
3 does not have an inverse  
8 does not have an inverse 
Option A:
It is not closed under multiplication. After multiplication modulo (10) we get ‘0’. The ‘0’ is not present in the set.
(2*5)%10 ⇒ 10%10 = 0
Option B:
2 does not have an inverse such as
(2*x)%10 ≠ 1
Option C:
3 have an inverse such that
(3*7)%10 = 1
Option D:
8 does not have an inverse such that
(8*x)%10 ≠ 1
Question 201 
A relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v. Then R is: Then R is:
Neither a Partial Order nor an Equivalence Relation  
A Partial Order but not a Total Order  
A Total Order  
An Equivalence Relation 
i) Symmetric
ii) Reflexive
iii) Transitive
If a relation is partial order relation then it must be
i) Reflexive
ii) Antisymmetric
iii) Transitive
If a relation is total order relation then it must be
i) Reflexive
ii) Symmetric
iii) Transitive
iv) Comparability
Given ordered pairs are (x,y)R(u,v) if (x
Here <, > are using while using these symbol between (x,y) and (y,v) then they are not satisfy the reflexive relation. If they uses (x<=u) and (y>=u) then reflexive relation can satisfies.
So, given relation cannot be a Equivalence. Total order relation or partial order relation.
Question 202 
We are given a set X = {x_{1}, .... x_{n}} where xi = 2^{i}. A sample S ⊆ X is drawn by selecting each x_{i} independently with probability p_{i} = 1/2. The expected value of the smallest number in sample S is:
1/n  
2  
√n  
n 
The given probability P_{i} is for selection of each item independently with probability 1/2.
Now, Probability for x_{1} to be smallest in S = 1/2
Now, Probability for x_{2} to be smallest in S = Probability of x_{1} not being in S × Probability of x_{2} being in S
= 1/2 × 1/2
Similarly, Probability x_{i} to be smallest = (1/2)^{i}
Now the Expected value is
Question 203 
For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is:
No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nC_{n}*(1/2)^{n}*(1/2)^{n} = 2nC_{n}/4^{n}
Question 204 
Let E, F and G be finite sets.
Let X = (E∩F)  (F∩G) and Y = (E  (E∩G))  (EF).
Which one of the following is true?
X ⊂ Y  
X ⊃ Y  
X = Y  
X  Y ≠ ∅ and Y  X ≠ ∅ 
E = {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)}
F = {(1,1), (2,2), (3,3)}
G = {(1,3), (2,1), (2,3), (3,1)}
X = (E∩F)  (F∩G)
= {(1,1), (2,2), (3,3)  ∅}
= {(1,1), (2,2), (3,3)} (✔️)
Y = (E  (E∩G)  (E  F))
= (E  {(1,3), (2,3), (3,1)}  {(1,2), (1,3), (2,3), (3,1)})
= {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)}  {(1,3), (2,2), (2,1)}  (1,2), (1,3), (2,3), (3,1)}
= {(1,1), (1,2), (2,2), (3,3)}  {(1,2), (1,3), (2,3), (3,1)}
= {(1,1), (2,2), (3,3)} (✔️)
X = Y
X = (E∩F)  (F∩G) = {2,5}  {5} = {2}
Y = (E  (E∩G)  (E  F))
= {(1,2,4,5)  (4,5)  (1,4)}
= {(1,2)  (1,4)}
= {2}
X = Y
Question 205 
F is an n×n real matrix. b is an n×1 real vector. Suppose there are two n×1 vectors, u and v such that, u≠v and Fu=b, Fv=b. Which one of the following statements is false?
Determinant of F is zero  
There are an infinite number of solutions to Fx=b  
There is an x≠0 such that Fx=0  
F must have two identical rows 
Fu = Fv
Fu  Fv = 0
F(u  v) = 0
Given u ≠ v
F = 0 (i.e., Singular matrix, so determinant is zero)
Option A is true.
⇾ Fx = b; where F is singular
It can have no solution (or) infinitely many solutions.
Option B is true.
⇾ x ≠ 0 such that Fx = 0 is True because F is singular matrix (“stated by singular matrix rules). Option C is true.
⇾ F can two identical columns and rows.
Option D is false.
Question 206 
Given a set of elements N = {1, 2, ..., n} and two arbitrary subsets A⊆N and B⊆N, how many of the n! permutations π from N to N satisfy min(π(A)) = min(π(B)), where min(S) is the smallest integer in the set of integers S, and π(S) is the set of integers obtained by applying permutation π to each element of S?
(nA ∪ B) A B  
(A^{2}+B^{2})n^{2}
 
n!(A∩B/A∪B)  
Two arbitrary subsets A⊆N and B⊆N.
Out of n! permutations π from N to N, to satisfy
min(π(A)) = min (π(B))
*) π(S) is the set of integers obtained by applying permutation π to each element of S.
If min(π(A)) =min (π(B)), say y = π(x) is the common minimum.
Since the permutation π is a 1to1 mapping of N,
x ∈ A∩B
∴ A∩B cannot be empty.
⇒ y = π(x)
= π(A∩B) is the minimum of π(A∪B) is the minimum of π(A) and π(B) are to be same.
You can think like
*) If the minimum of π(A) and π(B) are same [min π(A)] = min [π(B)]
then min(π(A∩B)) = min(π(A∪B))
∴ Total number is given by n! A∩B/A∪B
*) Finally
Considering all possible permutations, the fraction of them that meet this condition π(A∩B) / π(A∪B)
[The probability of single permutation].
Ex: N = {1, 2, 3, 4} A = {1, 3} B = {1, 2, 4}
Since π is one to one mapping
π(A∩B) = A∩B
∴ π(A) = {1, 2}
π(B) = {1, 4, 3}
π(A∩B) = {1}
π(A∪B) = {1, 2, 3, 4}
4! × 1/4 = 6
Question 207 
Let S = {1,2,3,....,m}, m > 3. Let x_{1}, x_{2},....x_{n} be the subsets of S each of size 3. Define a function f from S to the set of natural numbers as, f(i) is the number of sets of X_{j} that contain the element i. That is f(i) = {ji ∈ X_{j}}.
Then is
3m  
3n  
2m+1  
2n+1 
Question 208 
Which one of the first order predicate calculus statements given below correctly express the following English statement?
Tigers and lions attack if they are hungry or threatened.
∀x [(tiger(x) ∧ lion(x)) → {(hungry(x) ∨ threatened(x))
→ attacks(x)}]  
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x))
∧ attacks(x)}]  
∀x [(tiger(x) ∨ lion(x)) → {(attacks(x) → (hungry (x)) ∨ threatened (x))}]  
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]

Here we have two cases.
i) If Tiger is hungry (or) threaten that will attack.
ii) If Lion is hungry (or) threaten that will attack.
If Tiger is hungry (or) threaten then both lion and tiger will not attack only Tiger will attack and viceversa.
Then answer is
∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
Note: Don’t confuse with the statement Tiger and Lion.
Question 209 
Consider the following propositional statements:
 P1 : ((A ∧ B) → C)) ≡ ((A → C) ∧ (B → C))
P2 : ((A ∨ B) → C)) ≡ ((A → C) ∨ (B → C))
Which one of the following is true?
P1 is a tautology, but not P2  
P2 is a tautology, but not P1  
P1 and P2 are both tautologies  
Both P1 and P2 are not tautologies 
Both P1 and P2 are not Tautologies.
Question 210 
A logical binary relation ⊙,is defined as follows:
Let ~ be the unary negation (NOT) operator, with higher precedence than ⊙. Which one of the following is equivalent to A∧B ?
(~A⊙B)  
~(A⊙~B)  
~(~A⊙~B)  
~(~A⊙B) 
Question 211 
The 2^{n} vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.
The number of vertices of degree zero in G is:
1  
n  
n+1  
2^{n} 
= no. of subsets with size less than or equal to 1
= n+1, because in question it is given that the two vertices are connected if and only if the corresponding sets intersect in exactly two elements.
Question 212 
The 2^{n} vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.
The maximum degree of a vertex in G is:
2^{n2}  
2^{n3} × 3  
2^{n1} 
(k(_{c}))_{2} 2^{nk}
∴ We need to find 'k' value such that, the value will be maximum.[k should be an integer].
If you differentiate (k(_{c}))_{2} 2^{nk} w.r.t. k and equal to 0.
You will get k = 2/(log_{e})2 which is not an integer.
So you can see it like
∴ The maximum degree 3⋅2^{n3} at k=3 or k=4.
Question 213 
The 2^{n} vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.
The number of connected components in G is:
n  
n+2  
2^{n/2}  
2^{n} / n 
While other nodes are connected so that total number of connected components is (n+1)+1
(here we are adding 1 because it is connected corresponding remaining vertices)
= n+2
Question 214 
Let A, B and C be nonempty sets and let X = (A  B)  C and Y = (A  C)  (B  C). Which one of the following is TRUE?
X = Y  
X ⊂ Y  
Y ⊂ X  
None of these 
B = {1, 3, 4, 5}
C = {2, 4, 5, 6}
X = (A  B)  C
X = {2, 6}  {2, 4, 5, 6}
= ∅
Y = (A  C)  (B  C)
= {1, 3}  { 1, 3}
= ∅
X = Y
X = (A  B)  C
= (1, 5)  (5, 7, 4, 3)
= (1)
Y = (A  C)  (B  C)
= (1, 4)  (2, 4)
= (1)
X = Y
Question 215 
Let G be a simple connected planar graph with 13 vertices and 19 edges. Then, the number of faces in the planar embedding of the graph is:
6  
8  
9  
13 
F = E  V + 2 [From Euler's formula i.e., F + V  E = 2]
F = 19  13 +2
F = 8
Question 216 
Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is
12  
8  
Less than 8  
More than 12

Edges = 100
Minimum cover of vertex G is = 8
Maximum Independent set of G = No. of vertices  Minimum cover of vertex G
= 20  8
= 12
Question 217 
Let f(x) be the continuous probability density function of a random variable X. The probability that a < X ≤ b, is:
f(b  a)
 
f(b)  f(a)  
Then the probablity be area of the corresponding curve i.e.,
Question 218 
The set {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively:
3 and 13  
2 and 11  
4 and 13  
8 and 14 
Inverse of 4 = m; Inverse of 7 = n
(4×m)%15=1; (7*n)%15=1
Option A: m=3 n=13
12%15≠1 (✖️) 91%15=1 (✔️)
Option B: m=2 n=11
8%15≠1 (✖️) 11%15≠1 (✖️)
Option C: m=4 n=13
16%15=1(✔️) 91%15=1 (✔️)
Option D: m=8 n=14
120%15≠1(✖️) 98%15≠1(✖️)
Question 219 
Let P, Q and R be three atomic prepositional assertions. Let X denote (P ∨ Q) → R and Y denote (P → R) ∨ (Q → R). Which one of the following is a tautology?
X ≡ Y  
X → Y  
Y → X  
¬Y → X 
⇒ ∼(P∨Q) ∨ R
⇒ (∼P∧∼Q) ∨ R
⇒ (∼P∨R) × (∼Q∨R)
⇒ (P→R) ∧ (Q→R)
Option B: X→Y
[(P→R) × (Q→R)] → [(P→R) ∨ (Q→R)]
∼[(P→R) × (Q→R) ∨ (P→R) ∨ (Q→R)]
[∼(P→R) ∨ ∼(Q→R)] ∨ [(P→R) ∨ (Q→R)]
[∼(P→R) ∨ (P→R)] ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] ∨ [∼(Q→R) ∨ (Q→R)]
T ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] V T
T (✔️)
Question 220 
What is the first order predicate calculus statement equivalent to the following?
Every teacher is liked by some student
∀(x) [teacher(x) → ∃ (y) [student(y) → likes (y, x)]]  
∀(x) [teacher(x) → ∃ (y) [student(y) ∧ likes (y, x)]]  
∃(y) ∀(x) [teacher(x) → [student(y) ∧ likes (y, x)]]  
∀(x) [teacher(x) ∧ ∃ (y)[student(y) → likes (y, x)]] 
Option B: If x is a teacher, then there exists some y, who is a student and like x. (✔️)
Option C: There exists a student who likes all teachers.
Option D: If x is a teacher and then there exists some y, if y is a student then y likes x.
Question 221 
Let R and S be any two equivalence relations on a nonempty set A. Which one of the following statements is TRUE?
R∪S, R∩S are both equivalence relations.  
R∪S is an equivalence relation.  
R∩S is an equivalence relation.  
Neither R∪S nor R∩S is an equivalence relation. 
Let (a,b) present in R and (b,c) present in S and (a,c) is not present in either of them. Then R∪S will contain (a,b) and (b,c) but not (a,c) and hence not transitive.
And equivalence relation must satisfy 3 property:
(i) Reflexive
(ii) Symmetric
(iii) Transitive
But as we have seen that for R∪S, Transitivity is not satisfied.
Question 222 
Let f: B → C and g: A → B be two functions and let h = f∘g. Given that h is an onto function. Which one of the following is TRUE?
f and g should both be onto functions
 
f should be onto but g need not be onto  
g should be onto but f need not be onto  
both f and g need not be onto 
f: B→C and g: A→B are two functions.
h = f∘g = f(g(x))
→ If his onto function, that means for every value in C, there must be value in A.
→ Here, we are mapping C to A using B, that means for every value in C there is a value in B then f is onto function.
→ But g may (or) may not be the onto function i.e., so values in B which may doesn't map with A.
Question 223 
What is the minimum number of ordered pairs of nonnegative numbers that should be chosen to ensure that there are two pairs (a,b) and (c,d) in the chosen set such that
a ≡ c mod 3 and b ≡ d mod 5
4  
6  
16  
24 
That means a = 0,1,2 ⇒ 3
b = dmod5
That means b = 0,1,2,3,4 ⇒ 4
→ Total no. of order pairs = 3 * 5 = 15
→ Ordered pair (c,d) has 1 combination.
Then total no. of combinations = 15+1 = 16
Question 224 
Consider the set H of all 3 × 3 matrices of the type
a f e 0 b d 0 0 c
where a, b, c, d, e and f are real numbers and abc ≠ 0. Under the matrix multiplication operation, the set H is
a group  
a monoid but not a group
 
a semi group but not a monoid  
neither a group nor a semi group

The algebraic structure is a group because the given matrix can have inverse and the inverse is nonsingular.
Question 225 
Which one of the following graphs is NOT planar?
G_{1}  
G_{2}  
G_{3}  
G_{4} 
which is planar
G_{3} can also be drawn as
which is planar
G_{4} can also be drawn as
which is planar
But G_{1} cannot be drawn as planar graph.
Hence, option (A) is the answer.
Question 226 
Consider the following system of equations in three real variables x_{l}, x_{2} and x_{3}
2x_{l}  x_{2} + 3x_{3} = 1 3x_{l} 2x_{2} + 5x_{3} = 2 x_{l} + 4x_{2} + x_{3} = 3
This system of equations has
no solution  
a unique solution  
more than one but a finite number of solutions  
an infinite number of solutions

2(2  20) +1(3 + 5) + 3(12  2)
= 44 + 8 + 30
= 6 ≠ 0
→ A ≠ 0, we have Unique Solution.
Question 227 
What are the eigenvalues of the following 2 × 2 matrix?
2 1 4 5
1 and 1  
1 and 6  
2 and 5  
4 and 1 
A = (2  λ)(5  λ)  (4) = 0
10  7λ+ λ^{2}  4 = 0
λ^{2}  7λ + 6 = 0
λ^{2}  6λ  λ + 6 = 0
(λ  6) 1(λ  6) = 0
λ = 1 (or) 6
Question 228 
Let , where x<1. What is g(i)?
i  
i+1  
2i  
2^{i} 
Put g(i) = i+1
S = 1 + 2x + 3x^{2} + 4x^{3} + .....
Sx = 1x + 2x^{2} + 3x^{3} + 4x^{4} + ......
S  Sx = 1 + x + x^{2} + x^{3} + .....
[Sum of infinite series in GP with ratio < 1 is a/1r]
S  Sx = 1/(1x)
S(1x) = 1/(1x)
S = 1/(1x)^{2}
Question 229 
Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows:
(i) Select a box (ii) Choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q are (1/3) and (2/3), respectively.
Given that a ball selected in the above process is a red ball, the probability that it came from the box P is
4/19  
5/19  
2/9  
19/30 
Q → 3 red, 1 blue
The probability of selecting a red ball is
(1/3)(2/5) + (2/3)(3/4)
2/15 + 1/2 = 19/30
The probability of selecting a red ball from P
(1/3) * (2/5) = 2/15
→ The colour of ball is selected is to be red and that is taken from the box P.
⇒ Probability of selecting a red ball from P/Probability of selecting a red ball
⇒ (2/15)/(19/30)
⇒ 4/19
Question 230 
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is:
1/2^{n}  
1  1/n  
1/n!  
1(1/2^{n}) 
Hence Probability = (2^{n}  1) /2^{n} = 1  1/2^{n}
Question 231 
Identify the correct translation into logical notation of the following assertion.
"Some boys in the class are taller than all the girls"
Note: taller(x,y) is true if x is taller than y.
(∃x) (boy(x) → (∀y) (girl(y) ∧ taller(x,y)))  
(∃x) (boy(x) ∧ (∀y) (girl(y) ∧ taller(x,y)))  
(∃x) (boy(x) → (∀y) (girl(y) → taller(x,y)))  
(∃x) (boy(x) ∧ (∀y) (girl(y) → taller(x,y))) 
'∧' → predicts statements are always true, no matter the value of x.
'→' → predicts there is no need of left predicate to be true always, but whenever it becomes true, then right predicate must be true.
Option D:
There exists a some boys who are taller than of all girls y.
Question 232 
Consider the binary relation:
S = {(x,y)y = x+1 and x,y ∈ {0,1,2, ...}}
The reflexive transitive closure of S is
{(x, y)y > x and x, y ∈ {0, 1, 2, ... }}  
{(x, y)y ≥ x and x, y ∈ {0, 1, 2, ... }}
 
{(x, y)y < x and x, y ∈ {0, 1, 2, ... }}  
{(x, y)y ≤ x and x, y ∈ {0, 1, 2, ... }}

Answer is option B.
{(x, y)y ≥ x and x, y ∈ {0, 1, 2, ... }}
Question 233 
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?
3/8  
1/2  
5/8  
3/4 
Then total number of possibilities = 2^{4} = 16
No. of possibilities getting 2 heads and 2 tails is
HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6
Probability of getting 2 heads and 2 tails is
= No. of possibilities/Total no. of possibilities = 6/16 = 3/8
Question 234 
The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power(2,x) is same as 2^{x})
power (2,n)  
power (2,n^{2})  
power (2, (n^{2} + n)/2)
 
power (2, (n^{2}  n)/2)

A[i][j] = A[j][i]
So, we have only two choices, they are either upper triangular elements (or) lower triangular elements.
No. of such elements are
n + (n1) + (n2) + ... + 1
n(n+1)/2
We have two choices, thus we have
2^{(n(n+1)/2)} = 2^{((n2+n)/2) choices i.e., Power (2, (n2+n)/2). }
Question 235 
Let A, B, C, D be n × n matrices, each with nonzero determinant. If ABCD = 1, then B^{1} is
D^{1}C^{1}A^{1}  
CDA
 
ADC  
Does not necessarily exist 
ABCD = I
Pre multiply A^{1} on both sides
A^{1}ABCD = A^{1}⋅I
BCD = A^{1}
Pre multiply B^{1} on both sides
B^{1}BCD = B^{1}A^{1}
CD = B^{1}A^{1}
Post multiply A on both sides
CDA = B^{1}A^{1}⋅A
∴ CDA = B^{1}(I)
∴ CDA = B^{1}
Question 236 
The following propositional statement is (P → (Q v R)) → ((P v Q) → R)
satisfiable but not valid
 
valid  
a contradiction  
None of the above 
If P=T; Q=T; R=T
(P→(T∨T)) → ((T∨T)→R)
(P→T) → (T→R)
(T→T) → (T→T)
T→T
T(Satisfiable)
Question 237 
How many solutions does the following system of linear equations have?
x + 5y = 1 x  y = 2 x + 3y = 3
infinitely many  
two distinct solutions  
unique  
none 
rank = r(A) = r(AB) = 2
rank = total no. of variables
Hence, unique solution.
Question 238 
The following is the incomplete operation table a 4element group.
The last row of the table is
c a e b  
c b a e  
c b e a  
c e a b 
The last row is c e a b.
Question 239 
The inclusion of which of the following sets into
S = {{1,2}, {1,2,3}, {1,3,5}, (1,2,4), (1,2,3,4,5}}
is necessary and sufficient to make S a complete lattice under the partial order defined by set containment?
{1}  
{1}, {2, 3}  
{1}, {1, 3}  
{1}, {1, 3}, (1, 2, 3, 4}, {1, 2, 3, 5} 
For the set {1, 2, 3, 4, 5} there is no supremum element i.e., {1}.
Then clearly we need to add {1}, then it is to be a lattice.
Question 240 
An examination paper has 150 multiplechoice questions of one mark each, with each question having four choices. Each incorrect answer fetches 0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is:
0  
2550  
7525  
9375 
Probability of selecting a wrong answer = 3/4
For correct answer +1, for wrong answer0.25;
Expected marks for each question = (1/4) × 1 + (3/4) (0.25)
= 1/4 + (3/16)
= 43/16
= 1/16
= 0.0625
Expected marks for 150 questions = 150 × 0.625 = 9.375
The sum total of expected marks obtained by 1000 students is = 1000×9.375 = 9375
Question 241 
Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colourpairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement?
9  
8  
7  
6 
Each is printed twice the no. of letters = 26×2 = 52
If Mala has k colours, she can have k pairs of same colours.
She also can have ^{k}C_{2} different pairs in which each pair is having different colours.
So total no. of pairs that can be coloured = k+^{k}C_{2}
k+^{k}C_{2} ≥ 26
k+k(k1)/2 ≥ 26
k(k+1)/2 ≥ 26
k(k+1) ≥ 52
k(k+1) ≥ 7*8
k≥7
Question 242 
In an M×N matrix such that all nonzero entries are covered in a rows and b columns. Then the maximum number of nonzero entries, such that no two are on the same row or column, is
≤ a+b  
≤ max(a, b)  
≤ min(Ma, Nb)  
≤ min(a, b) 
→ Such that a row can have maximum of a elements and no row has separate element and for b also same.
→ By combining the both, it should be ≤ (a,b).
Question 243 
The minimum number of colours required to colour the following graph, such that no two adjacent vertices are assigned the same colour, is
2  
3  
4  
5 
→ a, b, c, d = 4
→ The minimum no. of colours required to colour a graph = 4 (no two adjacent vertices have same colours)
Question 244 
Two n bit binary strings, S_{1} and S_{2}, are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is
^{n}C_{d} /2^{n}
 
^{n}C_{d} / 2^{d}  
d/2^{n}  
1/2^{d} 
Total no. of cases where n positions have any binary bit = 2^{n}
The probability of 'd' bits differ = ^{n}C_{d} / 2^{n}
Question 245 
How many graphs on n labeled vertices exist which have at least (n^{2}  3n)/2 edges?
Maximum no. of vertices = n(n1)/2 = v
No. of graphs with minimum b edges is
= C(v,e) + C(v,e+1) + C(v,e+2) + ... + C(v,v)
= C((v,ve) + C(v,v(e+1)) + C(v,v(e+2)) + ... + C(v,0)
= C(a,n) + C(a,n1) + C(a,n2) + ... + C(a,0) (since ab=n)
= C(n(n1)/2,n) + C(n(n1)/2,n1) + ... + C(n(n1)/2,0)
Question 246 
A point is randomly selected with uniform probability in the XY plane within the rectangle with corners at (0,0), (1,0), (1,2) and (0,2). If p is the length of the position vector of the point, the expected value of p^{2} is
2/3  
1  
4/3  
5/3 
Above diagram shows the scenario of our question.
The length p of our position vector (x,y) is
p = √x^{2} + y^{2}
p^{2} = x^{2} + y^{2}
E(p^{2}) = E(x^{2} + y^{2}) = E(x^{2}) + E(y^{2})
Now we need to calculate the probability density function of X and Y.
Since distribution is uniform,
X goes from 0 to 1, so PDF(x) = 1/10 = 1
Y goes from 0 to 2, so PDF(y) = 1/20 = 1/2
Now we evaluate,
E(p^{2}) = E(x^{2}) + E(y^{2}) = 5/3
Question 247 
Let G_{1} = (V,E_{1}) and G_{2} = (V,E_{2}) be connected graphs on the same vertex set V with more than two vertices. If G_{1} ∩ G_{2} = (V, E_{1} ∩ E_{2}) is not a connected graph, then the graph G_{1} U G_{2} = (V, E_{1} U E_{2})
cannot have a cut vertex
 
must have a cycle  
must have a cutedge (bridge)
 
has chromatic number strictly greater than those of G_{1} and G_{2}

(A)
False, since in G_{1}∪G_{2} 'C' is a cut vertex.
(B) True, for all conditions.
(C)
False. G_{1}∪G_{2} has no bridge.
D)
False. G_{1}∪G_{2}, G_{1}, G_{2} all the three graphs have chromatic number of 2.
Question 248 
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(AB) and P(BA) respectively are
1/4, 1/2  
1/2, 1/4  
1/2, 1  
1, 1/2 
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
Question 249 
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3^{n}
Question 250 
Consider the set ∑* of all strings over the alphabet ∑ = {0, 1}. ∑* with the concatenation operator for strings
does not form a group  
forms a noncommutative group  
does not have a right identity element  
forms a group if the empty string is removed from Σ*

→ To perform concatenation with the given set can result a Monoid and it follows the property of closure, associativity and consists of identity element.
Question 251 
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between
k and n  
k – 1 and k + 1  
k – 1 and n – 1  
k + 1 and n – k 
If a vertex is removed then it results that all the components are also be disconnected. So removal can create (n1) components.
Question 252 
Let (S, ≤) be a partial order with two minimal elements a and b, and a maximum element c. Let P: S → {True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y ∈ S satisfying x ≤ y, where ⇒ stands for logical implication. Which of the following statements CANNOT be true?
P(x) = True for all x ∈ S such that x ≠ b
 
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c  
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c  
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x 
a or b the minimal element in set.
P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.
Option D is False.
Question 253 
Which of the following is a valid first order formula? (Here α and β are first order formulae with x as their only free variable)
((∀x)[α] ⇒ (∀x)[β]) ⇒ (∀x)[α⇒β]
 
(∀x)[α] ⇒ (∃x)[α ∧ β]  
((∀x)[α ∨ β] ⇒ (∃x)[α] ⇒ (∀x)[α]  
(∀x)[α ⇒ β] ⇒ ((∀x)[α] ⇒ (∀x)[β]) 
Here, α, β are holding values of x. Then and RHS saying that α holding the value of x and β is holding value of x.
Then LHS ⇒ RHS.
Question 254 
Consider the following formula a and its two interpretations I_{1} and I_{2}
α: (∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒ (∀x)[¬Px] I_{1}: Domain: the set of natural numbers Px ≡ 'x is a prime number' Qxy ≡ 'y divides x' I_{2}: same as I_{1} except that Px = 'x is a composite number'.
Which of the following statements is true?
I_{1} satisfies α, I_{2} does not  
I_{2} satisfies α, I_{1} does not
 
Neither I_{2} nor I_{1} satisfies α
 
Both I_{1} and I_{2} satisfy α 
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Q_{yy}]] ⇒(∀x)[¬Px]
Q_{yy} is always true, because y divide y, then ¬Q_{yy} is false.
∀x[(P(x) ⇔ ∀y [Qxy ⇔ False]]
∀y [Qxy ⇔ False] can be written as ∀y[¬axy]
⇒(∀x)[P(x) ⇔ ∀y[¬Qxy]]
Here, ¬Qxy says that y doesnot divides x, which is not always be true.
For example, if x=y then it is false then ∀y[¬Qxy] is not true for all values of y.
⇒(∀x)[P(x) ⇔ False]
⇒(∀x)[¬P(x) = RHS]
LHS = RHS
⇒ Irrespective of x, whether x is prime of composite number I_{1} and I_{2} satisfies α.
Question 255 
m identical balls are to be placed in n distinct bags. You are given that m ≥ kn, where, k is a natural number ≥1. In how many ways can the balls be placed in the bags if each bag must contain at least k balls?
. So option (B) is correct.
Question 256 
How many perfect matching are there in a complete graph of 6 vertices ?
15  
24  
30  
60 
(2n)!/n!×2^{n}
Given, 2n = 6 ⇒ n = 3
So, finally, 6!/3!×2^{3} = 15
Question 257 
Let f : A → B be an injective (onetoone) function. Define g: 2^{A}→2^{B} as: g(C) = {f(x)x ∈ C}, for all subsets C of A.
Define h: 2^{B} → 2^{A} as: h(D) = {xx ∈ A, f(x) ∈ D}, for all subsets D of B.
Which of the following statements is always true?
g(h(D)) ⊆ D  
g(h(D)) ⊇ D  
g(h(D)) ∩ D = ɸ  
g(h(D)) ∩ (B—D) ≠ ɸ 
→ g: 2^{A}→2^{B} be also one to one function and g(C) = f(x)x∈C}, for all subsets C of A.
The range of this function is n(2^{A}).
→ h: 2^{B}→2^{A} it is not a one to one function and given h(D) = {xx∈A, f(x)∈D}, for all subsets D of B.
The range of this function is also n(2^{A}).
→ The function g(h(D)) also have the range n(2^{A}) that implies n(A)≤n(B), i.e., n(2^{A}) is less than n(2^{B}).
Then this result is g(h(D)) ⊆ D.
Question 258 
Consider the set {a, b, c} with binary operators + and × defined as follows:
+ a b c × a b c a b a c a a b c b a b c b b c a c a c b c c c b
For example, a + c = c, c + a = a, c × b = c and b × c = a. Given the following set of equations:
(a × x) + (a × y) = c (b × x) + (c × y) = c
The number of solution(s) (i.e., pair(s) (x, y)) that satisfy the equations is:
0  
1  
2  
3 
In those (x, y) = (b,c) & (c,b) are the possible solution for the corresponding equations.
(x, y) = (b,c) ⇒ (a*b)+(a*c) ⇒ (b*b)+(c*c)
⇒ (b) + (c) ⇒ c + b
⇒ c (✔️) ⇒ c (✔️)
(x,y) = (c,b) ⇒ (a*c)+(a*b) ⇒ (b*c)+(c*b)
⇒ c+b ⇒ a+c
⇒ c (✔️) ⇒ c (✔️)
Question 259 
Let ∑ = (a, b, c, d, e) be an alphabet. We define an encoding scheme as follows:
g(a) = 3, g(b) = 5, g(c) = 7, g(d) = 9, g(e) = 11.
Let p_{i} denote the ith prime number (p_{1}=2).
 For a nonempty string s=a_{1} ... a_{n}, where each a_{i}∈Σ, define
For a nonempty sequence 〈s_{j}...s_{n}〉 of strings from ∑^{+}, define
Which of the following numbers is the encoding h of a nonempty sequence of strings?
2^{7}3^{7}5^{7}  
2^{8}3^{8}5^{8}
 
2^{9}3^{9}5^{9}  
2^{10}5^{10}7^{10} 
And f(S) = 2^{3} = 8
So answer is 2^{8}3^{8}5^{8}.