EngineeringMathematics
Question 1 
If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)^{2}] equals _________.
A  54 
B  55 
C  56 
D  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 ________.
A  15 
B  16 
C  17 
D  18 
Question 3 
A  1 
B  Limit does not exist 
C  53/12 
D  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?
A  R_{2} only 
B  R_{1} and R_{2} 
C  Neither R_{1} and R_{2} 
D  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?
A  I implies II; II does not imply I. 
B  II implies I; I does not imply II. 
C  I and II are equivalent statements. 
D  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
A  n! 
B  1 
C  (n1)! 
D 
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?
A  Only II 
B  Only I 
C  Neither I nor II 
D  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) _____.
A  0.3 
B  0.9 
C  0.1 
D  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?
A  S1 and S3 
B  S1, S2 and S3 
C  S2 and S3 
D  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 ______.
A  12 
B  17 
C  10 
D  8 
Question 11 
The largest eigenvalue of A is ________
A  3 
B  4 
C  5 
D  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 __________.
A  0.021 
B  0.022 
C  0.023 
D  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 _______.
A  1 
B  2 
C  3 
D  4 
Question 14 
Let G be a finite group on 84 elements. The size of a largest possible proper subgroup of G is _________.
A  41 
B  42 
C  43 
D  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, …?
A  3/(1x)^{2} 
B  3x/(1x)^{2} 
C  2x/(1x)^{2} 
D  3x/(1x)^{2} 
Question 16 
A  0.289 
B  0.298 
C  0.28 
D  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
A  F_{1}F_{2} and F_{3}F_{4} only

B  F_{2}F_{3} only 
C  F_{3}F_{4} only 
D  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 
φ ≡ ∃s∃t∃u∀v∀w∀x∀y ψ(s,t,u,v,w,x,y)
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?
A  There exists at least one model of φ with universe of size less than or equal to 3. 
B  There exists no model of φ with universe of size less than or equal to 3.

C  There exists no model of φ with universe of size greater than 7. 
D  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 _______ .
A  0.60 
B  0.61 
C  0.62 
D  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?
A  Q and S only 
B  P and S only 
C  P and R only 
D  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?
A  Only I and III are necessarily true 
B  Only II is necessarily true 
C  Only I and II are necessarily true 
D  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 = ___________.
A  109 
B  110 
C  111 
D  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
A  I only 
B  I and IV only 
C  II only 
D  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))
A  IV only 
B  I and IV only 
C  II only 
D  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  a unique solution at x = J_{n} where J_{n} denotes a ndimensional vector of all 1 
B  no solution 
C  infinitely many solutions 
D  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.A  4 and 15 respectively 
B  3 and 14 respectively 
C  4 and 14 respectively 
D  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 __________.
A  0 
B  1 
C  2 
D  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 
A  is 0 
B  is 1 
C  is 1 
D  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  a tautology 
B  a contradiction 
C  always TRUE when p is FALSE 
D  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?
A  2 
B  1/2 
C  1 
D  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?
A  Both (I) and (II) 
B  (I) only 
C  (II) only 
D  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 _________.
A  271 
B  272 
C  273 
D  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.
A  
B  
C  
D 
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
A  (¬p ∧ r) ∧ (¬r → (p ∧ q)) 
B  (¬p ∧ r) ∧ ((p ∧ q) → ¬r) 
C  (¬p ∧ r) ∨ ((p ∧ q) → ¬r) 
D  (¬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 _________.
A  0 
B  1 
C  2 
D  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 _________.
A  2 
B  3 
C  4 
D  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 ___________.
A  16 
B  17 
C  18 
D  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
A  
B  
C  
D 
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
A  Nβ(1 – β) 
B  Nβ 
C  N(1 – β) 
D  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 ________.
A  5 
B  6 
C  7 
D  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 _________.
A  11 
B  12 
C  13 
D  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  a_{n} = a_{(n1)} + 2a_{(n2)} 
B  a_{n} = a_{(n1)} + a_{(n2)} 
C  a_{n} = 2a_{(n1)} + a_{(n2)} 
D  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 
A  4 
B  3 
C  2 
D  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 _________.
A  0.7 
B  0.6 
C  0.5 
D  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 __________.
A  18 
B  15 
C  17 
D  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 _________.
A  10 
B  11 
C  12 
D  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 ___________.
A  198 
B  199 
C  200 
D  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 __________.
A  2 
B  3 
C  4 
D  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) ________.
A  0.33 
B  0.34 
C  0.35 
D  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 _________.
A  4 
B  5 
C  6 
D  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))