Engineering-Mathematics
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(X2) - (E(X))2 = 5
E(X2) = 5 + (E(X))2 = 5 + 25 = 30
So, E[(X + 2)2] = E[X2 + 4 + 4X]
= E(X2) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Question 2 |
If the ordinary generating function of a sequence is
, then a3 - a0 is equal to ________.
15 | |
16 | |
17 | |
18 |

Question 3 |

1 | |
Limit does not exist | |
53/12 | |
108/7 |

Question 4 |
Let G be an arbitrary group. Consider the following relations on G:
-
- R
-
- : ∀a,b ∈ G, aR
-
- b if and only if ∃g ∈ G such that a = g
-
- bg
-
- R
-
- : ∀a,b ∈ G, aR
-
- b if and only if a = b
Which of the above is/are equivalence relation/relations?
R2 only | |
R1 and R2 | |
Neither R1 and R2 | |
R1 only |
Consider Statement R1:
Reflexive:
aR1a
⇒ a = g-1ag
Left multiply both sides by g
⇒ ga = gg-1ag
Right multiply both sides by g-1
⇒ gag-1 = gg-1agg-1
⇒ gag-1 = a [∴ The relation is reflexive]
Symmetric:
If aR1b, then ∃g ∈ G such that gag-1 = b then a = g-1bg, which is Correct.
⇒ So, given relation is symmetric.
Transitive:
The given relation is Transitive.
So, the given relation R1 is equivalence.
R2:
The given relation is not reflexive.
So, which is not equivalence relation.
Such that a ≠ a-1.
So, only R1 is true.
Question 5 |
Let X be a square matrix. Consider the following two statements on X.
-
- I. X is invertible.
- II. Determinant of X is non-zero.
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 non-zero.
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 | |
(n-1)! | |
The total number of hamiltonian cycles in a complete graph are
(n-1)!/2, where n is number of vertices.
Question 7 |
Let U = {1,2,...,n}. Let A = {(x,X)|x ∈ X, X ⊆ U}. Consider the following two statements on |A|.

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 3x2 + 6xY + 3Y + 6 has only real roots is (rounded off to 1 decimal place) _____.
0.3 | |
0.9 | |
0.1 | |
0.8 |
3x2 + 6xY + 3Y + 6
= 3x2 + (6Y)x + (3Y + 6)
which is in the form: ax2 + bx + c
For real roots: b2 - 4ac ≥ 0
⇒ (6Y)2 - 4(3)(3Y + 6) ≥ 0
⇒ 36Y2 - 36Y - 72 ≥ 0
⇒ Y2 - Y - 2 ≥ 0
⇒ (Y+1)(Y-2) ≥ 0
Y = -1 (or) 2
The given interval is (1,6).
So, we need to consider the range (2,6).
The probability = (1/(6-1)) * (6-2) = 1/5 * 4 = 0.8
Question 9 |
Consider the first order predicate formula φ:
- ∀x[(∀z z|x ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z z|w ⇒ ((w = z) ∨ (z = 1)))]
Here 'a|b' denotes that 'a divides b', where a and b are integers. Consider the following sets:
-
S1. {1, 2, 3, ..., 100}
S2. Set of all positive integers
S3. Set of all integers
Which of the above sets satisfy φ?
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
⇒ λ2 - 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 equi-probable 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 {an}, where an = 2n+3 for all n = 0, 1, 2, …?
3/(1-x)2 | |
3x/(1-x)2 | |
2-x/(1-x)2 | |
3-x/(1-x)2 |

Question 16 |

0.289 | |
0.298 | |
0.28 | |
0.29 |

Question 17 |
Assume that multiplying a matrix G1 of dimension p×q with another matrix G2 of dimension q×r requires pqr scalar multiplications. Computing the product of n matrices G1G2G3…Gn can be done by parenthesizing in different ways. Define GiGi+1 as an explicitly computed pair for a given parenthesization if they are directly multiplied. For example, in the matrix multiplication chain G1G2G3G4G5G6 using parenthesization(G1(G2G3))(G4(G5G6)), G2G3 and G5G6 are the only explicitly computed pairs.
Consider a matrix multiplication chain F1F2F3F4F5, where matrices F1, F2, F3, F4 and F5 are of dimensions 2×25, 25×3, 3×16, 16×1 and 1×1000, respectively. In the parenthesization of F1F2F3F4F5 that minimizes the total number of scalar multiplications, the explicitly computed pairs is/ are
F1F2 and F3F4 only
| |
F2F3 only | |
F3F4 only | |
F1F2 and F4F5 only
|
→ Optimal Parenthesization is:
((F1(F2(F3 F4)))F5)
→ But according to the problem statement we are only considering F3, F4 explicitly computed pairs.
Question 18 |
Consider the first-order logic sentence
where ψ(s,t,u,v,w,x,y) is a quantifier-free first-order 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(HG) denote the probability that Guwahati has high temperature. Similarly, P(MG) and P(LG) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(HD), P(MD) and P(LD) 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 (HG) then the probability of Delhi also having a high temperature (HD) is 0.40; i.e., P(HD ∣ HG) = 0.40. Similarly, the next two entries are P(MD ∣ HG) = 0.48 and P(LD ∣ HG) = 0.12. Similarly for the other rows.
If it is known that P(HG) = 0.2, P(MG) = 0.5, and P(LG) = 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 (HG ) then the probability that Delhi also having a high temperature (HD ) is 0.40.
P (HD / HG ) = 0.40
We need to find out the probability that Guwahati has high temperature.
Given that Delhi has high temperature (P(HG / HD )).

P (HD / HG ) = P(HG ∩ HD ) / P(HD )
= 0.2×0.4 / 0.2×0.4+0.5×0.1+0.3×0.01
= 0.60
Question 20 |
Let N be the set of natural numbers. Consider the following sets,
-
- 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 a matrix P whose only eigenvectors are the multiples of
Consider the following statements.
-
- (I)
-
- does not have an inverse
-
- (II)
-
- has a repeated eigenvalue
-
- (III)
- 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 |
The statement (¬p) ⇒ (¬q) is logically equivalent to which of the statements below?
-
- 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 |
Consider the first-order logic sentence F: ∀x(∃yR(x,y)). Assuming non-empty logical domains, which of the sentences below are implied by F?
-
- 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 c1, cn be scalars not all zero. Such that the following expression holds:

where ai is column vectors in Rn. Consider the set of linear equations.
Ax = B.where A = [a1.......an] and
Then, Set of equations has
a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1 | |
no solution | |
infinitely many solutions | |
finitely many solutions |
AX = B

As given that

means c0a0 + c1a1 + ...cnan = 0, represents that a0, a1... an 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, 'Σai = b',
so for c1c2...cn = {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 σ2. 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 L-Hospital rule

Substitute x=1
⇒ (7(1)6-10(1)4)/(3(1)2-6(1)) = (7-10)/(3-6) = (-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 R2 whose Euclidean norms satisfy ||u||=2||v||. 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 R2, 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(u-v)+2α-4 = 0
4-2α = Cosθ(4-2α)
(4-2α)(Cosθ-1) = 0
4-2α = 0

Question 31 |
Let A be m×n real valued square symmetric matrix of rank 2 with expression given below.

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 engenvalues 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.

a2+b2+c2+d2 = 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,
|A-λI| = 0 (The characteristic equation)

-λ(-λ)-25 = 0
λ2-25 = 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|+C-|A∩B|-|A∩C|-|B∩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|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C|
= 166+100+71-33-28-14+4
= 271
Question 33 |
If f(x) = Rsin(πx/2) + S, f'(1/2) = √2 and , then the constants R and S are, respectively.
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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 |


R2→R2+R1

The number of non-zero rows of P + Q = 2,
So Rank = 2
Note: “Rank” is the number of independent vectors.
Method-1:
Each vector is a row in matrix.
Echelon form of a matrix has no. of zeroes increasing each rows.
The total non-zero rows left give the rank.
Method-2:
Find determinant of matrix, for 3×5, if determinant is ‘0’, the max rank can be 2.
If determinant of any n×n is non-zero, then rows proceed with (n-1)×(n-1).
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| = 2|E|
The relation between max and min degree of graph are
m ≤ 2|E| / |v| ≤ M
Given minimum degree = 3
So, 3 ≤2 |E| / |v|
3|v| ≤ 2|E|
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
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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)/(1-P(q)) = (2⁄3)/(1-1/6) = (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)=pj, pj≥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 gy(Z)=(1-β+βz)N. The expectation of Y is
Nβ(1 - β) | |
Nβ | |
N(1 - β) | |
Not expressible in terms of N and β alone |
Given gy (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(xj,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 λ3 - 4λ2 + 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 |
λ3 - 4λ2 + aλ + 30 = 0 ⇾ (1)
One eigen value is ‘2’, so substitute it
23 - 4(2)2 + a(2) + 30 = 0
8 - 16 + 2a + 30 = 0
2a = -22
a = -11
Substitute in (1),
λ3 - 4λ2 - 11 + 30 = 0

So, (1) can be written as
(λ - 2)(λ2 - 2λ - 15) = 0
(λ - 2)(λ2 - 5λ + 3λ - 15) = 0
(λ - 2)(λ - 3)(λ - 5) = 0
λ = 2, 3, 5
Max λ=5
Question 41 |
Let p,q,r,s represent the following propositions.
-
- 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 an be the number of n-bit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for an?
an = a(n-1) + 2a(n-2) | |
an = a(n-1) + a(n-2) | |
an = 2a(n-1) + a(n-2) | |
an = 2a(n-1) + 2a(n-2) |
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 a3 = a1 + a2
Similarly, an = an-1 + an-2
Question 43 |

4 | |
3 | |
2 | |
1 |

Question 44 |
A probability density function on the interval [a,1] is given by 1/x2 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/x2 , 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 2-i.
As per the theorems, the determinant of the matrix is the product of the eigen values.
So the determinant is (2+i)*(2-i)*3 = 15.
Question 46 |
The coefficient of x12 in (x3 + x4 + x5 + x6 + ...)3 is _________.
10 | |
11 | |
12 | |
13 |
⇒ [x3(1 + x + x2 + x3 + ...)]3
= x9(1 + x + x2 + x3 + ...)3
First Reduction:
As x9 is out of the series, we need to find the co-efficient of x3 in (1 + x + x2 + ⋯)3

Here, m=3, k=3, the coefficient

= 5C3 = 5!/2!3! = 10
Question 47 |
Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = K × 104. The value of K is ___________.
198 | |
199 | |
200 | |
201 |
Replace a(n-1)

⇒ an = 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯ a1
Given that a1 = 8, replace it
⇒ an = 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯8
= 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯ + 6(1)2 + 2(1)

= 6(n2 + (n-1)2 + (n-2)2 + ⋯ + 22 + 12) + 2(n + (n-1) + ⋯1)
Sum of n2 = (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 a99 = k×104
a99 = 2(99)(100)2 = 198 × 104
∴k = 198
Question 48 |
A function f:N+ → N+, defined on the set of positive integers N+, satisfies the following properties:
-
- 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 |
Consider the following experiment.
Step1.-
- Flip a fair coin twice.
-
- If the outcomes are (TAILS, HEADS) then output
-
- and stop.
-
- If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output
-
- and stop.
- 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)/(1-1/4)
= 1/3
= 0.33
Question 50 |
Consider the following expressions:
-
- (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 vertex-colour 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 |
Consider the systems, each consisting of m linear equations in n variables.
-
- 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(A|B)
then there will be no solution.
ii) False, because if R(A) = R(A|B),
then there will be solution possible.
iii) True, if R(A) = R(A|B),
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/ AT = 1/|A| = 1/8 = 0.125
Question 56 |
A binary relation R on ℕ × ℕ is defined as follows: (a,b)R(c,d) if a≤c or b≤d. Consider the following propositions:
-
- P:
-
- is reflexive
-
- Q:
- 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 well-formed 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 |
Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements:
-
- I. Each compound in
-
- reacts with an odd number of compounds.
-
- II. At least one compound in
-
- reacts with an odd number of compounds.
-
- III. Each compound in
- reacts with an even number of compounds.
Which one of the above statements is 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

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 1399(mod 17), in the range 0 to 16, is ________.
4 | |
5 | |
6 | |
7 |
a(p-1) ≡ 1 mod p (p is prime)
From given question,
p = 17
a(17-1) ≡ 1 mod 17
a16 ≡ 1 mod 17
1316 ≡ 1 mod 17
Given:
1399 mod 17

133 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 l22 of L is
5 | |
6 | |
7 | |
8 |

l11 = 2 -----(1)
l11u12 = 2
u12 = 2/2
u12 = 1 ----- (2)
l21 = 4 ----- (3)
l21u12+l22 = 9
l22 = 9 - l21u12 = 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/(1-x)2 |
Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1-h(x)=1-x/x-1=-1/x-1
h(g(x))=g(x)/g(x)-1=1-x/-x
⇒ g(h(x))/h(g(x))=x/(x-1)(1-x)=(x/x-1)/1-x=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 L3 = {(x,y,z): x, y, z ∈ L} be the set of all ordered triplets of the elements of L. Let pr be the probability that an element (x,y,z) ∈ L3 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)∈ L3 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) = 125-6 = 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=a-4b
⇒ a=5 ⇒ -7=5-4b
⇒ 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
∴3|R| = 2|E| ⇒ |R| = 2/3|E| ------(3)
Substituting (2), (3) in (1), we get
10 + 2/3|E| = |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 |

= 2-1/1(2)+3-2/2(3)+4-3/3(4)+…+100-99/99(100)
= 1/1-1/2+1/2-1/3+1/3…+1/98-1/99+1/99-1/100
= 1-1/100
= 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 |

⇒ λ2 - 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 |
= 22+3×52×7 (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 C1, C3 are proportional i.e., C3 = 15C1)
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 self-complementary if it is isomorphic to its complement for all self-complementary 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 xa and xb for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if f(xb) is very small and then xb 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: xa, xb, ε, N // ε = convergence indicator fb = f(xb) i = 0 while (i < N and |fb| > ε) do i = i + 1 // update counter xt = ? // missing expression for // intermediate value xa = xb // reset xa xb = xt // reset xb fb = f(xb) // function value at new xb end while if |fb| > ε then // loop is terminated with i = N write “Non-convergence” else write “return xb” end if
xb – (fb–f(xa))fb /(xb–xa) | |
xa – (fa–f(xa))fa /(xb–xa) | |
xb – (xb–xa)fb /(fb–f(xa)) | |
xa – (xb–xa) fa /(fb–f(xa)) |

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 X-axis, 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 one-to-one is ______.
0.95 | |
0.96 | |
0.97 | |
0.98 |
Number of functions from X to Y is 202 i.e., 400 and number of one-one functions from X to Y is 20P2 i.e., 20×19 = 380
∴ Probability of a function f being one-one 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 non-decreasing 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 4-digit 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 220 bytes, block size of 16 bytes and a direct mapped cache having 212 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 212 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 non-zero 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) + a2 = a/x - 25a ----- (3)
abf(1/x) + b2 = bk - 25b ----- (4)
Subtract (3) - (4), we get
(a2 - b2) f(x) = a/x- 25a - bx + 25b
f(x) = 1/(a2 - b2) (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 |

= 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 non-trivial solution,
- px + qy + rz = 0
qx + ry + pz = 0
rx + py + qz = 0
then which one of the following options is True?
p-q+r = 0 or p = q = -r | |
p+q-r = 0 or p = -q = r | |
p+q+r = 0 or p = q = r | |
p-q+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 p-s = q-r
⇒ r-q = s-p
⇒ (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?
G1=(V,E1) where E1={(u,v)|(u,v)∉E} | |
G2=(V,E2 )where E2={(u,v)│(u,v)∈E} | |
G3=(V,E3) where E3={(u,v)|there is a path of length≤2 from u to v in E} | |
G4=(V4,E) where V4 is the set of vertices in G which are not isolated |
→ It strongly connected.
(A) G1=(V,E1) where E1={(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) G2=(V,E2 )where E2={(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) G3=(V,E3) where E3={(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) G4=(V4,E) where V4 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
G4 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 4-by-4 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

(II) There exists such that

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(2-y) | |
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(2-y) |
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 six-sided 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
x1 + x2 + x3 + x4 = 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+x1+x2 = 22
x1+x2 = 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 n-pennant is a sequence of numbers with sum equal to n. For example, (1,1,2) is a 4-pennant. The set of all possible 1-pennants is {(1)}, the set of all possible 2-pennants is {(2), (1,1)}and the set of all 3-pennants 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 10-pennants 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 log2log2N is ______.
16 | |
17 | |
18 | |
19 |
{0,1}4 = {0,1}×{0,1}×{0,1}×{0,1} = 16
|S| = 216
N = 2|S|
loglogN=loglog2|S| = log |S| = log216 = 16
Question 100 |
Consider an undirected graph where self-loops 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 (d1, d2, ..., dn) with d1 ≥ d2 ≥ ... dn is called graphic if there exists a simple undirected graph with n vertices having degrees d1, d2, ..., dn respectively. Which of the following 6-tuples 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’ non-working 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 non-working 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 non-zero 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) = (x-1)(x-2)(x-3)
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 Newton-Raphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2 … is obtained for the function
0.75x3 – 2x2 – 2x + 4 = 0
Consider the statements
(I) x3 = 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 non-zero 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

x1 + x5 = λx1 ---------- (1)
x1 + x5 = λx5 ---------- (2)
(1) + (2) ⇒ 2(x1 + x5) = λ(x1 + x5) ⇒ λ1 = 2
x2 + x3 + x4 = λ∙x2 -------- (4)
x2 + x3 + x4 = λ∙x3 -------- (5)
x2 + x3 + x4 = λ∙x4 -------- (6)
(4)+(5)+(6) = 3(x2 + x3 + x4) = λ(x2 + x3 + x4 ) ⇒ λ2 = 3
Product = λ1 × λ2 = 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) = 1-P(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 n-3 vertices ( one is self and two other vertices in actual graph).
As per given question,
n-3 =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 one-one & 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 V1 and V2 are 4-dimensional subspace of a 6-dimensional vector space V, then the smallest possible dimension of V1∩V2 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
V1 ∩ V2 = V1 + V2 - V1 ∪ V2 = 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 | |
n-k+1 |
Option 1, 2 will give answer 1. (i.e. one edge among them),
Option 3: n-k = 0 edges.
Option 4: n-k+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 :
n-3n/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 = x2 + y2. 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 = x2 + y2 --------(1)
Replace x, y in (1)
y⊕x = y2 + x2 which is same as (1), so this is cumulative
(x⊕y)⊕z = (x2 + y2) ⊕ z
= (x2 + y2) + z2
= x2 + y2 + z2 + 2x2y2 ----------(2)
x⊕(y ⊕ z) = x ⊕ (y2 + z2)
= x2 + (y2 + z2)2
= x2 + y2 + z2 + 2y2z2 ----------- (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/(2e3) | |
9/(2e3) | |
17/(2e3) | |
26/(2e3) |
P(x : λ) = (e-λ λx)/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 30)/0!+(e-3 31)/1!+(e-3 32)/2!
= e-3+e-3∙3+(e-3)∙9)/2
= (17e-3<)/2
= 17/(2e3 )
Question 131 |
Which one of the following does NOT equal to
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Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C2 → C1 + C2



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+) = x-1 = 3-1 = 2
∴ f(x) is continuous.
Option B:
At x = 3, f(x) = 4
For x ≠ 3, f(x) = f(3+) = f(3-) = 8-x = 8-3 = 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 I1 and I2 are correct inferences | |
I1 is correct but I2 is not a correct inference | |
I1 is not correct but I2 is a correct inference | |
Both I1 and I2 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.
I2: 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 a11 = a12 = a21 = +1 and a22 = -1. Then the eigenvalues of the matrix A19 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, An has eigen value as λn.
Eigen value of

|A-λI| = 0

-(1-λ)(1+λ)-1 = 0
-(1-λ2 )-1 = 0
-1 = 1-λ2
λ2 = 2
λ = ±√2
A19 has (√2)19 = 29×√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 |
v-e+f = 2
Given 10 vertices & 15 edges
10-15+f = 2
f = 2+15-10
f = 7
There will be an unbounded face always. So, number of faces = 6.
Question 141 |
Which of the following graphs is isomorphic to

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(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) = x4 - x3 - x2 - 4 in the interval [1,9]. The method converges to a solution after ________ iterations.
1 | |
3 | |
5 | |
7 |
Question 143 |
Suppose a fair six-sided 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 n-element (n ≥ 2) set to a 2-element set?
2n | |
2n-1 | |
2n-2 | |
2(2n– 2) |

Onto function is possible if m ≥ n. So, no. of onto functions possible is,
nm - nC1 (n-1)m + nC2 (n-2)m + .......
Here in Question,
m = n, n = 2
So, the final answer will be,
= 2n - 2C1 (2-1)n + 2C2 (2-2)n
= 2n - 2 × 1 + 0
= 2n - 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 6C4 => 15.
From each set of 4 vertices, suppose a set {a, b, c, d} we can have cycles like
a-b-c-d
a-b-d-c
a-c-b-d
a-c-d-b
a-d-b-c
a-d-c-b (Total 6, which is equal to number of cyclic permutations (n-1)! )
As they are labelled you can observe, a-b-c-d and a-d-c-b 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[X2]) 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 = [x1, x2, …, xn]. 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 yi = a * xi + b, where a and b are positive constants. Let μy be the mean and σy 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. | |
μy = aμx + b | |
σy = aσx + b |
(σy)2 is variance so,
yi = a * xi + b
(σy)2 = a2 (σx)2
⇒ σy = a σx
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,
5C1 × 4C1 = 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| = 2|T| | |
|S| = |T| - 1 | |
|S| = |T| | |
|S| = |T| + 1 |

id= 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| = 2|E|
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 |
Newton-Raphson method is used to compute a root of the equation x2 - 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?
210 | |
215 | |
220 | |
225 |
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 non-diagonal 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 (n2-n)non-diagonal elements (i.e., 2n2-n)
Ex:
{(1,1)(2,2)(3,3)} ----- ‘0’ non-diagonal element
{(1,1)(2,2)(3,3)(1,2)} ----- ‘1’ non-diagonal 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)} (n2-n) diagonal elements
____________________
Total: 2n2-n
For the given question n = 5.
The number of reflexive relations = 2(25-5) = 220
Question 155 |
Consider the set S = {1, ω, ω2}, where ω and ω2 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, ω2} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w2' and inverse of 'w2' 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 1099 is a multiple of 1096?
1/625 | |
4/625 | |
12/625 | |
16/625 |
We can write 1099 as 1096×103
So, (1099)/(1096) to be a whole number, [1096×103/1096] ➝ (1)
We can observe that every divisor of 103 is a multiple of 1096
So number of divisor of 103 to be found first
⇒ 103 = (5×2)3 = 23×53
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 1099 are 1099 = 299×599 = 100×100 = 10000
Probability that divisor of 1099 is a multiple of 1096 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. d1, d2, d3... dn
⇾ Discard d1, subtrack ‘1’ from the next 'd1' 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 n-vertex simple connected graph which does not contain any odd length cycle? Assume n ≥ 2.
2 | |
3 | |
n-1 | |
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,....n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) 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 well-formed 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) ∪ (Pc∩Q∩R) ∪ Qc ∪ Rc is
Qc ∪ Rc
| |
P ∪ Qc ∪ Rc | |
Pc ∪ Qc ∪ Rc | |
U |
(P∩Q∩R)∪(Pc∩Q∩R)∪Qc∪Rc
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
-
x1 + x2 + 2x3 = 1
x1 + 2x2 + 3x3 = 2
x1 + 4x2 + ax3 = 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 a-5 = 0, then rank(A) = rank[A|B]<3,
So infinite number of solutions.
But, it is given that the given system has unique solution i.e., rank(A) = rank[A|B] = 3 will be retain only if a-5 ≠ 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 Newton-Raphson 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 vertex-cover 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 K4 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+...+(2k-1)
= (2-1)+(4-1)+(6-1)+(8-1)+...+(2k-1)
= (2+4+6+8+...+2k)+(-1+-1+-1+k times)
= Q-(1+1+...+k times)
= Q-k
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 3x4 - 16x3 + 24x2 + 37 is
0 | |
1 | |
2 | |
3 |
f’(x) = 12x3 + 48x2 + 48x = 0
12x(x2 - 4x + 4) = 0
x=0; (x-2)2 = 0
x=2
f’’(x) = 36x2 - 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 σy 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 | |
n2 and n | |
n2 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
= 32
= n2 (for set of n elements)
Question 187 |
Let G be the non-planar 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 ≤ 3n-6 (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 non-planar graph.
iv) e=10, n=6
10 ≤ 3(6) - 6
10 ≤ 18 - 6
10 ≤ 12
Yes, it is planar.
Question 188 |
How many different non-isomorphic Abelian groups of order 4 are there?
2 | |
3 | |
4 | |
5 |
4 = 22
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 k-regular 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 n-1/2 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 |

|(A-λI)2-I| = 0 [a2-b2 = (a+b)(a-b)]
|(A-λI+I)(A-λI-I) = 0
|(A-(λ-I)I)(A-(λ+I)I| = 0
Let us assume:
λ-1=k & λ +1=k
λ =k+1 λ =k-1
⇓ ⇓
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 π1, π2, π3, π4 on Let p be the partial order on the set of partitions S' = {π1, π2, π3, π4} defined as follows: πi p πj if and only if πi refines πj. The poset diagram for (S', p) is:
![]() | |
![]() | |
![]() | |
![]() |
And, neither π2 refines π3, nor π3 refines π2.
Here, only π1 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 ∈ R3| x1+x2+x3 = 0, where xT = [x1,x2,x3]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 R3 | |
None of the above
|
Question 195 |
Consider the series obtained from the Newton-Raphson method. The series converges to
1.5 | |
√2 | |
1.6 | |
1.4 |
xn+1 = xn/2+9/8xn ⟶ (I); x0 = 0.5
Equation based on Newton-Raphson is
xn+1 = xn-f(xn)/f'(xn) ⟶ (II)
Equate I and II
xn-f(xn)/f'(xn) = xn/2+9/8xn
xn-f(xn)/f'(xn) = xn-xn/2+9/8xn
xn-f(xn)/f'(xn) = xn-(4xn2-9)/8xn
So, f(x) = 4xn2-9
4x2-9 = 0
4x2 = 9
x2 = 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)?
![]() | |
220 | |
210 | |
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)?
29 | |
219 | |
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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) = a0 + a1x + a2x2 + a3x3, where ai ≠ 0, ∀i. The minimum number of multiplications needed to evaluate p on an input x is:
3 | |
4 | |
6 | |
9 |
p(x) = a0 + a1x + a2x2 + a3x3 where ai≠0
This can be written as
p(x) = a0 +x( a1 + a2x + a3x2)=a0+(a1+(a2+a3x)x)x
Total no. of multiplications required is 3
i.e., a3x = K.....(i)
(a2+K)x = M..... (ii)
(a1+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:
Z2xy | |
Z×2xy | |
Z2x+y | |
2xyz |
A set ‘P’ consists of m elements and ‘Q’ consists of n elements then total number of function from P to Q is mn.
⇒ E be the no. of subsets of W = 2|w| = 2|xxy| = 2xy
No. of function from Z to E is = (2xy)z = (2xy)z = 2xyz
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 |
We are given a set X = {x1, .... xn} where xi = 2i. A sample S ⊆ X is drawn by selecting each xi independently with probability pi = 1/2. The expected value of the smallest number in sample S is:
1/n | |
2 | |
√n | |
n |
The given probability Pi is for selection of each item independently with probability 1/2.
Now, Probability for x1 to be smallest in S = 1/2
Now, Probability for x2 to be smallest in S = Probability of x1 not being in S × Probability of x2 being in S
= 1/2 × 1/2
Similarly, Probability xi to be smallest = (1/2)i
Now the Expected value is

Question 202 |
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:
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No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nCn*(1/2)n*(1/2)n = 2nCn/4n
Question 203 |
Let E, F and G be finite sets.
Let X = (E∩F) - (F∩G) and Y = (E - (E∩G)) - (E-F).
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 204 |
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 205 |
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?
(n-|A ∪ B|) |A| |B| | |
(|A|2+|B|2)n2
| |
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 1-to-1 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 206 |
Let S = {1,2,3,....,m}, m > 3. Let x1, x2,....xn 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 Xj that contain the element i. That is f(i) = |{j|i ∈ Xj|}|.
Then is
3m | |
3n | |
2m+1 | |
2n+1 |

Question 207 |
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 vice-versa.
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 208 |
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 209 |
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 210 |
The 2n 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 | |
2n |
= 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 211 |
The 2n 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:
![]() | |
2n-2 | |
2n-3 × 3 | |
2n-1 |
(k(c))2 2n-k
∴ We need to find 'k' value such that, the value will be maximum.[k should be an integer].
If you differentiate (k(c))2 2n-k w.r.t. k and equal to 0.
You will get k = 2/(loge)2 which is not an integer.
So you can see it like

∴ The maximum degree 3⋅2n-3 at k=3 or k=4.
Question 212 |
The 2n 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 | |
2n/2 | |
2n / 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 213 |
Let A, B and C be non-empty 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 214 |
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 215 |
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 216 |
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) | |
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Then the probablity be area of the corresponding curve i.e.,

Question 217 |
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 218 |
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 219 |
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 220 |
Let R and S be any two equivalence relations on a non-empty 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 221 |
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 222 |
What is the minimum number of ordered pairs of non-negative 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 223 |
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 non-singular.
Question 224 |
Which one of the following graphs is NOT planar?

G1 | |
G2 | |
G3 | |
G4 |

which is planar
G3 can also be drawn as

which is planar
G4 can also be drawn as

which is planar
But G1 cannot be drawn as planar graph.
Hence, option (A) is the answer.
Question 225 |
Consider the following system of equations in three real variables xl, x2 and x3
2xl - x2 + 3x3 = 1 3xl- 2x2 + 5x3 = 2 -xl + 4x2 + x3 = 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 226 |
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+ λ2 - 4 = 0
λ2 - 7λ + 6 = 0
λ2 - 6λ - λ + 6 = 0
(λ - 6) -1(λ - 6) = 0
λ = 1 (or) 6
Question 227 |
Let , where |x|<1. What is g(i)?
i | |
i+1 | |
2i | |
2i |

Put g(i) = i+1

S = 1 + 2x + 3x2 + 4x3 + .....
Sx = 1x + 2x2 + 3x3 + 4x4 + ......
S - Sx = 1 + x + x2 + x3 + .....
[Sum of infinite series in GP with ratio < 1 is a/1-r]
S - Sx = 1/(1-x)
S(1-x) = 1/(1-x)
S = 1/(1-x)2
Question 228 |
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 229 |
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:
1/2n | |
1 - 1/n | |
1/n! | |
1-(1/2n) |
Hence Probability = (2n - 1) /2n = 1 - 1/2n
Question 230 |
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 231 |
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 232 |
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 = 24 = 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 233 |
The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power(2,x) is same as 2x)
power (2,n) | |
power (2,n2) | |
power (2, (n2 + n)/2)
| |
power (2, (n2 - 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 + (n-1) + (n-2) + ... + 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 234 |
Let A, B, C, D be n × n matrices, each with non-zero determinant. If ABCD = 1, then B-1 is
D-1C-1A-1 | |
CDA
| |
ADC | |
Does not necessarily exist |
ABCD = I
Pre multiply A-1 on both sides
A-1ABCD = A-1⋅I
BCD = A-1
Pre multiply B-1 on both sides
B-1BCD = B-1A-1
CD = B-1A-1
Post multiply A on both sides
CDA = B-1A-1⋅A
∴ CDA = B-1(I)
∴ CDA = B-1
Question 235 |
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 236 |
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(A|B) = 2
rank = total no. of variables
Hence, unique solution.
Question 237 |
The following is the incomplete operation table a 4-element 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 238 |
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 239 |
An examination paper has 150 multiple-choice 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 answer-0.25;
Expected marks for each question = (1/4) × 1 + (3/4) -(0.25)
= 1/4 + (-3/16)
= 4-3/16
= 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 240 |
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 colour-pairs 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 kC2 different pairs in which each pair is having different colours.
So total no. of pairs that can be coloured = k+kC2
k+kC2 ≥ 26
k+k(k-1)/2 ≥ 26
k(k+1)/2 ≥ 26
k(k+1) ≥ 52
k(k+1) ≥ 7*8
k≥7
Question 241 |
In an M×N matrix such that all non-zero entries are covered in a rows and b columns. Then the maximum number of non-zero entries, such that no two are on the same row or column, is
≤ a+b | |
≤ max(a, b) | |
≤ min(M-a, N-b) | |
≤ 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 242 |
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 243 |
Two n bit binary strings, S1 and S2, 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
nCd /2n
| |
nCd / 2d | |
d/2n | |
1/2d |
Total no. of cases where n positions have any binary bit = 2n
The probability of 'd' bits differ = nCd / 2n
Question 244 |
How many graphs on n labeled vertices exist which have at least (n2 - 3n)/2 edges?
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Maximum no. of vertices = n(n-1)/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,v-e) + C(v,v-(e+1)) + C(v,v-(e+2)) + ... + C(v,0)
= C(a,n) + C(a,n-1) + C(a,n-2) + ... + C(a,0) (since a-b=n)
= C(n(n-1)/2,n) + C(n(n-1)/2,n-1) + ... + C(n(n-1)/2,0)

Question 245 |
A point is randomly selected with uniform probability in the X-Y 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 p2 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 = √x2 + y2
p2 = x2 + y2
E(p2) = E(x2 + y2) = E(x2) + E(y2)
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) = 1/1-0 = 1
Y goes from 0 to 2, so PDF(y) = 1/2-0 = 1/2
Now we evaluate,

E(p2) = E(x2) + E(y2) = 5/3
Question 246 |
Let G1 = (V,E1) and G2 = (V,E2) be connected graphs on the same vertex set V with more than two vertices. If G1 ∩ G2 = (V, E1 ∩ E2) is not a connected graph, then the graph G1 U G2 = (V, E1 U E2)
cannot have a cut vertex
| |
must have a cycle | |
must have a cut-edge (bridge)
| |
has chromatic number strictly greater than those of G1 and G2
|
(A)

False, since in G1∪G2 'C' is a cut vertex.
(B) True, for all conditions.
(C)

False. G1∪G2 has no bridge.
D)

False. G1∪G2, G1, G2 all the three graphs have chromatic number of 2.
Question 247 |
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A|B) and P(B|A) 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 248 |
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
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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
= 3n
Question 249 |
Consider the set ∑* of all strings over the alphabet ∑ = {0, 1}. ∑* with the concatenation operator for strings
does not form a group | |
forms a non-commutative 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 250 |
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 (n-1) components.
Question 251 |
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 252 |
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 253 |
Consider the following formula a and its two interpretations I1 and I2
α: (∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒ (∀x)[¬Px] I1: Domain: the set of natural numbers Px ≡ 'x is a prime number' Qxy ≡ 'y divides x' I2: same as I1 except that Px = 'x is a composite number'.
Which of the following statements is true?
I1 satisfies α, I2 does not | |
I2 satisfies α, I1 does not
| |
Neither I2 nor I1 satisfies α
| |
Both I1 and I2 satisfy α |
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒(∀x)[¬Px]
Qyy is always true, because y divide y, then ¬Qyy 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 I1 and I2 satisfies α.
Question 254 |
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?
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. So option (B) is correct.
Question 255 |
How many perfect matching are there in a complete graph of 6 vertices ?
15 | |
24 | |
30 | |
60 |
(2n)!/n!×2n
Given, 2n = 6 ⇒ n = 3
So, finally, 6!/3!×23 = 15
Question 256 |
Let f : A → B be an injective (one-to-one) function. Define g: 2A→2B as: g(C) = {f(x)|x ∈ C}, for all subsets C of A.
Define h: 2B → 2A as: h(D) = {x|x ∈ 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: 2A→2B 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(2A).
→ h: 2B→2A it is not a one to one function and given h(D) = {x|x∈A, f(x)∈D}, for all subsets D of B.
The range of this function is also n(2A).
→ The function g(h(D)) also have the range n(2A) that implies n(A)≤n(B), i.e., n(2A) is less than n(2B).
Then this result is g(h(D)) ⊆ D.
Question 257 |
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 258 |
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 pi denote the i-th prime number (p1=2).
- For a non-empty string s=a1 ... an, where each ai∈Σ, define

For a non-empty sequence 〈sj...sn〉 of strings from ∑+, define

Which of the following numbers is the encoding h of a non-empty sequence of strings?
273757 | |
283858
| |
293959 | |
210510710 |
And f(S) = 23 = 8
So answer is 283858.
Question 259 |
A graph G = (V,E) satisfies |E|≤ 3|V|-6. The min-degree of G is defined as