Computer Science And Information Technology - CS 2017 GATE Paper - 2 (Practice Test) - GATE MCQ

Let us consider all choices :

V is to the West of W,

V

W

Z is to the East of X and the West of V.

X

Z

V

W

W is to West of Y.

X

Z

V

W

Y

Therefore, V is in middle. So option A is correct.

With adjectives and adverbs, enough comes after adjectives and adverbs. 
With nouns, enough comes before noun.
In the given question, enough is used with bright which is an adjective, so enough will come after the adjective.
So bright enough is the correct option. 

Ways to collect both Red = ³C₂

Ways to collect both Green = ³C₂

Ways to collect both Blue = ⁴C₂

Total Ways = ¹⁰C₂

P(Socks of Same Colour) = (³C₂ + ⁴C₂+ ³C₂ ) /  ¹⁰C₂ = 4/15

• Yield - produce or provide
• Resistance - the refusal to accept or comply with something. Rest are all synonyms. Hence, (D) is correct.

The given function is a continuous function in range [-5, 5]. The Intermediate Value Theorem states if a continuous function has values of opposite sign inside an interval, then it has a root in that interval. The function changes sign twice in range [-5, 5].

1. It becomes positive to negative in range [-5, 0] when e^x + 0.5x² becomes less than 2.
2. It becomes positive to negative in range [0, 5] when e^x + 0.5x² becomes more than 2.

In given question there are three boxes incorrectly labeled:

• Box 3 labeled as Apples and Oranges - Original Content can be Apple or Orangle (This box will have only one fruit, so whichever we pick would  be the correct label). Let's say the picked item was Apple, therefore the box labeled Orange will have Apples and Oranges.)
• Box 2 - Oranges - This box will definitely not have orange (as its mislabeled) and it can not contain only apple (As it is in box 3) - So this will have Original Content- both oranges and Apples
• Box 1 - Apples - This is the leftover box and it can now have Original Content - Orange

Therefore option B is correct

X = 4777....(till 30 digits)
or X = 4.7777⋯ ∗ 10²⁹
X²⁹ = (4.777⋯∗10²⁹)³
  = (4.777…)³ ∗ 10 ⁸⁷
So, (X³ )³  requires 3 + 87 = 90 digits.

Region R has maximum air pressure variation as it  which has maximum number of lines crossing the area. Therefore option C is correct 

Alternate Solution
We have to see for the region which has maximum change in air pressure for thunderstorm.
In region P – the air pressure changes only once by 0.05 as only 2 lines are crossing the region. The first line with the air pressure 0.95 and the next one is with pressure 0.9. 
In region Q – No line crosses the region which implies that the air pressure doesn’t change.
In region R – the air pressure changes at 4 places resulting in maximum change in air pressure in the whole region. The minimum pressure in R is 0.65 and the maximum is 0.8.
In region S – the air pressure in the region in 0.95 and other contouring line is outside the region thus no change in air pressure within the region.
Thus the probability of thunderstorm is maximum in region R. 

P - iii Syntax tree is input to semantic analyser Q - iv Character stream is input to lexical analyser. R - i Intermediate code is input to code generator S -ii  Token stream is input to syntax analyser. Therefore option C

In contiguous allocation there is always possibility of external fragmentation. Both indexed and linked allocation are free from external fragmentation.

Therefore, option D

In one to many or many to one participation, relation is merged in table which is on many side and if there is total participation then relation is merged in that side. Therefore, option C is correct

• Tower of Hanoi - Ɵ(2ⁿ)
• Heap sort worst case - Ɵ(n log n)
• Binary Search - Ɵ(log n)
• Addition of two nxn matrices - Ɵ (n²)

Therefore Option C is correct

Record Route is an optional field used to record address. 
In IPv4 header, 40 bytes are reserved for OPTIONS. For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40, 37 bytes are left. 
An IP4 address takes 32 bits or 4 bytes. 
We can store at most floor(37/4) = 9 router addresses. 

Alternate Solution
In IPV4 frame format there are 40 bytes of option field, out of which only 38 bytes can be used, 2 bytes are reserved . Also one IPV4 adress is of 4 bytes 
Thus maxmimum IPV4 address that can be hold <= 38/4 = 9

Given,

• Q -> R(Primary Key)
• S -> P (Primary Key)

Entry to be deleted - P (3) and Q(8)

• Q can be deleted directly
• Now, S - > P but the relationship given is on delete set NULL, Therefore when  we delete 3 from  T1 ,the  entry in T2 having 3 will be NULL.

Therfore, Option A  - Answer is 0 entries

In BFS, we print a starting node, then its adjacent, then adjacent of adjacent, and so on..

I. A connected UDP socket can be used to communicate with multiple peers simultaneously is WRONG : Even if UDP is connectionless, every UDP socket has a single specified IP number and port number to connect to. Therefore we can not use same socket to connect to multiple peers simultaneously. II. A process can successfully call connect function again for an already connected UDP socket is RIGHT : We can call connect again to connect to a new peer and disconnect previous peer if implementation allows

Context free language are closed under union and difference with regular language.
It is not closed under complementation and intersection . Complement of CFL is recursive language .

• static char var; -> A variable located in data section of memory as it is static in nature
• m = malloc(10); m = null; ->This is a lost memory which cannot be freed as m=NULL
• Char *Ptr[10]; -> 10 memory locations of char type are allocated to store addresses
• register int var1;-> Request to allocate a CPU register to store data

S --> XY
X --> aX | a // This produces only "a"
Y --> aYb | ∈ // This produces and "a" for every "b"

Option (A) and (B) are wrong because n can be zero also due to epsilon in Y Option (D) is wrong because Y-->aYb produces equal number of a's and b's.
Since there is one variable X which produces at least one a.
Therefore numbers of a's are always greater than numbers of b's.

As per handshaking lemma,

Sum of degree of all vertices < = 2 * no. of edges.

Let v be number of vertices in graph.
==> 3*v <= 2*25
==> Minimum value of v = 16

In 32-bit IEEE-754 format

1st bit represent sign
2-9th bit represent exponent and 10-32 represent Mantissa (Fraction part)

Sign = 0, so positive 
2-9 bits — 01111100 when subtracted by 01111111 i.e., 126 decimal value gives -> 0000 0011 
Which is -3.(negative as the value is less than 126) 
As number is less than 126 it is subtracted otherwise 126 would have been subtracted from it in 32 bit representation. 
Mantissa is normal ,hence, 1.M can be used .Which is 1.1101101. 
Thus, 
Data + 1.1101101 * 2^-3 (±M * B^(±e)) 
Mantissa shift right 3 times -> 
+0.0011101101 
= 0.228 
= 2.28 * 10^-1 
Thus, option c is correct.