GATE Computer Science Engineering(CSE) 2027 Practice Test: Engineering

canvas – a type of cloth
canvass – a campaign or surve

imminent – about to happen, to take place very soon
eminent – remarkable, distinguished

Irritate and Nettle are synonyms. Both mean to excite a feeling of anger. Rest all are antonyms.

From (I), selling price per article = Rs. 10
From (II), cost price per article = profit per article
Now, selling price = cost price + profit
So, cost price per article = profit per article = Rs. 5
Total 200 articles are sold.
Thus, total profit = Rs. 1000
Therefore, both (I) and (II) are together sufficient, but neither alone.

Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Desired options are HHH, HHT, HTH, THH, HTT, THT, TTH.
Thus, P(atmost two tails) = 7/8

Skeleton crew means the least number of employees required to keep a work running

I is true as city T must be having lower temperature than city S. That is the reason why the people of city S have decided to move to city T.

Since 2 black balls were drawn, the number of black balls may be either of 2,3,4,5.
Let A2 = bag contains 2 black balls
A3 = bag contains 3 black balls
A4 = bag contains 4 black balls
A5 = bag contains 5 black balls
E = 2 black balls are drawn

P(E / A2) = ²C₂ / ⁵C₂ = 1/10
P(E / A3) = ³C₂ / ⁵C₂ = 3/10
P(E / A4) = ⁴C₂ / ⁵C₂ = 3/5
P(E / A5) = ⁵C₂ / ⁵C₂ = 1

Since the number of black balls is unknown, all events A2, A3, A4, A5 are equally likely.
So, P(A2) = P(A3) = P(A4) = P(A5) = 1/4

By Bayes’ theorem,
P(A5 / E) = {P(A5) X P(E / A5) } / [{P(A2) X P(E / A2) } + {P(A3) X P(E / A3) } + {P(A4) X P(E / A4) } + {P(A5) X P(E / A5) }]

P(A5 / E) = {1/4 X 1} / [{1/4 X 1/10} + {1/4 X 3/10} + {1/4 X 3/5} + {1/4 X 1}]

P(A5 / E) = 1/2 = 0.5

Number of post graduate employees in company H = 80% of 3360 = 2688
Number of male post–graduate employees in company H = 1800
Therefore, number of female post–graduate employees in company H = 2688 – 1800 = 888
Also, Total female employees in company H = (5 / 14) * 3360 = 1200
So, number of female non post–graduate employees in company H = 1200 – 888 = 312
Hence, required percentage = (312 / 1200) * 100 = 26 %.

F = 2788 / 17 = 164
H = 3360 / 21 = 160
B = 2880 / 18 = 160
C = 2310 / 14 = 165

From option C, f(g(z)) = f(z+2) = (z+2)² + 3(z+2) – 8 = h(z)

There is a property in limits that lim_x–>0 (cos x – 1) / x = 0.

Selection sort has a complexity O ( n² ) in every case.
Insertion sort has the best case when the input is already sorted. Every insert requires constant time.
Heap sort has a complexity O (n log(n)) in every case.

T(n) = 2 T(n–1) + c, which is O(2ⁿ).
Since there can be atmost O(n) elements in function call stack, extra space required is O(n).

The function represents code for a variation of subset sum problem.
Here we find if there is a subsequences with sum of squares of elements equal to X or not.

DFS – ACEFHGBD
BFS – ACFBEHGD
So, A,C and D have the same position.
Thus, 3 is the required answer.

The height of a binary search tree is the maximum distance of a leaf node from the root.

FIRST (S) = {i, a}
FIRST (E) = {e, eps}
FIRST (C) = {b}

FOLLOW (S) = {$, e}
FOLLOW (E) = {$, e}
FOLLOW (C) = {t}

When we make the LL(1) parse table, we get a conflict in E row.
Thus, the given grammar is not LL(1).

No. of elements in Set 1 = 6. So, No. of elements in the power set of Set 1 = 2⁶ = 64
No. of elements in Set 2 = 3. So, No. of elements in the power set of Set 1 = 2³ = 8
No. of elements in Set 3 = 5. So, No. of elements in the power set of Set 1 = 2⁵ = 32
No. of elements in Set 4 = 2. So, No. of elements in the power set of Set 1 = 2² = 4

The Johnson counter resets the input, i.e., reaches the initial input in ‘2n’ clock cycles, where ‘n’ is the number of bits. So, for 4 – bit counter, the counter will reset in 2 X 4 = 8 clock cycles.

In register reference instructions, the first 12 bits (0 – 11) specify the register operation.
The next three bits are 111 and specify opcode and the last bit of the instruction is mode which is 0.
Therefore, last 4 bits are always 1110.

Port numbers 1 – 1023 belong to the ‘well – known’ ports.
Port numbers 1024 – 49151 belong to Registered ports
Port numbers 49152 – 65535 belong to Private ports.

The length of fragment offset field is 13 bits. Also, fragment offset is measured in 8 – byte blocks.
So, maximum offset is (2¹³ – 1) X 8 = 65528 bytes.

HTTPS – 443
SNMP – 161
SMB – 445
POP3 – 110

Making a DFA is all that is required to get the correct answer.

It takes 4 bits to reference 12 logical pages (2⁴ = 16, so maximum 16 pages can be referenced in 4 – bits). Now, we are left with 16 – 4 = 12 bits, which can be used for pages.
So, page size = 2¹² = 4K

For calculating ‘m’, any non – zero value is considered TRUE. In conditional operator, ++i makes i = –1, which makes it go to the TRUE case of the conditional operator. Now, ––k will return 0 and since there is AND after that, it will not be calculated any further as False && False or False && True will always return False.
So, i = –1, j = 2, k = 0, m = 0.

Both the queries will remove all the rows from the table. But, query 1 can be undone by using rollback whereas query 2 is permanent. Also, DELETE fires all the associated triggers whereas TRUNCATE does not.

So, the sum of all elements is 9.