GATE Shift 1 Computer Science Question Paper 1st February 2025: The Indian Institute of Technology (IIT) Roorkee commenced the GATE 2025 Examination today, February 1, 2025. As per the official timetable, Computer Science Shift 1 (CS1) Agriculture Engineering (AG), and Mathematics (MA) examinations were held in Forenoon Sessions from 9:30 AM to 12:30 PM. Candidates who appeared for the Computer Science Engineering examination can check the memory-based GATE Shift 1 Computer Science Question Paper 1st February 2025 and the unofficial answer key.
GATE 2025 CS Paper Analysis, Question Paper and Answer Key |
GATE 2025 CS Paper Expected Cutoff |
GATE Shift 1 Computer Science Question Paper 1st February 2025: PDF Link
The GATE 2025 Shift 1 CS memory-based question paper PDF is in the table below.
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GATE Shift 1 Computer Science Question Paper 1st February 2025 PDF Link (to be added soon) |
GATE Shift 1 Computer Science Question Paper 1st February 2025: Memory-Based Questions
The GATE CS Shift 1 of Feb 1 memory-based questions are given below.
- void foo(int*p, int x)
{
*p=x;
}
int main()
{
int *z;
int a=20, b=25;
z=&a;
foo(z,b);
printf("%d",a);
return 0;
} - The height of the rooted tree is defined as the maximum number of edges in the path from the root head to any key node. Suppose a Min Heap-T stores 32 keys, the height of T is: _____. [NAT]
- BST n distinct [MSQ]
(a) finding O(log 2n)
(b) Inorder-->Sorted sequence
(c) Maximum length f from root node to any other node
(d) BST is also min heap - int main()
{
int a[]={0,1,2,2,2,0,0,1,1};
printf("%d",fw(4,9));
return 0;
}
int fw(int s[], int size)
{
if (size==0) return 0;
if (size==1) return 1;
if (size[0]!==s[1])
1 + fw (A, size-1)
else
return( s+1, size-1)
fw(A,9)
} - -6 in 4 bit in 2's complement 1010
- Pseudo code:
- If A matrix [1 -1] then the eigen values of A13 = _________.
[1 1] - A box contains 5 coins, where 4 are real coins and 1 is a fake coin. Now, the fake coin has head on both the sides. If one coin is removed and then tossed twice, then find the probability of getting the fake coin.
- Which of the following techniques used in compiler code optimizations uses live variable analysis?
(a) constant folding
(b) runtime function call management
(c) register allocation to variables
(d) strength reduction - Match the following
1. Network layer A. Packet Handling 2. Transport layer B. Frame and Error Handling 3. DLL C. Host to Host Communication - A dice is thrown three times. Find the probability of getting one head.
- f(A,B,C,D)=∑(0,2,4,8,10,12)
- T(n)=2T(n-1)+n*2n where n>0
(a) O(n2*2n)
(b) O(logn)2*2n
(c) O(nX2n)
(d) O(4n) - If 5 bit are transmitted and probability of flipping the bit is 0.01, then what is the probability that the message received is error free?
- A Language L is accepted by NFA with n states:
(a) Every DFA that accepts L > 2n states
(b) L may have NFA < n states
(c) There exists DFA with less than or equal to 2n states for L
(d) L may have DFA < n states
GATE Shift 1 Computer Science Question Paper 1st February 2025: Answer Key
Candidates can check the answers to the memory-based questions below.
- void foo(int*p, int x)
{
*p=x;
}
int main()
{
int *z;
int a=20, b=25;
z=&a;
foo(z,b);
printf("%d",a);
return 0;
}
Answer: 25 - The height of the rooted tree is defined as the maximum number of edges in the path from the root head to any key node. Suppose a Min Heap-T stores 32 keys, the height of T is: _____.
Answer: 5 - BST n distinct
(a) finding O(log 2n)
(b) Inorder-->Sorted sequence
(c) Maximum length f from root node to any other node
(d) BST is also min heap
Answer: (b) and (c) - int main()
{
int a[]={0,1,2,2,2,0,0,1,1};
printf("%d",fw(4,9));
return 0;
}
int fw(int s[], int size)
{
if (size==0) return 0;
if (size==1) return 1;
if (size[0]!==s[1])
1 + fw (A, size-1)
else
return( s+1, size-1)
fw(A,9)
}
Answer: To Be Added - -6 in 4 bit in 2's complement 1010
Answer: 1111111111111010 - Pseudo code:
Answer: 435 - If A matrix [1 -1] then the eigen values of A13 = _________.
[1 1]
Answer: -64√2 e - A box contains 5 coins, where 4 are real coins and 1 is a fake coin. Now, the fake coin has head on both the sides. If one coin is removed and then tossed twice, then find the probability of getting the fake coin.
Answer: 1/2 - Which of the following techniques used in compiler code optimizations uses live variable analysis?
(a) constant folding
(b) runtime function call management
(c) register allocation to variables
(d) strength reduction
Answer: (c) - Match the following
Answer: 1(A), 2(C), 3(B)1. Network layer A. Packet Handling 2. Transport layer B. Frame and Error Handling 3. DLL C. Host to Host Communication - A dice is thrown three times. Find the probability of getting one head.
Answer: 25/72 - f(A,B,C,D)=∑(0,2,4,8,10,12)
Answer: C'D' + AB'C + B'D' - T(n)=2T(n-1)+n*2n where n>0
(a) O(n2*2n)
(b) O(logn)2*2n
(c) O(nX2n)
(d) O(4n)
Answer: (a) - If 5 bit are transmitted and probability of flipping the bit is 0.01, then what is the probability that the message received is error free?
Answer: 0.95099 - A Language L is accepted by NFA with n states:
(a) Every DFA that accepts L > 2n states
(b) L may have NFA < n states
(c) There exists DFA with less than or equal to 2n states for L
(d) L may have DFA < n states
Answer: (b), (c), (d)
The GATE Shift 1 Computer Science Question Paper 1st February 2025 is a memory-based one and IIT Roorkee will release the official GATE 2025 Question Paper 2025 soon after the exam concludes on February 16, 2025. Candidates are advised to visit the official website of GATE 2025 for more information.
GATE 2025 Subject-wise Expected Cutoff |
- GATE AG Expected Cutoff Marks 2025: Agricultural Engineering Qualifying Marks
- GATE MA Expected Cutoff Marks 2025: Mathematics Qualifying Marks
- GATE NM Expected Cutoff Marks 2025: Naval Architecture and Marine Engineering Qualifying Marks
- GATE MT Expected Cutoff Marks 2025: Metallurgical Engineering Qualifying Marks
- GATE TF Expected Cutoff Marks 2025: Textile Engineering and Fibre Science Qualifying Marks
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