CS and IT Gate 2026 Set-2 Questions with Answer

The correct answer is 0.5.
This is a conditional probability problem involving 6 independent fair coin tosses (T1 through T6). We need to find P(E₁ | E₂), where E₁ and E₂ are defined over overlapping but not identical subsets of tosses.
E₁ - among tosses 2, 4, 6, there are at least 2 heads. E₂ - among tosses 1, 2, 3, 5, there are equal heads and tails (exactly 2 heads, 2 tails).
Key Insight - Shared toss T2: Toss 2 is the only toss common to both events. Tosses T4 and T6 are entirely outside E₂'s scope, and tosses T1, T3, T5 are entirely outside E₁'s scope. Since all tosses are independent, T4 and T6 remain unconditionally fair (P = 1/2 each) even given E₂.
Conditional distribution of T2 given E₂: By symmetry of the 4 tosses in E₂, each individual toss is equally likely to be a head. So P(T2 = H | E₂) = 2/4 = 1/2 - the same as the unconditional probability.
Computing P(E₁ | E₂): Given E₂, the three tosses {T2, T4, T6} each independently show heads with probability 1/2. Therefore, the number of heads in {T2, T4, T6} given E₂ follows a Binomial(3, 1/2) distribution - identical to the unconditional case.
P(E₁ | E₂) = P(at least 2 heads in 3 fair tosses) = P(X=2) + P(X=3) = C(3,2)·(1/2)³ + C(3,3)·(1/2)³ = 3/8 + 1/8 = 4/8 = 0.5.

Propagation delay one-way = 3000 × 1000 m × 5 ns/m = 15 ms, so RTT = 30 ms.
Transmission time per byte = 8 bits ÷ 10⁸ bps = 80 ns.
For max utilization, window size W ≥ 1 + 2a, where a = 15 ms ÷ 80 ns = 187,500 → W = 375,001 → needs ⌈log₂(375,001)⌉ = 19 bits.
For no wrap-around in 60 seconds, bytes sent = 10⁸ ÷ 8 × 60 = 750,000,000 → needs ⌈log₂(750,000,000)⌉ = 30 bits.
N must satisfy both conditions, so N = max(19, 30) = 30.
The correct answer is N = 30.

The correct answer is N = 30.
Given:
Wrap-around time = 60 seconds
Bandwidth = 10⁸ bits per second = 10⁸/8 bytes per second
Each transmitted data byte gets a unique sequence number.
Key insight: Since every byte has a unique sequence number, we need enough sequence numbers to cover all bytes transmitted during the wrap-around period of 60 seconds - so sequence numbers don''t repeat within that window.
Number of bytes transmitted in 60 seconds:
Bytes/sec = 10⁸ / 8
Total bytes in 60s = 60 × (10⁸ / 8) = 750,000,000 bytes
Minimum bits required:
2^N ≥ 750,000,000
N ≥ log₂(60 × 10⁸/8) = log₂(750,000,000) ≈ 29.48
N = ⌈29.48⌉ = 30
Note - the propagation delay is used to determine the window size for maximum utilization, but for the sequence number field size, the binding constraint is the wrap-around time condition. With 30 bits, we get 2³⁰ ≈ 1.07 billion unique sequence numbers - enough to cover all bytes in 60 seconds without any wrap-around.

e correct answer is Option A - Accelerate.
All the given words - Expedite, Hasten, and Hurry - have a similar meaning. They all refer to making something happen faster or increasing the speed of an action.
Expedite means to make a process happen more quickly.
Hasten means to cause something to happen sooner.
Hurry means to act with speed.
Accelerate also means to increase the speed of something, so it matches the meaning of the given words.
Why other options are incorrect:
Option B - Retard: means to slow down or delay, which is the opposite meaning.
Option C - Provide: means to give or supply something, which is unrelated.
Option D - Disable: means to make something unable to function, which is completely different.

The correct answer is Option A — 1/4.
This is a classic Binomial probability problem. Each day independently has probability p = 0.5 of being cloudy and q = 0.5 of being sunny. We need exactly 3 cloudy days out of 4, which follows the binomial formula:
P(X = k) = C(n, k) × p^k × q^(n−k)
Here n = 4, k = 3, p = 0.5, q = 0.5:
P(X = 3) = C(4, 3) × (0.5)³ × (0.5)¹ = 4 × (1/8) × (1/2) = 4/16 = 1/4.
C(4, 3) = 4 accounts for the 4 different ways to choose which 3 of the 4 days are cloudy (CCCS, CCSC, CSCC, SCCC). Each such arrangement has probability (0.5)³ × (0.5)¹ = 1/16. Multiplying gives 4/16 = 1/4.
Why not Option D (3/8)? A common mistake is computing C(4,3) × (0.5)³ = 4/8 = 3/8, forgetting to multiply by (0.5)¹ for the one sunny day. Since the sunny day also has probability 0.5, it must be included in the product — giving (0.5)⁴ total, not (0.5)³.
Why not Option A being confused with 1/4 by intuition? One might guess 1/4 simply because 1 out of 4 days is sunny, but the actual derivation confirms the answer rigorously through the binomial formula.

The correct answer is Option A — Rs. 0.
To find the total profit, we calculate the number of shares purchased, the selling value, and the net gain or loss for each stock separately, then combine them.
Stock A: The investor buys at Rs. 50 per share and invests Rs. 100, acquiring 100 ÷ 50 = 2 shares. The next day, Stock A is worth Rs. 55 per share, so the selling value is 2 × 55 = Rs. 110. Profit from Stock A = Rs. 110 − Rs. 100 = +Rs. 10.
Stock B: The investor buys at Rs. 80 per share and invests Rs. 80, acquiring 80 ÷ 80 = 1 share. The next day, Stock B falls to Rs. 70 per share, so the selling value is 1 × 70 = Rs. 70. Loss from Stock B = Rs. 70 − Rs. 80 = −Rs. 10.
Total profit = (+Rs. 10) + (−Rs. 10) = Rs. 0.
The gain on Stock A is exactly cancelled out by the loss on Stock B, leaving the investor with zero net profit. The total amount invested was Rs. 180 (100 + 80) and the total selling value is also Rs. 180 (110 + 70), confirming the answer.

The correct answer is Option C — When peacocks are not dancing, it is not raining.
The given statement, "When it is raining, peacocks dance", is a conditional statement of the form R → D (Raining implies Dancing). In formal logic, only one transformation of a conditional is guaranteed to preserve its truth: the contrapositive.
The contrapositive of R → D is ¬D → ¬R — "If peacocks are not dancing, then it is not raining." This is logically equivalent to the original statement and is therefore necessarily true. This is Option C.
Option A — "Peacocks dance only when it is raining" (D → R): This is the converse of the original. The converse is not logically equivalent — peacocks could dance for reasons other than rain. Not necessarily true.
Option B — "When peacocks dance, it is raining" (D → R): Also the converse, same as Option A. Not necessarily true.
Option C — "When peacocks are not dancing, it is not raining" (¬D → ¬R): This is the contrapositive of the original, which is always logically equivalent to the original conditional. Necessarily true. Correct.
Option D — "When it is not raining, peacocks do not dance" (¬R → ¬D): This is the inverse of the original. Like the converse, the inverse is not logically equivalent — peacocks might still dance even when it is not raining. Not necessarily true.
The simple rule: of the four related conditionals (original, converse, inverse, contrapositive), only the contrapositive always shares the same truth value as the original.

The correct answer is Option A — P = Thirst; Q = Hunger.
This is a word analogy question asking us to identify the consistent relationship between Water and P, and between Food and Q. The analogy reads: Water is to P as Food is to Q.
The correct relationship is "substance : the biological craving or need for that substance." Thirst is the body''s need for water, and Hunger is the body''s need for food. Both pairs follow the exact same logical pattern, making the analogy complete and parallel.
Option A — P = Thirst, Q = Hunger: Water : Thirst and Food : Hunger — both are "substance : craving for that substance." The relationship is identical across both pairs. Correct.
Option B — P = Drink, Q = Hunger: "Drink" is an action performed with water, while "Hunger" is a craving. The relationship types are inconsistent across the two pairs — one is an action, the other a physiological state. Incorrect.
Option C — P = Thirst, Q = Satiated: "Thirst" is a craving that exists before consuming water, while "Satiated" is the state of fullness that exists after consuming food. The directions of the relationship are opposite, breaking the parallel. Incorrect.
Option D — P = Wet, Q = Critic: "Wet" is a property caused by water contact, and "Critic" has no meaningful relationship with food in this context. Both words are unrelated to each other in any consistent way. Incorrect.
The key to analogy questions is ensuring the same type of relationship holds for both pairs — here, "need/craving for" is the consistent link throughout.

The correct answer is Option C — The salesperson can make the trip on both networks (i) and (ii).
The salesperson''s trip is a Hamiltonian Cycle problem — a closed path that visits every city (vertex) exactly once and returns to the starting city. We check if such a cycle exists in each network.
Figure (i) — 4×4 Grid Network (16 cities): For a grid graph G(m×n), a Hamiltonian cycle exists if and only if m×n is even and both m ≥ 2, n ≥ 2. Here 4×4 = 16 (even) and both dimensions exceed 1. The Hamiltonian cycle exists.
Figure (ii) — Irregular 5-city Network: All 5 vertices have degree ≥ 2, and by tracing the edges a valid Hamiltonian cycle can be found — visiting all 5 cities exactly once before returning to the start.
Since a Hamiltonian cycle exists in both networks, the salesperson can complete the required trip on both highway systems.
Key distinction: this is a Hamiltonian cycle (every city visited once), not an Eulerian circuit (every road traversed once). For grid graphs, the even-vertex rule directly determines Hamiltonicity.

A dice is rolled twice, so the total number of equally likely outcomes is 6 × 6 = 36. The sum of two dice can range from 2 to 12. Among these, the prime numbers are 2, 3, 5, 7, and 11. Note that 4, 6, 8, 9, 10 and 12 are not prime, and sums less than 2 or greater than 12 are not possible.
Now counting all ordered pairs (first roll, second roll) that give a prime sum:
Sum = 2 (prime): (1,1) → 1 way
Sum = 3 (prime): (1,2), (2,1) → 2 ways
Sum = 5 (prime): (1,4), (4,1), (2,3), (3,2) → 4 ways
Sum = 7 (prime): (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) → 6 ways
Sum = 11 (prime): (5,6), (6,5) → 2 ways
Total favourable outcomes = 1 + 2 + 4 + 6 + 2 = 15
Probability = ¹⁵/₃₆
Correct answer: C — 15/36 ✓