Free GATE DA mock test — 65 questions, GATE marking 🇮🇳

8 original practice questions written by Myndaq in the GATE DA style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.

1. Question 1General AptitudeOne correct answer

   In a group of 60 students, 35 use Python, 28 use R and 10 use neither language. How many students use exactly one of the two languages?

   1. A. 13
   2. B. 22
   3. C. 37
   4. D. 50
   Show answer and explanation

   Answer: C

   Students using at least one language = 60 - 10 = 50. By inclusion-exclusion, the number using both = 35 + 28 - 50 = 13.

   Exactly one language = (at least one) - (both) = 50 - 13 = 37. Check: Python only = 35 - 13 = 22 and R only = 28 - 13 = 15, and 22 + 15 = 37.

   13 is the count using both, 22 is Python only, and 50 is the count using at least one, so none of those answers the question.

2. Question 2Probability and StatisticsType a numeric answer

   An email filter has learned that 20% of incoming emails are spam. The word 'winner' appears in 60% of spam emails and in 5% of non-spam emails. Given that an email contains the word 'winner', the probability that it is spam is ______ (rounded off to two decimal places).

   Show answer and explanation

   Answer: 0.75

   Let S be the event that the email is spam and W the event that it contains 'winner'. Then P(S) = 0.2, P(W | S) = 0.6 and P(W | not S) = 0.05.

   By the law of total probability, P(W) = 0.2 × 0.6 + 0.8 × 0.05 = 0.12 + 0.04 = 0.16.

   By Bayes' theorem, P(S | W) = 0.12 / 0.16 = 0.75. A common slip is to answer 0.6, which is P(W | S), the reverse conditional.

3. Question 3Linear AlgebraOne or more correct answers

   Let A be a 3 × 3 real matrix whose rank is 2. Which of the following statements is/are NECESSARILY true?

   1. A. det(A) = 0
   2. B. The null space of A has dimension 1
   3. C. 0 is an eigenvalue of A
   4. D. A has exactly two non-zero eigenvalues, counted with algebraic multiplicity
   Show answer and explanation

   Answer: A, B, C

   Rank 2 is less than 3, so A is singular and det(A) = 0 (A is true). By the rank-nullity theorem, dim(null space) = 3 - 2 = 1 (B is true). A non-zero vector x with Ax = 0 = 0·x exists, so 0 is an eigenvalue (C is true).

   D is false. The matrix N with rows (0, 1, 0), (0, 0, 1), (0, 0, 0) has rank 2, but it is upper triangular with zero diagonal, so all three of its eigenvalues are 0. The rank counts non-zero eigenvalues only for special classes such as symmetric (diagonalizable) matrices, not in general.

4. Question 4Calculus and OptimizationType a numeric answer

   Let f(x, y) = x² + 2y² - 2x + 8y + 5. Starting from (x0, y0) = (0, 0), one step of gradient descent with step size η = 0.1 produces the point (x1, y1). The value of f(x1, y1) is ______ (rounded off to two decimal places).

   Show answer and explanation

   Answer: -0.48

   The gradient is ∇f = (2x - 2, 4y + 8). At (0, 0) it equals (-2, 8).

   Update: (x1, y1) = (0, 0) - 0.1 × (-2, 8) = (0.2, -0.8).

   f(0.2, -0.8) = 0.04 + 2(0.64) - 0.4 - 6.4 + 5 = 0.04 + 1.28 - 0.4 - 6.4 + 5 = -0.48.

   Check by completing squares: f = (x - 1)² + 2(y + 2)² - 4, so f(0.2, -0.8) = 0.64 + 2(1.44) - 4 = -0.48. The step lowered f from f(0, 0) = 5 towards the minimum value -4 at (1, -2); answering -4 skips the single-step condition.

5. Question 5Programming, Data Structures and AlgorithmsOne correct answer

   Consider the following Python program.

   def g(a): if len(a) <= 1: return a m = len(a) // 2 return g(a[m:]) + g(a[:m])

   print(g([1, 2, 3, 4, 5]))

   What is printed when the program runs?

   1. A. [3, 4, 5, 1, 2]
   2. B. [1, 2, 3, 4, 5]
   3. C. [5, 4, 3, 2, 1]
   4. D. [2, 1, 5, 4, 3]
   Show answer and explanation

   Answer: C

   For [1, 2, 3, 4, 5], m = 2, so the result is g([3, 4, 5]) + g([1, 2]).

   g([3, 4, 5]): m = 1, giving g([4, 5]) + g([3]). g([4, 5]) = g([5]) + g([4]) = [5, 4], so g([3, 4, 5]) = [5, 4, 3]. Similarly g([1, 2]) = [2] + [1] = [2, 1]. The output is [5, 4, 3, 2, 1]: the function reverses the list by recursively swapping its halves.

   Option A swaps the halves only once and ignores the recursion. Option D reverses each half but keeps the halves in their original order. Option B assumes the swaps cancel out. (The running time satisfies T(n) = 2T(n/2) + O(n) because of slicing and concatenation, so it is O(n log n).)

6. Question 6Database Management and WarehousingOne correct answer

   The relation Emp(eid, dept, salary) contains exactly six tuples: (1, 'A', 50), (2, 'A', 70), (3, 'B', 40), (4, 'B', 40), (5, 'C', 90) and (6, 'C', NULL).

   The following SQL query is run on it:

   SELECT dept, AVG(salary) FROM Emp GROUP BY dept HAVING COUNT(*) > 1 AND AVG(salary) > 45;

   Which option lists exactly the tuples returned by the query (in any order)?

   1. A. ('A', 60) only
   2. B. ('A', 60) and ('C', 90)
   3. C. ('A', 60), ('B', 40) and ('C', 90)
   4. D. ('A', 60) and ('C', 45)
   Show answer and explanation

   Answer: B

   Group A: salaries 50 and 70, COUNT(*) = 2, AVG = 60, so both conditions hold. Group B: COUNT(*) = 2 but AVG = 40, which fails AVG(salary) > 45. Group C: COUNT(*) counts rows, including the one with a NULL salary, so it is 2; AVG ignores NULL, so AVG = 90. Group C qualifies.

   The result is ('A', 60) and ('C', 90). Option A would follow only if NULL were treated as 0 (average 45, which fails > 45) or if COUNT(*) skipped the NULL row; SQL does neither. Option C ignores the HAVING condition on the average, and option D is impossible because a group with average 45 cannot pass AVG(salary) > 45.

7. Question 7Machine LearningOne or more correct answers

   Which of the following statements is/are CORRECT?

   1. A. In k-nearest-neighbour classification, increasing k generally reduces the variance of the classifier
   2. B. L1 (lasso) regularization in linear regression can drive some coefficients to exactly zero
   3. C. Adding more training examples always reduces the training error of an ordinary least-squares linear regression model
   4. D. A decision tree grown until every leaf is pure typically has high bias and low variance
   Show answer and explanation

   Answer: A, B

   A is correct: a larger k averages over more neighbours, giving a smoother decision boundary that changes less from one training sample to another (lower variance, at the cost of higher bias). B is correct: the L1 penalty has corners on the coordinate axes, so the optimum often lies where some coefficients are exactly 0, which is why lasso performs feature selection.

   C is wrong: a line fits two points with zero training error, but adding a third point that is not collinear makes the training error positive; training error usually rises, not falls, as data grows. D is wrong: a fully grown tree fits the training data almost perfectly, so it has low bias and high variance.

8. Question 8Artificial IntelligenceType a numeric answer

   A two-player game tree has a MAX node at the root with three MIN children B, C and D, explored in that order. Each MIN node has three leaf children, also explored left to right, with these utility values: under B the leaves are 5, 9, 7; under C they are 4, 10, 3; under D they are 8, 2, 6.

   If minimax search with alpha-beta pruning is applied with the given left-to-right ordering, the number of leaf nodes that are NEVER evaluated is ______.

   Show answer and explanation

   Answer: 3

   B is searched fully: min(5, 9, 7) = 5, so the root's alpha becomes 5.

   At C the first leaf is 4, so C's value is at most 4, which is already no better than alpha = 5 for MAX. The remaining leaves 10 and 3 are pruned (2 leaves).

   At D the first leaf 8 gives an upper bound of 8 > 5, so search continues. The second leaf 2 makes D's value at most 2 ≤ 5, so the last leaf 6 is pruned (1 leaf).

   Total pruned leaves = 2 + 1 = 3, and the root's minimax value is 5 (the full values are B = 5, C = 3, D = 2).