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100+ Free PAU Applied Mathematics for Social Sciences II (Cantabria) Practice Questions

Cantabria PAU Applied Mathematics for Social Sciences II 2026 (Pruebas de Acceso a la Universidad, Universidad de Cantabria) practice questions are available now; exam metadata is being verified.

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Key Facts: PAU Applied Mathematics for Social Sciences II (Cantabria) Exam

90 min

Exam time duration for the PAU modality-subject paper

University of Cantabria (UNICAN) PAU Organising Commission

5 Blocks

Core syllabus areas: Matrices, Linear Programming, Analysis, Integrals, Probability

2º Bachillerato Curriculum (LOMLOE), Ministerio de Educación

0–10 scale

Scoring system (minimum 4.0 required in Access Phase)

Cantabria PAU Regulations

EUR 71.09

Base registration fee for PAU Access Phase in Cantabria

University of Cantabria (UNICAN)

100

High-quality practice questions in this OpenExamPrep subject bank

OpenExamPrep

Master the 2026 Cantabria PAU Applied Mathematics for Social Sciences II exam with 100 practice questions covering matrix algebra, linear programming, calculus, integration, and probability. This English-language MCQ bank is a study adaptation of the theory and calculation methods behind the official written exam, not a format simulation.

Sample PAU Applied Mathematics for Social Sciences II (Cantabria) Practice Questions

Try these sample questions to test your PAU Applied Mathematics for Social Sciences II (Cantabria) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Given matrices A = [[3, -1], [2, 4]] and B = [[1, 5], [-2, 0]], calculate 2A - 3B.
A.[[3, -17], [10, 8]]
B.[[5, 13], [-2, 8]]
C.[[3, -7], [2, 8]]
D.[[4, 4], [0, 4]]
Explanation: First compute 2A = [[6, -2], [4, 8]] and 3B = [[3, 15], [-6, 0]]. Subtracting componentwise yields 2A - 3B = [[6-3, -2-15], [4-(-6), 8-0]] = [[3, -17], [10, 8]].
2Compute the matrix product AB for A = [[2, 1], [0, 3]] and B = [[4, -2], [1, 5]].
A.[[8, -2], [0, 15]]
B.[[9, 1], [3, 15]]
C.[[8, 1], [3, 15]]
D.[[9, 1], [1, 15]]
Explanation: Multiply rows of A by columns of B: row 1 gives [2(4)+1(1), 2(-2)+1(5)] = [9, 1]; row 2 gives [0(4)+3(1), 0(-2)+3(5)] = [3, 15]. Thus AB = [[9, 1], [3, 15]].
3Calculate the determinant of the matrix A = [[6, -3], [4, 2]].
A.0
B.-24
C.24
D.18
Explanation: For a 2x2 matrix [[a, b], [c, d]], det(A) = ad - bc. Here det(A) = (6)(2) - (-3)(4) = 12 - (-12) = 24.
4Given A = [[1, 2], [3, 4]], compute the matrix sum A + A^T.
A.[[2, 4], [6, 8]]
B.[[1, 5], [5, 4]]
C.[[2, 6], [4, 8]]
D.[[2, 5], [5, 8]]
Explanation: The transpose A^T is [[1, 3], [2, 4]]. Adding A and A^T entrywise yields [[1+1, 2+3], [3+2, 4+4]] = [[2, 5], [5, 8]].
5A company has two stores. Store 1 sells 10 units of product X and 20 of product Y. Store 2 sells 15 of X and 25 of Y. Prices are €5 for X and €8 for Y. Which matrix multiplication represents total revenue per store?
A.[[10, 20], [15, 25]] * [[5], [8]] = [[210], [275]]
B.[[10, 20], [15, 25]] * [[8], [5]] = [[180], [245]]
C.[[10, 15], [20, 25]] * [[5], [8]] = [[200], [275]]
D.[[5, 8]] * [[10, 20], [15, 25]] = [[170, 300]]
Explanation: Represent sales by matrix M = [[10, 20], [15, 25]] (rows = stores, columns = products) and price by vector P = [[5], [8]]. Revenue vector MP = [[10(5)+20(8)], [15(5)+25(8)]] = [[210], [275]] euros.
6Find the determinant of the lower triangular matrix A = [[3, 0, 0], [1, 4, 0], [5, -2, -2]].
A.24
B.-24
C.0
D.-6
Explanation: The determinant of a triangular matrix equals the product of its main diagonal entries: det(A) = 3 * 4 * (-2) = -24.
7Find the inverse matrix A^(-1) for A = [[4, 3], [1, 1]].
A.[[-1, 3], [1, -4]]
B.[[1, 1], [3, 4]]
C.[[1, -3], [-1, 4]]
D.[[4, -3], [-1, 1]]
Explanation: First calculate det(A) = 4(1) - 3(1) = 1. Using the 2x2 inverse formula A^(-1) = (1/det(A)) * [[d, -b], [-c, a]], we get [[1, -3], [-1, 4]].
8Calculate the determinant of the 3x3 matrix A = [[1, 2, 0], [3, -1, 4], [2, 1, 1]] using Sarrus's rule.
A.15
B.-5
C.10
D.5
Explanation: Sarrus's rule: det(A) = (1*-1*1 + 2*4*2 + 0*3*1) - (0*-1*2 + 1*4*1 + 2*3*1) = (-1 + 16 + 0) - (0 + 4 + 6) = 15 - 10 = 5.
9Solve the matrix equation A X = B for vector X, where A = [[2, 1], [3, 2]] and B = [[5], [7]].
A.[[3], [-1]]
B.[[-3], [1]]
C.[[17], [29]]
D.[[1], [3]]
Explanation: det(A) = 2(2) - 1(3) = 1. The inverse A^(-1) = [[2, -1], [-3, 2]]. Multiply X = A^(-1) B = [[2(5) - 1(7)], [-3(5) + 2(7)]] = [[10-7], [-15+14]] = [[3], [-1]].
10Determine the rank of the matrix A = [[1, 1, 1], [2, 2, k], [3, 3, 3]] as a function of the parameter k.
A.Rank is always 3 for any real k
B.Rank is 1 if k = 2, and rank is 2 if k ≠ 2
C.Rank is 2 if k = 2, and rank is 3 if k ≠ 2
D.Rank is 0 if k = 2, and rank is 1 if k ≠ 2
Explanation: Row 3 is 3 times Row 1, so rows 1 and 3 are linearly dependent. If k = 2, Row 2 is 2 times Row 1, making all rows proportional (rank 1). If k ≠ 2, Row 2 is independent of Row 1, so rank is 2.

About the PAU Applied Mathematics for Social Sciences II (Cantabria) Practice Questions

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