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2026 Statistics
Key Facts: Castilla-La Mancha PAU Applied Mathematics for Social Sciences II (Matemáticas Aplicadas a las Ciencias Sociales II - UCLM 2026) Exam
90 min
PAU Math CCSS II exam duration
Tribunal PAU Universidad de Castilla-La Mancha
0–10
Grading scale (minimum 4.0 to weight)
Normativa Académica UCLM
EUR 52.99
Standard PAU UCLM Access Phase registration fee
Universidad de Castilla-La Mancha 2026
100
Practice calculation questions in this assessment bank
OpenExamPrep
Prepare for the UCLM 2026 PAU Applied Mathematics for Social Sciences II exam with 100 worked practice questions on matrices, linear programming, calculus, and statistics. This English-language multiple-choice bank is a study adaptation of the official written PAU paper, not a replica of it.
Sample Castilla-La Mancha PAU Applied Mathematics for Social Sciences II (Matemáticas Aplicadas a las Ciencias Sociales II - UCLM 2026) Practice Questions
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1Given the objective function Z = 3x + 4y to be optimized over the feasible region with vertices A(0,0), B(5,0), C(3,4), and D(0,6), what is the maximum value of Z?
A.25
B.24
C.15
D.21
Explanation: Evaluating the objective function Z = 3x + 4y at each vertex: Z(0,0) = 0; Z(5,0) = 3(5)+4(0) = 15; Z(3,4) = 3(3)+4(4) = 9 + 16 = 25; Z(0,6) = 3(0)+4(6) = 24. The maximum value is 25 at the point (3,4).
2Calculate the area of the feasible region bounded by the constraints x ≥ 0, y ≥ 0, and x + y ≤ 6.
A.36 sq units
B.18 sq units
C.12 sq units
D.9 sq units
Explanation: The feasible region is a right triangle bounded by the coordinate axes and vertices (0,0), (6,0), and (0,6). The base b = 6 and height h = 6, so the area is A = (b · h)/2 = (6 · 6)/2 = 18 square units.
3To minimize the function Z = 2x + 5y over the region defined by the vertices (1,5), (2,2), (6,1), and (4,7), at which vertex is the minimum achieved?
A.(6,1)
B.(1,5)
C.(2,2)
D.(4,7)
Explanation: Evaluating Z = 2x + 5y at the vertices: Z(1,5) = 2(1)+5(5) = 27; Z(2,2) = 2(2)+5(2) = 14; Z(6,1) = 2(6)+5(1) = 17; Z(4,7) = 2(4)+5(7) = 43. The minimum value of Z is 14 and is reached at vertex (2,2).
4Find the coordinates of the intersection vertex of the boundary lines x + y = 10 and 2x - y = 5.
A.(3, 7)
B.(4, 6)
C.(6, 4)
D.(5, 5)
Explanation: Adding both equations to eliminate y: (x + y) + (2x - y) = 10 + 5 => 3x = 15 => x = 5. Substituting x = 5 into the first equation: 5 + y = 10 => y = 5. The intersection vertex is (5, 5).
5Given the profit function F(x,y) = 5x + 2y defined over a polygon with vertices A(0,4), B(3,5), C(6,1), and D(2,0), calculate the maximum profit.
A.32
B.25
C.10
D.8
Explanation: Evaluating F(x,y) = 5x + 2y at each vertex: F(0,4) = 8; F(3,5) = 5(3)+2(5) = 25; F(6,1) = 5(6)+2(1) = 32; F(2,0) = 5(2)+2(0) = 10. The maximum profit is 32, obtained at point C(6,1).
6Determine whether the point (3,2) belongs to the feasible region defined by the inequalities 2x + y ≤ 8 and x + 3y ≥ 7.
A.No, because it violates the first inequality.
B.Yes, because it satisfies both inequalities simultaneously.
C.No, because it violates the second inequality.
D.No, because it violates both constraints.
Explanation: For (3,2): 1st inequality: 2(3) + 2 = 6 + 2 = 8 ≤ 8 (Satisfied). 2nd inequality: 3 + 3(2) = 3 + 6 = 9 ≥ 7 (Satisfied). Since both constraints are satisfied, the point (3,2) belongs to the feasible region.
7Maximize Z = x + 3y subject to x ≥ 0, y ≥ 0, x ≤ 4, y ≤ 5.
A.19
B.4
C.15
D.12
Explanation: The feasible region is the rectangle with vertices (0,0), (4,0), (4,5), and (0,5). Evaluating Z = x + 3y at the vertices yields Z(0,0)=0, Z(4,0)=4, Z(4,5)=4+3(5)=19, and Z(0,5)=15. The maximum value of 19 is achieved at (4,5).
8A small company manufactures product X and product Y. Product X requires 2 hours of labor and 1 kg of raw material per unit. Product Y requires 1 hour of labor and 3 kg of raw material per unit. The company has up to 100 hours of labor and 150 kg of raw material available. Which set of inequalities defines the feasible region for production (x ≥ 0, y ≥ 0)?
A.2x + y ≤ 100; x + 3y ≤ 150
B.x + 2y ≤ 100; 3x + y ≤ 150
C.2x + 3y ≤ 100; x + y ≤ 150
D.2x + y ≥ 100; x + 3y ≥ 150
Explanation: Labor constraint: 2x + 1y ≤ 100. Raw material constraint: 1x + 3y ≤ 150. Non-negativity constraints: x ≥ 0, y ≥ 0. The correct system of inequalities is 2x + y ≤ 100 and x + 3y ≤ 150.
9Find the maximum of Z = 10x + 15y in an unbounded feasible region defined by x ≥ 0, y ≥ 0, and x + y ≥ 4.
A.60
B.40
C.It does not exist (the function is unbounded above in the region).
D.0
Explanation: The region extends indefinitely into the first quadrant as x and y increase (e.g., points like (100, 100) are feasible). Since the objective function has positive coefficients (10 and 15), Z can increase without bound. Therefore, no finite maximum exists.
10Consider the objective function Z = ax + y. For what value of parameter a does the function achieve its maximum simultaneously at the vertices (2,4) and (5,1) of a feasible region?
A.a = 1
B.a = -1
C.a = 2
D.a = 0.5
Explanation: For the maximum to be achieved at both vertices, the value of Z must be equal at both points: Z(2,4) = Z(5,1) => 2a + 4 = 5a + 1 => 3a = 3 => a = 1.
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