AP EAMCET 2025 previous year paper practice · PYQ-pattern set

AP EAMCET 2025 PYQ-pattern questions

Showing 10 of the 14 PYQ-pattern questions in this set — freshly worded in that year's pattern, not the original questions. Answers and solutions open under each question.

1. Q1 · Trigonometric Ratios up to Transformations

   If sin θ + cos θ = √2, then the value of sin θ · cos θ is

   • A. 1/4
   • B. 1/2
   • C. 1
   • D. √2
   Answer and solution

   Answer: B

   Squaring both sides: sin²θ + cos²θ + 2 sin θ cos θ = 2. Since sin²θ + cos²θ = 1, we get 1 + 2 sin θ cos θ = 2, so sin θ cos θ = 1/2.

2. Q2 · Product of Vectors

   If vectors **a** = 2**i** + 3**j** - **k** and **b** = **i** - 2**j** + 4**k**, then the projection of **a** on **b** is

   • A. 4/√21
   • B. -4/√21
   • C. 8/√21
   • D. -8/√21
   Answer and solution

   Answer: D

   Dot product a·b = 2-6-4 = -8. Magnitude of b = √(1+4+16)=√21. Projection of a on b = (a·b)/|b| = -8/√21.

3. Q3 · Transformation of Axes

   When the origin is shifted to the point (2, -3), the transformed equation of the curve x² + y² - 4x + 6y - 12 = 0 is

   • A. x² + y² = 25
   • B. x² + y² = 16
   • C. x² + y² = 20
   • D. x² + y² = 9
   Answer and solution

   Answer: A

   Substitute x = X + 2, y = Y - 3 into the original equation: (X+2)² + (Y-3)² - 4(X+2) + 6(Y-3) - 12 = 0. Expanding and simplifying gives X² + Y² = 25.

4. Q4 · Parabola

   If the normal to the parabola y² = 16x at the point (4, 8) meets the parabola again at point Q, then the coordinates of Q are

   • A. (36, -24)
   • B. (16, -16)
   • C. (25, -20)
   • D. (9, -12)
   Answer and solution

   Answer: A

   For y² = 4ax with a = 4, point (4, 8) corresponds to parameter t₁ = 1. The normal meets the parabola again at parameter t₂ = -t₁ - 2/t₁ = -1 - 2 = -3. Thus Q = (at₂², 2at₂) = (36, -24).

5. Q5 · Limits and Continuity

   Evaluate: lim(x→3) [(x² - 9)/(x - 3)]

   • A. 3
   • B. 6
   • C. 9
   • D. 0
   Answer and solution

   Answer: B

   Factoring the numerator: (x² - 9) = (x - 3)(x + 3). The limit becomes lim(x→3) (x + 3) = 6.

6. Q6 · Differential Equations

   The general solution of the differential equation dy/dx + y tan(x) = sec(x) is:

   • A. y sec(x) = tan(x) + C
   • B. y cos(x) = sin(x) + C
   • C. y sec(x) = x + C
   • D. y sin(x) = cos(x) + C
   Answer and solution

   Answer: A

   This is a linear first-order DE. Integrating factor: e^∫tan(x)dx = sec(x). Multiplying and integrating: y sec(x) = tan(x) + C.

7. Q7 · Motion in a Straight Line

   A particle starts from rest and accelerates uniformly at 3 m/s² for 6 seconds. What is the distance covered by the particle in this time?

   • A. 36 m
   • B. 54 m
   • C. 72 m
   • D. 108 m
   Answer and solution

   Answer: B

   Using s = ut + ½at², with u = 0, a = 3 m/s², t = 6 s: s = 0 + ½(3)(6)² = 54 m.

8. Q8 · Oscillations

   A simple pendulum has a time period of 2 seconds on Earth. What will be its time period on a planet where the acceleration due to gravity is 4 times that on Earth?

   • A. 0.5 s
   • B. 1 s
   • C. 2 s
   • D. 4 s
   Answer and solution

   Answer: B

   Time period T ∝ 1/√g. If g becomes 4g, then T' = T/√4 = 2/2 = 1 s.

9. Q9 · Mechanical Properties of Fluids

   A cylindrical tank of base area 0.5 m² is filled with water to a height of 4 m. A small circular hole of area 2 cm² is made at the bottom of the tank. The initial speed of water flowing out of the hole is (g = 10 m/s²)

   • A. 4√2 m/s
   • B. 8√2 m/s
   • C. 4√5 m/s
   • D. 2√10 m/s
   Answer and solution

   Answer: C

   Using Torricelli's theorem, the velocity of efflux v = √(2gh) = √(2×10×4) = √80 = 4√5 m/s.

10. Q10 · Ray Optics and Optical Instruments

    A convex lens of focal length 20 cm forms a real image at a distance of 60 cm from the lens. The object distance is

    • A. 15 cm
    • B. 20 cm
    • C. 30 cm
    • D. 40 cm
    Answer and solution

    Answer: C

    Using lens formula 1/f = 1/v - 1/u, we have 1/20 = 1/60 - 1/u. Solving: 1/u = 1/60 - 1/20 = (1-3)/60 = -2/60, so u = -30 cm. The magnitude of object distance is 30 cm.