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100+ Free TS ECET Practice Questions

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Key Facts: TS ECET Exam

200 MCQs

Total questions in the official TS ECET lateral entry exam

TSCHE Exam Pattern

180 Minutes

Total time allowed to answer all questions

TS ECET Notification

₹900 / ₹500

Registration fee for general and reserved categories

TS ECET Brochure

50 Marks

Minimum qualifying score required for general candidates (25%)

TSCHE Guidelines

The TS ECET lateral entry exam consists of 200 MCQs to be completed in 180 minutes, with no negative marking. The application fee is ₹900 (₹500 for SC/ST). Candidates must score at least 25% (50 marks) to qualify, with no minimum mark required for SC/ST candidates.

Sample TS ECET Practice Questions

Try these sample questions to test your TS ECET exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Find the determinant of the 3x3 matrix: [1 2 3] [0 1 4] [5 6 0]
A.-9
B.1
C.9
D.-1
Explanation: To find the determinant, expand along the first row: det(A) = 1 * (1*0 - 4*6) - 2 * (0*0 - 4*5) + 3 * (0*6 - 1*5) = 1 * (-24) - 2 * (-20) + 3 * (-5) = -24 + 40 - 15 = 1.
2What are the eigenvalues of the upper triangular matrix A = [2 3] [0 5]?
A.2 and 5
B.3 and 5
C.2 and 3
D.0 and 5
Explanation: For any triangular or diagonal matrix, the eigenvalues are simply the diagonal entries. Here, the diagonal entries are 2 and 5, so the eigenvalues are 2 and 5.
3Find the rank of the 3x3 matrix A = [1 2 3] [2 4 6] [3 6 9]
A.3
B.2
C.1
D.0
Explanation: Observe that the second row is twice the first row (R2 = 2*R1) and the third row is three times the first row (R3 = 3*R1). Performing row operations yields a matrix with two zero rows. Since there is only one non-zero row remaining, the rank is 1.
4If tan A = 1/2 and tan B = 1/3, find the value of tan(A + B).
A.1/6
B.5/6
C.1
D.5/5
Explanation: Using the compound angle formula: tan(A + B) = (tan A + tan B) / (1 - tan A * tan B) = (1/2 + 1/3) / (1 - (1/2)*(1/3)) = (5/6) / (1 - 1/6) = (5/6) / (5/6) = 1.
5Which of the following is a fundamental hyperbolic identity equivalent to 1?
A.cosh^2(x) - sinh^2(x)
B.cosh^2(x) + sinh^2(x)
C.sinh^2(x) - cosh^2(x)
D.1 - cosh^2(x)
Explanation: The fundamental identity for hyperbolic functions is cosh^2(x) - sinh^2(x) = 1, analogous to the circular trigonometric identity cos^2(x) + sin^2(x) = 1.
6Find the amplitude (principal argument) of the complex number z = -1 - i * sqrt(3).
A.pi/3
B.-2*pi/3
C.-pi/3
D.2*pi/3
Explanation: The complex number z lies in the third quadrant since both real and imaginary parts are negative. The reference angle is theta = arctan(|-sqrt(3)|/|-1|) = arctan(sqrt(3)) = pi/3. The principal argument in the third quadrant is given by -pi + theta = -pi + pi/3 = -2*pi/3.
7Find the perpendicular distance from the point (2, 3) to the straight line 3x + 4y - 8 = 0.
A.1
B.3
C.2
D.4
Explanation: The formula for the perpendicular distance from (x1, y1) to ax + by + c = 0 is d = |a*x1 + b*y1 + c| / sqrt(a^2 + b^2). Substituting the values: d = |3(2) + 4(3) - 8| / sqrt(3^2 + 4^2) = |6 + 12 - 8| / 5 = 10 / 5 = 2.
8What is the radius of the circle given by the equation x^2 + y^2 - 4x - 6y - 12 = 0?
A.4
B.5
C.12
D.sqrt(13)
Explanation: Comparing with the general equation x^2 + y^2 + 2gx + 2fy + c = 0: 2g = -4 => g = -2; 2f = -6 => f = -3; c = -12. The radius r = sqrt(g^2 + f^2 - c) = sqrt((-2)^2 + (-3)^2 - (-12)) = sqrt(4 + 9 + 12) = sqrt(25) = 5.
9Determine the eccentricity of the ellipse 9x^2 + 25y^2 = 225.
A.3/5
B.4/5
C.16/25
D.5/4
Explanation: Dividing the equation by 225: x^2 / 25 + y^2 / 9 = 1. This is of the form x^2/a^2 + y^2/b^2 = 1, where a^2 = 25 and b^2 = 9. The eccentricity e = sqrt(1 - b^2/a^2) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5.
10Evaluate the limit: lim (x -> 0) [sin(5x) / x].
A.5
B.1
C.0
D.1/5
Explanation: Using the standard limit lim (theta -> 0) [sin(theta)/theta] = 1, we multiply and divide the expression by 5: lim (x -> 0) [5 * sin(5x) / (5x)] = 5 * 1 = 5. Alternatively, using L'Hopital's Rule: lim (x -> 0) [5 cos(5x) / 1] = 5.

About the TS ECET Exam

The TS ECET is a state-level entrance exam conducted for lateral entry into second-year B.E. / B.Tech. and B.Pharmacy courses in colleges across Telangana. Designed for diploma holders and B.Sc. Mathematics graduates, this practice test contains exactly 100 high-quality MCQs distributed proportionally: 25 in Mathematics, 13 in Physics, 12 in Chemistry, and 50 in Engineering Fundamentals. Detailed explanations are provided for every question.

Questions

200 scored questions

Time Limit

180 minutes

Passing Score

25% (50 marks out of 200) for general candidates, no minimum qualifying mark for SC/ST

Exam Fee

₹900 for General/OBC, ₹500 for SC/ST (Telangana State Council of Higher Education (TSCHE))

TS ECET Exam Content Outline

25%

Mathematics

Matrices, Trigonometry, Analytical Geometry, Calculus, Differential Equations, Laplace Transforms, and Fourier Series.

13%

Physics

Analytical mechanics, vector forces, friction, thermodynamics, SHM, sound properties, and modern physics principles.

12%

Chemistry

Atomic structures, water treatments, electrochemistry, corrosion mechanics, polymers, fuels, and environmental pollutants.

50%

Engineering Paper

Core concepts and syllabus fundamentals corresponding to Civil, Mechanical, Electrical, and Computer Science Engineering branches.

How to Pass the TS ECET Exam

What You Need to Know

  • Passing score: 25% (50 marks out of 200) for general candidates, no minimum qualifying mark for SC/ST
  • Exam length: 200 questions
  • Time limit: 180 minutes
  • Exam fee: ₹900 for General/OBC, ₹500 for SC/ST

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

TS ECET Study Tips from Top Performers

1Dedicate ample practice time to Mathematics as it accounts for 50 marks and contains highly scoreable calculation-based topics.
2Work systematically through standard numericals on matrices, Laplace transforms, and integration to build strong speed.
3Thoroughly memorize fundamental physics equations regarding kinematics, simple harmonic motion, and ideal gases.
4Review organic polymers and water treatment techniques, which are frequently asked and relatively easy scoring areas in Chemistry.
5Practice previous year branch-specific questions for the Engineering paper, focusing on the core principles of your respective engineering discipline.

Frequently Asked Questions

What is the minimum qualifying score for TS ECET?

For general and OBC candidates, the qualifying score is 25% of the total marks, which is 50 marks out of 200. There is no minimum qualifying mark requirement for SC and ST candidates.

Is there a negative marking policy in TS ECET?

No, there is no negative marking in the TS ECET exam. Candidates receive 1 mark for each correct answer, and no marks are deducted for incorrect or unattended questions.

Who is eligible to apply for TS ECET lateral entry?

Candidates holding a Diploma in Engineering/Technology/Pharmacy from the State Board of Technical Education, or a B.Sc. degree with Mathematics as a subject with at least 45% marks (40% for reserved categories) are eligible.

What is the application fee structure?

The application registration fee is ₹900 for general and OBC category candidates, and ₹500 for candidates belonging to SC and ST categories.

Which university conducts the TS ECET?

The TS ECET is conducted by Osmania University (or another state university appointed by TSCHE) on behalf of the Telangana State Council of Higher Education (TSCHE).