PracticeVideosBlogFlashcardsEspañol
All Practice Exams

100+ Free CG PET Practice Questions

Pass your Chhattisgarh Pre Engineering Test exam on the first try — instant access, no signup required.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
Merit List Placement Pass Rate
100+ Questions
100% Free

Loading practice questions...

Same family resources

Explore More India State-Level Admissions and Polytechnic Entrance Exams

Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.

AP EdCET — Andhra Pradesh Education Common Entrance Test

Practice Questions

AP POLYCET — Andhra Pradesh Polytechnic Common Entrance Test

Practice Questions

Bihar Combined Entrance Competitive Examination (BCECE)

Practice Questions

Bihar Polytechnic (DCECE) Entrance Exam (India)

Practice Questions

CG PPT — Chhattisgarh Pre-Polytechnic Test (CG Vyapam, India)

Practice Questions

JEMAT — Joint Entrance Management Aptitude Test (MAKAUT, West Bengal, India)

Practice Questions
2026 Statistics

Key Facts: CG PET Exam

150

Total Questions

CG Vyapam

3 Hours

Exam Duration

CG Vyapam

None

Negative Marking

CG Vyapam

Free

For CG Domiciles

State Policy

45%

PCM 12th Req.

CG DTE

10%

Min. Merit Score

CG Vyapam

CG PET (Chhattisgarh Pre Engineering Test) is conducted offline by CG Vyapam. The exam comprises 150 multiple-choice questions across Physics, Chemistry, and Mathematics (50 questions each) with no negative marking. Qualifying candidates are ranked in the merit list to gain admission into undergraduate engineering programs in Chhattisgarh.

Sample CG PET Practice Questions

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

1Evaluate the limit: \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}.
A.3
B.2
C.4
D.0
Explanation: Using factoring: (x^3 - 8)/(x^2 - 4) = ((x-2)(x^2 + 2x + 4)) / ((x-2)(x+2)). For x != 2, this simplifies to (x^2 + 2x + 4)/(x+2). Taking the limit as x -> 2 gives (4 + 4 + 4)/(2+2) = 12/4 = 3.
2Find the derivative dy/dx of the parametric curve defined by x = a(t - \sin t) and y = a(1 - \cos t) at t = \pi/2.
A.0
B.1
C.-1
D.1/2
Explanation: We use the parametric derivative formula: dy/dx = (dy/dt) / (dx/dt). Here, dy/dt = a\sin t and dx/dt = a(1 - \cos t). Thus, dy/dx = \sin t / (1 - \cos t). At t = \pi/2, dy/dx = \sin(\pi/2) / (1 - \cos(\pi/2)) = 1 / (1 - 0) = 1.
3An open box is to be made from a square sheet of cardboard of side 12 cm by cutting equal squares of side x from the corners and folding up the sides. Find the value of x (in cm) that maximizes the volume of the box.
A.1
B.3
C.2
D.4
Explanation: The volume V of the box is given by V(x) = x(12 - 2x)^2 = 4x^3 - 48x^2 + 144x. To find the maximum, we differentiate with respect to x: dV/dx = 12x^2 - 96x + 144. Setting dV/dx = 0 yields 12(x^2 - 8x + 12) = 0, which factors to 12(x-2)(x-6) = 0. Since x = 6 would reduce the side length to 0, the feasible critical point is x = 2. The second derivative is d^2V/dx^2 = 24x - 96, which is negative (-48) at x = 2, confirming a maximum.
4Evaluate the definite integral: \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx.
A.\pi/2
B.\pi/4
C.\pi
D.0
Explanation: Using the integral property \int_a^b f(x) dx = \int_a^b f(a+b-x) dx, let I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx. Then I = \int_{0}^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx. Adding the two equations: 2I = \int_{0}^{\pi/2} 1 dx = \pi/2. Therefore, I = \pi/4.
5Find the area of the region bounded by the curves y^2 = 4x and x^2 = 4y.
A.16/3
B.8/3
C.4/3
D.32/3
Explanation: The intersection points are found by substituting y = x^2/4 into y^2 = 4x, giving (x^2/4)^2 = 4x => x^4 = 64x => x(x^3 - 64) = 0. The points are x = 0 and x = 4. The area A is given by \int_{0}^{4} (\sqrt{4x} - x^2/4) dx = [\frac{4}{3}x^{3/2} - \frac{x^3}{12}]_0^4 = (\frac{4}{3}(8) - \frac{64}{12}) = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}.
6Solve the differential equation: dy/dx + y \tan x = \sec x.
A.y \sec x = \tan x + C
B.y = \sin x + C \cos x
C.y \cos x = \sin x + C
D.y = \tan x + C \sec x
Explanation: This is a linear first-order differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = \tan x and Q(x) = \sec x. The integrating factor is IF = e^{\int \tan x dx} = e^{\ln(\sec x)} = \sec x. The general solution is y * IF = \int Q(x) * IF dx => y \sec x = \int \sec^2 x dx = \tan x + C. Dividing by \sec x gives y = (\sin x / \cos x) / (1 / \cos x) + C \cos x = \sin x + C \cos x.
7If A is a 3x3 non-singular matrix such that |A| = 5, find the determinant of the adjoint of A, i.e., |adj(A)|.
A.5
B.125
C.25
D.1
Explanation: For any n x n matrix A, the determinant of its adjoint is given by the formula |adj(A)| = |A|^{n-1}. Since A is a 3x3 matrix (n = 3) and |A| = 5, we have |adj(A)| = 5^{3-1} = 5^2 = 25.
8For what value of k does the system of equations: x + y + z = 1, x + 2y + 4z = k, x + 4y + 10z = k^2 have a solution?
A.k = 1 or k = 2
B.k = 1 or k = 3
C.k = 2 or k = 3
D.k = 0 or k = 1
Explanation: We can write the system as an augmented matrix and apply row operations. R2 -> R2 - R1 gives: y + 3z = k - 1. R3 -> R3 - R1 gives: 3y + 9z = k^2 - 1. Now apply R3 -> R3 - 3*R2 to get: 0 = (k^2 - 1) - 3(k - 1) = k^2 - 3k + 2. For the system to have a solution, this equation must hold, so k^2 - 3k + 2 = 0 => (k-1)(k-2) = 0 => k = 1 or k = 2.
9If \omega is a complex cube root of unity, evaluate the value of (1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5.
A.32
B.-32
C.64
D.-64
Explanation: We use the properties of cube roots of unity: 1 + \omega + \omega^2 = 0. Therefore, 1 + \omega^2 = -\omega and 1 + \omega = -\omega^2. Substituting these into the expression: (-\omega - \omega)^5 + (-\omega^2 - \omega^2)^5 = (-2\omega)^5 + (-2\omega^2)^5 = -32\omega^5 - 32\omega^{10}. Since \omega^3 = 1, we have \omega^5 = \omega^2 and \omega^{10} = \omega. The expression becomes -32(\omega^2 + \omega) = -32(-1) = 32.
10If the roots of the quadratic equation x^2 - px + q = 0 differ by unity, which of the following relations is correct?
A.p^2 = 4q + 1
B.p^2 = 4q - 1
C.q^2 = 4p + 1
D.q^2 = 4p - 1
Explanation: Let the roots be \alpha and \beta. We are given |\alpha - \beta| = 1. We know that (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta. The sum of the roots is \alpha + \beta = p and the product is \alpha\beta = q. Substituting these gives 1^2 = p^2 - 4q, which simplifies to p^2 = 4q + 1.

About the CG PET Exam

The Chhattisgarh Pre Engineering Test (CG PET) is a state-level entrance exam for B.E. / B.Tech courses in government and private colleges of Chhattisgarh. Learn and prepare with our exam-style questions.

Questions

150 scored questions

Time Limit

3 hours

Passing Score

10% for General/OBC, 5% for SC/ST/PH

Exam Fee

₹200 (CG Vyapam)

CG PET Exam Content Outline

33.33%

Physics

50 questions covering Mechanics, Thermodynamics, Waves, Electromagnetism, Optics, and Modern Physics

33.33%

Chemistry

50 questions covering Physical Chemistry, Inorganic Chemistry, and Organic Chemistry

33.33%

Mathematics

50 questions covering Algebra, Calculus, Coordinate Geometry, Vectors, and Trigonometry

How to Pass the CG PET Exam

What You Need to Know

  • Passing score: 10% for General/OBC, 5% for SC/ST/PH
  • Exam length: 150 questions
  • Time limit: 3 hours
  • Exam fee: ₹200

Keys to Passing

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

CG PET Study Tips from Top Performers

1Since there is no negative marking, ensure you attempt all 150 questions on the exam.
2Focus heavily on Mathematics (Calculus and Coordinate Geometry) and Physics (Mechanics and Electromagnetism) as they tend to be highly scoring.
3Solve previous years' question papers of CG PET and JEE Main to understand the difficulty level and types of questions asked.
4Learn and memorize key reactions in organic chemistry, and practice numerical problems in physical chemistry and physics.
5Master the use of shortcut formulas to solve math questions quickly, as you only have an average of 1.2 minutes per question.

Frequently Asked Questions

What is the exam pattern for CG PET 2026?

CG PET 2026 is an offline (pen-and-paper) test lasting 3 hours. It contains 150 multiple-choice questions: 50 in Physics, 50 in Chemistry, and 50 in Mathematics. Each correct answer carries 1 mark, and there is no negative marking.

Is there any application fee for Chhattisgarh residents?

As per the policy of the Chhattisgarh state government, local native residents of Chhattisgarh are exempted from paying application fees for exams conducted by CG Vyapam. Non-Chhattisgarh residents must pay the standard application fee (₹200 for General, ₹150 for OBC, ₹100 for SC/ST/PH).

What is the minimum qualifying score for CG PET?

To be placed on the merit list, General and OBC candidates must score at least 10% (15 out of 150 marks). SC, ST, and PwD candidates must score at least 5% (8 out of 150 marks). However, actual admission depends on college-specific cutoff ranks.

What are the eligibility criteria for CG PET?

Candidates must have passed or be appearing in the 10+2 exam with Physics and Mathematics as compulsory subjects, plus one optional subject (like Chemistry, Biology, or Biotech). A minimum aggregate of 45% in PCM is required for General category candidates, and 40% for SC/ST/OBC/PH/Female candidates. Candidates should also be at least 17 years old.

Is CG PET open to non-Chhattisgarh candidates?

Yes, candidates from other states can apply, but they are only eligible for the management quota or seats specifically designated for non-domiciled candidates in private colleges. Government college seats are reserved for Chhattisgarh residents.