Free Practice Questions for BUET Undergraduate Admission Main Written Examination

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Exam pass rate: Extremely selective; 1,309 seats filled from 10,351 GPA-shortlisted written candidates in the 2025-26 session (9,051 appeared, 87.44% attendance)

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Key Facts: BUET Undergraduate Admission Main Written Examination Exam

400 Marks

Total Written Marks (Module A)

BUET Admission Circular

3 Hours

Module A Duration

BUET Admission Committee (10 January 2026 sitting)

40 Questions

Written Problem Count

BUET Admission Guidelines

~1,309

Total Undergraduate Seats

BUET Academic Council

3 Subjects

Math, Physics, Chemistry

BUET Admission Syllabus

Single Attempt

First-Time Only Policy

BUET Admission Regulations

The BUET Undergraduate Admission Main Written Examination (Module A) is a 3-hour, 400-mark test of written engineering problems across Higher Mathematics, Physics, and Chemistry, sat by roughly the top 10,000 GPA-shortlisted applicants for 1,309 seats. This practice bank provides 100 rigorous analytical questions adapted into an English-language MCQ study companion with comprehensive explanations.

Sample BUET Undergraduate Admission Main Written Examination Practice Questions

Try these sample questions to review concepts for the BUET Undergraduate Admission Main Written Examination exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Solve the initial value problem: (x² + 1)(dy/dx) + 2xy = 4x³, with y(0) = 2. What is the value of y(1)?

A.2.0

B.1.5

C.3.0

D.2.5

Explanation: Rewriting the ODE in standard linear form: dy/dx + [2x/(x²+1)]y = 4x³/(x²+1). The integrating factor is IF = e^(∫ 2x/(x²+1) dx) = e^(ln(x²+1)) = x²+1. Multiplying gives d/dx [y(x²+1)] = 4x³, so y(x²+1) = x⁴ + C. Using y(0) = 2 yields 2(1) = 0 + C => C = 2. At x = 1, y(1+1) = 1⁴ + 2 = 3, so y(1) = 3/2 = 1.5.

2Find the general solution to the homogeneous differential equation dy/dx = (y² - x²)/(2xy).

A.x² + y² = Cx

B.x² - y² = Cy

C.x² + y² = Cy

D.y² - 2x² = Cx

Explanation: Substitute y = vx, dy/dx = v + x(dv/dx). The ODE becomes v + x(dv/dx) = (v² - 1)/(2v) => x(dv/dx) = -(v² + 1)/(2v). Separating variables gives -2v/(v² + 1) dv = dx/x. Integrating yields -ln(v² + 1) = ln|x| - ln|C| => ln[(v²+1)x] = ln|C| => (y²/x² + 1)x = C => (x² + y²)/x = C, which simplifies to x² + y² = Cx.

3Evaluate the limit: L = lim(x -> 0) [(tan x - sin x) / x³].

A.1/2

B.1/3

C.1/6

D.1

Explanation: Rewrite tan x - sin x = sin x (1/cos x - 1) = sin x (1 - cos x) / cos x. As x -> 0, cos x -> 1, sin x ~ x, and (1 - cos x) = 2 sin²(x/2) ~ 2(x/2)² = x²/2. Thus, the numerator is asymptotically x(x²/2) = x³/2. Dividing by x³ gives lim(x -> 0) (x³/2)/x³ = 1/2.

4A cylindrical open water tank of fixed volume V = 27π m³ is to be constructed with a circular base and vertical cylindrical walls. What base radius r minimizes the total surface area of sheet metal required (base + lateral wall)?

A.3 m

B.2 m

C.6 m

D.3√2 m

Explanation: The volume of the open cylinder is V = πr²h = 27π => h = 27/r². The total surface area is S = πr² + 2πrh = πr² + 2πr(27/r²) = πr² + 54π/r. Taking the derivative with respect to r: dS/dr = 2πr - 54π/r². Setting dS/dr = 0 gives 2πr = 54π/r² => r³ = 27 => r = 3 m. The second derivative d²S/dr² = 2π + 108π/r³ > 0 confirms a minimum.

5Find the equation of the normal to the curve y = x³ - 3x + 2 at the point where x = 2.

A.x + 9y - 38 = 0

B.9x + y - 22 = 0

C.x - 9y + 34 = 0

D.9x - y - 14 = 0

Explanation: At x = 2, y = (2)³ - 3(2) + 2 = 8 - 6 + 2 = 4, so the point of tangency is (2, 4). The derivative is dy/dx = 3x² - 3. At x = 2, the tangent slope is m_t = 3(4) - 3 = 9. The normal slope is m_n = -1/9. Using point-slope form: y - 4 = (-1/9)(x - 2) => 9(y - 4) = -(x - 2) => 9y - 36 = -x + 2 => x + 9y - 38 = 0.

6If y = (sin⁻¹ x)², prove that (1 - x²) y₂ - x y₁ equals a constant. What is the value of this constant?

A.2

B.1

C.-2

D.0

Explanation: Differentiating y = (sin⁻¹ x)² with respect to x gives y₁ = 2(sin⁻¹ x) / √(1 - x²). Multiplying by √(1 - x²) and squaring: (1 - x²)(y₁)² = 4(sin⁻¹ x)² = 4y. Differentiating implicitly with respect to x: -2x(y₁)² + (1 - x²) 2 y₁ y₂ = 4 y₁. Dividing through by 2 y₁ (for y₁ ≠ 0) yields (1 - x²) y₂ - x y₁ = 2.

7Evaluate the definite integral: I = ∫[0 to π/2] [sin³ x / (sin³ x + cos³ x)] dx.

A.π/4

B.π/2

C.π/3

D.1

Explanation: Using King's property of definite integrals, ∫[0 to a] f(x) dx = ∫[0 to a] f(a - x) dx. Here, replacing x with (π/2 - x) gives I = ∫[0 to π/2] [cos³ x / (cos³ x + sin³ x)] dx. Adding the two expressions: 2I = ∫[0 to π/2] [(sin³ x + cos³ x) / (sin³ x + cos³ x)] dx = ∫[0 to π/2] 1 dx = π/2. Therefore, I = π/4.

8Evaluate the definite integral: I = ∫[0 to π] [x sin x / (1 + cos² x)] dx.

A.π²/4

B.π²/2

C.π/4

D.π²

Explanation: Let I = ∫[0 to π] [x sin x / (1 + cos² x)] dx. Applying the property ∫[0 to a] f(x) dx = ∫[0 to a] f(a - x) dx: I = ∫[0 to π] [(π - x) sin(π - x) / (1 + cos²(π - x))] dx = ∫[0 to π] [(π - x) sin x / (1 + cos² x)] dx. Adding both equations gives 2I = π ∫[0 to π] [sin x / (1 + cos² x)] dx. Let u = cos x, du = -sin x dx. When x = 0, u = 1; when x = π, u = -1. Then 2I = π ∫[-1 to 1] du/(1 + u²) = π [tan⁻¹(1) - tan⁻¹(-1)] = π [π/4 - (-π/4)] = π(π/2) = π²/2. Hence, I = π²/4.

9Calculate the area of the region bounded between the parabola y² = 4ax and the straight line y = mx.

A.8a² / (3m³)

B.4a² / (3m³)

C.16a² / (3m³)

D.8a² / (m³)

Explanation: Intersection points: (mx)² = 4ax => m²x² - 4ax = 0 => x = 0 and x = 4a/m². The area is A = ∫[0 to 4a/m²] (√(4ax) - mx) dx = [ 2√a · (2/3) x^(3/2) - (m/2) x² ] from 0 to 4a/m². At the upper limit: (4/3)√a · (4a/m²)^(3/2) - (m/2) (16a²/m⁴) = (4/3)√a · (8a^(3/2)/m³) - 8a²/m³ = (32/3)(a²/m³) - (24/3)(a²/m³) = (8/3)(a²/m³) = 8a² / (3m³).

10Evaluate the indefinite integral: ∫ [dx / (x(x⁴ + 1))].

A.(1/4) ln |x⁴ / (x⁴ + 1)| + C

B.(1/4) ln |(x⁴ + 1) / x⁴| + C

C.ln |x / (x⁴ + 1)| + C

D.(1/2) ln |x² / (x⁴ + 1)| + C

Explanation: Multiply numerator and denominator by x³: ∫ [x³ dx / (x⁴(x⁴ + 1))]. Let u = x⁴, then du = 4x³ dx => x³ dx = du/4. The integral becomes (1/4) ∫ [du / (u(u + 1))] = (1/4) ∫ [1/u - 1/(u + 1)] du = (1/4) [ln|u| - ln|u + 1|] + C = (1/4) ln|u / (u + 1)| + C = (1/4) ln|x⁴ / (x⁴ + 1)| + C.

About the BUET Undergraduate Admission Main Written Examination Exam

The BUET Undergraduate Admission Written Examination is Bangladesh's most prestigious and rigorous engineering entrance examination. Administered annually by the Bangladesh University of Engineering and Technology (BUET), the test evaluates candidates across Advanced Mathematics, Advanced Physics, and Advanced Chemistry. Candidates solve complex multi-step problems requiring deep theoretical understanding, mathematical derivation, and rapid precision under strict time limits. This practice bank adapts these classic engineering problems into English-language multiple-choice questions with step-by-step solutions.

Exam sponsor: BUET Undergraduate Admission Committee. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Single-phase written admission test (2025-26 session; the earlier preliminary MCQ round was scrapped by the Admission Test Organising Committee on 25 October 2025). Roughly the top 10,351 applicants are shortlisted on SSC/HSC GPA merit, including GPA-5.00 individually in Physics, Chemistry, and Mathematics. Module A (Engineering departments + URP) is a 400-mark written paper in Higher Mathematics, Physics, and Chemistry over 3 hours; Module B (adding Architecture) supplements it with a 1.5-hour freehand drawing and visual-spatial intelligence test. Final admission fills 1,309 total seats by written-exam merit.

Time Limit

3 hours (9:00 AM-12:00 PM, first shift) for Module A; 1.5 hours (2:30-4:00 PM, second shift) for Module B drawing/aptitude

Passing Score

Merit-based admission ranking based on raw marks obtained in the written exam

Exam / Certification Fees

BDT 1,000 (Module A) / BDT 1,200 (Module B with Architecture); subject to yearly admission notification.

Exam sponsor website

Reported exam pass rate: Extremely selective; 1,309 seats filled from 10,351 GPA-shortlisted written candidates in the 2025-26 session (9,051 appeared, 87.44% attendance). Admission is purely merit-based determined by the aggregate score across Physics, Chemistry, and Mathematics written sections. This describes exam candidates, not OpenExamPrep users or results from using our resources. Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

35%

Advanced Higher Mathematics

Differential and integral calculus, ordinary differential equations, coordinate geometry of lines, circles and conics, vectors, statics & dynamics, complex numbers, polynomials, and probability & series.

33%

Advanced Physics

Newtonian mechanics, rotational kinematics and dynamics, gravitation, fluid dynamics and elasticity, thermodynamics and entropy, electrostatics, current electricity and AC circuits, electromagnetic induction, physical optics, and modern physics & quantum theory.

32%

Advanced Chemistry

Quantitative chemistry and multi-step stoichiometry, chemical equilibrium, buffer solutions & pH calculations, solubility product (Ksp), chemical kinetics rate laws, thermochemistry, redox titrations, coordination chemistry, and organic reaction mechanisms and synthesis.

Preparing for the BUET Undergraduate Admission Main Written Examination Exam

What You Need to Know

Passing score: Merit-based admission ranking based on raw marks obtained in the written exam
Assessment: Single-phase written admission test (2025-26 session; the earlier preliminary MCQ round was scrapped by the Admission Test Organising Committee on 25 October 2025). Roughly the top 10,351 applicants are shortlisted on SSC/HSC GPA merit, including GPA-5.00 individually in Physics, Chemistry, and Mathematics. Module A (Engineering departments + URP) is a 400-mark written paper in Higher Mathematics, Physics, and Chemistry over 3 hours; Module B (adding Architecture) supplements it with a 1.5-hour freehand drawing and visual-spatial intelligence test. Final admission fills 1,309 total seats by written-exam merit.
Time limit: 3 hours (9:00 AM-12:00 PM, first shift) for Module A; 1.5 hours (2:30-4:00 PM, second shift) for Module B drawing/aptitude
Exam / certification fees: BDT 1,000 (Module A) / BDT 1,200 (Module B with Architecture); subject to yearly admission notification. Official sources

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BUET Undergraduate Admission Main Written Examination: Suggested Study Strategy

1Focus on multi-step derivation and boundary-condition analysis rather than memorizing formula shortcuts, as BUET problems frequently test edge cases and non-standard physical configurations.

2Cultivate rapid calculator proficiency and dimensional verification to eliminate arithmetic errors in multi-part numerical evaluations.

3In Higher Mathematics, prioritize definite integral transformations, differential equation modeling, conic section standard forms, and vector statics/dynamics equilibrium.

4In Physics, master rotational dynamics, thermodynamic work integrals, LCR AC resonant circuits, and wave optics path differences.

5In Chemistry, practice rigorous simultaneous ionic equilibria, buffer capacity calculations, redox electron-transfer balancing, and organic reaction mechanisms.

Frequently Asked Questions

What is the structure of the BUET Written Admission Examination?

The BUET Written Admission Test (Module A - Engineering Faculties and URP) consists of 40 written analytical questions carrying 10 marks each, totaling 400 marks over a duration of 3 hours (9:00 AM to 12:00 PM in the 10 January 2026 sitting). The questions are distributed across Higher Mathematics, Physics, and Chemistry (~13-14 questions per subject). Candidates must write calculations and final answers within designated answer boxes on the exam script.

Is this practice bank identical to the official exam format?

This practice bank is an English-language study adaptation designed to master the analytical depth and numerical problem-solving techniques tested on BUET's written examination. While the official exam uses subjective written answer boxes, our 100 multiple-choice questions retain full mathematical rigor, multi-step problem structures, and complete pedagogical explanations.

Are non-programmable scientific calculators allowed in the BUET written test?

Yes, BUET permits approved non-programmable scientific calculators (such as standard fx-991ES/EX series, subject to current BUET admission committee guidelines specified on the admit card). Graphical and programmable calculators are strictly prohibited.

Can second-time candidates sit for the BUET admission test?

Under current BUET admission regulations, second-time test attempts are not permitted. Only candidates who passed their HSC (or equivalent GCE A-Level) examination in the immediate preceding academic cycle are eligible to apply.

What is the difference between Module A and Module B in BUET admission?

Module A is for Engineering faculties and the Department of Urban and Regional Planning (URP), compulsory for all engineering aspirants and consisting of the 400-mark Mathematics, Physics, and Chemistry written examination (3 hours, morning shift). Module B candidates, which include applicants to the Department of Architecture, additionally sit a 1.5-hour afternoon Freehand Drawing and Visual-Spatial Intelligence test (2:30-4:00 PM in the 2025-26 session).