240+ Free JEE Main Paper 2B — B.Planning Practice Questions

Prepare for the Joint Entrance Examination (Main) — Paper 2B (B.Planning) exam with instant access — no signup required.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately

240+ Questions

100% Free

Loading practice questions...

Same family resources

Explore More JEE Main Architecture Entrance

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

JEE Main Paper 2A — B.Arch (National Testing Agency, India)

Practice Questions100 questions

2026 Statistics

Key Facts: JEE Main Paper 2B — B.Planning Exam

100 questions

Total questions (25 Math, 50 Aptitude, 25 Planning)

NTA JEE Main Bulletin

400 marks

Maximum possible score (+4 for correct, -1 for incorrect)

NTA marking scheme

3 hours

Total duration of the computer-based exam

NTA

3 SPA institutes

Schools of Planning and Architecture (Delhi, Bhopal, Vijayawada)

JoSAA Seat Matrix

JEE Main Paper 2B (B.Planning) is a 3-hour computer-based test of 100 questions. It consists of Mathematics, General Aptitude, and specialized Planning-based questions, determining admissions to prestigious B.Plan courses in India.

Sample JEE Main Paper 2B — B.Planning Practice Questions

Try these sample questions to test your JEE Main Paper 2B — B.Planning exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 240+ question experience with AI tutoring.

1Find the equation of the line passing through the point (2, 3) and parallel to the line 3x - 4y + 5 = 0.

A.3x - 4y + 6 = 0

B.3x - 4y - 6 = 0

C.4x + 3y - 17 = 0

D.3x + 4y - 18 = 0

Explanation: Since the line is parallel to 3x - 4y + 5 = 0, its equation is of the form 3x - 4y + c = 0. Substituting the point (2, 3): 3(2) - 4(3) + c = 0 => 6 - 12 + c = 0 => c = 6. Thus, the equation is 3x - 4y + 6 = 0.

2Evaluate the limit: lim (x -> 0) [sin(3x) / tan(2x)].

A.3/2

B.2/3

C.1

D.0

Explanation: We can rewrite the limit as lim (x -> 0) [(sin(3x)/(3x)) * (3x) / ((tan(2x)/(2x)) * (2x))]. Since lim (theta -> 0) sin(theta)/theta = 1 and lim (theta -> 0) tan(theta)/theta = 1, the expression simplifies to lim (x -> 0) [1 * 3x / (1 * 2x)] = 3/2.

3Two cards are drawn sequentially without replacement from a standard pack of 52 cards. What is the probability that both are aces?

A.1/221

B.1/169

C.1/26

D.3/676

Explanation: The probability of drawing the first ace is 4/52 = 1/13. Since there is no replacement, the probability of drawing the second ace is 3/51 = 1/17. The combined probability is (1/13) * (1/17) = 1/221.

4If A is a symmetric matrix and B is a skew-symmetric matrix of the same order, then the matrix (AB - BA) is always a:

A.Symmetric matrix

B.Skew-symmetric matrix

C.Diagonal matrix

D.Identity matrix

Explanation: Let C = AB - BA. Taking transpose: C^T = (AB - BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T. Since A is symmetric (A^T = A) and B is skew-symmetric (B^T = -B), we get C^T = (-B)A - A(-B) = -BA + AB = AB - BA = C. Since C^T = C, the matrix is symmetric.

5If alpha and beta are the roots of the quadratic equation x^2 - 5x + 6 = 0, find the quadratic equation whose roots are (alpha + 1) and (beta + 1).

A.x^2 - 7x + 12 = 0

B.x^2 - 5x + 12 = 0

C.x^2 - 7x + 6 = 0

D.x^2 + 7x + 12 = 0

Explanation: Let y = x + 1 be the new roots, so x = y - 1. Substituting this into the original equation: (y - 1)^2 - 5(y - 1) + 6 = 0 => y^2 - 2y + 1 - 5y + 5 + 6 = 0 => y^2 - 7y + 12 = 0. Replacing y back with x gives x^2 - 7x + 12 = 0.

6Find the scalar projection of the vector a = 2i + 3j + k on the vector b = i + 2j + 2k.

A.10/3

B.10

C.8/3

D.4/3

Explanation: The scalar projection of a on b is given by (a . b) / |b|. First, compute the dot product a . b = (2)(1) + (3)(2) + (1)(2) = 2 + 6 + 2 = 10. Next, compute the magnitude of b: |b| = sqrt(1^2 + 2^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3. Thus, scalar projection = 10/3.

7Find the shortest distance between the parallel lines r = (1,0,0) + lambda(1,1,0) and r = (0,1,1) + mu(1,1,0).

A.sqrt(3)

B.sqrt(2)

C.sqrt(6)

D.1/sqrt(3)

Explanation: Take the displacement between one point on each line as (-1,1,1) and the common direction as (1,1,0). Their cross product is (-1,1,-2), with magnitude sqrt(6). Dividing by the direction magnitude sqrt(2) gives shortest distance sqrt(6)/sqrt(2) = sqrt(3).

8Solve the first-order linear differential equation: dy/dx + y/x = x^2 (for x > 0).

A.y = (x^3 / 4) + C / x

B.y = x^3 + C

C.y = (x^2 / 3) + C

D.y = (x^3 / 3) + C / x

Explanation: This is a linear differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = 1/x and Q(x) = x^2. The integrating factor is IF = e^(integral (1/x) dx) = e^(ln x) = x. Multiplying the equation by x gives: x(dy/dx) + y = x^3 => d/dx (xy) = x^3. Integrating both sides with respect to x: xy = (x^4 / 4) + C => y = (x^3 / 4) + C/x.

9In a town of 100 families, 65 subscribe to newspaper A, 45 subscribe to newspaper B, and 30 subscribe to both. How many families subscribe to neither newspaper?

A.20

B.10

C.15

D.25

Explanation: Let A and B be the sets of families subscribing to newspapers A and B respectively. We are given n(A) = 65, n(B) = 45, and n(A intersect B) = 30. Using the formula: n(A union B) = n(A) + n(B) - n(A intersect B) = 65 + 45 - 30 = 80. The number of families subscribing to neither is Total - n(A union B) = 100 - 80 = 20.

10Simplify the expression: sin(15 degrees) + cos(15 degrees).

A.sqrt(3/2)

B.sqrt(2)

C.sqrt(3)

D.1/sqrt(2)

Explanation: Let S = sin(15) + cos(15). Squaring both sides: S^2 = (sin(15) + cos(15))^2 = sin^2(15) + cos^2(15) + 2 sin(15)cos(15) = 1 + sin(30) = 1 + 1/2 = 3/2. Since sin(15) and cos(15) are both positive, S = sqrt(3/2).

About the JEE Main Paper 2B — B.Planning Exam

JEE Main Paper 2B is the national level entrance exam conducted by the National Testing Agency (NTA) for admission to undergraduate Bachelor of Planning (B.Plan) courses at premier institutions in India, including School of Planning and Architecture (SPA) Delhi, Bhopal, and Vijayawada. The exam assesses mathematical foundations, spatial and general aptitude, and basic understanding of planning principles, urban development, and civic infrastructure.

Questions

100 scored questions

Time Limit

3 hours (180 minutes); 4 hours for eligible PwD/PwBD candidates using compensatory time

Passing Score

No fixed pass mark; admission is determined by NTA percentile ranks and Joint Seat Allocation Authority (JoSAA) counseling

Exam Fee

Per 2026 session at an India centre: INR 1,000 General male; INR 900 Gen-EWS/OBC-NCL male; INR 800 General/Gen-EWS/OBC-NCL female; INR 500 SC/ST/PwD/PwBD or Third Gender, plus processing charges and GST (National Testing Agency (NTA))

JEE Main Paper 2B — B.Planning Exam Content Outline

25%

Part I: Mathematics

Algebra, matrices, determinants, coordinate geometry, calculus, vector algebra, 3D geometry, statistics, and probability from Class 11-12 syllabus.

50%

Part II: General Aptitude

Awareness of persons, buildings, materials, objects, and textures related to architecture and the built environment; 2D-to-3D visualization; visual, numerical, and verbal mental ability; spatial perception; scale and proportion; and 2D/3D form and composition.

25%

Part III: Planning-based Questions

General awareness of prominent cities, development issues, and government programs; social sciences, including history, geography, civics, economics, population, and settlements; and thinking skills such as comprehension, map reading, critical reasoning, data interpretation, statistics, and quantitative reasoning.

How to Pass the JEE Main Paper 2B — B.Planning Exam

What You Need to Know

Passing score: No fixed pass mark; admission is determined by NTA percentile ranks and Joint Seat Allocation Authority (JoSAA) counseling
Exam length: 100 questions
Time limit: 3 hours (180 minutes); 4 hours for eligible PwD/PwBD candidates using compensatory time
Exam fee: Per 2026 session at an India centre: INR 1,000 General male; INR 900 Gen-EWS/OBC-NCL male; INR 800 General/Gen-EWS/OBC-NCL female; INR 500 SC/ST/PwD/PwBD or Third Gender, plus processing charges and GST

Keys to Passing

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

JEE Main Paper 2B — B.Planning Study Tips from Top Performers

1Revise NCERT Class 11 and 12 Mathematics thoroughly, focusing on coordinate geometry, statistics, and probability.

2Develop spatial visualization skills by practicing 3D object rotations, projections, and counting surfaces.

3Study basic urban planning terminology, green building guidelines (GRIHA/LEED), and civic sanitation systems.

4Practice map scale conversions and mathematical calculations for sound (decibels) and population projection.

5Keep updated with major Government of India urban development schemes like Smart Cities, AMRUT, and PMAY.

Frequently Asked Questions

What is the exam pattern of JEE Main Paper 2B?

Paper 2B contains 100 questions in total: 25 in Mathematics (20 MCQs + 5 compulsory numerical questions), 50 MCQs in General Aptitude, and 25 MCQs in Planning-based questions. The total marks are 400.

What is the marking scheme for the B.Planning paper?

Correct answers receive +4 marks. Incorrect answers (both MCQs and numerical questions) receive a penalty of -1 mark. Unattempted questions receive 0 marks.

Which colleges accept JEE Main Paper 2B scores?

Scores are accepted for admission to B.Plan programs at the three Schools of Planning and Architecture (SPA Delhi, SPA Bhopal, SPA Vijayawada), select NITs, and other Centrally Funded Technical Institutes (CFTIs).

Can I appear for both Paper 2A (B.Arch) and Paper 2B (B.Planning)?

Yes, candidates can choose to appear for Paper 2A, Paper 2B, or both. The exam sessions are scheduled such that candidates can write both papers if registered.