Free Practice Questions for IOE Entrance Nepal

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Key Facts: IOE Entrance Nepal Exam

100

Total MCQs

140

Total Marks

120 min

Exam Duration

10%

Negative Marking

2,000 NPR

Application Fee

The Tribhuvan University IOE Entrance Examination is a 2-hour computer-based test with 100 multiple-choice questions totalling 140 marks (60 one-mark and 40 two-mark items) for admission to BE and B.Arch programmes. Marks are split Mathematics 50, Physics 40, Chemistry 30, and English 20. Wrong answers carry 10% negative marking, and seats are allocated by merit rank. The application fee is NPR 2,000. OpenExamPrep provides free independent practice questions with worked solutions.

Sample IOE Entrance Nepal Practice Questions

Try these sample questions to review concepts for the IOE Entrance Nepal exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1What is the modulus and principal argument of the complex number z = (1 + i*sqrt(3)) / (1 + i)?

A.Modulus 2, argument pi/6

B.Modulus sqrt(2), argument 5pi/12

C.Modulus sqrt(2), argument pi/12

D.Modulus 1/sqrt(2), argument -pi/12

Explanation: To find the modulus and argument of the quotient z = z1 / z2, express both numerator and denominator in polar form. For the numerator z1 = 1 + i*sqrt(3), modulus |z1| = sqrt(1^2 + (sqrt(3))^2) = sqrt(4) = 2, and argument arg(z1) = arctan(sqrt(3)/1) = pi/3 (60 degrees). For the denominator z2 = 1 + i, modulus |z2| = sqrt(1^2 + 1^2) = sqrt(2), and argument arg(z2) = arctan(1/1) = pi/4 (45 degrees). The modulus of the quotient is |z| = |z1| / |z2| = 2 / sqrt(2) = sqrt(2). The argument of the quotient is arg(z) = arg(z1) - arg(z2) = pi/3 - pi/4 = (4pi - 3pi)/12 = pi/12 (15 degrees). Therefore, |z| = sqrt(2) and arg(z) = pi/12.

2For what condition will the roots of the quadratic equation ax^2 + bx + c = 0 (where a != 0) be reciprocal to each other?

A.b = 0

B.c = a

C.b^2 = 4ac

D.c = -a

Explanation: Let the roots of the quadratic equation ax^2 + bx + c = 0 be alpha and beta. If the roots are reciprocals of each other, then beta = 1/alpha, which implies that their product alpha * beta = alpha * (1/alpha) = 1. According to Vieta's formulas, the product of the roots is given by c / a. Setting c / a = 1 gives c = a. Thus, the necessary and sufficient condition for the roots to be reciprocal is c = a.

3If A is an invertible square matrix of order 3 such that det(A) = 4, what is the determinant of the adjugate matrix, det(adj(A))?

A.4

B.64

C.12

D.16

Explanation: For any n x n square matrix A, the fundamental property relating a matrix and its adjugate is A * adj(A) = det(A) * I_n. Taking the determinant on both sides: det(A * adj(A)) = det(det(A) * I_n). By determinant properties, det(A) * det(adj(A)) = (det(A))^n * det(I_n) = (det(A))^n * 1. Dividing both sides by det(A) yields det(adj(A)) = (det(A))^(n - 1). Here, the order is n = 3 and det(A) = 4, so det(adj(A)) = 4^(3 - 1) = 4^2 = 16.

4What is the coefficient of x^4 in the binomial expansion of (x/2 - 3/x^2)^10?

A.405 / 256

B.-405 / 256

C.135 / 128

D.405 / 512

Explanation: In the expansion of (x/2 - 3/x^2)^10, the general term is given by T_(r+1) = C(10, r) * (x/2)^(10-r) * (-3/x^2)^r = C(10, r) * (1/2)^(10-r) * (-3)^r * x^(10 - r - 2r) = C(10, r) * (1/2)^(10-r) * (-3)^r * x^(10 - 3r). We require the term containing x^4, so set the exponent of x to 4: 10 - 3r = 4 => 3r = 6 => r = 2. Now compute the coefficient for r = 2: C(10, 2) = (10 * 9) / 2 = 45. For the constants: (1/2)^(10 - 2) = (1/2)^8 = 1/256, and (-3)^2 = 9. Multiplying these gives: 45 * (1/256) * 9 = 405 / 256. Since (-3)^2 is positive, the coefficient is +405 / 256.

5An infinite geometric series has first term a = 6 and common ratio r = 1/3. What is the sum of this infinite series?

A.8

B.18

C.9

D.2

Explanation: For an infinite geometric progression with first term a and common ratio r where |r| < 1, the sum to infinity is given by S_infinity = a / (1 - r). Here, a = 6 and r = 1/3 (|1/3| < 1, so the series converges). Substituting these values: S_infinity = 6 / (1 - 1/3) = 6 / (2/3) = 6 * (3/2) = 18 / 2 = 9.

6In how many distinct ways can 6 engineering students be seated around a circular conference table?

A.720

B.60

C.24

D.120

Explanation: The number of distinct circular permutations of n distinct objects is (n - 1)!, because fixing the position of one object eliminates the rotational symmetry of the circle. For n = 6 students around a table, the total number of arrangements is (6 - 1)! = 5! = 5 * 4 * 3 * 2 * 1 = 120 ways.

7How many solutions exist for the trigonometric equation cos(2x) = sin(x) in the interval [0, 2*pi)?

A.2

B.3

C.4

D.1

Explanation: Using the double-angle identity cos(2x) = 1 - 2*sin^2(x), the equation becomes: 1 - 2*sin^2(x) = sin(x) => 2*sin^2(x) + sin(x) - 1 = 0. Factor this quadratic in sin(x): (2*sin(x) - 1)(sin(x) + 1) = 0. This yields two possibilities: 1) 2*sin(x) - 1 = 0 => sin(x) = 1/2. In the interval [0, 2*pi), sin(x) = 1/2 at x = pi/6 and x = 5pi/6 (2 solutions). 2) sin(x) + 1 = 0 => sin(x) = -1. In [0, 2*pi), sin(x) = -1 at x = 3pi/2 (1 solution). Combining both cases gives 2 + 1 = 3 distinct solutions in [0, 2*pi).

8What is the exact value of arctan(1/2) + arctan(1/3)?

A.pi / 4

B.pi / 6

C.pi / 3

D.pi / 2

Explanation: Use the addition formula for inverse tangent: arctan(x) + arctan(y) = arctan((x + y) / (1 - xy)), valid when xy < 1. Here, x = 1/2 and y = 1/3, so xy = (1/2)*(1/3) = 1/6 < 1. Calculating the numerator: x + y = 1/2 + 1/3 = 5/6. Calculating the denominator: 1 - xy = 1 - 1/6 = 5/6. Thus, (x + y) / (1 - xy) = (5/6) / (5/6) = 1. Therefore, arctan(1/2) + arctan(1/3) = arctan(1) = pi/4 radians (45 degrees).

9In a triangle ABC, the lengths of the sides are a = 7, b = 8, and c = 9. What is the value of cos(A)?

A.11 / 16

B.5 / 7

C.3 / 4

D.2 / 3

Explanation: According to the Law of Cosines, for any triangle with sides a, b, c, the cosine of angle A is given by cos(A) = (b^2 + c^2 - a^2) / (2bc). Substituting the given side lengths: a = 7 => a^2 = 49; b = 8 => b^2 = 64; c = 9 => c^2 = 81. Numerator = 64 + 81 - 49 = 145 - 49 = 96. Denominator = 2 * b * c = 2 * 8 * 9 = 144. Therefore, cos(A) = 96 / 144. Dividing numerator and denominator by 48 gives 2 / 3.

10What is the acute angle between the pair of straight lines represented by the homogeneous equation 2x^2 + 7xy + 3y^2 = 0?

A.pi / 3 (60 degrees)

B.pi / 6 (30 degrees)

C.pi / 4 (45 degrees)

D.arctan(3/5)

Explanation: The general homogeneous equation of second degree representing a pair of lines passing through the origin is ax^2 + 2hxy + by^2 = 0. Comparing 2x^2 + 7xy + 3y^2 = 0 with the standard form: a = 2, 2h = 7 => h = 7/2, and b = 3. The angle theta between the two lines is given by tan(theta) = |2 * sqrt(h^2 - ab) / (a + b)|. Calculate h^2 - ab: (7/2)^2 - 2 * 3 = 49/4 - 6 = 49/4 - 24/4 = 25/4. Thus, sqrt(h^2 - ab) = sqrt(25/4) = 5/2. Now substitute into the formula: tan(theta) = |2 * (5/2) / (2 + 3)| = |5 / 5| = 1. Since tan(theta) = 1, theta = pi/4 radians (45 degrees).

About the IOE Entrance Nepal Exam

The IOE BE/B.Arch Entrance Examination is Tribhuvan University's computer-based entrance test for admission to undergraduate engineering and architecture programmes in Nepal. The 2-hour examination has 100 multiple-choice questions totalling 140 marks (Mathematics 50, Physics 40, Chemistry 30, English 20), with 10% negative marking for wrong answers.

Exam sponsor: Institute of Engineering (IOE) Entrance Examination Board, Tribhuvan University. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Computer-based test of 100 multiple-choice questions totalling 140 marks (60 one-mark and 40 two-mark questions), 2 hours, with 10% negative marking for wrong answers. Marks: Mathematics 50, Physics 40, Chemistry 30, English 20.

Time Limit

120 minutes

Passing Score

Admission by merit rank; see the current IOE notice for any qualifying cutoff

Exam / Certification Fees

2,000 NPR

Exam sponsor website

Reported exam pass rate: Not published; merit rank determines seat allocation. High merit ranks are required to obtain regular subsidized seats at premier constituent campuses including Pulchowk Campus, Thapathali Campus, Western Campus (Pokhara), and Eastern Campus (Dharan). 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.

50 of 140 marks

Mathematics

Sets, functions, algebra, matrices, complex numbers, trigonometry, coordinate geometry, vectors, calculus and differential equations.

40 of 140 marks

Physics

Mechanics, heat and thermodynamics, waves and optics, electricity and magnetism, and modern physics.

30 of 140 marks

Chemistry

Physical, inorganic and organic chemistry.

20 of 140 marks

English

Grammar, usage and vocabulary.

Preparing for the IOE Entrance Nepal Exam

What You Need to Know

Passing score: Admission by merit rank; see the current IOE notice for any qualifying cutoff
Assessment: Computer-based test of 100 multiple-choice questions totalling 140 marks (60 one-mark and 40 two-mark questions), 2 hours, with 10% negative marking for wrong answers. Marks: Mathematics 50, Physics 40, Chemistry 30, English 20.
Time limit: 120 minutes
Exam / certification fees: 2,000 NPR Official sources

Using Our Practice Resources

Work through all 100 available questions
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IOE Entrance Nepal: Suggested Study Strategy

1Mathematics (50 marks) and Physics (40 marks) carry 90 of the 140 marks, so prioritise problem solving in these two subjects.

2Two-mark questions are worth double, so practise multi-step numericals in calculus, mechanics and physical chemistry.

3Factor in the 10% negative marking: avoid blind guessing, especially on two-mark items.

4Practise full 100-question sets in 2 hours to build pace for the computer-based format.

Frequently Asked Questions

What is the examination pattern and marking scheme for IOE BE/B.Arch Entrance?

The examination is a 2-hour (120-minute) computer-based test (CBT) comprising 100 multiple-choice questions totaling 140 marks. It includes 60 one-mark questions and 40 two-mark questions. Incorrect answers incur a 10% negative marking penalty (-0.1 mark for 1-mark questions, -0.2 marks for 2-mark questions), while unattempted questions receive 0 marks.

What is the subject distribution of questions in the IOE entrance paper?

Marks are distributed as Mathematics 50, Physics 40, Chemistry 30, and English 20, for a total of 140 marks across 100 questions.

How are IOE entrance results used for admission?

Candidates are ranked by entrance score, and seats in constituent and affiliated engineering colleges are allocated in merit order. Check the current IOE notice for any qualifying cutoff.

Where is the IOE entrance examination conducted?

The entrance is a computer-based test held at centres announced by the IOE Entrance Examination Board in its admission notice.

Is OpenExamPrep affiliated with Tribhuvan University or IOE?

No. OpenExamPrep is an independent educational platform and is not affiliated with, endorsed by, or connected to Tribhuvan University, the Institute of Engineering (IOE), or the IOE Entrance Examination Board. All practice materials and solutions are created independently for study practice.