Free Practice Questions for IOQM

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Key Facts: IOQM Exam

3 hours

Exam Duration

MTA(I)

OMR Integer Questions

HBCSE

10%

Minimum for Certificate

MTA(I)

The IOQM consists of 30 integer-answer questions (answers from 00 to 99) in a 3-hour offline format. Scoring at least 10% grants a certificate, while higher regional merit cutoffs determine qualification for the Regional Mathematical Olympiad (RMO).

Sample IOQM Practice Questions

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

1Let $P(x)$ be a polynomial with integer coefficients such that $P(1) = 4$ and $P(2) = 11$. Find the remainder when $P(x)$ is divided by $(x-1)(x-2)$, evaluated at $x = 10$.

A.57

B.63

C.67

D.73

Explanation: Let the remainder be $R(x) = ax + b$ since the divisor $(x-1)(x-2)$ is of degree 2. By the Remainder Theorem, $P(1) = R(1) = a + b = 4$ and $P(2) = R(2) = 2a + b = 11$. Subtracting the first equation from the second gives $a = 7$. Substituting back, we find $b = 4 - 7 = -3$. Thus, the remainder is $R(x) = 7x - 3$. Evaluating this remainder at $x = 10$ yields $R(10) = 7(10) - 3 = 67$.

2Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 - 10x + 18 = 0$. Find the value of $\alpha^2 + \beta^2$.

A.54

B.64

C.74

D.82

Explanation: By Vieta's formulas, the sum of the roots is $\alpha + \beta = 10$ and the product of the roots is $\alpha\beta = 18$. We want to find $\alpha^2 + \beta^2$. Using the algebraic identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, we substitute the values to get: $\alpha^2 + \beta^2 = 10^2 - 2(18) = 100 - 36 = 64$.

3If $x + 2y + 3z = 24$ and $3x + 2y + z = 16$, find the value of $5x + 6y + 7z$.

A.48

B.56

C.64

D.72

Explanation: We search for constants $A$ and $B$ such that $A(x + 2y + 3z) + B(3x + 2y + z) = 5x + 6y + 7z$. Equating coefficients: for $x$, $A + 3B = 5$; for $y$, $2A + 2B = 6 \implies A + B = 3$; for $z$, $3A + B = 7$. From $A + B = 3$, we have $B = 3 - A$. Substituting this into the first equation: $A + 3(3 - A) = 5 \implies -2A + 9 = 5 \implies A = 2$. Thus, $B = 1$. Checking the $z$ coefficient: $3(2) + 1 = 7$, which matches. Thus, the expression equals $2(x + 2y + 3z) + 1(3x + 2y + z) = 2(24) + 16 = 48 + 16 = 64$.

4In an arithmetic progression, the sum of the first, third, and fifth terms is 15, and the sum of the second, fourth, and sixth terms is 24. Find the seventh term of this progression.

A.14

B.17

C.20

D.23

Explanation: Let the terms of the arithmetic progression be $a_n$ with common difference $d$. The sum of the first, third, and fifth terms is $a_1 + a_3 + a_5 = 3a_3 = 15 \implies a_3 = 5$. The sum of the second, fourth, and sixth terms is $a_2 + a_4 + a_6 = 3a_4 = 24 \implies a_4 = 8$. The common difference is $d = a_4 - a_3 = 8 - 5 = 3$. Therefore, the seventh term is $a_7 = a_4 + 3d = 8 + 3(3) = 17$.

5If $x^2 - 3x + 1 = 0$, find the value of $x^4 + \frac{1}{x^4}$.

A.43

B.47

C.49

D.51

Explanation: Since $x \neq 0$, we can divide $x^2 - 3x + 1 = 0$ by $x$ to get $x - 3 + \frac{1}{x} = 0 \implies x + \frac{1}{x} = 3$. Squaring both sides: $(x + \frac{1}{x})^2 = 3^2 \implies x^2 + 2 + \frac{1}{x^2} = 9 \implies x^2 + \frac{1}{x^2} = 7$. Squaring once more: $(x^2 + \frac{1}{x^2})^2 = 7^2 \implies x^4 + 2 + \frac{1}{x^4} = 49 \implies x^4 + \frac{1}{x^4} = 47$.

6Let $a$ and $b$ be the roots of the equation $x^2 - p x + 12 = 0$. If $a^2 + b^2 = 25$ and $p > 0$, find the value of $p^2$.

A.25

B.37

C.49

D.61

Explanation: By Vieta's formulas, $a + b = p$ and $ab = 12$. We are given $a^2 + b^2 = 25$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we substitute the knowns: $25 = p^2 - 2(12) \implies 25 = p^2 - 24 \implies p^2 = 49$.

7The sum of the first $n$ terms of a sequence is given by $S_n = 2n^2 + 3n$. Find the value of the 10th term, $a_{10}$.

A.37

B.39

C.41

D.43

Explanation: The 10th term $a_{10}$ is given by $S_{10} - S_9$. We calculate: $S_{10} = 2(10)^2 + 3(10) = 200 + 30 = 230$, and $S_9 = 2(9)^2 + 3(9) = 162 + 27 = 189$. Thus, $a_{10} = 230 - 189 = 41$. Alternatively, since $S_n = 2n^2 + 3n$ represents the sum of an arithmetic progression, the $n$-th term is $a_n = S_n - S_{n-1} = 2n^2 + 3n - [2(n-1)^2 + 3(n-1)] = 4n + 1$. Thus $a_{10} = 4(10) + 1 = 41$.

8For any positive real numbers $a$ and $b$ such that $a + b = 10$, find the maximum possible integer value of $ab$.

A.20

B.24

C.25

D.26

Explanation: By the AM-GM Inequality, for positive real numbers $a$ and $b$, $\sqrt{ab} \le \frac{a+b}{2}$. Substituting $a+b=10$ gives $\sqrt{ab} \le 5 \implies ab \le 25$. The maximum value of $25$ is achieved when $a = b = 5$. Since 25 is an integer, this is the maximum integer value.

9Let $\alpha, \beta, \gamma$ be the roots of the cubic polynomial $P(x) = x^3 - 5x^2 + 7x + 13$. Find the value of $(3-\alpha)(3-\beta)(3-\gamma)$.

A.12

B.16

C.20

D.24

Explanation: Since $\alpha, \beta, \gamma$ are the roots of $P(x)$, we can factor the polynomial as $P(x) = (x - \alpha)(x - \beta)(x - \gamma)$. Evaluating $P(x)$ at $x = 3$ gives $P(3) = (3 - \alpha)(3 - \beta)(3 - \gamma)$. We compute $P(3) = 3^3 - 5(3^2) + 7(3) + 13 = 27 - 45 + 21 + 13 = 16$.

10Let $\alpha, \beta, \gamma$ be the roots of $x^3 - 7x^2 + 5x - 1 = 0$. Find the value of $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}$.

A.9

B.11

C.13

D.15

Explanation: By Vieta's formulas, the elementary symmetric sums are $\alpha + \beta + \gamma = 7$, $\alpha\beta + \beta\gamma + \gamma\alpha = 5$, and $\alpha\beta\gamma = 1$. We rewrite the desired expression: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{(\alpha\beta\gamma)^2}$. The numerator is $\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) = 5^2 - 2(1)(7) = 25 - 14 = 11$. The denominator is $(\alpha\beta\gamma)^2 = 1^2 = 1$. Thus, the value is $\frac{11}{1} = 11$.

About the IOQM Exam

The Indian Olympiad Qualifier in Mathematics (IOQM) is the first stage of the Mathematics Olympiad selection process in India. It is a highly challenging examination testing concepts in algebra, geometry, combinatorics, and number theory for students in classes 8-12.

Exam sponsor: Mathematics Teachers' Association (India) [MTA(I)] and Homi Bhabha Centre for Science Education [HBCSE]. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Questions

30 questions

Time Limit

3 hours

Passing Score

Scoring 10% or more (10 marks out of 100) receives a certificate

Exam / Certification Fees

₹240 (₹150 for Kendriya Vidyalaya students, ₹125 for Jawahar Navodaya Vidyalaya students)

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.

25-30%

Algebra

Solve problems involving polynomials, Vieta's relations, algebraic inequalities (AM-GM, Cauchy-Schwarz), functional equations, and sequences.

25-30%

Geometry

Solve problems using cyclic quadrilaterals, Power of a Point, Ptolemy's Theorem, Ceva and Menelaus, and properties of incircles and circumcircles.

20-25%

Number Theory

Apply modular arithmetic, divisibility rules, Diophantine equations, Fermat's Little Theorem, and Euler's totient function.

20-25%

Combinatorics

Master permutations, combinations, bijections, the pigeonhole principle, grid path counting, recurrence relations, and double counting.

Preparing for the IOQM Exam

What You Need to Know

Passing score: Scoring 10% or more (10 marks out of 100) receives a certificate
Exam length: 30 questions
Time limit: 3 hours
Exam / certification fees: ₹240 (₹150 for Kendriya Vidyalaya students, ₹125 for Jawahar Navodaya Vidyalaya students) Official sources

Using Our Practice Resources

Work through all 100 available questions
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IOQM: Suggested Study Strategy

1Focus heavily on geometry and number theory, as they represent a substantial portion of the exam and have highly structured solution patterns.

2Get comfortable with integer-only answers; practice checking your work to avoid small arithmetic errors that lead to a completely incorrect answer.

3Solve past IOQM, PRMO, and RMO question papers under timed conditions to build stamina for the 3-hour exam.

4Master LaTeX mathematical concepts like modular arithmetic equivalence and similar triangle ratios to speed up diagram and algebraic manipulations.

Frequently Asked Questions

What is the format of the IOQM exam?

IOQM is a pen-and-paper exam featuring 30 questions. The answers are integers in the range 00-99 and must be filled on an OMR sheet. There is no negative marking.

Who is eligible for IOQM?

Students in classes 8, 9, 10, 11, and 12 who are Indian citizens (or meet OCI criteria) and born on or after a specified date (usually October 30, 2006, for the 2026 cycle) are eligible.

What is the fee for registering for the IOQM?

The general registration fee is ₹240. However, Kendriya Vidyalaya (KV) students pay ₹150, and Jawahar Navodaya Vidyalaya (JNV) students pay ₹125.

What score is required to pass the IOQM?

Scoring 10% or more (10 marks out of 100) makes a student eligible for an IOQM certificate. However, qualification to the next stage (RMO) is based on regional cutoffs and ranks.