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100+ Free Odisha BSE HSC Mathematics Practice Questions

Odisha HSC Class 10 Mathematics (BSE Odisha) practice questions are available now; exam metadata is being verified.

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Overall HSC pass percentages are published with each BSE Odisha result; no fixed standing subject-wise Mathematics pass-rate figure is used here. Pass Rate
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2026 Statistics

Key Facts: Odisha BSE HSC Mathematics Exam

100 marks

Maximum Score

BSE Odisha HSC Pattern of Syllabus / Question 2026

2 h 45 m

Exam Duration

BSE Odisha Pattern of Syllabus 2026

50 + 50

Objective MCQ marks + Subjective marks

BSE Odisha Pattern of Question 2026 (50 × 1 MCQ + 5 × 10 subjective)

~30/100

Typical Subject Pass Mark

Secondary sources citing BSE Odisha HSC rules (confirm current circular)

~33%

Typical Aggregate Pass Level

Secondary sources citing BSE Odisha HSC rules (confirm current circular)

MCQ study aid

Local bank is a 100-item English MCQ adaptation — not a full official paper simulation

OpenExamPrep practice disclosure

Odisha BSE HSC Mathematics is a Class 10 board paper scored out of 100 marks over 2 hours 45 minutes. Expect a mixed paper (50 MCQ marks + 50 subjective marks), not pure open study MCQ. Pass is typically about 30 marks in the subject and ~33% aggregate—confirm the latest circular. Fee is part of the annual HSC form-fill-up fee notified by BSE Odisha. This site offers free English MCQ practice with worked solutions aligned to the Class X syllabus.

Sample Odisha BSE HSC Mathematics Practice Questions

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

1Using Euclid's division algorithm, find the HCF of 196 and 38220.
A.196
B.98
C.14
D.28
Explanation: Apply Euclid's algorithm: 38220 = 196 × 195 + 0. The remainder is 0, so HCF(196, 38220) = 196. Equivalently, 196 divides 38220 exactly.
2The HCF of 96 and 404 is 4. Their LCM is:
A.9696
B.2424
C.38784
D.1616
Explanation: For two positive integers, HCF × LCM = product of the numbers. LCM = (96 × 404) / 4 = 38784 / 4 = 9696.
3If a = x³y² and b = xy³ where x and y are primes, then LCM(a, b) is:
A.xy
B.x³y³
C.x³y²
D.xy³
Explanation: LCM takes the highest power of each prime: max(3, 1) = 3 for x and max(2, 3) = 3 for y. So LCM(a, b) = x³y³.
4The decimal expansion of 17/8 terminates after how many places?
A.1
B.2
C.3
D.Does not terminate
Explanation: 17/8 = 17/(2³). Completing the denominator to powers of 10 needs 5³, so the decimal terminates after 3 places: 2.125.
5Which of the following is an irrational number?
A.√4
B.√9
C.√2
D.0.333... (3 repeating)
Explanation: √2 cannot be written as a ratio of integers and is irrational. √4 = 2 and √9 = 3 are integers; 0.333... = 1/3 is rational.
6The LCM of 12, 15 and 21 is:
A.420
B.210
C.84
D.105
Explanation: 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. LCM = 2² × 3 × 5 × 7 = 420.
7If HCF(336, 54) = 6, then LCM(336, 54) equals:
A.3024
B.2016
C.1512
D.1008
Explanation: LCM = (336 × 54) / 6 = 18144 / 6 = 3024.
8If one zero of the quadratic polynomial x² + 3x + k is 2, then k is:
A.−10
B.10
C.−2
D.2
Explanation: Substitute x = 2: 4 + 6 + k = 0 ⇒ k = −10.
9The zeros of the polynomial x² − 2x − 8 are:
A.4 and −2
B.−4 and 2
C.4 and 2
D.−4 and −2
Explanation: x² − 2x − 8 = (x − 4)(x + 2). Zeros are 4 and −2. Sum is 2 and product is −8, matching the coefficients.
10If α and β are zeros of x² + 7x + 12, then α + β equals:
A.−7
B.7
C.12
D.−12
Explanation: For ax² + bx + c, sum of zeros = −b/a. Here a = 1, b = 7, so α + β = −7.

About the Odisha BSE HSC Mathematics Practice Questions

Verified exam format metadata for Odisha HSC Class 10 Mathematics (BSE Odisha) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.