Class 9 CBSE Mathematics — Chapter 5: I'm Up and Down, and Round and Round MCQs with Answers
90 practice questions · 30 Easy · 30 Medium · 30 Hard
Practise the most important Class 9 CBSE Mathematics questions from Chapter 5, "I'm Up and Down, and Round and Round" — 9 separately timed quizzes built from 90 NCERT-aligned multiple-choice questions, with answers and explanations. The set is split into 30 Easy, 30 Medium and 30 Hard, so you can warm up on the fundamentals and then push into the exam-level problems that separate top scorers in CBSE Board exams and the JEE & NEET foundation years.
"I'm Up and Down, and Round and Round" is a chapter that rewards problem-solving speed, formula recall and step-by-step reasoning. Each MCQ on this chapter is timed and uses exam-grade marking (+4 correct, −1 wrong, 0 skipped), training the accuracy-under-pressure that real papers demand. Every question carries a short explanation, so a wrong answer becomes a quick lesson rather than a dead end — the fastest way to close gaps before a test.
Use this chapter as targeted revision: attempt the Easy set first to confirm your basics on I'm Up and Down, and Round and Round, then move to Medium and Hard to test application and problem-solving. Your accuracy, streaks and XP save automatically, and the chapter feeds into your overall Class 9 Mathematics mastery score. 14 sample questions are solved in full below — answer and worked explanation — and signing in free opens all 90.
Key concepts: I'm Up and Down, and Round and Round (Class 9 Mathematics)
Everything in a circle follows from one fact: every point on it is the same distance from the centre. From that single property come the chord theorems, the relationship between the angle at the centre and the angle at the circumference, and the rule that opposite angles of a cyclic quadrilateral are supplementary.
- Circle
- The set of all points at a fixed distance (the radius) from a fixed point (the centre). Every theorem in the chapter is a consequence of this one definition.
- Chord
- A segment joining two points on the circle. The longest possible chord is the diameter, which passes through the centre.
- Arc, sector and segment
- An arc is a piece of the circle itself; a sector is the region between two radii and an arc; a segment is the region between a chord and its arc.
- Perpendicular from centre to a chord
- The perpendicular dropped from the centre bisects the chord, and conversely the line from the centre to a chord's midpoint is perpendicular to it.
Class 9 Mathematics I'm Up and Down, and Round and Round MCQs with answers
14 solved questions across the difficulty levels this chapter is graded on — answer and explanation shown for each. The remaining 76 are timed and scored when you sign in.
- Q1Easy
The longest chord of a circle is called its:
A.sectorB.radiusC.arcD.diameter✓ CorrectAnswer: D. diameter
Explanation: The diameter is a chord that passes through the centre, and among all chords it is the longest — any chord not through the centre is shorter than it, never longer.
- Q2Easy
A circle has radius 7 cm. Its diameter is:
A.7 cmB.21 cmC.14 cm✓ CorrectD.3.5 cmAnswer: C. 14 cm
Explanation: Diameter = 2 × radius = 2 × 7 = 14 cm. A student who forgets to double the radius might pick 7 cm, mistaking the radius itself for the diameter.
- Q3Easy
The region of a circle enclosed between a chord and its corresponding arc is called a:
A.segment✓ CorrectB.quadrantC.sectorD.semicircleAnswer: A. segment
Explanation: A chord splits the circle into two regions called segments, each bounded by the chord and an arc. A sector, by contrast, is bounded by two radii and an arc, not a chord.
- Q4Easy
A line that intersects a circle in two distinct points is called a:
A.tangentB.radiusC.secant✓ CorrectD.chordAnswer: C. secant
Explanation: A secant is a line that cuts the circle at two distinct points; a tangent touches the circle at only one point, so it is not the answer here.
- Q5Easy
The region enclosed by two radii of a circle and the arc between them is called a:
A.arcB.segmentC.chordD.sector✓ CorrectAnswer: D. sector
Explanation: A sector is the region enclosed by two radii and the arc between their endpoints — like a pizza slice. A segment, by contrast, is bounded by a chord and an arc, not two radii.
- Q6Easy
The diameter of a circle is 20 cm. Its radius is:
A.40 cmB.10 cm✓ CorrectC.20 cmD.5 cmAnswer: B. 10 cm
Explanation: Radius = diameter ÷ 2 = 20 ÷ 2 = 10 cm. Halving gives 10 cm; doubling the diameter to 40 cm mixes up which operation converts diameter to radius.
- Q7Medium
O is the centre of a circle and A, B are on it with ∠AOB = 110°. If C is a point on the major arc AB, then ∠ACB is:
A.110°B.55°✓ CorrectC.250°D.70°Answer: B. 55°
Explanation: The angle at the circumference is half the central angle: ∠ACB = 110° ÷ 2 = 55°. Using the full 110° would ignore the halving rule for a point on the major arc.
- Q8Medium
In cyclic quadrilateral ABCD, ∠C is 40° more than ∠A. Then ∠A equals:
A.110°B.90°C.70°✓ CorrectD.55°Answer: C. 70°
Explanation: ∠A and ∠C are supplementary and ∠C = ∠A + 40°, so 2∠A + 40° = 180°, giving ∠A = 70°. Treating ∠A as equal to ∠C directly would miss the 40° offset.
- Q9Medium
A chord of length 16 cm lies at a distance of 6 cm from the centre of a circle. The radius of the circle is:
A.10 cm✓ CorrectB.8 cmC.12 cmD.14 cmAnswer: A. 10 cm
Explanation: Half the chord = 8 cm, so by the Pythagorean theorem radius = √(8² + 6²) = √100 = 10 cm. Adding 8 + 6 = 14 cm directly skips the required squaring step.
- Q10Medium
AB is a diameter and C is a point on the circle. If ∠BAC = 25°, then ∠ABC equals:
A.25°B.65°✓ CorrectC.90°D.115°Answer: B. 65°
Explanation: Since AB is a diameter, ∠ACB = 90° (angle in a semicircle); the triangle's angles sum to 180°, so ∠ABC = 180° − 90° − 25° = 65°, not simply 25° carried over unchanged.
- Q11Medium
O is the centre of a circle and A, B lie on it with ∠AOB = 100°. Then ∠OAB equals:
A.100°B.50°C.80°D.40°✓ CorrectAnswer: D. 40°
Explanation: Triangle OAB is isosceles since OA = OB (both radii), so the base angles are equal: ∠OAB = (180° − 100°) ÷ 2 = 40°. Treating ∠OAB as equal to the full central angle 100° ignores that it is only one base angle.
- Q12Hard
AB is a diameter of a circle and C is a point on it. The bisector of ∠ACB meets the circle again at D. Then ∠ABD equals:
A.90°B.45°✓ CorrectC.30°D.60°Answer: B. 45°
Explanation: ∠ACB = 90° (angle in a semicircle), so its bisector makes ∠ACD = 45°; since ∠ABD and ∠ACD both subtend arc AD from the same segment, ∠ABD = 45° too, not the full 90°.
- Q13Hard
ABCD is a cyclic quadrilateral with centre O and ∠BCD = 120°. The central angle ∠BOD (on the side of A) is:
A.120°✓ CorrectB.60°C.240°D.100°Answer: A. 120°
Explanation: ∠BAD = 180° − 120° = 60° (opposite angle); central angle = 2 × inscribed angle, so ∠BOD = 2 × 60° = 120°, not 240° (doubling ∠BCD directly).
- Q14Hard
In a circle with centre O, points B and C lie on it and A is on the major arc BC with ∠BAC = 40°. Then ∠OBC equals:
A.40°B.50°✓ CorrectC.80°D.70°Answer: B. 50°
Explanation: ∠BOC = 2 × ∠BAC = 80° (central angle theorem); since OB = OC, triangle OBC is isosceles, so ∠OBC = (180° − 80°) ÷ 2 = 50°, not simply half of 80° (40°).
Unlock the full chapter — free
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Start this chapter free →I'm Up and Down, and Round and Round — FAQs
What are the key concepts in Class 9 Mathematics I'm Up and Down, and Round and Round?+
Everything in a circle follows from one fact: every point on it is the same distance from the centre. From that single property come the chord theorems, the relationship between the angle at the centre and the angle at the circumference, and the rule that opposite angles of a cyclic quadrilateral are supplementary. Key ideas include Circle, Chord, Arc, sector and segment, Perpendicular from centre to a chord.
What does Class 9 Mathematics Chapter 5 (I'm Up and Down, and Round and Round) cover on XamBaaz?+
It covers 90 NCERT-aligned MCQs on "I'm Up and Down, and Round and Round" — 30 Easy, 30 Medium and 30 Hard — making 9 separately timed quizzes you can sit without ever repeating the same set, each with an instant explanation, suitable for CBSE Board exams and the JEE & NEET foundation years.
Are these "I'm Up and Down, and Round and Round" questions free to practise?+
Yes — sign in with Google to practise "I'm Up and Down, and Round and Round" free. Full unlimited access is ₹999/year (limited-time launch price), with no per-chapter charges.
How should I revise "I'm Up and Down, and Round and Round" for the exam?+
Start with the Easy quiz to confirm your fundamentals, then attempt Medium and Hard for application-level practice. There are 9 separately timed quizzes on this chapter, so you can come back and get a fresh set rather than re-sitting one you have seen. Review each explanation, retry the questions you miss, and track your accuracy on this chapter until it is consistently high.
Are these "I'm Up and Down, and Round and Round" MCQs available with answers?+
Yes. 14 sample questions are shown here in full, each with the correct option and a step-by-step "Why" explanation. Sign in free with Google to practise all 90 questions with instant scoring.
Is there negative marking in the "I'm Up and Down, and Round and Round" quizzes?+
Yes — the timed quizzes use exam-grade marking: +4 for a correct answer, −1 for a wrong one and 0 for a skipped question. Note that MHT-CET and the CBSE board papers themselves carry no negative marking — our mocks for those are scored their way, not this way.
Are these important questions for I'm Up and Down, and Round and Round?+
The set is curated to the NCERT syllabus and weighted toward the question patterns that actually appear in CBSE Board exams and the JEE & NEET foundation years, across Easy, Medium and Hard — so it doubles as an "important questions" revision list for "I'm Up and Down, and Round and Round".