Half of the CBSE Class 10 paper is competency-based, and the hardest part usually isn't the arithmetic — it's recognising which mathematical tool a real situation calls for when nobody tells you the chapter. This set contains 15 scenarios and 50 linked sub-questions, deliberately spread across the chapters that competency questions draw from most, and every numerical answer here was computed and independently verified before publishing.
How to use this page: Read the full scenario first, then attempt all its sub-questions on paper before revealing any answer — the sub-questions are designed to build on each other, so checking one early gives away the next.
For the format itself — what competency questions are and the five types CBSE uses — see our Competency-Based Questions guide. This page is the practice that follows from it.
Chapters Covered
tanks, wells, combined solids
mean, median, mode from grouped data
real sampling situations
maps, layouts, section formula
tangents in design contexts
similarity for indirect measurement
HCF and LCM in scheduling and tiling
surveying, savings patterns
Case 1 · Surface Areas & Volumes
The Village Water Tank
(i) Calculate the volume of the tank in cubic metres.
Reveal Answer
Formula: V = πr²h = (22/7) × (1.4)² × 2.5
(22/7) × 1.96 = 6.16; then 6.16 × 2.5 = 15.4
Answer: ✓ Volume = 15.4 m³
(ii) Express the tank's capacity in litres.
Reveal Answer
1 m³ = 1000 litres, so 15.4 × 1000
Answer: ✓ Capacity = 15,400 litres
(iii) Find the cost of painting the curved surface at ₹50 per m².
Reveal Answer
Formula: CSA = 2πrh = 2 × (22/7) × 1.4 × 2.5 = 22 m²
Answer: ✓ Cost = 22 × 50 = ₹1,100
Note that only the curved surface is painted here — a common trap is including the base or top when the scenario doesn't call for it.
(iv) If each household needs 70 litres per day, how many households can the full tank serve for one day?
Reveal Answer
15,400 ÷ 70 = 220
Answer: ✓ 220 households
Case 2 · Combination of Solids
Designing a Wooden Toy
(i) Find the slant height of the conical part.
Reveal Answer
Formula: l = √(h² + r²) = √(144 + 12.25) = √156.25
Answer: ✓ Slant height = 12.5 cm
(ii) Calculate the total surface area to be polished.
Reveal Answer
Curved surface of cone = πrl = (22/7)(3.5)(12.5) = 137.5 cm²
Curved surface of hemisphere = 2πr² = 2(22/7)(12.25) = 77 cm²
Answer: ✓ Total polished area = 137.5 + 77 = 214.5 cm²
The flat circular face where the cone meets the hemisphere is hidden inside the toy, so it is never included in the exposed surface area.
(iii) Find the volume of the conical part alone.
Reveal Answer
Formula: V = (1/3)πr²h = (1/3)(22/7)(12.25)(12)
Answer: ✓ Volume of cone = 154 cm³
Case 3 · Statistics
A School Screen-Time Survey
(i) Calculate the mean daily screen time.
Reveal Answer
Class midpoints: 1, 3, 5, 7. Products fx: 8, 45, 90, 63 → Σfx = 206
Formula: Mean = Σfx / Σf = 206 / 50
Answer: ✓ Mean = 4.12 hours
(ii) Find the median class and calculate the median.
Reveal Answer
Cumulative frequencies: 8, 23, 41, 50. Since n/2 = 25, the median class is 4–6.
Formula: Median = l + [(n/2 − cf)/f] × h = 4 + [(25 − 23)/18] × 2
Answer: ✓ Median ≈ 4.22 hours
(iii) Calculate the mode.
Reveal Answer
Modal class is 4–6 (highest frequency, 18). Here f₁=18, f₀=15, f₂=9.
Formula: Mode = 4 + [(18−15)/(36−15−9)] × 2 = 4 + (3/12)×2
Answer: ✓ Mode = 4.5 hours
(iv) The school wants to report a single "typical" screen time to parents. Which measure would you recommend, and why?
Reveal Answer
Answer: ✓ All three measures here fall close together (4.12, 4.22, 4.5), which indicates the data is fairly symmetric and any of them would represent it reasonably. The mean is the most commonly reported and uses every observation, making it a defensible choice — though mentioning that all three agree closely is itself a strong observation, since it shows the summary isn't being distorted by extreme values.
Questions like this have no single "correct" number — marks come from justifying your choice with reference to the actual data, not from naming a measure.
Case 4 · Probability
The School Lucky Draw
(i) Find the probability of drawing a red ball.
Reveal Answer
Total balls = 5 + 8 + 7 = 20
Answer: ✓ P(red) = 5/20 = 1/4
(ii) Find the probability that the ball drawn is not green.
Reveal Answer
Either count the non-green balls (5 + 8 = 13), or use P(not green) = 1 − P(green) = 1 − 7/20
Answer: ✓ P(not green) = 13/20
(iii) A student claims that since there are three colours, the probability of each colour is 1/3. Is this correct? Justify.
Reveal Answer
Answer: ✓ No, this is incorrect. Probability is only equally divided when all outcomes are equally likely, which requires an equal number of balls of each colour. Here the counts differ (5, 8, 7), so the probabilities differ accordingly: 1/4, 2/5, and 7/20 respectively. The number of categories does not determine probability — the number of favourable outcomes does.
Case 5 · Coordinate Geometry
Mapping a Neighbourhood
(i) Find the distance from the school to the library.
Reveal Answer
Formula: d = √[(6−2)² + (6−3)²] = √(16 + 9) = √25
Answer: ✓ Distance = 5 km
(ii) Find the distances home-to-school and home-to-library.
Reveal Answer
Home to school: √[(2−2)² + (6−3)²] = √9 = 3 km
Home to library: √[(6−2)² + (6−6)²] = √16 = 4 km
Answer: ✓ 3 km and 4 km respectively
(iii) Show that these three locations form a right-angled triangle, and identify where the right angle is.
Reveal Answer
Check the converse of Pythagoras: 3² + 4² = 9 + 16 = 25, and 5² = 25.
Answer: ✓ Since HS² + HL² = SL², the triangle is right-angled at H (the home) — the two shorter sides meet there.
The right angle is always at the vertex between the two shorter sides, not at the end of the longest side.
Case 6 · Coordinate Geometry — Section Formula
Placing a Streetlight
(i) Find the coordinates of the point dividing AB in the ratio 1 : 2 (measured from A).
Reveal Answer
Formula: x = (1×8 + 2×2)/(1+2) = 12/3 = 4; y = (1×12 + 2×3)/3 = 18/3 = 6
Answer: ✓ The point is (4, 6)
(ii) Find the midpoint of AB.
Reveal Answer
Formula: Midpoint = ((2+8)/2, (3+12)/2)
Answer: ✓ Midpoint = (5, 7.5)
(iii) If a second streetlight is placed at the midpoint, which of the two lights is closer to junction A? Explain without calculating distances.
Reveal Answer
Answer: ✓ The point dividing AB in the ratio 1:2 is closer to A. A ratio of 1:2 means the point sits one-third of the way along AB from A, whereas the midpoint sits one-half of the way along. Since one-third is less than one-half, the first light is nearer to A.
Case 7 · Circles — Tangents
A Circular Park's Boundary Path
(i) What is the geometrical name for the walkway PT, and what is the angle between OT and PT (where O is the centre)?
Reveal Answer
Answer: ✓ PT is a tangent to the circle. The radius drawn to the point of contact is always perpendicular to the tangent, so ∠OTP = 90°.
(ii) Calculate the length of the walkway PT.
Reveal Answer
Since ∠OTP = 90°, triangle OTP is right-angled at T.
Formula: PT = √(OP² − OT²) = √(13² − 5²) = √(169 − 25) = √144
Answer: ✓ PT = 12 m
(iii) If a second walkway is built from P touching the circle at another point T′, what can you say about its length, and why?
Reveal Answer
Answer: ✓ It will also be 12 m long. The two tangents drawn from an external point to a circle are always equal in length, because both form congruent right triangles with the same hypotenuse OP and the same radius.
Case 8 · Triangles — Similarity
Measuring a Tower Without Climbing It
(i) Explain why the pole and the tower form similar triangles with their shadows.
Reveal Answer
Answer: ✓ Both the pole and the tower stand vertically, so each makes a 90° angle with the ground. Since the measurements are taken at the same moment, the sun's rays strike both at the same angle of elevation. With two pairs of equal angles, the triangles are similar by the AA (angle-angle) criterion.
(ii) Calculate the height of the tower.
Reveal Answer
Formula: Corresponding sides are proportional: 6/4 = h/28
h = (6 × 28)/4 = 168/4
Answer: ✓ Tower height = 42 m
(iii) Why is it essential that both shadows are measured at the same time of day?
Reveal Answer
Answer: ✓ Because the sun's angle of elevation changes continuously through the day, which changes shadow lengths. If the two measurements were taken at different times, the angles would differ, the triangles would no longer be similar, and the proportion used to find the height would be invalid.
Case 9 · Real Numbers — HCF & LCM
Temple Bells and a Tiled Floor
(i) At what time will the three bells next ring together?
Reveal Answer
This requires the LCM of 9, 12, and 15.
Formula: 9 = 3², 12 = 2²×3, 15 = 3×5 → LCM = 2² × 3² × 5 = 180 minutes
Answer: ✓ 180 minutes = 3 hours, so they ring together again at 9:00 a.m.
Recurring-event questions use LCM; questions about dividing into largest equal groups use HCF. Identifying which one the scenario calls for is exactly what this question type tests.
(ii) Find the largest possible side length of a square tile that covers the floor exactly.
Reveal Answer
This requires the HCF of 525 and 450.
Formula: 525 = 3 × 5² × 7, 450 = 2 × 3² × 5² → HCF = 3 × 5² = 75
Answer: ✓ Largest tile side = 75 cm
(iii) How many such tiles are needed?
Reveal Answer
Formula: Number of tiles = (525 × 450) / (75 × 75) = 236250 / 5625
Answer: ✓ 42 tiles
Case 10 · Surface Areas & Volumes
Plastering a Village Well
(i) Calculate the inner curved surface area to be plastered.
Reveal Answer
Diameter 4 m means radius r = 2 m.
Formula: CSA = 2πrh = 2 × (22/7) × 2 × 14
Answer: ✓ Area = 176 m²
(ii) Find the cost of plastering at ₹40 per m².
Reveal Answer
Answer: ✓ Cost = 176 × 40 = ₹7,040
(iii) Why is only the curved surface plastered, and not the base?
Reveal Answer
Answer: ✓ The base of a well is the water-bearing floor, which is left unplastered so groundwater can seep in. Plastering the base would seal the well off from its water source, defeating its purpose. The curved walls are plastered to prevent soil erosion and collapse.
This is the kind of sub-question that carries no calculation at all — it tests whether you understand what the mathematics is describing.
Case 11 · Trigonometry
Drone Survey of a Field
(i) Find the horizontal distance from the point directly below the drone to the farther marker.
Reveal Answer
A smaller angle of depression corresponds to the farther object, so the 30° sighting is the farther marker.
Formula: tan 30° = 60/d → d = 60/tan30° = 60√3
Answer: ✓ Distance ≈ 103.92 m
(ii) Find the horizontal distance to the nearer marker.
Reveal Answer
Formula: tan 60° = 60/d → d = 60/√3 = 20√3
Answer: ✓ Distance ≈ 34.64 m
(iii) Calculate the distance between the two markers.
Reveal Answer
Formula: 60√3 − 20√3 = 40√3
Answer: ✓ Distance between markers = 40√3 ≈ 69.28 m
Case 12 · Arithmetic Progressions
A Monthly Savings Plan
(i) Explain why this savings pattern forms an arithmetic progression, and state its first term and common difference.
Reveal Answer
Answer: ✓ The amount saved increases by a constant ₹50 each month, and a sequence in which consecutive terms differ by a fixed amount is by definition an AP. Here a = 200 and d = 50.
(ii) How much is saved in the 12th month?
Reveal Answer
Formula: aₙ = a + (n−1)d = 200 + 11 × 50
Answer: ✓ ₹750 in the 12th month
(iii) Find the total amount saved over the first 12 months.
Reveal Answer
Formula: Sₙ = n/2 × (first + last) = 12/2 × (200 + 750) = 6 × 950
Answer: ✓ Total saved = ₹5,700
When you already know the last term, S = n/2 × (first + last) is faster than the full formula — and less error-prone.
Case 13 · Probability
The Numbered Card Game
(i) Find the probability that the number drawn is a multiple of 5.
Reveal Answer
Multiples of 5: 5, 10, 15, 20, 25 — that's 5 outcomes.
Answer: ✓ P = 5/25 = 1/5
(ii) Find the probability that the number drawn is prime.
Reveal Answer
Primes up to 25: 2, 3, 5, 7, 11, 13, 17, 19, 23 — that's 9 outcomes. (Note that 1 is not prime.)
Answer: ✓ P = 9/25
Forgetting that 1 is not a prime number is one of the most common errors in this exact question type.
(iii) Find the probability that the number drawn is not a perfect square.
Reveal Answer
Perfect squares up to 25: 1, 4, 9, 16, 25 — that's 5 outcomes, so P(perfect square) = 5/25 = 1/5.
Formula: P(not a perfect square) = 1 − 1/5
Answer: ✓ P = 4/5
Case 14 · Statistics
Comparing Two Class Sections
(i) Calculate the mean mark for Section A.
Reveal Answer
Midpoints 5, 15, 25, 35, 45. Σfx = 20 + 90 + 250 + 420 + 360 = 1140
Formula: Mean = 1140/40
Answer: ✓ Mean (Section A) = 28.5 marks
(ii) Calculate the mean mark for Section B.
Reveal Answer
Σfx = 10 + 120 + 350 + 350 + 270 = 1100
Formula: Mean = 1100/40
Answer: ✓ Mean (Section B) = 27.5 marks
(iii) Find the modal class for Section A and calculate its mode.
Reveal Answer
Highest frequency in A is 12, so the modal class is 30–40. Here f₁=12, f₀=10, f₂=8.
Formula: Mode = 30 + [(12−10)/(24−10−8)] × 10 = 30 + (2/6)×10
Answer: ✓ Mode ≈ 33.33 marks
(iv) A teacher concludes that "Section A is clearly the stronger section." Evaluate whether the data fully supports this claim.
Reveal Answer
Answer: ✓ The data only partially supports it. Section A's mean is higher, but only by 1 mark out of 50 — a small difference given 40 students per section. Section A also has more students in the lowest band (4 versus 2), meaning its marks are more spread out, while Section B's scores cluster more tightly around the middle. A one-mark difference in means is too slim to describe either section as "clearly" stronger without also considering the spread of the data.
Evaluative sub-questions reward careful hedging backed by specific figures. An answer that simply agrees or disagrees without citing the data earns far less than one that quantifies why the difference is or isn't meaningful.
Case 15 · Surface Areas & Volumes
A Conical Grain Heap
(i) Find the volume of grain in the heap.
Reveal Answer
Formula: V = (1/3)πr²h = (1/3) × (22/7) × 12.25 × 12
Answer: ✓ Volume = 154 m³
(ii) Calculate the area of canvas needed to cover the heap.
Reveal Answer
First find the slant height: l = √(12² + 3.5²) = √156.25 = 12.5 m
Formula: Canvas area = curved surface area = πrl = (22/7)(3.5)(12.5)
Answer: ✓ Canvas needed = 137.5 m²
The base of the heap rests on the ground and needs no canvas — so the curved surface area, not the total surface area, is what the situation calls for.
(iii) A supplier sells canvas only in whole square metres. How much canvas should be purchased, and why might even this be insufficient in practice?
Reveal Answer
Answer: ✓ Rounding up to whole square metres gives 138 m². In practice even this may be insufficient, because a canvas sheet needs overlap at seams and extra material to be secured at the base — the calculated surface area is a mathematical minimum, not a practical purchase quantity.
Modelling, Solving, and Interpreting Are Three Separate Skills
Across these 15 scenarios, notice how different the sub-questions are from each other. Some ask you to set up the mathematics, some to compute, and several — like the well's base, the teacher's claim, or the canvas purchase — involve no calculation at all. A student strong in one of these can be weak in another, and a single chapter score hides that completely.
What a Genelis weak area map looks like after competency practice
Next session: interpretation and evaluation (28%) — the sub-questions with no calculation at all, and the ones most often left blank. Genelis tracks these as distinct skills rather than one combined topic score.
Genelis is an AI-powered personalized learning platform built on Adaptive Personalized Intelligence. The Genelis learning system generates fresh competency scenarios across every chapter and tracks modelling, computation, and interpretation separately — so a strong calculation score never hides a weakness in reading the situation correctly. Every wrong answer is logged to your wrong-question notebook for reattempt.
Learn smarter. Practice deeper. Improve continuously.
Genelis combines Adaptive Personalized Intelligence, AI-generated notes, targeted practice, mock tests, analytics, and personalised revision to help students improve every study session.
Questions Students Commonly Ask
Quick answers to the most common questions related to this guide.
How is a competency-based Maths question different from a normal word problem?
A normal word problem usually asks for one final answer from one given situation. A competency-based question presents a single realistic scenario followed by two to four linked sub-questions that build on each other — typically asking you to first model the situation mathematically, then solve it, and then apply the result to something further such as a cost, a comparison, or a judgement about whether the answer is reasonable. The mathematics itself is often not harder; what is tested is whether you can translate a real situation into the right mathematical tool without being told which one to use.
Which Class 10 Maths chapters appear most often in competency-based questions?
Surface Areas and Volumes appears very frequently because real objects naturally combine shapes and lead to cost calculations. Statistics and Probability lend themselves to survey and data scenarios. Coordinate Geometry appears through maps and layouts, Trigonometry through height and distance measurement, Real Numbers through scheduling and tiling problems, and Arithmetic Progressions through savings, seating and production patterns. Triangles and Circles appear through similarity-based measurement and tangent-based design situations.
Why do competency questions often ask whether an answer is reasonable?
Because interpreting a result in context is a distinct skill from calculating it. A calculation can be arithmetically perfect and still produce an answer that makes no sense in the real situation — a negative length, an impossible age, or a tank capacity far beyond what the described container could hold. Questions that ask you to comment on reasonableness are testing whether you actually understand what your number represents, rather than whether you can follow a formula.