Learn

Class 10 Maths competency-based questions 2026-27: what CBSE asks, with 8 of our own

CMDLearn · 3 November 2026

A competency-based maths question gives you a situation, a table or a claim and asks you to use a method you know to decide something: which plan is cheaper, whether a ladder reaches, whether a company's claim is true. You still use the textbook's formulas. What changes is that the question does not tell you which formula to use. This guide explains what CBSE says about these questions for 2026-27, shows where they sit in the official Class 10 Maths sample paper, and then gives 8 practice questions of our own, two each from four chapters, with worked answers and the NCERT page each answer rests on.

The 8 questions are our own, drafted with AI help. Every number was re-computed by a script and every method was checked against the NCERT chapter, but no practising teacher has reviewed them yet. The data in the questions is made up for practice. If you find a mistake, please tell us.

What CBSE means by competency-based

CBSE's secondary curriculum for 2026-27 says board papers set out to test whether students can apply, analyse and evaluate, and that at present about half of the questions are competency-focused. The kinds it names include questions built on a case or a source, questions that ask you to read data, and situation or application questions, often mixing more than one topic. The rest are multiple-choice and other answer types (Secondary Curriculum 2026-27, Part 1, section 3.1, PDF page 24).

The Class 10 Maths curriculum says the same in its own words. Its opening page says the course moves away from learning by rote towards building competencies, with more weight on real-life applications (Maths curriculum 2026-27, PDF page 1). Its question paper design splits the 80 marks by the kind of thinking each question needs:

(Maths curriculum 2026-27, PDF pages 8 and 9.) So a Standard paper puts about 37 of its 80 marks on applying, analysing and evaluating. Basic puts 20 there. Competency questions are not a separate chapter. They are a way of asking about the same seven units.

Where they sit in the 2026-27 sample paper

The Standard sample paper has 38 questions in five sections (sample paper, PDF page 1). Its footnote says the question paper design and assessment pattern have not changed for 2026-27. The competency questions show up in four places:

The marking scheme gives marks step by step in the long answers and the cases, often in half marks, while Section A has no part marks (marking scheme, PDF pages 1 and 10-12). A correct method with a slip can still earn most of the marks in a case, but not in an MCQ.

How to answer them

• Turn the words into maths first. Name your unknowns ("let x be the number of meals"), then write the equation, the AP or the right triangle. Most of the marks start here.
• Answer the question that was asked. If it asks "which plan is cheaper?", the last line must name the plan, not only show two numbers.
• Check a claim with numbers. "Ravi says..." or "the company claims..." questions want a calculation, then a clear yes or no, then one line on why.
• Use exact values, then round once. Keep √3 or a fraction until the last step, then use √3 = 1.732 if the question allows it. Rounding early is the most common way to lose the last half mark.

Pair of Linear Equations in Two Variables (Chapter 3)

Question 1. A school canteen offers two monthly lunch plans. Plan A costs ₹300 for the month plus ₹20 for each meal. Plan B has no monthly charge and costs ₹35 for each meal. (a) Write the monthly cost y for x meals under each plan. (b) For how many meals do the two plans cost the same, and what is that cost? (c) Meera eats 22 school lunches in a month. Which plan should she pick, and how much does she save? (d) Her brother eats only 15 lunches. Which plan suits him? (e) If both plans are drawn as lines on one graph, what does their meeting point tell you?

Answer: (a) Plan A: y = 300 + 20x; Plan B: y = 35x; (b) 20 meals, ₹700; (c) Plan A, saving ₹30; (d) Plan B, saving ₹75; (e) the number of meals at which the two plans cost the same. (a) Each plan is a linear equation in two variables, x (meals) and y (cost in rupees). (b) Substitute y = 35x into the other equation: 35x = 300 + 20x, so 15x = 300 and x = 20. Then y = 35 × 20 = ₹700. Substitution is one of the chapter's two algebraic methods (pages 30 and 37). (c) Plan A: 300 + 20 × 22 = ₹740. Plan B: 35 × 22 = ₹770. Plan A is cheaper by ₹30. (d) Plan A: 300 + 20 × 15 = ₹600. Plan B: 35 × 15 = ₹525. Plan B is cheaper by ₹75. Below 20 meals Plan B wins; above 20 meals Plan A wins. (e) The two lines intersect at one point, (20, 700), which is the unique solution of the pair (page 37). On that graph you can read the answer to (c) and (d) without working: to the right of x = 20, the Plan A line is the lower one.

Question 2 (assertion and reason, with a follow-up). Kabir says: "At the stationery shop, 3 notebooks and 5 pens cost ₹190." His sister Tara says: "At the same prices, 6 notebooks and 10 pens cost ₹360."

Assertion (A): Kabir and Tara cannot both be right. Reason (R): For the pair 3x + 5y = 190 and 6x + 10y = 360, the ratio of the x-coefficients equals the ratio of the y-coefficients, but not the ratio of the constant terms.

Choose: (A) both true and R explains A; (B) both true but R does not explain A; (C) A true, R false; (D) A false, R true. Then answer: if Tara had said ₹380 instead, could you find the price of one notebook? Why or why not?

Answer: (A). With ₹380, no: there would be infinitely many possible prices. Write the pair as 3x + 5y − 190 = 0 and 6x + 10y − 360 = 0. Then a₁/a₂ = 3/6 = 1/2 and b₁/b₂ = 5/10 = 1/2, but c₁/c₂ = 190/360 = 19/36, which is not 1/2. When the first two ratios are equal and the third is different, the pair is inconsistent: the lines are parallel and there is no solution (page 37). No prices could make both statements true, so A is true, R is true, and R is the reason for A. Another way to see it: 6 notebooks and 10 pens are exactly twice 3 notebooks and 5 pens, so they must cost twice ₹190, which is ₹380, not ₹360. With ₹380 all three ratios would be 1/2, the lines would coincide, and the pair would be dependent, with infinitely many solutions (page 37). Tara's statement would add no new information, so one notebook's price cannot be found.

Arithmetic Progressions (Chapter 5)

Question 3. Neha is training for a 6 km school run. She runs 1.5 km on day 1 and adds 0.25 km each day. (a) How far does she run on day 15? (b) On which day does she first run 6 km? (c) What is her total distance over the first 19 days? (d) Her coach says her total for the first three weeks (21 days) must not go above 80 km. Does the plan keep to that? If not, for how many days can she follow it before the total would go over 80 km?

Answer: (a) 5 km; (b) day 19; (c) 71.25 km; (d) no, the 21-day total is 84 km; she can follow it for 20 days (77.5 km). The daily distances 1.5, 1.75, 2, ... form an AP with a = 1.5 and d = 0.25. (a) aₙ = a + (n − 1)d (page 72), so a₁₅ = 1.5 + 14 × 0.25 = 5 km. (b) 1.5 + (n − 1) × 0.25 = 6 gives (n − 1) × 0.25 = 4.5, so n − 1 = 18 and n = 19. (c) The sum of a finite AP is S = n/2 × (a + l), with l the last term (page 72): S₁₉ = 19/2 × (1.5 + 6) = 19 × 3.75 = 71.25 km. (d) Sₙ = n/2 × [2a + (n − 1)d] (page 72): S₂₁ = 21/2 × (3 + 20 × 0.25) = 21/2 × 8 = 84 km, which is more than 80. S₂₀ = 20/2 × (3 + 19 × 0.25) = 10 × 7.75 = 77.5 km, which is within the limit. So the plan works for 20 days, and day 21 would push the total over.

Question 4. For a science fair, a class builds a tower of paper cups. The bottom row has 40 cups, and each row above has 3 cups fewer than the row below it. Ravi says: "Row 10 from the bottom will have 40 − 10 × 3 = 10 cups." (a) Is Ravi right? If not, find the correct number and say what he did wrong. (b) What is the greatest number of rows the tower can have? How many cups are in its top row? (c) How many cups does that full tower need? (d) The class has only 250 cups. How many complete rows can they build from the bottom, and how many cups are left over?

Answer: (a) No, row 10 has 13 cups; (b) 14 rows, with 1 cup on top; (c) 287 cups; (d) 8 rows, with 14 cups left over. The rows form an AP with a = 40 and d = −3. (a) aₙ = a + (n − 1)d (page 72), so a₁₀ = 40 + 9 × (−3) = 13. Ravi took away 3 cups ten times, but row 1 already has 40 cups: only 9 steps of −3 separate row 1 from row 10. (b) A row needs at least 1 cup: 40 − 3(n − 1) > 0 gives n − 1 < 13.33, so n is at most 14. Row 14 has 40 − 39 = 1 cup; a row 15 would need −2 cups, which is impossible. (c) S = n/2 × (a + l) (page 72): S₁₄ = 14/2 × (40 + 1) = 7 × 41 = 287 cups. (d) Sₙ = n/2 × [80 − 3(n − 1)]. S₈ = 4 × 59 = 236 cups, and S₉ = 9/2 × 56 = 252 cups, which is more than 250. So they can build 8 complete rows and have 250 − 236 = 14 cups left, not enough for row 9, which needs 40 − 24 = 16 cups.

Some Applications of Trigonometry (Chapter 9)

Question 5. A fire engine carries a ladder 20 m long. Its foot is fixed on the truck, 2 m above the ground. A person is waiting at a window 18 m above the ground. Take √3 = 1.732 and √2 = 1.414. (a) If the ladder is raised to 60° with the horizontal, how high does its top reach above the ground? Does it reach the window? (b) How far from the wall must the foot of the ladder be for the top to rest on the wall at that angle? (c) Because of parked cars, the truck can only use an angle of 45°. Does the ladder still reach? (d) At 45°, how long would the ladder need to be?

Answer: (a) about 19.32 m, so yes; (b) 10 m; (c) no, it reaches only about 16.14 m; (d) about 22.62 m. The ladder, the horizontal through its foot and the wall form a right triangle with the ladder as the hypotenuse. (a) sin 60° = √3/2 (Table 8.1, page 125), so the rise above the foot is 20 × √3/2 = 10√3 ≈ 17.32 m. Add the 2 m of the truck: about 19.32 m, which is above 18 m. (b) cos 60° = 1/2 (page 125), so the horizontal distance is 20 × 1/2 = 10 m. (c) sin 45° = 1/√2 (page 125), so the rise is 20/√2 = 10√2 ≈ 14.14 m, and the top is at about 16.14 m, below the window. (d) The ladder must rise 18 − 2 = 16 m. At 45°, length × 1/√2 = 16, so length = 16√2 ≈ 22.62 m. Each part uses one trigonometric ratio of a right triangle, which is how the chapter finds heights and distances (pages 133 and 143).

Question 6. A mobile tower is 30 m tall and stands on level ground. Seen from a spot P on that ground, its top is at an angle of elevation of 60°. Ishaan says: "If I walk back to twice that distance, the angle of elevation will halve to 30°." (a) How far is P from the foot of the tower? (b) Find tan of the angle at twice the distance, and decide whether Ishaan is right. (c) Between which two of the angles 30°, 45° and 60° does the new angle lie? (d) At what distance is the angle of elevation exactly 30°? How many times the distance from P is that?

Answer: (a) 10√3 ≈ 17.32 m; (b) tan of the new angle is √3/2, not 1/√3, so Ishaan is wrong; (c) between 30° and 45°; (d) 30√3 ≈ 51.96 m, three times the distance from P. Let d be the distance from the foot. (a) tan 60° = 30/d and tan 60° = √3 (page 125), so d = 30/√3 = 10√3 ≈ 17.32 m. (b) At 20√3 m, tan θ = 30/(20√3) = √3/2 ≈ 0.866. But tan 30° = 1/√3 ≈ 0.577 (page 125), so the angle is not 30°. Doubling the distance halves tan θ, not the angle itself. (c) tan 30° ≈ 0.577 and tan 45° = 1, and 0.866 lies between them, so θ is between 30° and 45° (tan grows as the angle grows from 0° to 90°). (d) tan 30° = 30/d gives d = 30√3 ≈ 51.96 m, which is 3 × 10√3: three times as far, not twice. The height of an object or the distance to it can be found from one known side and one angle in a right triangle (page 143).

Statistics (Chapter 13)

Question 7. A food delivery app recorded how long its last 50 orders in one town took to arrive (made-up data).

The app's advert makes two claims: "Our average delivery takes under 35 minutes" and "Half of all orders arrive within 40 minutes." (a) Find the mean delivery time. (b) Find the median. (c) Find the mode. (d) Check each claim against the data.

Answer: (a) 39.8 minutes; (b) 40 minutes; (c) about 40.56 minutes; (d) the first claim is false, the second is true for this data. (a) Class marks are 20, 30, 40, 50 and 60. Σfᵢxᵢ = 120 + 330 + 640 + 600 + 300 = 1,990 and Σfᵢ = 50, so the mean is 1,990 ÷ 50 = 39.8 minutes (direct method, page 200). (b) n/2 = 25. Cumulative frequencies: 6, 17, 33, 45, 50. 25 falls in the 35-45 class: the first cumulative frequency above 25 is 33 (page 193). Median = l + [(n/2 − cf)/f] × h = 35 + [(25 − 17)/16] × 10 = 35 + 5 = 40 minutes (page 201). (c) The modal class is 35-45 (16 orders). Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 35 + [(16 − 11)/(32 − 11 − 12)] × 10 = 35 + 50/9 ≈ 40.56 minutes (page 201). (d) The mean, 39.8 minutes, is not under 35, so the first claim is false. The median, 40 minutes, means about half the orders took less than 40 minutes and half took more (the chapter reads a median this way on page 190), so the second claim holds for this data.

Question 8. A walking club recorded the average number of steps each of its 40 members walked per day in a month (made-up data). Work in thousands of steps.

(a) Find the mean by the step-deviation method. (b) Find the median. (c) Find the mode. (d) The club's newsletter says: "At least half our members walk 9,000 steps or more a day." Is that true? (e) The club wants one number to describe its typical member. Which of the three would you report, and why?

Answer: (a) 8,550 steps; (b) 8,800 steps; (c) about 9,571 steps; (d) no; (e) the median, explained below. (a) Class marks: 4, 6, 8, 10, 12. Take a = 8 and h = 2, so u = (x − 8)/2 is −2, −1, 0, 1, 2. Σfᵢuᵢ = −8 − 7 + 0 + 12 + 14 = 11. Mean = a + h × Σfᵢuᵢ/Σfᵢ = 8 + 2 × 11/40 = 8.55 thousand (page 201). (b) n/2 = 20. Cumulative frequencies: 4, 11, 21, 33, 40, so the median class is 7,000-9,000 (page 193). Median = 7 + [(20 − 11)/10] × 2 = 8.8 thousand (page 201). (c) The modal class is 9,000-11,000. Mode = 9 + [(12 − 10)/(24 − 10 − 7)] × 2 = 9 + 4/7 ≈ 9.571 thousand (page 201). (d) The median, 8,800, is below 9,000. Directly from the table, 4 + 7 + 10 = 21 of the 40 members walk under 9,000 steps, so fewer than half walk 9,000 or more. The claim is false. (e) The median: it is the middle member's value, so half the club walks less and half more. The mode (about 9,571) describes only the most crowded class, and the mean is pulled down by the few members in the 3,000-5,000 class. The chapter says the choice of measure depends on what the situation needs (pages 185-186); for "a typical member" the middle value is the fairest single number here.

Practise more, and check your marks

• The official paper, section by section: our CBSE Class 10 Maths sample paper 2026-27 guide. What is and is not examined: the Class 10 Maths syllabus 2026-27 guide.
• More questions with marks labels, chapter by chapter: Class 10 Maths important questions 2026-27, and the formulas in one place: Class 10 Maths formula sheet 2026-27.
• Practice for the four chapters here: Pair of Linear Equations, Arithmetic Progressions, Some Applications of Trigonometry and Statistics.
• The same kind of questions for Science: Class 10 Science competency-based questions 2026-27.

Where CMDLearn fits

CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.

Sources

All opened and read on 3 November 2026. CBSE page numbers are PDF pages; NCERT page numbers are the printed page numbers of the textbook.

• CBSE, Secondary School Curriculum 2026-27, Part 1, Introduction, section 3.1 "Nature of Question Papers" (about 50% competency-focused questions; the types named), PDF page 24: cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/Curriculum_SecP1_2026-27.pdf
• CBSE, Mathematics curriculum, Class X, 2026-27 (codes 041 and 241): PDF page 1 (the shift to competency-based education), PDF page 8 (Standard question paper design: 43, 19 and 18 marks), PDF page 9 (Basic: 60, 12 and 8 marks): cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/Maths_SecP1X_2026-27.pdf
• CBSE, Mathematics Standard (041) Sample Question Paper, Class X, 2026-27: general instructions and Q3 (PDF page 1), Q19-Q20 (page 2), Q32 and Q35 (page 4), Q36-Q38 (pages 5-6), footnote on every page: cbseacademic.nic.in/web_material/SQP/ClassX_2026_27/MathsStandard-SQP.pdf
• CBSE, Mathematics Standard (041) Marking Scheme, Class X, 2026-27: Section A rule (PDF page 1), Q36-Q38 (pages 10-12): cbseacademic.nic.in/web_material/SQP/ClassX_2026_27/MathsStandard-MS.pdf
• NCERT, Mathematics, Textbook for Class X (Reprint 2026-27), Chapter 3 "Pair of Linear Equations in Two Variables": algebraic methods, page 30; summary of the graphical meaning and the ratio conditions, page 37: ncert.nic.in/textbook/pdf/jemh103.pdf
• NCERT, Mathematics, Textbook for Class X, Chapter 5 "Arithmetic Progressions": summary (nth term, sum of n terms, sum with last term), page 72: ncert.nic.in/textbook/pdf/jemh105.pdf
• NCERT, Mathematics, Textbook for Class X, Chapter 8 "Introduction to Trigonometry": Table 8.1, values of the ratios at 0°, 30°, 45°, 60° and 90°, page 125: ncert.nic.in/textbook/pdf/jemh108.pdf
• NCERT, Mathematics, Textbook for Class X, Chapter 9 "Some Applications of Trigonometry": angle of elevation, page 133; summary, page 143: ncert.nic.in/textbook/pdf/jemh109.pdf
• NCERT, Mathematics, Textbook for Class X, Chapter 13 "Statistics": choosing a measure, pages 185-186; what a median tells you, page 190; median class, page 193; summary (mean by the three methods, mode, median), pages 200-201: ncert.nic.in/textbook/pdf/jemh113.pdf

The board-pattern facts are from CBSE's documents above, and the method in each answer is from the textbook pages named. The situations, data, questions and worked answers are our own, and every number was re-computed by script. This guide is general information; follow your teacher, the textbook and CBSE's own notices.

Practise this class: Class 10 worksheets, chapter by chapter · free practice, marked instantly · timed revision papers