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Class 10 maths is one paper, but its five sections reward different habits. A student who treats a 1-mark question like a 5-mark proof runs out of time; one who treats a 5-mark question like a quick MCQ loses step marks. This post goes through the 2026-27 CBSE Mathematics Standard sample question paper section by section, with worked answers from its questions and a time plan you can test at home.
How is the paper divided?
The sample paper's general instructions set out 38 compulsory questions in five sections, for 80 marks in 3 hours:
Section A: 18 multiple-choice and 2 assertion-reason questions, 1 mark each (20 marks).
Section B: 5 very short answer questions, 2 marks each (10 marks).
Section C: 6 short answer questions, 3 marks each (18 marks).
Section D: 4 long answer questions, 5 marks each (20 marks).
Section E: 3 case-study questions, 4 marks each, with sub-parts of 1, 1 and 2 marks (12 marks).
There is no overall choice, but there is internal choice in 2 questions each in Sections B, C and D, and in the 2-mark sub-part of every case study. Calculators are not allowed, and you take π = 22/7 unless told otherwise. Each page notes that there is no change in the question paper design for 2026-27.
Which chapters carry the marks, and how is Basic different?
The 2026-27 maths curriculum gives the unit marks: Number Systems 6, Algebra 20, Coordinate Geometry 6, Geometry 15, Trigonometry 12, Mensuration 10, and Statistics and Probability 11.
The same document shows how Standard and Basic differ in what they test. In Standard, about 43 marks (54%) are for remembering and understanding, 19 (24%) for applying, and 18 (22%) for analysing, evaluating and creating. In Basic, the split is 60 (75%), 12 (15%) and 8 (10%). So Basic puts far fewer marks on questions of the analyse-and-create kind. If you are still choosing, read our guide to Maths Standard or Basic.
Section A: how do you avoid traps in 1-mark questions?
Section A rewards a quick, careful set-up. Q1 asks for the greatest number that divides 134 and 188 leaving remainders 4 and 6.
Remove the remainders: 134 − 4 = 130 and 188 − 6 = 182.
Factorise: 130 = 2 × 5 × 13 and 182 = 2 × 7 × 13.
HCF = 2 × 13 = 26, option (B).
The trap is taking the HCF of 134 and 188 directly, which gives 2. For assertion-reason questions, judge each statement separately before you ask whether the reason explains the assertion.
Sections B to D: how do you choose between internal options?
Spend 30 seconds on each pair before you write. Pick the option where you can see the whole method, not the one that looks shorter. A good check is whether you can say the first two lines out loud.
Q30 in Section C is a good model of a full 3-mark answer. In what ratio does the x-axis divide the segment joining (−4, −6) and (−1, 7)?
Let the ratio be k : 1. A point on the x-axis has y = 0.
Section formula for y: (7k − 6)/(k + 1) = 0, so 7k = 6 and k = 6/7. The ratio is 6 : 7.
x = (6 × (−1) + 7 × (−4))/(6 + 7) = (−6 − 28)/13 = −34/13. The point is (−34/13, 0).
Common mistakes: putting the ratio on the wrong point, and writing −3² when you mean (−3)². Our post on Class 10 maths board answer mistakes has more on how step marks are given in Sections B to D.
Section E: how should you read a case study?
Each case study has three sub-parts of 1, 1 and 2 marks. The first two are usually set-up questions, and the 2-mark part uses them. So answer (i) and (ii) carefully even if they feel easy.
Q36 is a sanitizer company. Small bottles cost ₹10 and large ₹15; in June it sold 1000 bottles for ₹12,750.
(i) The sale for June: 10x + 15y = 12,750, where x and y are the numbers of small and large bottles.
(ii) In July both prices rise by ₹2, so the sale for July is 12x + 17y = 34,250.
(iii)(A) Now bring in the bottle count, x + y = 1000. Multiply by 10 to get 10x + 10y = 10,000, and subtract it from 10x + 15y = 12,750: 5y = 2,750, so y = 550 and x = 450. In June it sold 450 small and 550 large bottles.
Notice the trap in (i): the question asks for the sale, so the answer is the rupee equation, not x + y = 1000. The marking scheme gives the mark for 10x + 15y = 12,750. For (iii)(B), July had 2500 bottles: 12x + 12y = 30,000, so 5y = 4,250, y = 850 and x = 1,650. Check: 12 × 1,650 + 17 × 850 = 19,800 + 14,450 = 34,250.
Q37 is a circular garden of radius 9 m with P at 18 m from the centre. The radius meets the tangent at 90°, so ∠OAP = 90°. Then PA = √(18² − 9²) = √243 = 9√3 m. Since sin ∠APO = 9/18, ∠APO = 30°, the tangents meet at 60°, and ∠AOB = 180° − 60° = 120°. See our guide to circles and tangents for the theorems behind this.
How should you spend the 3 hours?
With 80 marks in 180 minutes, budget about 2 minutes per mark, and test this plan on a timed paper:
Reading and choosing, 5 minutes.
Section A, 25 minutes for 20 marks. Mark any doubtful question and move on.
Section B, 15 minutes and Section C, 30 minutes.
Section D, 45 minutes for 20 marks. Draw the figure first.
Section E, 25 minutes for 12 marks.
Checking, 35 minutes: marked questions first, then signs and units.
Some students prefer to do Section E before Section D while their reading is fresh. Try both orders in practice.
A weekly practice routine
Do one section of the sample paper on each of five days, timed to the plan above, and mark it with the official marking scheme.
Keep a list of marks lost by section. If Section A costs you more than it should, practise traps; if Section D does, practise full layouts.
On day six, mix five questions from five different chapters, without looking at chapter names.
If you cannot see where a method went wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Print the sample paper today and time yourself on Section E alone.
