The UP Board Class 10 Mathematics 2026 question paper, code 822(FX), had a 20-mark multiple-choice section and a 50-mark descriptive section. The paper covered algebra, geometry, trigonometry, mensuration, statistics and probability. Below are selected questions with worked answers, an explanation of the section choices and a link to the complete question paper PDF and available solutions.

This article covers the English version of paper 822(FX). Students comparing answers should match the paper code and the question itself, since another set may use a different order. The examination was held on 27 February 2026.

UP Board Class 10 Mathematics 2026 paper pattern

The five main questions in Section B contained several parts. Its first question required all the parts, while the second asked students to attempt any five parts. The remaining questions offered alternatives marked OR. These instructions explain why the paper has more individual problems than its count of main questions suggests.

Questions, Answers and Section Analysis

Section A: short calculations and concept checks

The objective section included LCM and HCF, zeroes of a polynomial, quadratic roots, arithmetic progressions and coordinate geometry. Trigonometry questions tested standard angle values and ratios in a right triangle. Probability, grouped data and volumes of combined solids also appeared.

Some answers needed only one calculation, but the choice of formula still mattered. For example, the intersection of a parallelogram's diagonals is found with the midpoint formula. For a combined solid, the volumes of both parts must be included. The data questions also distinguished the median class from the modal class.

Section B: proofs, working and word problems

The descriptive section included an irrationality proof, coordinate calculations, a trigonometric identity, an arc-length problem and grouped-data calculations. Other parts asked for geometry proofs, probabilities, the sum of an arithmetic progression and the solution of simultaneous equations.

The longer problems turned written information into equations or a diagram. They included a two-digit number, a right triangle, heights and distances, circle segments and a combined cone-and-cylinder solid. Showing how the given measurements connect is part of explaining these answers; writing only the final number leaves out the method.

Some of the Questions of the Paper

The questions below are shortened without changing their meaning. References follow the sections and question labels in the linked paper.

Intersection of a parallelogram's diagonals (Section A, Question 6)

Question: A parallelogram has vertices A(6, 1), B(8, 2), C(9, 4) and D(7, 3), taken in order. Where do its diagonals meet?

Answer: (15/2, 5/2).

The diagonals of a parallelogram bisect each other. Their intersection is therefore the midpoint of AC:

((6 + 9)/2, (1 + 4)/2) = (15/2, 5/2).

The midpoint of BD gives the same result: ((8 + 7)/2, (2 + 3)/2). This provides a quick check using the other diagonal. Finding the length of a diagonal would not answer the question, which asks for a point.

Finding a trigonometric ratio (Section A, Question 9)

Question: In triangle ABC, angle B is 90 degrees, AB is 5 cm and BC is 12 cm. Find sin A.

Answer: 12/13.

AC is the hypotenuse because it is opposite the right angle. By Pythagoras' theorem, AC² = 5² + 12² = 169, so AC = 13 cm. The side opposite angle A is BC. Thus, sin A = BC/AC = 12/13.

The fraction 5/13 uses the side adjacent to A and gives cos A instead. Labelling the sides from the angle named in the question prevents this mix-up.

Volume of a cone on a hemisphere (Section A, Question 18)

Question: A cone stands on a hemisphere. Both have radius 2 cm, and the cone's height equals its radius. Find the total volume.

Answer: 8π cm³.

The cone has radius 2 cm and height 2 cm. Its volume is (1/3)πr²h = 8π/3 cm³. The hemisphere's volume is (2/3)πr³ = 16π/3 cm³.

Adding them gives (8π + 16π)/3 = 8π cm³. The shared circular base is a surface, not an overlapping volume, so nothing needs to be subtracted. This differs from finding the exposed surface area of the same solid.

Finding the median class (Section A, Question 19)

Question: The class intervals are 0-5, 5-10, 10-15, 15-20, 20-25 and 25-30. Their frequencies are 2, 8, 10, 5, 4 and 1. Find the median class.

Answer: 10-15.

The total frequency is 30, so N/2 = 15. The cumulative frequencies are 2, 10, 20, 25, 29 and 30. The first cumulative frequency greater than 15 is 20, corresponding to the class 10-15.

The question asks for the class containing the median, not the numerical median. The modal class is found by the highest frequency. Both happen to be 10-15 in this example, but the two methods are different and need not give the same class in another table.

Proving a number is irrational (Section B, Question 1(a))

Question: Prove that 2√2 is irrational.

Answer: A rational value of 2√2 would imply that √2 is rational, which leads to a contradiction.

Suppose √2 = p/q, where p and q are integers with no common factor and q is not zero. Squaring gives p² = 2q². Thus p², and hence p, is even. Write p = 2k. Substituting gives q² = 2k², so q is also even.

This contradicts the choice of p and q with no common factor. Therefore √2 is irrational. If 2√2 were rational, dividing it by the non-zero rational number 2 would make √2 rational. Hence 2√2 is irrational too.

Throwing a die twice (Section B, Question 2(c))

Question: A die is thrown twice. Find the probability that six appears neither time, and the probability that it appears at least once.

Answer: 25/36 and 11/36, respectively.

For a fair die, the probability of not getting six on one throw is 5/6. The throws are independent, so the probability of no six in both throws is (5/6) × (5/6) = 25/36.

The event "at least one six" is the complement of "no six". Its probability is therefore 1 - 25/36 = 11/36. It includes a six on the first throw, the second throw or both. Simply adding 1/6 and 1/6 counts the double-six outcome twice.

A number and its reversed digits (Section B, Question 3)

Question: A two-digit number and the number formed by reversing its digits add up to 110. Its tens digit is six more than its units digit. Find the number.

Answer: 82.

Let the tens digit be x and the units digit be y. The numbers are 10x + y and 10y + x. Their sum gives 11x + 11y = 110, or x + y = 10. The second condition gives x - y = 6.

Adding the equations gives 2x = 16, so x = 8 and y = 2. The original number is 82, not 28, because its tens digit must be the larger digit. The check is 82 + 28 = 110 and 8 - 2 = 6.

Height and distance between buildings (Section B, Question 4)

Question: From the top of a taller building, the angles of depression to the top and bottom of a 6 m building are 30 degrees and 60 degrees. Find the taller building's height and the distance between the buildings.

Answer: Height = 9 m; distance = 3√3 m.

Let the taller building's height be H metres and the horizontal distance be d metres, with both buildings on the same level ground. The angle of depression to the bottom gives H/d = tan 60° = √3.

The vertical difference between the two tops is H - 6, so (H - 6)/d = tan 30° = 1/√3. Substituting H = √3d gives √3d - 6 = d/√3. Multiplying by √3 gives 3d - 6√3 = d, hence d = 3√3 and H = 9.

The 30-degree triangle uses the difference in heights, not the full height of the taller building. Checking this distance in both tangent ratios confirms the result.

Important points from this paper

The questions required different kinds of answers: a coordinate pair, a dimensionless trigonometric ratio, a volume with cubic units, a class interval and a proof. Giving an answer in the requested form matters. The descriptive section also required attention to internal choices rather than treating every printed alternative as compulsory.

Several problems offered a simple way to check the working. Both diagonals gave the same midpoint, the two probabilities added to one, and the number 82 satisfied both conditions. These checks come directly from the question and can reveal an error without starting again.

Frequently asked questions

Which UP Board Mathematics paper is explained here?

The article covers Class 10 Mathematics 2026, code 822(FX), using the English paper linked above.

How were the 70 marks divided?

Section A had 20 one-mark multiple-choice questions. Section B carried the remaining 50 marks through descriptive questions.

Did the time limit include reading time?

Yes. The 3 hours 15 minutes included the first 15 minutes for reading the paper.

Are all the questions solved in this article?

No. The article explains selected questions from both sections. The linked complete-paper page provides the paper and an additional solution PDF.