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CBSE Class 10 Maths Standard 2026 Question Paper 30/5/2 with Solutions

All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/5/2 (2026), with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.

Set 30/1/1Set 30/1/2Set 30/1/3Set 30/2/1Set 30/2/2Set 30/2/3Set 30/3/1Set 30/3/2Set 30/3/3Set 30/4/1Set 30/4/2Set 30/4/3Set 30/5/1Set 30/5/2Set 30/5/3
Q11 markMCQTriangles

Devansh proved that using SAS similarity criteria. If he found , then which of the following was proved true ?

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (D)
  1. In SAS similarity, the sides including the equal angles must be proportional.
  2. The sides including are AC and BC; the sides including are PR and QR.
  3. So , i.e. .
Q21 markMCQCircles

In the given figure, PQ is tangent to the circle with centre O. S is a point on the circle such that . The m is

Diagram for CBSE 2026 Class 10 Maths question 2
  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (B)
  1. (radius is perpendicular to tangent), so .
  2. (radii), so .
  3. .
  4. In : .
Also asked in: 2026 Standard 30/5/1

In an A.P., if , then the common difference of the A.P. is

  1. (A)6
  2. (B)4
  3. (C)
  4. (D)3
Show answer & solution
Answer: (B) 4

The value of p for which roots of the quadratic equation are rational, is

  1. (A)1
  2. (B)
  3. (C)25
  4. (D)
Show answer & solution
Answer: (B)
  1. Roots are rational when is a perfect square (and p is rational).
  2. : (no real roots).
  3. : , a perfect square; gives , which are rational.
  4. : , not a perfect square. : .
  5. So .
Q51 markMCQTriangles

In the given figure, such that XP : PY = 2 : 3. If PQ = 5 cm, then YZ equals

Diagram for CBSE 2026 Class 10 Maths question 5
  1. (A)12.5 cm
  2. (B)10 cm
  3. (C)15 cm
  4. (D)7.5 cm
Show answer & solution
Answer: (A) 12.5 cm
  1. , so (AA).
  2. , so cm
Also asked in: 2026 Standard 30/5/1

For an acute angle , if , then equals

  1. (A)
  2. (B)0
  3. (C)
  4. (D)1
Show answer & solution
Answer: (C)
Q71 markMCQProbability

A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (B)
  1. Number of tens = 4, number of kings = 4, so favourable outcomes = 8.
  2. P(ten or king)
Also asked in: 2026 Standard 30/5/1

The line segment joining the points and is divided by y-axis in the ratio

  1. (A)2 : 5
  2. (B)1 : 2
  3. (C)2 : 1
  4. (D)5 : 2
Show answer & solution
Answer: (A) 2 : 5
  1. Let the y-axis divide PQ in the ratio .
  2. The point of division has x-coordinate , which is 0 on the y-axis.
  3. , so .
  4. Ratio = 2 : 5

Simplest form of is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. (for acute A)

A wire is attached from a point A on the ground to the top of a pole BC, making an angle of elevation as . If AB = m, then length of the wire is

  1. (A)10 m
  2. (B) m
  3. (C)15 m
  4. (D) m
Show answer & solution
Answer: (B) m
  1. In right ,
  2. m
Q111 markMCQPolynomials

If sum and product of zeroes of a polynomial are and respectively, then a polynomial is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (B)
  1. Required polynomial
  2. With :
  3. Check: sum , product .
Q121 markMCQProbability

Meena calculates that the probability of her winning the first prize in a lottery is 0.08. If total 800 tickets were sold, the number of tickets bought by her, is

  1. (A)64
  2. (B)640
  3. (C)100
  4. (D)10
Show answer & solution
Answer: (A) 64
  1. P(winning)
  2. Tickets bought

A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use )

Diagram for CBSE 2026 Class 10 Maths question 13
  1. (A)
  2. (B)314
  3. (C)
  4. (D)
Show answer & solution
Answer: (A)
  1. The largest cone has radius cm and height cm (radius of hemisphere).
  2. Slant height cm
  3. CSA
Q141 markMCQStatistics

While calculating mean of a grouped frequency distribution, step deviation method was used . It was found that , h = 5 and a = 62.5. The value of is

  1. (A)0.5
  2. (B)1.5
  3. (C)0.3
  4. (D)7.5
Show answer & solution
Answer: (C) 0.3
  1. , so

The area of a sector of a circle of radius 10 cm is . The value of central angle is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (D)

A camping tent in hemispherical shape of radius 1.4 m, has a door opening of area 0.50 . Outer surface area of the tent is

  1. (A)11.78
  2. (B)12.32
  3. (C)11.82
  4. (D)12.86
Show answer & solution
Answer: (C) 11.82
  1. CSA of hemisphere
  2. Outer surface area
Q171 markMCQProbability

Which of the following can not be the probability of an event ?

  1. (A)
  2. (B)
  3. (C)
  4. (D)10%
Show answer & solution
Answer: (C)
  1. Probability of an event lies between 0 and 1.
  2. , so it cannot be a probability.
  3. The others are 0.39, 0.00005 and 0.1, all between 0 and 1.

The value of k for which the equation has real and equal roots, is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. For real and equal roots, .
Also asked in: 2026 Standard 30/5/1

Assertion (A): The system of linear equations and is inconsistent.
Reason (R): When two linear equations don’t have unique solution, they always represent parallel lines.

  1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. (C)Assertion (A) is true, but Reason (R) is false.
  4. (D)Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (C) Assertion (A) is true, but Reason (R) is false.
  1. , ,
  2. , so the lines are parallel and the system is inconsistent. A is true.
  3. If there is no unique solution, the lines may be parallel or coincident (infinitely many solutions). So R is false.
Q201 markAssertion–ReasonReal Numbers

Assertion (A): H.C.F. , where m is a prime number.
Reason (R): H.C.F. of two numbers is always less than or equal to the smaller number.

  1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. (C)Assertion (A) is true, but Reason (R) is false.
  4. (D)Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  1. and .
  2. HCF (18m divides ). A is true.
  3. The HCF divides both numbers, so it cannot exceed the smaller one. R is true.
  4. But R does not explain why the HCF here equals 18m; that follows from 18m being a factor of . So the answer is (B).
Q212 marksVery Short AnswerTriangles

In the given figure, and . Show that . If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, then find the length DE.

Diagram for CBSE 2026 Class 10 Maths question 21
Show answer & solution
Answer: DE = 14 cm
  1. , so (corresponding angles, with transversal EF).
  2. , so (corresponding angles).
  3. Hence (AA similarity).
  4. cm
  5. , so cm
Q222 marksVery Short AnswerIntroduction to Trigonometry

For acute angles A and B, if and , then find the values of A and B.

Show answer & solution
Answer: ,
  1. , so ... (1)
  2. , so ... (2)
  3. Adding: , so ; then .
OR
Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry

Evaluate :

Show answer & solution
Answer:
  1. , ,
  2. Numerator
  3. Value
Q232 marksVery Short AnswerReal Numbers

Prove that is an irrational number given that is irrational.

Show answer & solution
Answer: Proved.
  1. Suppose is rational. Let , where r is rational.
  2. Then , so .
  3. Since r is rational, is rational, so would be rational.
  4. This contradicts the fact that is irrational.
  5. Hence is irrational.
Q242 marksVery Short AnswerProbability

A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is 3/5, then find the number of yellow balls.

Show answer & solution
Answer: 10 yellow balls
  1. Number of green balls
  2. Number of yellow balls
Q252 marksVery Short AnswerCoordinate Geometry

Vertices of a right triangle ABC with are , and . Find the value of tan A.

Show answer & solution
Answer:
  1. In right with :
Also asked in: 2026 Standard 30/5/1
OR
Q25 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry

Using distance formula, prove that the points , and are collinear.

Show answer & solution
Answer: Proved.
  1. Hence A, B, C are collinear (C lies between A and B).
Also asked in: 2026 Standard 30/5/1
Q263 marksShort AnswerCoordinate Geometry

A circle centered at passes through the points and . Find the value(s) of K. Hence find length of chord AB.

Show answer & solution
Answer: K = 4 or K = −2; AB = units (K = 4) or units (K = −2)
  1. Let centre .
  2. :
  3. , so and
  4. or
  5. For :
  6. For :
OR
Q26 (OR) (OR)3 marksShort AnswerCoordinate Geometry

Prove that the point P dividing the line segment joining the points and in the ratio 3 : 2, lies on the line . Also find length of PA and PB.

Show answer & solution
Answer: P(2, 1) lies on the line; PA = units, PB = units
  1. Check: . So P lies on the line .
Q273 marksShort AnswerReal Numbers

The dimensions of a window are 156 cm × 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

Show answer & solution
Answer: Side = 12 cm; number of squares = 234
  1. ,
  2. HCF , so the side of each square is 12 cm.
  3. Number of squares

Use graphical method to solve the system of linear equations : and .

Show answer & solution
Answer: ,
  1. is a horizontal line through .
  2. For : . Points: , , .
  3. Plot both lines on the same axes.
  4. The lines intersect at .
  5. Solution: ,
Q293 marksShort AnswerArithmetic Progressions

In an A.P., term exceeds the term by 21. If sum of first 10 terms is 55, then form the A.P.

Show answer & solution
Answer:
  1. , so .
  2. , so .
  3. A.P.:
OR
Q29 (OR) (OR)3 marksShort AnswerArithmetic Progressions

The sum of first n terms of an A.P. is . Find its term and hence term.

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Answer: ;
Q303 marksShort AnswerAreas Related to Circles

A circle of diameter 20 cm is equally divided into five sectors. Find the area and perimeter of one of the sectors.

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Answer: Area (≈ 62.86 ); perimeter cm (≈ 32.57 cm)
  1. Radius cm; each sector has angle .
  2. Area of one sector ≈ 62.86
  3. Arc length cm
  4. Perimeter cm ≈ 32.57 cm
Also asked in: 2026 Standard 30/5/1
Q313 marksShort AnswerIntroduction to Trigonometry

Prove that :

Show answer & solution
Answer: Proved.
  1. Write in terms of and :
  2. LHS
  3. = RHS
Q325 marksLong AnswerQuadratic Equations

By selling an article for ₹ 48, a trader loses as much percent as half of the cost price of the article. Calculate the cost price and loss amount of the article.

Show answer & solution
Answer: Cost price ₹ 80 with loss ₹ 32, or cost price ₹ 120 with loss ₹ 72
  1. Let the cost price be ₹ x. Loss percent .
  2. Loss
  3. SP = CP − Loss:
  4. , so or .
  5. If CP = ₹ 80: loss = 40%, loss amount = ₹ 32 (80 − 32 = 48).
  6. If CP = ₹ 120: loss = 60%, loss amount = ₹ 72 (120 − 72 = 48).
  7. Both values satisfy the conditions.
Q335 marksLong AnswerCircles

PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

Diagram for CBSE 2026 Class 10 Maths question 33
Show answer & solution
Answer: AB = cm, PA = cm
  1. , so cm.
  2. C lies on OP and OC = 5 cm, so PC = 13 − 5 = 8 cm. at C.
  3. Let AC = x. Tangents from A are equal, so AQ = AC = x and PA = 12 − x.
  4. In right :
  5. , so and cm.
  6. PA cm
  7. By symmetry BC = AC, so AB = 2x cm.
Q345 marksLong AnswerTriangles

D is the mid-point of side BC of . CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that
(i) OBGC is a parallelogram.
(ii)
(iii)

Diagram for CBSE 2026 Class 10 Maths question 34
Show answer & solution
Answer: Proved.
  1. (i) BD = DC (D is the mid-point of BC) and OD = DG (given). So the diagonals BC and OG of quadrilateral OBGC bisect each other. Hence OBGC is a parallelogram.
  2. (ii) As OBGC is a parallelogram, and , i.e. and .
  3. In , , so by BPT .
  4. In , , so by BPT .
  5. Hence , and by the converse of BPT in , .
  6. (iii) In and : is common, and (corresponding angles, ).
  7. Hence (AA similarity).
OR
Q34 (OR) (OR)5 marksLong AnswerTriangles

Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

Diagram for CBSE 2026 Class 10 Maths question 34 (OR)
Show answer & solution
Answer: Proved.
  1. (i) In and : DQ = CQ (Q is the mid-point of CD), (vertically opposite), (alternate angles, ).
  2. So (ASA), hence AQ = QR.
  3. (ii) In and : and (alternate angles, ).
  4. So (AA), and .
  5. , so , i.e. AP = 2PQ.
  6. (iii) PR = PQ + QR = PQ + AQ (from (i)) = PQ + (AP + PQ) = AP + 2PQ = AP + AP = 2AP (using (ii)).
Q355 marksLong AnswerStatistics

Find the mean and mode of the following frequency distribution :

Class Interval :400-450450-500500-550550-600600-650650-700
Frequency :151820232212
Show answer & solution
Answer: Mean = 550; mode = 587.5
  1. N = 15 + 18 + 20 + 23 + 22 + 12 = 110
  2. Class marks: 425, 475, 525, 575, 625, 675. Take a = 575, h = 50;
  3. Mean
  4. Modal class 550-600: l = 550, , , , h = 50
  5. Mode
OR
Q35 (OR) (OR)5 marksLong AnswerStatistics

If the median of the following frequency distribution is 32.5 and sum of all frequencies is 40, then find the values of and :

Class Interval :0-1010-2020-3030-4040-5050-6060-70
Frequency :391262
Show answer & solution
Answer: ,
  1. , so
  2. Median 32.5 lies in 30-40: l = 30, f = 12, h = 10, cf ,
  3. , so and

Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is 54 metres away from the water tank.
From a window (W) of the building, the angle of elevation of top of the tank is and angle of depression of its foot is .
(i) Write a relation between d (the height of window) and y. (1)
(ii) Determine the value of h. (1)
(iii) (a) Determine height of the water tank. (2)
OR (iii) (b) Find the value of and height of the window above ground level. (2)

Diagram for CBSE 2026 Class 10 Maths question 36
Show answer & solution
Answer: (i) y = 2d (ii) h = 54 m (iii) (a) m ≈ 85.18 m OR (iii) (b) m, window height m ≈ 31.18 m
  1. XW = AC = 54 m, , , AX = WC = d.
  2. (i) In right : , so .
  3. (ii) In right : , so h = 54 m.
  4. (iii) (a) , so m.
  5. Height of tank AB m
  6. (iii) (b) , so m (≈ 76.37 m).
  7. Height of window m
Q374 marksCase StudyPolynomials

An arch of a railway bridge, built on Chenab riverbed, is shown in the above diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial .
Observe the diagram and based on above information, answer the following questions :
(i) Write the co-ordinates of point A. (1)
(ii) Find the span of the arch. (1)
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials. (2)
OR (iii) (b) Find the values of at and . Are they same ? (2)

Diagram for CBSE 2026 Class 10 Maths question 37
Show answer & solution
Answer: (i) A(0, 136) (ii) 467 m (iii) (a) Zeroes 228.5 and −238.5; sum = −10 = OR (iii) (b) p(100) = 108.5, p(−100) = 113.5; not the same
  1. (i) A lies on the y-axis, so x = 0 and . A = (0, 136).
  2. (ii) Span = PQ = 228.5 − (−238.5) = 467 m.
  3. (iii) (a) From the diagram the zeroes are 228.5 and −238.5.
  4. Sum of zeroes = 228.5 + (−238.5) = −10.
  5. . Hence sum of zeroes , verified.
  6. (iii) (b)
  7. The values are not the same.
Q384 marksCase StudySurface Areas and Volumes

A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter 7 cm is latched with a very thin rod. (Ignore the rod while making calculations.)
Dimensions of the cuboid are 24 cm × 12 cm × 17 cm.
(i) Find the surface area of the bulb. (1)
(ii) What could be the maximum diameter of the bulb if at least 1 cm space is left from each side ? (1)
(iii) (a) Find the area of the fabric used if there is a fold of 2 cm on top and bottom edges. (2)
OR (iii) (b) Find the space available inside the lamp. (2)

Diagram for CBSE 2026 Class 10 Maths question 38
Show answer & solution
Answer: (i) 154 (ii) 10 cm (iii) (a) 1512 OR (iii) (b) 4896 (lamp volume); 4716.33 after excluding the bulb
  1. (i) r = 3.5 cm. Surface area
  2. (ii) The smallest dimension of the lamp is 12 cm. Leaving 1 cm on each side, maximum diameter = 12 − 2 = 10 cm.
  3. (iii) (a) The fabric covers the four vertical faces (open top and bottom). Height with folds = 17 + 2 + 2 = 21 cm.
  4. Area
  5. (iii) (b) Volume of the lamp
  6. Volume of bulb
  7. Space available (excluding the bulb)
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