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CBSE Class 10 Maths Basic 2025 Question Paper 430/6/1 with Solutions

All 44 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/6/1 (2025), with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.

Set 430/1/1Set 430/1/2Set 430/1/3Set 430/2/1Set 430/2/2Set 430/2/3Set 430/3/1Set 430/3/2Set 430/3/3Set 430/4/1Set 430/4/2Set 430/4/3Set 430/5/1Set 430/5/2Set 430/5/3Set 430/6/1Set 430/6/2Set 430/6/3
Q11 markMCQPolynomials

In the given figure, graph of polynomial is shown. Number of zeroes of is

Diagram for CBSE 2025 Class 10 Maths question 1
  1. (A)3
  2. (B)2
  3. (C)1
  4. (D)4
Show answer & solution
Answer: (A) 3
  1. The number of zeroes equals the number of points where the graph meets the -axis.
  2. The graph meets the -axis at O and at two more points, i.e. 3 points.
  3. So has 3 zeroes.

term of the A.P. : , ........ is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. , .
  2. .
Also asked in: 2025 Basic 430/6/3

The line intersects – axis at

  1. (A)(0, )
  2. (B)(0, 3)
  3. (C)(, 0)
  4. (D)(3, 0)
Show answer & solution
Answer: (D) (3, 0)
  1. On the -axis, .
  2. , so . The point is (3, 0).
Also asked in: 2025 Basic 430/6/2

Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is

Diagram for CBSE 2025 Class 10 Maths question 4
  1. (A)8 cm
  2. (B) cm
  3. (C) cm
  4. (D)12 cm
Show answer & solution
Answer: (B) cm
  1. Height of one cone cm.
  2. The solid is two cones joined at their vertices, so its height cm.

The value of k for which the system of equations and is inconsistent, is

  1. (A)
  2. (B)
  3. (C)6
  4. (D)
Show answer & solution
Answer: (A)
  1. Inconsistent when .
  2. gives , and .
Q61 markMCQProbability

Two dice are rolled together. The probability of getting a sum more than 9 is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. Total outcomes = 36.
  2. Sum 10: (4, 6), (5, 5), (6, 4); sum 11: (5, 6), (6, 5); sum 12: (6, 6). That is 6 outcomes.
  3. P(sum more than 9) .
Also asked in: 2025 Basic 430/6/3

ABCD is a rectangle with its vertices at (2, ), (8, 4), (4, 8) and (, 2) taken in order. Length of its diagonal is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (D)
  1. Diagonal AC joins (2, ) and (4, 8).
  2. AC .
Q81 markMCQCircles

In the given figure, PA is tangent to a circle with centre O. If and OA = 2.5 cm, then OP is equal to

Diagram for CBSE 2025 Class 10 Maths question 8
  1. (A)2.5 cm
  2. (B)5 cm
  3. (C) cm
  4. (D)2 cm
Show answer & solution
Answer: (B) 5 cm
  1. OA PA, so is right-angled at A.
  2. , so and OP = 5 cm.
Q91 markMCQProbability

If probability of happening of an event is 57%, then probability of non-happening of the event is

  1. (A)0.43
  2. (B)0.57
  3. (C)53%
  4. (D)
Show answer & solution
Answer: (A) 0.43
  1. P(E) = 57% = 0.57.
  2. P(not E) = 1 0.57 = 0.43.

OAB is sector of a circle with centre O and radius 7 cm. If length of arc cm, then is equal to

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. Arc length .
  2. .
  3. .
Q111 markMCQTriangles

In , DE BC. If AE = cm, EC = 4 cm, AD = cm and DB = 3 cm, then value of is

Diagram for CBSE 2025 Class 10 Maths question 11
  1. (A)1
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (B)
  1. By BPT, , so .
  2. , so and .
Q121 markMCQProbability

Three coins are tossed together. The probability that exactly one coin shows head, is

  1. (A)
  2. (B)
  3. (C)1
  4. (D)
Show answer & solution
Answer: (D)
  1. Total outcomes = 8.
  2. Exactly one head: HTT, THT, TTH, i.e. 3 outcomes.
  3. Probability .
Also asked in: 2025 Basic 430/6/2
Q131 markMCQCircles

In two concentric circles centred at O, a chord AB of the larger circle touches the smaller circle at C. If OA = 3.5 cm, OC = 2.1 cm, then AB is equal to

Diagram for CBSE 2025 Class 10 Maths question 13
  1. (A)5.6 cm
  2. (B)2.8 cm
  3. (C)3.5 cm
  4. (D)4.2 cm
Show answer & solution
Answer: (A) 5.6 cm
  1. OC AB (radius tangent), and the perpendicular from the centre bisects the chord, so AB = 2AC.
  2. AC cm.
  3. AB = 2 2.8 = 5.6 cm.

If , then value of is

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (D)
  1. .
  2. So .
Also asked in: 2025 Basic 430/6/3
Q151 markMCQStatistics

To calculate mean of a grouped data, Rahul used assumed mean method. He used d = , where A is assumed mean. Then is equal to

  1. (A)A +
  2. (B)A + h
  3. (C)h (A + )
  4. (D)A h
Show answer & solution
Answer: (A) A +
  1. In the assumed mean method with , .
  2. No class size factor h appears because d is not divided by h.

If the sum of first n terms of an A.P. is given by , then the first term of the A.P. is

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

In , . If , then cos C is equal to

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. Since , AC is the hypotenuse and AB is opposite to C.
  2. , so .
  3. .
Also asked in: 2025 Basic 430/6/2

The volume of air in a hollow cylinder is 450 cm. A cone of same height and radius as that of cylinder is kept inside it. The volume of empty space in the cylinder is

Diagram for CBSE 2025 Class 10 Maths question 18
  1. (A)225 cm
  2. (B)150 cm
  3. (C)250 cm
  4. (D)300 cm
Show answer & solution
Answer: (D) 300 cm
  1. Volume of cylinder cm.
  2. Volume of cone cm.
  3. Empty space cm.
Q191 markAssertion–ReasonReal Numbers

Assertion (A) : is a rational number, where a and b are positive integers.
Reason (R) : Product of two irrationals is always rational.

  1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for 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. , an integer, so it is rational. A is true.
  2. R is false: e.g. is irrational.
Q201 markAssertion–ReasonTriangles

Assertion (A) : such that , . Hence .
Reason (R) : Sum of all angles of a triangle is .

  1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for 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: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  1. R is true (angle sum property).
  2. In , by R.
  3. Corresponding angles of similar triangles are equal, so . A is true.
  4. R is used to find the angle, so R explains A.
Q212 marksVery Short AnswerQuadratic Equations

Solve the equation , using quadratic formula.

Show answer & solution
Answer: ,
  1. , , .
  2. .
  3. .
Also asked in: 2025 Basic 430/6/2
OR
Q21 (OR) (OR)2 marksVery Short AnswerQuadratic Equations

Find the nature of roots of the equation .

Show answer & solution
Answer: Real and equal roots
  1. , , .
  2. .
  3. Since D = 0, the roots are real and equal.
Also asked in: 2025 Basic 430/6/2
Q222 marksVery Short AnswerTriangles

In a trapezium ABCD, AB DC and its diagonals intersect at O. Prove that .

Show answer & solution
Answer: Proved.
  1. In and :
  2. (alternate angles, AB DC)
  3. (alternate angles)
  4. So (AA similarity).
  5. Hence (corresponding sides of similar triangles are proportional).
Q232 marksVery Short AnswerProbability

A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that
(i) it bears a 2– digit number
(ii) the number is a perfect square.

Show answer & solution
Answer: (i) (ii)
  1. Total outcomes = 120.
  2. (i) 2-digit numbers are 10 to 99, i.e. 90 numbers. P = .
  3. (ii) Perfect squares up to 120: 1, 4, 9, ..., 100, i.e. 10 numbers (). P .
Q242 marksVery Short AnswerIntroduction to Trigonometry

Evaluate :

Show answer & solution
Answer:
  1. , , .
  2. Denominator .
  3. Value .
OR
Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry

Verify that , for .

Show answer & solution
Answer: Verified: both sides equal .
  1. LHS .
  2. RHS .
  3. LHS = RHS, hence verified.
Q252 marksVery Short AnswerReal Numbers

Using prime factorisation, find the HCF of 180, 140 and 210.

Show answer & solution
Answer: HCF = 10
  1. HCF = product of the smallest powers of common primes .
Also asked in: 2025 Basic 430/6/3
Q263 marksShort AnswerPolynomials

If , are zeroes of the polynomial , then form a quadratic polynomial in whose zeroes are and .

Show answer & solution
Answer: (or any non-zero multiple )
  1. , .
  2. Sum of new zeroes .
  3. Product of new zeroes .
  4. Required polynomial .
Also asked in: 2025 Basic 430/6/2
OR
Q26 (OR) (OR)3 marksShort AnswerPolynomials

Find the zeroes of the polynomial and verify the relationship between zeroes and its coefficients.

Show answer & solution
Answer: Zeroes: and ; relationship verified.
  1. .
  2. Zeroes: , .
  3. Sum and .
  4. Product and .
  5. Hence the relationship is verified.
Also asked in: 2025 Basic 430/6/2
Q273 marksShort AnswerQuadratic Equations

Find length and breadth of a rectangular park whose perimeter is 100 m and area is 600 m.

Show answer & solution
Answer: Length = 30 m, breadth = 20 m
  1. Let length = m. Then gives breadth m.
  2. Area: , so .
  3. , so or .
  4. Taking length as the longer side: length = 30 m, breadth = 20 m.
Also asked in: 2025 Basic 430/6/3
Q283 marksShort AnswerReal Numbers

Three measuring rods are of lengths 120 cm, 100 cm and 150 cm. Find the least length of a fence that can be measured an exact number of times, using any of the rods. How many times each rod will be used to measure the length of the fence ?

Show answer & solution
Answer: Least length = 600 cm (6 m); the rods are used 5, 6 and 4 times respectively.
  1. , , .
  2. LCM cm.
  3. Rod of 120 cm: times; rod of 100 cm: times; rod of 150 cm: times.
Q293 marksShort AnswerAreas Related to Circles

AB and CD are diameters of a circle with centre O and radius 7 cm. If , then find the area and perimeter of the shaded region.

Diagram for CBSE 2025 Class 10 Maths question 29
Show answer & solution
Answer: Area cm (about 25.67 cm); perimeter cm (about 35.33 cm)
  1. The shaded region is two sectors BOD and AOC, each of angle (, vertically opposite).
  2. Area cm.
  3. Each arc cm.
  4. Perimeter = 4 radii + 2 arcs cm.
Q303 marksShort AnswerIntroduction to Trigonometry

Prove that .

Show answer & solution
Answer: Proved.
  1. Let , so .
  2. LHS .
  3. .
  4. .
  5. So LHS = RHS.
Q313 marksShort AnswerArithmetic Progressions

Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.

Show answer & solution
Answer: A.P.: 4, 10, 16, 22, ...;
  1. , so .
  2. , so .
  3. A.P.: 4, 10, 16, 22, ...
  4. .
OR
Q31 (OR) (OR)3 marksShort AnswerArithmetic Progressions

Find the sum of first 20 terms of an A.P. whose term is given by . Can 52 be a term of this A.P. ?

Show answer & solution
Answer: ; No, 52 is not a term.
  1. , .
  2. .
  3. If , then , which is not a natural number.
  4. So 52 is not a term of the A.P.

Solve the following pair of linear equations by graphical method :
and

Show answer & solution
Answer: , (the lines intersect at (4, 1))
  1. For : points (0, 9), (3, 3), (4, 1).
  2. For : points (2, 0), (0, ), (4, 1).
  3. Plot both lines on the same axes.
  4. The lines intersect at (4, 1).
  5. Check: and . So , .
OR
Q32 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables

Nidhi received simple interest of ₹ 1,200 when invested ₹ at 6% p.a. and ₹ y at 5% p.a. for 1 year. Had she invested ₹ at 3% p.a. and ₹ y at 8% p.a. for that year, she would have received simple interest of ₹ 1,260. Find the values of and y.

Show answer & solution
Answer: = ₹ 10,000, y = ₹ 12,000
  1. , so ... (1)
  2. , so ... (2)
  3. (2) 2: ... (3)
  4. (3) (1): , so .
  5. From (1): , so .
Q335 marksLong AnswerCircles

The given figure shows a circle with centre O and radius 4 cm circumscribed by . BC touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.

Diagram for CBSE 2025 Class 10 Maths question 33
Show answer & solution
Answer: AE cm (about 5.82 cm)
  1. Tangents from an external point are equal: BF = BD = 6, CE = CD = 10, AE = AF = (say).
  2. Sides: BC = 16, CA = , AB = ; semi-perimeter .
  3. Area perimeter .
  4. By Heron: Area .
  5. So , i.e. .
  6. , so cm.
OR
Q33 (OR) (OR)5 marksLong AnswerCircles

PA and PB are tangents drawn to a circle with centre O.
If and OA = 10 cm, then
(i) Find . (1)
(ii) Find the perimeter of . (3)
(iii) Find the length of chord AB. (1)

Diagram for CBSE 2025 Class 10 Maths question 33 (OR)
Show answer & solution
Answer: (i) (ii) cm (about 47.3 cm) (iii) AB cm
  1. (i) , so . OP bisects , so .
  2. (ii) In right : , so OP = 20 cm; , so AP cm.
  3. Perimeter cm.
  4. (iii) is isosceles with OA = OB = 10 and . The perpendicular from O to AB bisects AB and .
  5. Half of AB , so AB cm.

The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are and respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use )

Show answer & solution
Answer: Height = 13.5 m; distance m (7.785 m)
  1. Let the multi-storeyed building be m high and the distance between the buildings be m.
  2. Foot of the 9 m building (depression ): , so .
  3. Top of the 9 m building (depression ): , so .
  4. , so and m.
  5. m.
Q355 marksLong AnswerStatistics

Find ‘mean’ and ‘mode’ of the following data :
Class: 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45
Frequency: 12, 10, 15, 11, 7, 5

Show answer & solution
Answer: Mean = 28; Mode
  1. Class marks: 17.5, 22.5, 27.5, 32.5, 37.5, 42.5; .
  2. .
  3. Mean .
  4. Modal class 25 – 30 (highest frequency 15): , , , , .
  5. Mode .
Q364 marksCase StudyTriangles

A triangular window of a building is shown above. Its diagram represents a with and AB = AC. Points P and R trisect AB and PQ RS AC.
Based on the above, answer the following questions :
(i) Show that . (1)
(ii) Prove that PQ AC. (1)
(iii) (a) If AB = 3 m, find length BQ and BS. Verify that BQ BS. (2)
OR
(iii) (b) Prove that BR + RS BC. (2)

Diagram for CBSE 2025 Class 10 Maths question 36
Show answer & solution
Answer: (i) Proved (AA similarity) (ii) Proved (iii) (a) BQ m, BS m, so BQ BS OR (b) Proved
  1. (i) In and : is common and (corresponding angles, PQ AC). So (AA).
  2. (ii) From (i), . P and R trisect AB with P nearer B, so BP AB. Hence PQ AC.
  3. (iii) (a) AC = AB = 3 m, so BC m. From the similarity, BQ BC m.
  4. Similarly with BR AB, so BS BC m. Thus BQ BS.
  5. (iii) (b) with ratio , so BR AB and RS AC.
  6. BR + RS (AB + AC) BC (Pythagoras in , right-angled at A).
Q374 marksCase StudySurface Areas and Volumes

A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm.
Based on the above, answer the following questions :
(i) Find the dimensions of the cuboidal box. (1)
(ii) Find the total outer surface area of the box. (1)
(iii) (a) Find the difference between the capacity of the bowl and the volume of the box. (use ) (2)
OR
(iii) (b) The inner surface of the bowl and the thickness is to be painted. Find the area to be painted. (2)

Diagram for CBSE 2025 Class 10 Maths question 37
Show answer & solution
Answer: (i) 21 cm 21 cm 10.5 cm (ii) 1764 cm (iii) (a) about 2537.17 cm OR (b) about 660.79 cm
  1. (i) The bowl just fits, so length = breadth = outer diameter = 21 cm and height = outer radius = 10.5 cm.
  2. (ii) TSA cm.
  3. (iii) (a) Capacity of bowl cm.
  4. Volume of box cm.
  5. Difference cm.
  6. (iii) (b) Inner curved surface cm.
  7. Rim (thickness) cm.
  8. Area to be painted cm.
Q384 marksCase StudyCoordinate Geometry

Gurveer and Arushi built a robot that can paint a path as it moves on a graph paper. Some co-ordinate of points are marked on it. It starts from (0, 0), moves to the points listed in order (in straight lines) and ends at (0, 0).
Arushi entered the points P(8, 6), Q(12, 2) and S( 6, 6) in order. The path drawn by robot is shown in the figure.
Based on the above, answer the following questions :
(i) Determine the distance OP. (1)
(ii) QS is represented by equation . Find the co-ordinates of the point where it intersects y – axis. (1)
(iii) (a) Point R(4.8, y) divides the line segment OP in a certain ratio, find the ratio. Hence, find the value of y. (2)
OR
(iii) (b) Using distance formula, show that . (2)

Diagram for CBSE 2025 Class 10 Maths question 38
Show answer & solution
Answer: (i) 10 units (ii) (iii) (a) 3 : 2, y = 3.6 OR (b) Shown: PQ , OS
  1. (i) OP units.
  2. (ii) On the y-axis : , . Point .
  3. (iii) (a) Let R divide OP in the ratio . Then , so , . Ratio .
  4. y .
  5. (iii) (b) PQ ; OS .
  6. .
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