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## CBSE Class 10 Maths Sample Paper PDF Free Download

### CBSE Class 10 Maths Sample Paper

The Central Board of Secondary Education (CBSE) has released the sample papers and marking scheme for all subjects of class 10.

The latest CBSE sample papers with marking schemes are the best resources for the annual exam preparation as the similar paper pattern and marks distribution will be followed in the CBSE Class 10 Board Exams 2023-24.

You can find here the CBSE Class 10 Mathematics (Standard) Sample Paper 2023-24. This sample paper is going to be extremely helpful in preparing for the year-end board exam in the right way from the beginning of the session.

CBSE Class 10 Maths Marking Scheme is helpful to know the step-wise solutions for questions given in the sample paper and understand how answers will be evaluated in a step-by-step manner in board exams.

### Sample Paper Class 10 Maths

The pattern for CBSE Class 10 Maths (Basic & Standard) has been discussed along with the Class 10 Maths Sample Paper 2023 which is as follows-

**For Basic Maths Syllabus- **

1. Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.

2. Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.

3. Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.

4. Section D has 4 Long Answer (LA) type questions carrying 5 marks each.

5. Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1, and 2 marks each respectively.

**For Standard Maths Syllabus- **

1. Section A has 20 MCQs carrying 1 mark each

2. Section B has 5 questions carrying 02 marks each.

3. Section C has 6 questions carrying 03 marks each.

4. Section D has 4 questions carrying 05 marks each.

5. Section E has 3 case-based integrated units of assessment (04 marks each) with subparts of the values of 1, 1, and 2 marks each respectively.

### Class 10 Maths Sample Paper 2023 PDF

To give students a clear vision for the Class 10 Mathematics 2023 Exam, CBSE has uploaded sample question papers for CBSE Class 10th Maths Sample Paper 2023 (Basic & Standard) on its official website https://cbseacademic.nic.in/index.html.

**CBSE Class 10 Maths Sample Paper with Solutions PDF** can be downloaded directly from here along with the answer key containing detailed solutions to the Class 10 Maths Sample Papers.

CBSE Class 10 Maths Sample Paper and Solutions 2023 | |

Class 10 Basic Maths Sample Paper 2023 | Solution Link |

Class 10 Standard Maths Sample Paper 2023 | Solution Link |

#### CBSE Class 10th Sample Paper and Solution 2023(Basic Maths)

As we know CBSE has recently released the sample paper for class 10th on the official website. Below are the sample questions from the sample paper for you to practice thoroughly so as to score good marks in the board examination.

#### Section A

**Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.**

**Question1: ** If two positive integers p and q can be expressed as p = ab² and q = a³b; a, b being prime numbers, then LCM (p, q) is

(a) ab

(b) a²b ²

(c) a³b²

(d) a³b³

**Solution: **(c) a³b²

**Question2**. What is the greatest possible speed at which a man can walk 52 km and 91 km in an exact number of hours?

(a) 17 km/hours

(b) 7 km/hours

(c) 13 km/hours

(d) 26 km/hours

**Solution: **(c) 13 km/hours

**Question3.** If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is

(a) 10

(b) -10

(c) 5

(d) –5

**Solution: **(b) -10

**Question4.** Graphically, the pair of equations given by

6x – 3y + 10 = 0

2x – y + 9 = 0 represents two lines that are

(a) intersecting at exactly one point.

(b) parallel.

(c) coincident.

(d) intersecting at exactly two points.

**Solution: **(b) Parallel.

**Question5.** If the quadratic equation x2 + 4x + k = 0 has real and equal roots, then

(a) k < 4

(b) k > 4

(c) k = 4

(d) k ≥ 4

**Solution: **(c) k = 4

**Question6.** The perimeter of a triangle with vertices (0, 4), (0, 0), and (3, 0) is

(a) 5 units

(b) 12 units

(c) 11 units

d) (7 + √5) units

**Solution: **(b) 12

**Question7. **If in triangles ABC and DEF, AB/DE =BC/FD, then they will be similar, when

(a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠A = ∠F

**Solution:**(c) ∠B = ∠D** **

**Question8.** In which ratio the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4)?.

(a) 1: 5

(b) 5: 1

(c) 1: 1

(d) 1: 2

**Solution: **(b) 5: 1

**Question9.** In the figure, if PA and PB are tangents to the circle with center O such that ∠APB = 50°, then ∠OAB is equal to

(a) 25°

(b) 30°

(c) 40°

(d) 50°

**Solution: **(a) 25°

**Question10.** If sin A =1/2, then the value of sec A is :

(a) √3/2

(b) 1/√3

(c) √3

(d) 1

**Solution: **(a) √3/2

**Question11. **√3 cos2A + √3 sin2A is equal to

(a) 1

(b) 1/√3

(c) √3

(d) 0

**Solution: **(c) √3

**Question12. **The value of cos1° cos2° cos3° cos4°…………..…..cos90° is

(a) 1

(b) 0

(c) – 1

(d) 2

**Solution: **(b) 0

**Question13. ** If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

(a) 22: 7

(b) 14: 11

(c) 7: 22

(d) 11: 14

**Solution: **(b) 14: 11

**Question14. ** If the radii of two circles are in the ratio of 4 : 3, then their areas are in the ratio of :

(a) 4 : 3

(b) 8 : 3

(c) 16: 9

(d) 9: 16

**Solution: **(c) 16: 9

**Question15. ** The total surface area of a solid hemisphere of radius 7 cm is :

(a) 447π cm²

(b) 239π cm²

(c) 174π cm²

(d) 147π cm²

**Solution: **(d) 147π cm2

**Question16. ** For the following distribution :

Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |

Frequency | 10 | 15 | 12 | 20 | 9 |

the upper limit of the modal class is

(a) 10

(b) 15

(c) 20

(d) 25

**Solution: **(c) 20

**Question17. ** If the mean of the following distribution is 2.6, then the value of y is

Variable (x) | 1 | 2 | 3 | 4 | 5 |

Frequency | 4 | 5 | y | 1 | 2 |

(a) 3

(b) 8

(c) 13

(d) 24

**Solution: **(b) 8

**Question18. ** A card is selected at random from a well-shuffled deck of 52 cards. The probability of its being a red face card is

(a)3/26

(b)3/13

(c)2/13

(d)1/2

Direction for questions 19 & 20: In questions numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

**Solution:**(a)3/26** **

**Question19. ** **Assertion: **If the HCF of 510 and 92 is 2, then the LCM of 510 & 92 is 32460

**Reason: **as HCF(a,b) x LCM(a,b) = a x b

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

**Solution: **(d) Assertion (A) is false but Reason (R) is true.

**Question20. **Assertion (A): The ratio in which the line segment joining (2, -3) and (5, 6) internally divided by the x-axis is 1:2.

Reason (R): as a formula for the internal division is (𝑚𝑥2 + 𝑛𝑥1/𝑚 + 𝑛, 𝑚𝑦2 + 𝑛𝑦1/𝑚 + 𝑛)

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

**Solution: **(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

#### Section B

**Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each. **

**Question21. ** For what values of k will the following pair of linear equations have infinitely many solutions?

kx + 3y – (k – 3) = 0

12x + ky – k = 0

**Solution: **For a pair of linear equations to have infinitely many solutions :

a1/a2 = b1/b2 = c1/c2

⇒ k/12 = 3/k = k−3/k

𝑘/12 = 3/𝑘

⇒ k² = 36 ⇒ k = ± 6

Also, 3/𝑘 = 𝑘−3/𝑘 ⇒ k² – 6k = 0 ⇒ k = 0, 6.

Therefore, the value of k, which satisfies both conditions, is k = 6.

**Question22. **In the figure, altitudes AD and CE of Δ ABC intersect each other at point P. Show that:

(i) ΔABD ~ ΔCBE

(ii) ΔPDC ~ ΔBEC

**Solution: **(i) In ΔABD and ΔCBE

∠ADB = ∠CEB = 90º

∠ABD = ∠CBE (Common angle)

⇒ ΔABD ~ ΔCBE (AA criterion)

(ii) In ΔPDC and ΔBEC

∠PDC = ∠BEC = 90º

∠PCD = ∠BCE (Common angle)

⇒ ΔPDC ~ ΔBEC (AA criterion)

**[OR]**

**In the figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC**

**Solution: **In ΔABC, DE || AC

BD/AD = BE/EC ………(i) (Using BPT)

In ΔABE, DF || AE

BD/AD = BF/FE ……..(ii) (Using BPT)

From (i) and (ii)

BD/AD = BE/EC = BF/FE

Thus, BF

FE = BE/EC

**Question23.** Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution:** Let O be the center of the concentric circle of radii 5 cm

and 3 cm respectively. Let AB be a chord of the larger circle touching the smaller circle at P

Then AP = PB and OP⊥AB

Applying Pythagoras’ theorem in △OPA, we have

OA²=OP²+AP²

⇒ 25 = 9 + AP2

⇒ AP²= 16 ⇒ AP = 4 cm

∴ AB = 2AP = 8 cm

**Question24. ** If cot θ =7/8 , evaluate (1 + sin θ) (1− sin θ)/(1 + cos θ) (1− cos θ)

**Solution: **Now,

(1 + sinθ)(1 − sinθ) /(1 + cosθ)(1 − cosθ)

= (1 – sin²θ)/ (1 – cos²θ)

= cos²θ/ sin²θ

= ( cosθ /sinθ )² = cot²θ

= ( 7 /8 )² = 49 /64

**Question25.** Find the perimeter of a quadrant of a circle of radius 14 cm.

**Solution: **Perimeter of quadrant = 2r + 1/4 × 2 π r

⇒ Perimeter = 2 × 14 + 1/2 × 22 /7 × 14

⇒ Perimeter = 28 + 22 =28+22 = 50 cm

**[OR]**

Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.

**Solution: **Area of the circle = Area of first circle + Area of the second circle

⇒ πR² = π (r1)² + π (r1)²

⇒ πR² = π (24)² + π (7)²

⇒ πR² = 576π +49π

⇒ πR² = 625π

⇒ R² = 625

⇒ R = 25

Thus, diameter of the circle = 2R = 50 cm.

#### Section C

**Section C consists of 6 questions of 3 marks each.**

**Question26. **Prove that √5 is an irrational number.

**Solution: **Let us assume to the contrary, that √5 is rational. Then we can find a and b ( ≠ 0) such that √5 = 𝑎/𝑏(assuming that a and b are co-primes).

So, a = √5 b ⇒ a² = 5b²

Here 5 is a prime number that divides a² then 5 divides an also

(Using the theorem, if a is a prime number and if a divides p², then a divides p, where a is a positive integer)

Thus 5 is a factor of a

Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c

We get (5c)² = 5b² ⇒ 5c² = b²

This means 5 divides b²so 5 divides b also (Using the theorem, if a is a prime number and if a divides p², then a divides p, where a is a positive integer).

Hence a and b have at least 5 as common factors.

But this contradicts the fact that a and b are coprime. This is the contradiction to our assumption that p and q are co-primes.

So, √5 is not a rational number. Therefore, the √5 is irrational.

**Question27. ** Find the zeroes of the quadratic polynomial 6x² – 3 – 7x and verify the relationship between the zeroes and the coefficients.

**Solution: **6x² – 7x – 3 = 0

⇒ 6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(3x + 1) = 0

⇒ 2x – 3 = 0 & 3x + 1 = 0 x = 3/2 & x

= -1/3 Hence, the zeros of the quadratic polynomials are 3/2 and -1/3.

For verification Sum of zeros = – coefficient of x /coefficient of x²

⇒ 3/2 + (-1/3) = – (-7) / 6 ⇒ 7/6 = 7/6

Product of roots = constant/ coefficient of x²

⇒ 3/2 x (-1/3) = (-3) / 6 ⇒ -1/2 = -1/2 Therefore, the relationship between zeros and their coefficients is verified.

**Question28. **A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 16 for a book kept for four days. Find the fixed charges and the charge for each extra day.

**Solution: **Let the fixed charge by Rs x and the additional charge by Rs y per day

Number of days for Latika = 6 = 2 + 4

Hence, Charge x + 4y = 22

x = 22 – 4y ………(1)

Number of days for Anand = 4 = 2 + 2

Hence, Charge x + 2y = 16

x = 16 – 2y ……. (2)

On comparing equations (1) and (2), we get,

22 – 4y = 16 – 2y ⇒ 2y = 6 ⇒ y = 3

Substituting y = 3 in equation (1), we get,

x = 22 – 4 (3) ⇒ x = 22 – 12 ⇒ x = 10

Therefore, fixed charge = Rs 10 and additional charge = Rs 3 per day

**[OR]**

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

**Solution: **AB = 100 km. We know that, Distance = Speed × Time.

AP – BP = 100 ⇒ 5x − 5y = 100 ⇒ x−y=20…..(i)

AQ + BQ = 100 ⇒ x + y = 100….(ii)

Adding equations (i) and (ii), we get,

x − y + x + y = 20 +100 ⇒ 2x = 120 ⇒ x = 60

Substituting x = 60 in equation (ii), we get, 60 + y = 100 ⇒ y = 40

Therefore, the speed of the first car is 60 km/hr, and the speed of the second car

is 40 km/hr.

**Question29. **In the figure, PQ is a chord of length 8 cm of a circle. radius 5 cm. The tangents at P and Q intersect at a point

T. Find the length TP.

**Solution: **Since OT is the perpendicular bisector of PQ.

Therefore, PR=RQ=4 cm

Now, OR = √𝐎𝐏² − 𝐏𝐑²

= √𝟓² − 𝟒² =3cm

Now, ∠TPR + ∠RPO = 90° (∵TPO=90°)

& ∠TPR + ∠PTR = 90° (∵TRP=90∘)

So, ∠RPO = ∠PTR

So, ⍙TRP ~ ⍙PRO [By A-A Rule of similar triangles]

So, TP/PO = RP/RG

⇒ TP/5 = 4/ 3

⇒ TP =m20/3cm

**Question30. **Prove that tan θ /1 − cot θ +cot θ/1 − tan θ = 1 + sec θ cosec θ

**Solution: **LHS = tan θ/1−cot θ + cot θ/1−tan θ

= tan θ/1−1/tanθ + 1/ tanθ/ 1−tan θ

=tan²θ/tan θ−1 + 1/tan θ (1−tan θ)

= tan³θ−1/tan θ (tan θ−1)

=(tanθ −1) (tan³θ + tanθ+1 )/tan θ (tan θ−1)

=(tan³θ + tanθ+1 )/tan θ

= tanθ + 1 + sec = 1 + tanθ + secθ

= 1 + sin θ/cos θ + cos θ/sin θ

= 1 + sin²θ+ cos²θ/sin θ cos θ

= 1 + 1/sin θ cos θ

= 1 + sec θ cosec θ

**[OR]**

**If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1**

**Solution: **sin θ + cos θ = √3

⇒ (sin θ + cos θ)² = 3

⇒ sin²θ + cos²θ + 2sin θ cos θ = 3

⇒ 1 + 2sin θ cos θ = 3 ⇒ 1 sin θ cos θ = 1

Now tanθ + cotθ = sin θ/cos θ + cos θ/sin θ

= sin²θ+ cos²θ/sin θ cos θ

=1/ sin θ cos θ

=1/1

= 1

**Question31.** Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8? and (ii) 13? (iii) less than or equal to 12?

**Solution: **(i) P(8 ) = 5 /36 (ii) P(13 ) = 0 /36 = 0 (iii) P(less than or equal to 12) = 1

Language | English |

No. of Pages | 9 |

PDF Size | 1.5 MB |

Category | Education |

Source/Credits | cbseacademic.nic.in |

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