CBSE Class 10 Maths Sample Paper 2023

CBSE Class 10 Maths Sample Paper 2023: CBSE has released the CBSE Class 10 Maths Sample Paper 2023 Exam for giving ease to the 10th class students to understand and be familiar with the new pattern to be followed. The duration of the test will be 3 hours (180 minutes) and it will cover the entire syllabus. The CBSE Class 10 Maths Sample Paper 2022-2023 are now available at CBSE’s academic website i.e. cbseacademic.nic.in or you can also download Class 10 Maths Sample Papers for Basic & Standard Mathematics from the direct links provided in the article. 

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 2023Solution Link
Class 10 Standard Maths Sample Paper 2023 Solution Link 

CBSE Class 10 Maths Syllabus 2023- Click to Check

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 - 55 - 1010 - 1515 - 20 20 - 25 
Frequency1015 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)  123 4 5
Frequency  4 5  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

Section D

Section D consists of 4 questions of 5 marks each. 

Question32. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains. 

Solution: Let the average speed of the passenger train = x km/h. 

and the average speed of the express train = (x + 11) km/h.

As per the given data, the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. 

Therefore, 132 /𝑥 − 132 /𝑥+11 = 1 ⇒ 132 (𝑥+11−𝑥) /𝑥 (𝑥+11) = 1 

⇒ 132 (𝑥 +11- 𝑥) /𝑥(𝑥+11) = 1 

⇒ 132 × 11 = x(x + 11) 

⇒ x² + 11x – 1452 = 0 

⇒ x² + 44x -33x -1452 = 0 

⇒ x (x + 44) -33(x + 44) = 0 

⇒ (x + 44)(x – 33) = 0 

⇒ x = – 44, 33 

As the speed cannot be negative, the speed of the passenger train will be 33 km/h and the speed of the express train will be 33 + 11 = 44 km/h.

[OR] 

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution: Let the speed of the stream be x km/hr 

So, the speed of the boat in upstream = (18 - x) km/hr & 

the speed of the boat in downstream = (18 + x) km/hr 

ATQ, distance upstream speed - distance downstream speed = 1 ⇒ 24 /18 − 𝑥 − 24 /18 + 𝑥 = 1 

⇒ 24 [ 1 /18 − 𝑥 − 1 /18 + 𝑥 ] = 1 

⇒ 24 [ 18 + 𝑥−(18−𝑥) (18 − 𝑥)/(18 + 𝑥) ] = 1 

⇒ 24 [ 2𝑥 (18 − 𝑥)/(18 + 𝑥) ] = 1 

⇒ 24 [ 2𝑥 (18 − 𝑥)/(18 + 𝑥) ] = 1 

⇒ 48x = 324 - x²

⇒ x ² + 48x - 324 = 0 

⇒ (x + 54)(x - 6) = 0

⇒ x = -54 or 6 As speed to stream can never be negative, the speed of the stream is 6 km/hr

Question33.  Prove that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In the figure, find EC if AD DB = AE EC using the above theorem.

Solution: Figure Given, 

To prove, 

constructions Proof Application ----

Question34.  A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution: Volume of one conical depression = 1/3 x π r² h = 1 3 x 22 7 x 0.5 ² x 1.4 cm³ 

= 0.366 cm³ Volume of 4 conical depression 

= 4 x 0.366 cm³ 

= 1.464 cm³ Volume of cuboidal box 

= L x B x H = 15 x 10 x 3.5 cm³ 

= 525 cm³ Remaining volume of the box 

= Volume of cuboidal box – Volume of 4 conical depressions = 525 cm³ − 1.464 cm³ = 523.5 cm³

[OR]

Ramesh made a bird bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath

Solution: Let h be the height of the cylinder, and r the common radius of the cylinder and hemisphere. 

Then, the total surface area = CSA of cylinder + CSA of hemisphere = 2𝜋rh + 2𝜋r² 

= 2𝜋 r (h + r) 

= 2 x 22 7 x 30 (145 + 30) cm² 

= 2 x 22 7 x 30 x 175 cm²

= 33000 cm² 

= 3.3 m² 

Question35. A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons aged 18 years onwards but less than 60 years.

Age (in years)            Number of policyholders 
Below 20 2
20-254
25-3018 
30-3521
 35-40 33
40-4511 
45-503
50-55 6
55-602

Solution:

Age (in years)            Number of policyholders Cumulative Frequency (cf) 
Below 20 22
20-2546
25-3018 24
30-352145
 35-40 3378
40-4511 89
45-50392
50-55 698
55-602100

n = 100 ⇒ n/2 = 50, 

Therefore, median class = 35 – 40, 

Class size, h = 5, Lower limit of median class, l = 35, frequency f = 33, cumulative frequency cf = 45 

⇒Median = l + [ n 2 − cf f ] × h 

⇒Median = 35 + [ 50 − 45 33 ] × 5 = 35 + 25 33 = 35 + 0.76 = 35.76 Therefore, median age is 35.76 years

Section E 

Case study-based questions are compulsory.

Question36 Case Study – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in the 4th year and 2600 cars in the 8th year. Assuming that the production increases uniformly by a fixed number every year. Based on the above information answer the following questions.

I. Find the products in the 1st year. 1

II. Find the products in the 12th year. 1

III. Find the total production in the first 10 years.

Solution: I. Since the production increases uniformly by a fixed number every year, the number of Cars manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.

So, a + 3d = 1800 & a + 7d = 2600

So d = 200 & a = 1200

II. t12 = a + 11d ⇒ t30 = 1200 + 11 x 200

⇒ t12 = 3400

III.  Sn =𝑛/2 [2𝑎 + (𝑛 − 1)𝑑] ⇒ S10 = 10/2 [2 𝑥 1200 + (10 − 1) 200]

⇒ S10 =13/ 2 [2 𝑥 1200 + 9 x 200]

⇒ S10 = 5 x [2400 + 1800 ]

⇒ S10 = 5 x 4200= 21000

[OR]

In which year the total production will reach 15000 cars?

Solution:[OR]

Let in n years the production will reach to 31200

Sn = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑] = 31200 ⇒ 𝑛/2 [2 𝑥 1200 + (𝑛 − 1)200] = 31200

⇒𝑛/2 [2 x 1200 + (𝑛 − 1)200] = 31200 ⇒ 𝑛 [ 12 + (𝑛 − 1) ] = 312

⇒ n/2 + 11n -312 = 0

⇒ n/2 + 24n - 13n -312 = 0

⇒ (n +24)(n -13) = 0

⇒ n = 13 or – 24. As n can’t be negative. So n = 13 

Question 37. Case Study – 2 

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

Based on the above information answer the following questions using the coordinate geometry.

I.Find the distance between Lucknow (L) to Bhuj(B).
II.If Kota (K), internally divides the line segment joining Lucknow (L) to Bhuj (B) into 3: 2 then find the coordinate of Kota (K).
III.

Name the type of triangle formed by the places Lucknow (L), Nashik (N), and Puri (P) 

[OR] 

Find a place (point) on the longitude (y-axis) that is equidistant from the points Lucknow (L) and Puri (P).

Solution: I.LB = √ (𝑥2 − 𝑥1 )² + (𝑦2 − 𝑦1 )² 

⇒ LB = √ (0 − 5)²+ (7 − 10) ² 

LB = √ (5) ² + (3) ² 

⇒ LB = √25 + 9 

LB = √34 

Hence the distance is 150 √34 km  

II. Coordinate of Kota (K) is ( 3 x 5 + 2 x 0 3 + 2 , 3 x 7 + 2 x 10 3 + 2 ) 

 = ( 15+0/5 , 21+20/5 )

= (3, 41 5 ) 

 III. L(5, 10), N(2,6), P(8,6) 

LN = √ (2 − 5) 2 + (6 − 10) 2 = √ (3) 2 + (4) 2 = √9 + 16 = √25 = 5 

NP = √ (8 − 2) 2 + (6 − 6) 2 = √ (4) 2 + (0) 2 = 4 

PL = √ (8 − 5) 2 + (6 − 10) 2 = √ (3) 2 + (4) 2 

⇒ LB = √9 + 16 = √25 = 5 

as LN = PL ≠ NP, so ∆ LNP is an isosceles triangle.

OR 

N Let A (0, b ) be a point on the y – axis then AL = AP

 ⇒ √ (5 − 0) ² + (10 − b) ²

 = √ (8 − 0) ² + (6 − b) ² 

⇒ (5)²+ (10 − b)² 

= (8)² + (6 − b)²

⇒ 25 + 100 − 20𝑏 + b² 

= 64 + 36 − 12𝑏 + b² 

⇒ 8b = 25 

⇒ b = 25/ 8 So, the coordinate on the y-axis is (0, 25/8 ) 

Question38.  Case Study – 3 

Lakshaman Jhula is located 5 kilometers northeast of the city of Rishikesh in the Indian state of Uttarakhand. The bridge connects the villages of Tapovan to Jonk. Tapovan is in Tehri Garhwal district, on the west bank of the river, while Jonk is in Pauri Garhwal district, on the east bank. Lakshman Jhula is a pedestrian bridge also used by motorbikes. It is a landmark of Rishikesh.

A group of Class X students visited Rishikesh in Uttarakhand on a trip. They observed from a point (P) on a river bridge that the angles of depression of opposite banks of the river are 60° and 30° respectively. The height of the bridge is about 18 meters from the river.

Based on the above information answer the following questions

I.Find the distance PA.
II.Find the distance PB
III.

Find the width AB of the river.

[OR] 

Find the height BQ if the angle of the elevation from P to Q is 30°.

Solution: I. sin 60° =PC/PA

⇒√3/2 = 18/PA

⇒ PA = 12 √3 m

II. sin 30° = PC/PB

⇒1/2

=18/PB

⇒ PB = 36 m

III. tan 60° =PC/AC

⇒ √3 =18/AC

⇒ AC = 6 √3 m

tan 30° =PC/CB

⇒1/√3

=18/CB

⇒ CB = 18 √3 m

Width AB = AC + CB = 6 √3 + 18 √3 = 24 √3 m

[OR]

RB = PC =18 m & PR = CB = 18 √3 m

tan 30° = QR/PR

⇒1/√3

=QR/18 √3

⇒ QR = 18 m

QB = QR + RB = 18 + 18 = 36m. Hence height BQ is 36m.

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

Below are the sample questions from the official sample paper recently released by CBSE recently. 

Section A

Section A has 20 MCQs carrying 1 mark each.

Question1. Let a and b be two positive integers such that a = p³q⁴ and b = p²q³, where p and q are prime numbers. If HCF(a,b) = pmqn and LCM(a,b) = pr qs, then (m+n)(r+s)=

(a) 15 (b) 30 (c) 35 (d) 72

Solution: (c) 35

Question2. Let p be a prime number. The quadratic equation has its roots as factors of p is

(a) x²–px +p=0 (b) x²–(p+1)x +p=0 (c) x²+(p+1)x +p=0 (d) x² –px+p+1=0

Solution: (b) x² –(p+1)x +p=0

Question3. If α and β are the zeros of a polynomial f(x) = px²– 2x + 3p and α + β = αβ, then p is

(a)-2/3 (b) 2/3 (c) 1/3 (d) -1/3

Solution: (b) 2/3

Question4. If the system of equations 3x+y =1 and (2k-1)x +(k-1)y =2k+1 is inconsistent, then k =

(a) -1 (b) 0 (c) 1 (d) 2

Solution: (d) 2

Question 5. If the vertices of a parallelogram PQRS taken in order are P(3,4), Q(-2,3), and R(-3,-2), then the coordinates of its fourth vertex S are

(a) (-2,-1) (b) (-2,-3) (c) (2,-1) (d) (1,2)

Solution: (c) (2,-1)

Question6. ∆ABC~∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and AB²: PQ² = 4 : 9, then AM: PN =

(a) 3:2 (b) 16:81 (c) 4:9 (d) 2:3 

Solution: (d) 2:3

Question7. If x tan 60° cos 60°= sin60° cot 60° , 

then x = (a) cos30° (b) tan30° (c) sin30° (d) cot30°  

Solution: (b) tan 30° 

Question8.  If sinθ + cosθ = √2, then tanθ + cot θ = (a) 1 (b) 2 (c) 3 (d) 4 

Solution: (b) 2

Question9.  In the given figure, DE ∥ BC, AE = a units, EC =b units, DE =x units and BC = y units. Which of the following is true?

(a) x= 𝑎+𝑏/𝑎𝑦 (b) y= 𝑎𝑥/ 𝑎+𝑏 (c) x= 𝑎𝑦 /𝑎+𝑏 (d) 𝑥 𝑦 = 𝑎 /b

Solution: (c) x= 𝑎𝑦 /𝑎+𝑏

Question10. ABCD is a trapezium with AD ∥ BC and AD = 4cm. If the diagonals AC and BD intersect each other at O such that AO/OC = DO/OB =1/2, then BC = 

(a) 6cm (b) 7cm (c) 8cm (d) 9cm  

Solution: (c) 8cm

Question11. If two tangents inclined at an angle of 60ᵒ are drawn to a circle of radius 3cm, then the length of each tangent is equal to 

(a) 3√3 2 cm (b) 3cm (c) 6cm (d) 3√3cm 

Solution: (d) 3√3cm

Question12.  The area of the circle that can be inscribed in a square of 6cm is 

(a) 36π cm² (b) 18π cm² (c) 12 π cm² (d) 9π cm²

Solution:  (d) 9π cm²

Question13.  The sum of the length, breadth and height of a cuboid is 6√3cm and the length of its diagonal is 2√3cm. The total surface area of the cuboid is 

(a) 48 cm² (b) 72 cm² (c) 96 cm² (d) 108 cm² 

Solution:(c) 96 cm²   

Question14.  If the difference of Mode and Median of a data is 24, then the difference of median and mean is 

(a) 8 (b) 12 (c) 24 (d) 36  

Solution: (b) 12  

Question15. The number of revolutions made by a circular wheel of radius 0.25m in rolling a distance of 11km is 

(a) 2800 (b) 4000 (c) 5500 (d) 7000

Solution: (d) 7000  

Question16.  For the following distribution

Class  0-55-10 10-15 15-20 20-25
Frequency10 15 12 20   9

 the sum of the lower limits of the median and modal class is 

(a) 15 (b) 25 (c) 30 (d) 35

Solution:  (b) 25 

Question17.  Two dice are rolled simultaneously. What is the probability that 6 will come up at least once? 

(a)1/6 (b) 7/36 (c) 11/36 (d) 13/36 

Solution: (c) 11/36  

Question18. If 5 tanβ =4, then 5 𝑠𝑖𝑛𝛽−2 cos𝛽 5 sin𝛽+2 cos𝛽 = 

(a) 1/3 (b) 2/5 (c) 3/5 (d) 6 

Solution:  (a) 1/3

Question19.  DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct option Statement 

A (Assertion): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340 

Statement R( Reason) : HCF is always a factor of LCM 

(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 and 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: (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)    

Question20.  Statement A (Assertion): If the co-ordinates of the mid-points of the sides AB and AC of ∆ABC are D(3,5) and E(-3,-3) respectively, then BC = 20 units

Statement R( Reason) : The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

(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 and 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 consists of 5 questions of 2 marks each. 

Question21 If 49x+51y= 499, 51 x+49 y= 501, then find the value of x and y

Solution: Adding the two equations and dividing by 10, we get : x+y = 10 Subtracting the two equations and dividing by -2, we get : x-y =1 Solving these two new equations, we get, x = 11/2 y = 9/2

Question22 In the given figure below, AD /AE = AC /BD and ∠1 = ∠2. Show that ∆ BAE~ ∆CAD

Solution: In ΔABC, ∠1 = ∠2 ∴ AB = BD ………………………(I) 

Given, AD/AE = AC/BD Using equation (I), 

we get AD/AE = AC/AB ……………….(ii)

In ΔBAE and ΔCAD, by equation (ii), AC/AB = AD/AE ∠A= ∠A (common) 

∴ ΔBAE ~ ΔCAD [By SAS similarity criterion]

Question23 In the given figure, O is the center of the circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB= 75°. 

Solution: ∠PAO = ∠ PBO = 90° ( angle b/w radius and tangent)

 ∠AOB = 105° (By angle sum property of a triangle) 

∠AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the angle subtended by the arc at the center)

Question24 The length of the minute hand of a clock is 6cm. Find the area swept by it when it moves from 7:05 p.m. to 7:40 p.m. 

Solution: We know that, in 60 minutes, the tip of the minute hand moves 360°

 In 1 minute, it will move =360°/60 = 6°

 ∴ From 7: 05 pm to 7: 40 pm i.e. 35 min, 

it will move through = 35 × 6° = 210° 

∴ Area swept by the minute hand in 35 min = Area of the sector with sectorial angle θ of 210° 

and radius of 6 cm = 210 /360 x π x 6 2 = 7 /12 x 22 /7 x 6 x 6 =66cm² 

OR

 In the given figure, arcs have been drawn of radius 7cm each with vertices A, B, C, and D of quadrilateral ABCD as centers. Find the area of the shaded region. 

Solution: Let the measure of ∠A, ∠B, ∠C, and ∠D be θ₁, θ₂, θ₃, and θ₄ respectively 

Required area = Area of the sector with center A + Area of the sector with center B + Area of the sector with center C + Area of the sector with center D 

 = 𝛉₁/360 x π x 7 ²+ 𝛉₂ /360 x π x 7 ²+ 𝛉₃ /360 x π x 7 ² + 𝛉₄ /360 x π x 7 ² 

= (𝛉₁ + 𝛉₂ + 𝛉₃ + 𝛉₄) /360 x π x 7 ² 

= (𝟑𝟔𝟎)/ 360 x 𝟐𝟐 /7 x 7x 7 ( By angle sum property of a triangle) 

= 154 cm²

Question25 If sin(A+B) =1 and cos(A-B)= √3/2, 0°< A+B ≤ 90° and A> B, then find the measures of angles A and B. 

Solution: sin(A+B) =1 = sin 90, so A+B = 90……………….(I)

 cos(A-B)= √3/2 = cos 30, so A-B= 30……………(ii) 

From (i) & (ii) ∠A = 60° And ∠B = 30° 

OR 

Find an acute angle θ when cosθ − sin θ /cosθ+sin θ = 1−√3/ 1+√3

Solution: cosθ − sin θ /cosθ+sin θ = 1−√3/1+√3 

Dividing the numerator and denominator of LHS by cosθ, we get 1 − tan θ /1+tan θ = 1−√3 /1+√3 

Which on simplification (or comparison) gives tanθ = √3 Or θ= 60°

Section C

Question26 Given that √3 is irrational, prove that 5 + 2√3 is irrational.  

Solution: Let us assume 5 + 2√3 is rational, 

then it must be in the form of p/q where p and q are co-prime integers and q ≠ 0 

i.e 5 + 2√3 = p/q So √3 = 𝑝−5𝑞 2𝑞 ……………………(I) 

Since p, q, 5 and 2 are integers and q ≠ 0, HS of equation (i) is rational. 

But the LHS of (i) is √3 which is irrational. 

This is not possible. This contradiction has arisen due to our wrong assumption that 5 + 2√3 is rational. 

So, 5 + 2√3 is irrational.

Question27 If the zeroes of the polynomial x² +px +q are double in value to the zeroes of the polynomial 2x² -5x -3, then find the values of p and q. 

Solution: Let α and β be the zeros of the polynomial 2x² -5x -3 

Then α + β = 5/2 And αβ = -3/2. 

Let 2α and 2β be the zeros x2 + px +q 

Then 2α + 2β = -p 2(α + β) = -p 2 x 5/2 =-p So p = -5 And 2α x 2β = q 4 αβ = q So q = 4 x-3/2 = -6

Question28 A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr; it would have taken 6 hours more than the scheduled time. Find the length of the journey. 

Solution: Let the actual speed of the train be x km/hr and let the actual time taken by y hours. 

Distance covered is xy km If the speed is increased by 6 km/hr, 

then the time of the journey is reduced by 4 hours 

i.e., when the speed is(x+6)km/hr, the time of the journey is(y−4) hours. 

∴ Distance covered =(x+6)(y−4) 

⇒xy=(x+6)(y−4) 

⇒−4x+6y−24=0

 ⇒−2x+3y−12=0 …………………………….(I) 

Similarly xy=(x−6)(y+6) 

⇒6x−6y−36=0 

⇒x−y−6=0 ………………………………………(ii) 

Solving (i) and (ii) we get x=30 and y=24 Putting the values of x and y in equation (I), 

we obtain Distance =(30×24)km =720km. 

Hence, the length of the journey is 720km. 

OR 

Anuj had some chocolates, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate of ₹1 per chocolate and got a total of ₹400. If he had sold the first lot at the rate of ₹1 per chocolate, and the second lot at the rate of ₹4 for 5 chocolates, his total collection would have been ₹460. Find the total number of chocolates he had.

Solution: Let the number of chocolates in lot A be x 

And let the number of chocolates in lot B be y 

∴ total number of chocolates =x+y Price of 1 chocolate = ₹ 2/3, so for x chocolates = 𝟐 /𝟑 x 

and price of y chocolates at the rate of ₹ 1 per chocolate =y. ∴ by the given condition 𝟐/ 𝟑 x +y=400

 ⇒2x+3y=1200 ..............(I)

 Similarly x+ 𝟒 /𝟓 y = 460 

⇒5x+4y=2300 ........ (ii) 

Solving (i) and (ii) we get x=300 and y=200 

∴x+y=300+200=500 

So, Anuj had 500 chocolates.

Question29 Prove the following that- tan³ θ + cot³θ = secθ cosecθ – 2 sinθ cosθ 1+ tan² θ 1+ cot²θ 

Solution: LHS : sin³ θ/ cos³ θ + cos³ θ/ sin³ θ 1+ sin² θ/cos²θ 1+ cos² θ/ sin² θ

= sin³ θ/ cos³ θ + cos ³ θ/ sin³ θ (cos² θ + sin² θ)/cos² θ (sin² θ + cos² θ)/ sin² θ 

= sin³ θ/ cosθ+ cos³ θ /sinθ 

= sin⁴ θ + cos⁴ θ /cosθsinθ 

= (sin² θ + cos² θ)2 – 2 sin² θcos² θ /cosθsinθ 

= 1 - 2 sin² θcos² θ /cosθsinθ 

= 1/cosθsinθ - 2 sin² θcos² θ /cosθsinθ 

= secθcosecθ – 2sinθcosθ 

= RHS

Question30 Prove that a parallelogram circumscribing a circle is a rhombus 

Solution: Let ABCD be the rhombus circumscribing the circle with center O, such that AB, BC, CD, and DA touch the circle at points P, Q, R and S respectively. We know that the tangents drawn to a circle from an exterior point are equal in length. 

∴ AP = AS………….(1) 

BP = BQ……………(2) 

CR = CQ …………...(3) 

DR = DS……………(4). 

Adding (1), (2), (3), and (4) we get AP+BP+CR+DR = 

AS+BQ+CQ+DS (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) ∴ AB+CD=AD+BC-----------(5)

Since AB=DC and AD=BC (opposite sides of parallelogram ABCD) 

putting in (5) we get, 2AB=2AD or AB = AD. 

∴ AB=BC=DC=AD Since a parallelogram with equal adjacent sides is a rhombus, ABCD is a rhombus 

OR

In the figure XY and X'Y' are two parallel tangents to a circle with center O and another tangent AB with the point of contact C interesting XY at A and X'Y' at B, what is the measure of ∠AOB.

Solution: Join OC

 In Δ OPA and Δ OCA 

OP = OC (radii of same circle) 

PA = CA (length of two tangents from an external point) 

AO = AO (Common) 

Therefore, Δ OPA ≅ Δ OCA (By SSS congruency criterion)

Hence, ∠ 1 = ∠ 2 (CPCT) 

Similarly ∠ 3 = ∠ 4 ∠PAB + ∠QBA =180°(co interior angles are supplementary as XY∥X’Y’) 

2∠2 + 2∠4 = 180° ∠2 + ∠4 = 90°-------------------------(1) 

∠2 + ∠4 +∠AOB = 180° (Angle sum property) 

Using (1), we get, ∠AOB = 90°

Question31 Two coins are tossed simultaneously. What is the probability of getting

 (i) At least one head? 

(ii) At most one tail? 

(iii) A head and a tail?

Solution: (i) P (At least one head) = 3 /4

 (ii) P(At most one tail) = 3 /4 

(iii) P(A head and a tail) = 2/ 4 = 1/ 2

Section D

Question32 To fill a swimming pool two pipes are used. If the pipe of a larger diameter is used for 4 hours and the pipe of a smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately if the pipe of a smaller diameter takes 10 hours more than the pipe of a larger diameter to fill the pool. 

Solution: Let the time taken by the larger pipe alone to fill the tank= x hours 

Therefore, the time taken by the smaller pipe = x+10 hours Water filled by the larger pipe running for 4 hours = 4 /𝑥 liters 

Water filled by a smaller pipe running for 9 hours = 9 /𝑥+10 liters

We know that 4 /𝑥 + 9/ 𝑥+10 = 1 2 

Which on simplification gives: x ²−16x−80=0 

x ²−20x + 4x−80=0 

x(x-20) + 4(x-20)= 0 

(x +4)(x-20)= 0 

x=- 4, 20 x cannot be negative. 

Thus, x=20 x+10= 30 Larger pipe would alone fill the tank in 20 hours and a smaller pipe would fill the tank alone in 30 hours.

OR 

In a flight of 600km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr from its usual speed and the time of the flight increased by 30 min. Find the scheduled duration of the flight. 

Solution: Let the usual speed of the plane be x km/hr 

and the reduced speed of the plane be (x-200) km/hr 

Distance =600 km [Given] 

According to the question, (time taken at reduced speed) - (Schedule time) = 30 minutes = 0.5 hours. 

600/𝑥−200 − 600 /𝑥 = 1 /2 

Which on simplification gives: x ²- 200x−240000=0

 x ² -600x + 400x −240000=0 

x(x- 600) + 400( x-600) = 0 

(x-600)(x+400) =0 

x=600 or x=−400

But speed cannot be negative.

 ∴ The usual speed is 600 km/hr and the scheduled duration of the flight is 600/ 600 = 1 hour 

Question33. Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. Using the above theorem proves that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the nonparallel sides in the same ratio. 

Solution: For the Theorem: Given, To prove, Construction and figure Proof 

Let ABCD be a trapezium DC∥AB and EF is a line parallel to AB and hence to DC. 

To prove: 𝐃𝐄 /𝐄𝐀 = 𝐂𝐅/ 𝐅𝐁 

Construction: Join AC, meeting EF in G.

 Proof: In △ABC, we have GF∥AB CG/GA=CF/FB [By BPT] ......(1) 

In △ADC, we have EG∥DC ( EF ∥AB & AB ∥DC) DE/EA

= CG/GA [By BPT] .....(2) From (1) & (2), 

we get, 𝐃𝐄/ 𝐄𝐀 = 𝐂𝐅/ 𝐅𝐁 

Question34 Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively decided to provide a place and canvas for 1500 tents and share the whole expenditure equally. The lower part of each tent is cylindrical with a base radius of 2.8 m and a height of 3.5 m and the upper part is conical with the same base radius but of height 2.1 m. If the canvas used to make the tents costs ₹120 per m², find the amount shared by each school to set up the tents. 

Solution: Radius of the base of cylinder (r) = 2.8 m = Radius of the base of the cone (r) 

Height of the cylinder (h)=3.5 m Height of the cone (H)=2.1 m. 

Slant height of conical part (l)=√r²+H² = √(2.8)²+(2.1)² = √7.84+4.41 = √12.25 = 3.5 m 

Area of canvas used to make tent = CSA of cylinder + CSA of cone = 2×π×2.8×3.5 + π×2.8×3.5 = 61.6+30.8 = 92.4m²

Cost of 1500 tents at ₹120 per sq.m = 1500×120×92.4 = 16,632,000 

Share of each school to set up the tents = 16632000/50 = ₹332,640

OR 

There are two identical solid cubical boxes of sides 7cm. From the top face of the first cube, a hemisphere of diameter equal to the side of the cube is scooped out. This hemisphere is inverted and placed on the top of the second cube’s surface to form a dome. Find 

(i) the ratio of the total surface area of the two new solids formed 

(ii) the volume of each new solid formed

Solution: (i) SA for first new solid (S₁): 6×7×7 + 2 π ×3.52 - π ×3.52 

= 294 + 77 – 38.5 = 332.5cm²

SA for second new solid (S₂): 6×7×7 + 2 π ×3.52 - π ×3.52 = 294 + 77 – 38.5 = 332.5 cm² 

So S₁: S₂ = 1:1 

(ii) Volume for first new solid (V₁)= 7×7×7 - 2 3 π ×3.53 = 343 - 539 6 = 1519 6 cm3 Volume for second new solid (V₂)= 7×7×7 + 2 3 π ×3.53 = 343 + 539 6 = 2597 6 cm

Question35 The median of the following data is 525. Find the values of x and y, if the total frequency is 100

Class interval         Frequency
0−100
100−2005
200−300 x
300−400 12 
400−50017 
500−60020
600−700 y
700−8009
800−900  7
 900−1000 4 

Solution: Median = 525, so Median Class = 500 – 600

Class interval         FrequencyCumulative Frequency
0−1002
100−20057
200−300 x7+x 
300−400 12 19+ x
400−50017 36+ x
500−6002056+ x
600−700 y56+x+y
700−800965+x +y
800−900  772+x+y
 900−1000 4 76+x+y

76+x+y=100⇒x+y=24 ….(i)

Median = l +n/2−cf/fx h

Since, l=500, h=100, f=20, cf=36+x and n=100

Therefore, putting the value in the Median formula, we get;

525 = 500 + 50−(36+x)/20x 100

so x = 9

y = 24 – x (from eq.i)

y = 24 – 9 = 15

Therefore, the value of x = 9

and y = 15

Section E 

Case study-based questions are compulsory. 

Question 36. A tiling or tessellation of a flat surface is the covering of a plane using one or more geometric shapes, called tiles, with no overlaps and no gaps. Historically, tessellations were used in ancient Rome and in Islamic art. You may find tessellation patterns on floors, walls, paintings, etc. Shown below is a tiled floor in the archaeological Museum of Seville, made using squares, triangles, and hexagons.

A craftsman thought of making a floor pattern after being inspired by the above design. To ensure accuracy in his work, he made the pattern on the Cartesian plane. He used regular octagons, squares and triangles for his floor tessellation pattern

Use the above figure to answer the questions that follow: 

(i) What is the length of the line segment joining points B and F? 

(ii) The centre ‘Z’of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z? 

(iii) What are the coordinates of the point on y-axis equidistant from A and G? 

Solution: (i) B(1,2), F(-2,9) BF² 

= ( -2-1)²+ ( 9-2)²

 = ( -3)²+ ( 7)² 

= 9 + 49 = 58 So, BF = √58 units

(ii)W(-6,2), X(-4,0), O(5,9), P(3,11) 

Clearly, WXOP is a rectangle 

The point of intersection of the diagonals of a rectangle is the midpoint of the diagonals. 

So the required point is mid point of WO or XP 

= (−6+5 2 , 2+9 2 ) = (−1 2 , 11 2 ) 

(iii) A(-2,2), G(-4,7) Let the point on y-axis be Z(0,y) AZ² = GZ²

( 0+2)² + ( y-2)² 

= ( 0+4)² + ( y-7)² ( 2)² + y² + 4 -4y

= (4)²+ y² + 49 -14y 8-4y = 65-14y 10y

= 57 So, 

y= 5.7 i.e. the required point is (0, 5.7) 

OR What is the area of Trapezium AFGH?

Solution: A(-2,2), F(-2,9), G(-4,7), H(-4,4) 

Clearly GH = 7-4=3units 

AF = 9-2=7 units 

So, the height of the trapezium AFGH = 2 units 

So, area of AFGH = 1 2 (AF + GH) x height = 1 2 (7+3) x 2 = 10 sq. units

Question 37. The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in a concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

(i) If the first circular row has 30 seats, how many seats will be there in the 10th row? 

(ii) For 1500 seats in the auditorium, how many rows need to be there? 

OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after the 10th row? 

(iii) If there were 17 rows in the auditorium, how many seats will be there in the middle row?  

Solution: (i) Since each row is increasing by 10 seats, it is an AP with the first term a= 30, and common difference d=10.

 So number of seats in 10th row = 𝑎10 = a+ 9d = 30 + 9×10 = 120 

(ii) Sn = n 2 ( 2a + (n-1)d) 1500 = n 2 ( 2 × 30 + (n-1)10) 3000 = 50n + 10n2 n 2 +5n -300 =0 n 2 + 20n -15n – 300 =0 (n+20) (n-15) =0 Rejecting the negative value, n= 15

OR 

No. of seats already put up to the 10th row = S10 S10 = 10 /2 {2 × 30 + (10-1)10)}

= 5(60 + 90) = 750

 So, the number of seats still required to be put are 1500 -750 = 750 

(iii) If no. of rows =17 then the middle row is the 9 th row 

𝑎8 = a+ 8d = 30 + 80 = 110 seats 

Question38. We all have seen airplanes flying in the sky but might have not thought of how they actually reached the correct destination. Air Traffic Control (ATC) is a service provided by ground-based air traffic controllers who direct aircraft on the ground and through a given section of controlled airspace and can provide advisory services to aircraft in non-controlled airspace. Actually, all this air traffic is managed and regulated by using various concepts based on coordinate geometry and trigonometry.

At a given instance, ATC finds that the angle of elevation of an airplane from a point on the ground is 60°. After a flight of 30 seconds, it is observed that the angle of elevation changes to 30°. The height of the plane remains constant as 3000√3 m. Use the above information to answer the questions that follow- 

(i) Draw a neatly labeled figure to show the above situation diagrammatically. 

(ii) What is the distance traveled by plane in 30 seconds? 

Solution: P and Q are the two positions of the plane flying at a height of 3000√3m. 

A is the point of observation. 

(ii) In △ PAB, tan60° =PB/AB Or √3 = 3000√3/ AB So AB=3000m tan30°= QC/AC 1/√3= 3000√3 / AC AC = 9000m distance covered = 9000- 3000 = 6000 m.

OR 

Keeping the height constant, during the above flight, it was observed that after 15(√3 -1) seconds, the angle of elevation changed to 45°. How much is the distance traveled in that duration? 

(iii) What is the speed of the plane in km/hr?

Solution: In △ PAB, tan60° =PB/AB Or √3 = 3000√3/ AB

 So AB=3000m tan45° = RD/AD 1= 3000√3 / AD 

AD = 3000√3 m distance covered = 3000√3 - 3000 = 3000(√3 -1)m.(iii)speed = 6000/ 30 = 200 m/s = 200 x 3600/1000 = 720km/h

Alternatively: speed = 3000(√3 −1) 15(√3 −1) = 200 m/s = 200 x 3600/1000 = 720km/hr

Steps to download CBSE Class 10 Maths Sample Paper 2023

Download the CBSE Maths Sample Paper Class 10, 2023 by following the below-mentioned steps

Step I- Visit the official website of CBSE Academic @ www.cbseacademic.nic.in. 

Step II- Click on the notification appearing in the academic section- “Sample Question Papers of Classes X for Board Exams 2023”.

Step III- Now click on the link mentioned under “Sample Papers Class X”.

Step IV- The list of all subjects “Class X Sample Question Paper & Marking Scheme for Exam 2023” appears on the screen. 

Step V- Click on “SQP” for “Maths” and download CBSE Class 10th Maths Sample Paper 2023 pdf along with Answer Key. 

Step VI- Check the marking scheme after attempting each subject CBSE Class 10th Maths Sample Paper 2023

Related Links
CBSE Class 10 English Syllabus 2023CBSE Class 10 Science Syllabus 2023
CBSE Class 10 Maths Syllabus 2023CBSE Class 10 Hindi Syllabus 2023
CBSE Class 10 Science Sample Paper 2023CBSE Class 10 Hindi Sample Paper 2023
CBSE Class 10 English Sample Paper 2023CBSE Class 10 Social Science Sample Paper 2023
CBSE Class 10 Sample Paper 2023

 

CBSE Class 10 Maths Sample Paper 2023: FAQs

Ans. CBSE has released single sample paper for each subject.

Ans. The marking scheme for CBSE Class 10 Maths Exam has been discussed in detail in the article, check here.

Ans. Yes, CBSE has released Class 10 Maths Sample Paper 2023 on CBSE Academic official website.

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