Maths Sample Paper Class 10 Term 2
CBSE Class 10 Term 2 Maths Sample Paper: CBSE has released the CBSE Class 10 Maths Sample Paper for Term 2 Exam for giving ease to the 10th class students to understand and be familiar with the new pattern to be followed. CBSE has scheduled CBSE Class 10 Maths Term 2 Exam for 05th May 2022 (Thursday) and students must prepare well for the examination. The duration of the test will be 120 minutes (2 hours) and it will cover only the rationalized syllabus of Term II only (i.e. approx. 50% of the entire syllabus). The CBSE Class 10 Maths Sample Paper 2021-22 are now available at CBSEβs academic website i.e. cbseacademic.nic.in or you can also download Class 10 Maths Term 2 Sample Papers for Basic & Standard Mathematics from the direct links provided in the article.
Sample Paper Class 10 Term 2 Maths
The pattern for CBSE Class 10 Maths Term 2 (Basic & Standard) has been discussed along with the Class 10 Maths Sample Paper which is as follows-
For Basic Maths Syllabus- Class 10 Term 2 Basic Maths Exam will consist of 14 questions divided into 3 sections A, B, C. Section A comprises of 6 questions of 2 marks each, Section B comprises of 4 questions of 3 marks each, and Section C comprises of 4 questions of 4 marks each.
For Standard Maths Syllabus- Class 10 Term 2 Standard Maths Exam will consist of 14 questions divided into 3 sections A, B, C. Section A comprises of 6 questions of 2 marks each, Section B comprises of 4questions of 3 marks each, and Section C comprises of 4 questions of 4 marks each.
Maths Class 10 Term 2 Answer Key 2022- Click Here
Class 10 Maths Term 2 Sample Paper PDF
To give students a clear vision for the Class 10 Mathematics Term 2 Exam, CBSE has uploaded sample question papers for CBSE Class 10th Maths Sample Paper 2022 Term 2 (Basic & Standard) on its official website https://cbseacademic.nic.in/index.html. CBSE Class 10 Maths Term 2 Sample Paper with Solutions PDF can be downloaded directly from here along with the answer key containing detailed solutions to the Class 10 Maths Term 2 Sample Papers.
CBSE Class 10 Term 2 Maths Syllabus-Click to Check
Maths Sample Paper Class 10 Term 2 with Solutions
Below are the questions that have been mentioned in the CBSE Class 10 Term-2 Maths Sample Paper with Solutions.
SECTION A
Question 1- Find the roots of the quadratic equation 3xΒ² β 7π₯ β 6 = 0.
Solution- 3xΒ² β 7π₯ β 6 = 0
β 3xΒ² β 9π₯ + 2π₯ β 6 = 0
β 3π₯(π₯ β 3) + 2(π₯ β 3) = 0
β (π₯ β 3)(3π₯ + 2) = 0
β΅ π₯ = 3, β 2/3
OR
Find the values of k for which the quadratic equation 3xΒ² + ππ₯ + 3 = 0 has real and equal roots.
Solution- Since the roots are real and equal,
β΄ π· = bΒ² β 4ππ = 0
β kΒ² β 4Γ3Γ3 = 0 (β΅ π = 3, π = π, π = 3)
β kΒ² = 36
β k = 6 ππ β6
Question 2- Three cubes each of volume 64cmΒ³ are joined end to end to form a cuboid. Find the total surface area of the cuboid so formed?
Solution- Let π be the side of the cube and L, B, H be the dimensions of the cuboid
Since bΒ³ = 64cmΒ³ β΄ π = 4 ππ
Total surface area of cuboid is 2[πΏπ΅ + π΅π» + π»πΏ], Where L=12, B=4 and H=4
=2(12 Γ 4 + 4 Γ 4 + 4 Γ 12) cmΒ² = 224 cmΒ²
Question 3- An inter-house cricket match was organized by a school. The distribution of runs made by the students is given below. Find the median runs scored.
Runs scored | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Number of students | 4 | 6 | 5 | 3 | 4 |
Solution-
Runs scored | Frequency | Cumulative Frequency |
0-20 | 4 | 4 |
20-40 | 6 | 10 |
40-60 | 5 | 15 |
60-80 | 3 | 18 |
80-100 | 4 | 22 |
Total frequency (N) = 22
π/2 = 11; So 40-60 is the median class.
Median = π + ( π/2 )βππ/π Γ β
= 40 + 11β10/5 x 20
= 44 runs
Question 4- Find the common difference of the AP 4,9,14,β¦ If the first term changes to 6 and the common difference remains the same then write the new AP.
Solution- The common difference is 9 - 4=5
If the first term is 6 and the common difference is 5,
then the new AP is, 6, 6+5, 6+10β¦
=6,11,16β¦
Question 5- The mode of the following frequency distribution is 38. Find the value of x.
Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 60-70 |
Frequency | 7 | 9 | 12 | 16 | x | 11 |
Solution- β΅ Mode = 38.
β΄ The modal class is 30-40.
Mode = π + π1β π0 /2π1βπ0βπ2 Γ β
=30 + 16β12/ (32β12βπ₯ )x10 = 38
4/ 20βπ₯ x 10 =8
8(20-x) = 40
20-x= 5
X= 15
Question 6- XY and MN are the tangents drawn at the endpoints of the diameter DE of the circle with centre O. Prove that XY || MN.

Solution- β΅XY is the tangent to the circle at the point D
β΄ OD β XY β γ₯ODX = 90Β° β γ₯EDX = 90Β°
Also, MN is the tangent to the circle at E
β΄ OE β MN β γ₯OEN = 90Β° β γ₯DEN = 90Β°
γ₯EDX = γ₯ DEN (πππβ 90Β°).
which are alternate interior angles.
β΄ XY ΰ₯₯ MN
OR
In the given figure, a circle is inscribed in the quadrilateral ABCD. Given AB=6 cm, BC=7 cm and CD=4 cm. Find AD.

Solution- β΅Tangent segments drawn from an external point to a circle are equal
β΄ BP=BQ
CR=CQ
DR=DS
AP=AS
β BP+CR+DR+AP = BQ+CQ+DS+AS
β AB+DC = BC+AD
β΄ AD= 10-7= 3 cm
SECTION B
Question 7- An AP 5, 8, 11β¦has 40 terms. Find the last term. Also find the sum of the last 10 terms.
Solution- First Term of the AP(a) = 5
Common difference (d) = 8-5=3
Last term = a40 = a+(40-1) d
= 5 + 39 Γ 3 = 122
Also π31 = π + 30π = 5 + 30 Γ 3 = 95
Sum of last 10 terms = π/2 (π31 + π40)
= 10/2 (95 + 122)
= 5 Γ 217 = 1085
Question 8- A tree is broken due to the storm in such a way that the top of the tree touches the ground and makes an angle of 30Β° with the ground. The length of the broken upper part of the tree is 8 meters. Find the height of the tree before it was broken.
Solution- Let, AB be the tree broken at C,
Also let π΄πΆ = π₯
In β CAD, sin30Β° = π΄πΆ π·πΆ
β 1/2 = π₯/8 β π₯ = 4 π
β the length of the tree is = 8+4 =12m
OR
Two poles of equal height are standing opposite each other on either side of the road 80m wide. From a point between them on the road the angles of elevation of the top of the two poles are respectively 60Β° and 30Β° . Find the distance of the point from the two poles.
Solution- Let AB and CD be two poles of height h meters also let P be a point between them on the road which is x meters away from foot of first pole AB, PD= (80-x) meters.
In βABP, π‘ππ60Β° = β/π₯ β β = π₯β3 .β¦(1)
In βCDP, π‘ππ 30Β° = β/80βπ₯ β β =80βπ₯/β3 β¦.(2)
π₯β3 = 80 β π₯/β3 [β΅ πΏπ»π(1) = πΏπ»π(2), π π πππ’ππ‘πππ π π»π]
β 3π₯ = 80 β π₯ β 4π₯ = 80 β π₯ = 20π
So, 80 β π₯ = 80 β 20 = 60π
Hence the point is 20m from one pole and 60 meters from the other pole.
Question 9- PA and PB are the tangents drawn to a circle with centre O. If PA= 6 cm and β APB=60Β° , then find the length of the chord AB.

Solution- PA = PB (Tangent segments drawn to a circle from an external point are equal)
β΄ In βπ΄ππ΅, γ₯PAB = γ₯PBA
Also, γ₯APB = 60Β°
In βπ΄ππ΅, sum of three angles is 180Β°
Therefore, γ₯PAB + γ₯PBA = 180Β°
β΄ γ₯APB= 180Β° β 60Β° = 120Β°
γ₯PAB = γ₯PBA = 60Β°(β΄ γ₯PAB = γ₯PBA)
β΅ βπ΄ππ΅ is an equilateral triangle.
So, π΄π΅ = 6cm
Question 10- The sum of the squares of three positive numbers that are consecutive multiples of 5 is 725. Find the three numbers.
Solution- Let the three consecutive multiples of 5 be 5x, 5x+5, 5x+10.
Their squares are (5π₯)Β², (5π₯ + 5)Β² and(5π₯ + 10)Β²
(5π₯)Β²+ (5π₯ + 5)Β²+ (5π₯ + 10)Β²= 725
β25π₯Β²+ 25π₯Β²+ 50x + 25 + 25π₯Β²+ 100x + 100 = 725
β 75π₯Β²+ 150π₯ β 600 = 0
β π₯Β² + 2π₯ β 8 = 0
β (π₯ + 4)(π₯ β 2) = 0
β π₯ = β4, 2
β π₯ = 2 (ignoring βve value)
So the numbers are 10, 15 and 20
Question 11- Construct two concentric circles of radii 3cm and 7cm. Draw two tangents to the smaller circle from a point P which lies on the bigger circle.
Solution- Draw two concentric circles with centre O and radii 3 cm and 7 cm respectively.
βJoin OP and bisect it at π β², so ππβ² = π β²π
βConstruct circle with centre π β² and radius π β²π
βJoin PA and PB

OR
Draw a pair of tangents to a circle of radius 6cm which are inclined to each other at an angle of 60Β° . Also, find the length of the tangent.
Solution- Draw a circle of radius 6cm
βDraw OA and Construct β π΄ππ΅ = 120Β°
βDraw β ππ΄π = β ππ΅π = 90Β°
βPA and PB are required tangents
βJoin OP and apply tanβ π΄ππ = tan 30Β° = 6 ππ΄ β Length of tangent = 6β3 cm

Question 12- The following age-wise chart of 300 passengers flying from Delhi to Pune is prepared by the Airlines staff. Find the mean age of the passengers.
Age | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 | Less than 60 | Less than 70 | Less than 80 |
Number of passengers | 14 | 44 | 82 | 134 | 184 | 245 | 287 | 300 |
Solution- Converting the cumulative frequency table into exclusive classes, we get:
Age | No of passengers (fi) | xi | fi xi |
0-10 | 14 | 5 | 70 |
10-20 | 30 | 15 | 450 |
20-30 | 38 | 25 | 950 |
30-40 | 52 | 35 | 1820 |
40-50 | 50 | 45 | 2250 |
50-60 | 61 | 55 | 3355 |
60-70 | 42 | 65 | 2730 |
70-80 | 13 | 75 | 975 |
β ππ = 300 | β πππ₯π =12600 |
Mean age = π₯Μ = β πππ₯π/ β ππ = 12600/300
π₯Μ = 42
Question 13- A lighthouse is a tall tower with light near the top. These are often built on islands, coasts or on cliffs. Lighthouses on water surface act as a navigational aid to the mariners and send warning to boats and ships for dangers. Initially wood, coal would be used as illuminators. Gradually it was replaced by candles, lanterns, electric lights. Nowadays they are run by machines and remote monitoring. Prongs Reef lighthouse of Mumbai was constructed in 1874-75. It is approximately 40 meters high and its beam can be seen at a distance of 30 kilometres. A ship and a boat are coming towards the lighthouse from opposite directions. Angles of depression of flash light from the lighthouse to the boat and the ship are 30Β° and 60Β° respectively.

i) Which of the two, the boat or the ship is nearer to the lighthouse. Find its distance from the lighthouse?
ii) Find the time taken by the boat to reach the lighthouse if it is moving at the rate of 20 km per hour.
Solution (i) The ship is nearer to the lighthouse as its angle of depression is greater.
In β ACB, tan 60Β° = AB/BC
ββ3 = 40/BC
β΄ BC = 40/β3 = 40β3/3 m

Solution (ii) In β ADB, tan 30Β° = AB/BC
β 1/ β3 = 40/DB
β΄ DB = 40β3m
Time taken to cover this distance = ( 60/ 2000 Γ 40β3) minutes
= 60β3/100 = 2.076 minutes
Question 14- Krishnanagar is a small town in the Nadia District of West Bengal. Krishnanagar clay dolls are unique in their realism and the quality of their finish. They are created by modeling coils of clay over a metal frame. The figures are painted in natural colours and their hair is made either from sheepβs wool or jute. Artisans make models starting from fruits, animals, God, goddesses, farmers, fishermen, weavers to Donald Duck and present comic characters. These creations are displayed in different national and international museums.
Here are a few images (not to scale) of some clay dolls of Krishnanagar.

The ratio of diameters of red spherical apples in Doll-1 to that of spherical oranges in Doll-2 is 2:3. In Doll-3, a male doll of blue colour has a cylindrical body and a spherical head. The spherical head touches the cylindrical body. The radius of both the spherical head and the cylindrical body is 3cm and the height of the cylindrical body is 8cm.
Based on the above information answer the following questions:
i) What is the ratio of the surface areas of red spherical apples in Doll-1 to that of spherical oranges in Doll-2.?
ii) The blue doll of Doll-3 is melted and its clay is used to make the cylindrical drum of Doll-4. If the radius of the drum is also 3cm, find the height of the drum.
Solution- (i) Let π1πππ π2 be respectively the radii of apples and oranges
β΅ 2π1: 2π2 = 2: 3 β π1: π2 = 2: 3
4ππ1Β²: 4ππ2Β² = (π1/π2)Β² = (2/3)Β²
= 4: 9
Solution- (ii) Let the height of the drum be h.
The volume of the drum = volume of the cylinder + volume of the sphere
Ο3Β²h = (Ο3Β²Γ 8 + 4/3 Ο3Β³) ππΒ³
β β = (8 + 4)ππ
β β = 12ππ
Class 10 Maths Term 2 Sample Paper- Standard Maths
Question 1- Find the value of a25 - a15 for the AP: 6, 9, 12, 15, β¦β¦β¦..
Solution- a = 6, d = 3 ; a25 = 6 + 24(3) = 78
a15 = 6 + 14(3) = 48 ; a25 - a15 = 78 - 48 = 30
OR
If 7 times the seventh term of the AP is equal to 5 times the fifth term, then find the value of its 12th term.
Solution- 7(π + 6π) = 5(π + 4π)
β2π + 22π = 0
β π + 11π = 0
β π‘12 = 0
Question 2- Find the value of π so that the quadratic equation ππ₯(5π₯ β 6) = 0 has two equal roots.
Solution- 5mxΒ² - 6mx + 9 = 0
bΒ² - 4ac = 0 β (-6m)Β² - 4(5m)(9) = 0
β 36m(m - 5) = 0
β m = 0, 5 ; rejecting m=0, we get m = 5
Question 3- From a point P, two tangents PA and PB are drawn to a circle C(0, r). If OP = 2r, then find β π΄ππ΅. What type of triangle is APB?

Solution- let β π΄ππ = π
ππππ = ππ΄/ ππ = 1/ 2 β π = 30Β°
β β π΄ππ΅ = 2π = 60Β°
Also β ππ΄π΅ = β ππ΅π΄ = 60Β°
(β΅ ππ΄ = ππ΅)
ββ³ π΄ππ΅ is equilateral
Question 4- The curved surface area of a right circular cone is 12320 cmΒ². If the radius of its base is 56 cm, then find its height.
Question 5- Mrs. Garg recorded the marks obtained by her students in the following table. She calculated the modal marks of the students of the class as 45. While printing the data, a blank was left. Find the missing frequency in the table given below
Marks Obtained | 0 β 20 | 20 β 40 | 40-60 | 60-80 | 80-100 |
Number of Students | 5 | 10 | β | 6 | 3 |
Question 6- If Ritu were younger by 5 years than what she really is, then the square of her age would have been 11 more than five times her present age. What is her present age?
OR
Solve for x: 9xΒ² - 6px + (pΒ² - qΒ²) = 0
Question 7- Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median
Distance (in m) | 0 β 1 | 1 β 2 | 2 β 3 | 3 β 4 | 4 β 5 |
Number of Students | 40 | 80 | 62 | 38 | 30 |
Question 8- Construct a pair of tangents to a circle of radius 4cm, which are inclined to each other at an angle of 60Β°.
Question 9- The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
Runs scored | 0 β 40 | 40 β 80 | 80 β 120 | 120 β 160 | 160 β 200 |
Number of batsmen | 12 | 20 | 35 | 30 | 23 |
Question 10- Two vertical poles of different heights are standing 20m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60Β° and angle of elevation of the top of the second pole from the foot of the first pole is 30Β°. Find the difference between the heights of two poles. (Take β3 = 1.73)
OR
A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60Β°. Calculate the height of the building. (Take β3 = 1.73)
Question 11- The internal and external radii of a spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of diameter 14cm, find the height of the cylinder. Also find the total surface area of the cylinder. (Take π = 22/7 )
Question 12- Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.
OR
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that β PTQ = 2β OPQ

Question 13- Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites. A guard, stationed at the top of a 240m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes(tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30Β°.

(Lighthouse of Mumbai Harbour. Picture credits - Times of India Travel) i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower. ii) After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(β3 - 1) m. He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower?
Question 14- Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms, and shoulders, support required from other muscles helps in toning up the whole body.

Nitesh wants to participate in the push-up challenge. He can currently make 3000 push-ups in one hour. But he wants to achieve a target of 3900 push-ups in 1 hour for which he practices regularly. With each day of practice, he is able to make 5 more push-ups in one hour as compared to the previous day. If on first day of practice he makes 3000 push-ups and continues to practice regularly till his target is achieved.
Keeping the above situation in mind answer the following questions:
i) Form an A.P representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished?
ii) Find the total number of push-ups performed by Nitesh up to the day his goal is achieved.
Steps to download CBSE Class 10 Maths Sample Paper 2021-22
Download the CBSE Maths Sample Paper 2022 Class 10 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- βSample Question Papers of Classes X for Term 2 Exams 2022β.
Step III- Now click on the link mentioned under βSample Papers Class Xβ.
Step IV- The list of all subjects βClass X Sample Question Paper & Marking Scheme for Exam 2022β appears on the screen.
Step V- Click on βSQPβ for βMathsβ and download CBSE Class 10th Maths Sample Paper Term 2 pdf along with Answer Key.
Step VI- Check the marking scheme after attempting each subject CBSE Class 10th Maths Sample Paper 2022 Term 2.
Class 10 Maths Term-1 Sample Paper | Solution |
Class 10 Basic Maths Term-1 Sample Paper | Solution Link |
Class 10 Standard Maths Term-1 Sample Paper | Solution Link |