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Term 2 Class 10 Maths Sample Paper 2022
Central Board of Secondary Education has published the CBSE Class 10 Maths Sample Paper 2021-22 on its official website on 19th January 2022 on the official website https://www.cbse.gov.in/. The Central Board of Secondary Education will conduct CBSE Term 2 examination for Class 10th and 12th in March-April 2022. The candidates who are going to appear in CBSE Term 2 examination must solve the CBSE Class 10 Maths Sample Paper 2021-22 given on this page. Through CBSE Class 10 Maths Sample Paper 2021-22, the candidates will get to know about the difficulty level of the CBSE Term 2 examination. In this article, we have given CBSE Class 10 Maths Sample Paper 2021-22. The candidates must solve the CBSE Class 10 Maths Sample Paper 2021-22 given on this page and bookmark this page to get all the updates.
Read: Tips to get 95+ in Term 2 Class 10 Maths
Read: HPBOSE 10th, 12th Term 1 Result 2022 is out at @hpbose.org.
Class 10 Maths Standard Term 2 Sample Paper 2022
Class 10 Maths Standard Sample Paper 2022 Term 2 Answer keys
CBSE Class 10 Term 2 Maths Basic Sample Paper 2022
Term 2 Class 10 Maths Basic Sample Paper 2021-2022 Answer keys
CBSE Class 10th Maths Sample Paper 2022 Term-2: Exam Pattern
Here we have discussed the CBSE Class 10 Maths Term-2 exam pattern. The candidates appearing in the CBSE Term 2 examination must check the exam pattern given below:
Exam Pattern for Basic Maths
- The Basic Maths exam of Class 10th Term 2 consists of 14 questions.
- The Basic Maths is divided into three sections namely A, B, C.
- Section A: In Section A, 6 questions will be asked. Each question carries 2 marks.
- Section B: In Section B, 4 questions will be asked. Each question carries 3 marks.
- Section C: In Section C, 4 questions will be asked. Each question carries 4 marks.
Exam Pattern for Standard Maths
- The Standard Maths exam of Class 10th Term 2 consists of 14 questions.
- The Standard Maths is divided into three sections namely A, B, C.
- Section A: In Section A, 6 questions will be asked. Each question carries 2 marks.
- Section B: In Section B, 4 questions will be asked. Each question carries 3 marks.
- Section C: In Section C, 4 questions will be asked. Each question carries 4 marks.
CBSE Term 2 Class 10 Maths Sample Paper 2022: Important Questions
Below we have given the CBSE Class 10th Maths Term-2 Sample Paper 2022. Solve the CBSE Class 10th Maths Term-2 Sample Paper 2022 and assess your preparation:
- Find the roots of the quadratic equation 3𝑥 2 − 7𝑥 − 6 = 0.
OR
Find the values of k for which the quadratic equation 3𝑥 2 + 𝑘𝑥 + 3 = 0 has real and equal roots.
Solution:
3𝑥 2 − 7𝑥 − 6 = 0
⇒ 3𝑥 2 − 9𝑥 + 2𝑥 − 6 = 0
⇒ 3𝑥(𝑥 − 3) + 2(𝑥 − 3) = 0
⇒ (𝑥 − 3)(3𝑥 + 2) = 0
∵ 𝑥 = 3, − 2/3
OR
Since the roots are real and equal,
∴ 𝐷 = 𝑏 2 − 4𝑎𝑐 = 0
⇒k 2 – 4×3×3 = 0 (∵ 𝑎 = 3, 𝑏 = 𝑘, 𝑐 = 3)
⇒k 2 = 36
⇒k = 6 𝑜𝑟 −6
2. Three cubes each of volume 64cm3 are joined end to end to form a cuboid. Find the total surface area of the cuboid so formed?
Solution:
Let 𝑙 be the side of the cube and L, B, H be the dimensions of the cuboid Since 𝑙 3 = 64 𝑐𝑚3 ∴ 𝑙 = 4 𝑐𝑚
Total surface area of cuboid is 2[𝐿𝐵 + 𝐵𝐻 + 𝐻𝐿], Where L=12, B=4 and H=4
=2(12 × 4 + 4 × 4 + 4 × 12) 𝑐𝑚2 = 224𝑐𝑚2
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
𝑁/2 = 11; So 40-60 is the median class.
Median = 𝑙 + ( 𝑁 2 )−𝑐𝑓 𝑓 × ℎ
= 40 + 11−10/ 5 x 20
= 44 runs
4. Find the common difference of the AP 4,9,14,… 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+10…
=6,11,16….
5. The mode of the following frequency distribution is 38. Find the value of x
| Runs scored | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 60-70 |
| Number of students | 7 | 9 | 12 | 16 | x | 11 |
Solution:
∵ Mode = 38.
∴ The modal class is 30-40.
Mode = 𝑙 + 𝑓1− 𝑓0 /2𝑓1−𝑓0−𝑓2 × ℎ
=30 + 16−12/ (32−12−𝑥 )x10 = 38
4/ 20−𝑥 x 10 =8
8(20-x) = 40
20-x= 5
X= 15
Q.6 XY and MN are the tangents drawn at the endpoints of the diameter DE of the circle with center O. Prove that XY MN.
OR
In the given figure, a circle is inscribed in the quadrilateral ABCD. Given AB=6cm, BC=7cm and CD=4cm. Find AD.
Solution:
∵XY is the tangent to the circle at the point D
∴ OD XY ODX = 900
EDX = 900
Also, MN is the tangent to the circle at E
∴ OE MN OEN = 900 DEN = 900
⇒ EDX = DEN (𝑒𝑎𝑐ℎ 900 ).
which are alternate interior angles.
∴ XY || MN
OR
∵Tangent segments drawn from an external point to a circle are equal
∴ BP=BQ
CR=CQ
DR=DS
AP=AS
⇒BP+CR+DR+AP = BQ+CQ+DS+AS
⇒ AB+DC = BC+AD
∴ AD= 10-7
= 3 cm
7. An AP 5, 8, 11…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 × 3 = 122
Also 𝑎31 = 𝑎 + 30𝑑 = 5 + 30 × 3 = 95
Sum of last 10 terms = 𝑛 2 (𝑎31 + 𝑎40) = 10
2 (95 + 122) = 5 × 217 = 1085
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 300 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.
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 600 and 300 . Find the distance of the point from the two poles.
Solution:
Let, AB be the tree broken at C,
Also let 𝐴𝐶 = 𝑥
In ∆ CAD, sin300 = 𝐴𝐶
𝐷𝐶 ⇒ 1 2 = 𝑥 8
⇒ 𝑥 = 4 𝑚
⇒the length of the tree is = 8+4 =12m
OR
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 ∆ABP, 𝑡𝑎𝑛60𝑜 = ℎ 𝑥
⇒ ℎ = 𝑥√3 .…(1)
In ∆CDP, 𝑡𝑎𝑛 30𝑜 = ℎ 80−𝑥
⇒ ℎ = 80−𝑥 √3 ….(2) 𝑥√3 = 80 − 𝑥 √3 [∵ 𝐿𝐻𝑆(1) = 𝐿𝐻𝑆(2), 𝑠𝑜 𝑒𝑞𝑢𝑎𝑡𝑖𝑛𝑔 𝑅𝐻𝑆] ⇒ 3𝑥 = 80 − 𝑥
⇒ 4𝑥 = 80
⇒ 𝑥 = 20𝑚 So, 80 − 𝑥 = 80 − 20 = 60𝑚
Hence the point is 20m from one pole and 60 meters from the other pole.
9. PA and PB are the tangents drawn to a circle with center O. If PA= 6 cm and APB=600, then find the length of the chord AB.
Solution:
PA = PB (Tangent segments drawn to a circle from an external point are equal)
∴ In ∆𝐴𝑃𝐵, PAB = PBA
Also, APB = 600
In ∆𝐴𝑃𝐵, sum of three angles is 1800 .
Therefore, PAB + PBA = 1800 – APB= 1800 – 600 = 1200 .
∴ PAB = PBA = 600 (∵ PAB = PBA)
∵ ∆𝐴𝑃𝐵 is an equilateral triangle.
So, 𝐴𝐵 = 6𝑐𝑚
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 (5𝑥) 2 , (5𝑥 + 5) 2and(5𝑥 + 10) 2 .
(5𝑥) 2 + (5𝑥 + 5) 2 + (5𝑥 + 10) 2 = 725
⇒25𝑥 2 + 25𝑥 2+ 50x + 25 + 25𝑥 2 + 100x + 100 = 725
⇒ 75𝑥 2 + 150𝑥 − 600 = 0
⇒ 𝑥 2 + 2𝑥 − 8 = 0
⇒ (𝑥 + 4)(𝑥 − 2) = 0
⇒ 𝑥 = −4, 2
⇒ 𝑥 = 2 (ignoring –ve value)
So the numbers are 10, 15 and 20
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.
OR
Draw a pair of tangents to a circle of radius 6cm which are inclined to each other at an angle of 600 . Also find the length of the tangent.
Solution:
Draw two concentric circles with center O and radii 3cm and 7cm respectively. Join OP and bisect it at 𝑂 ′ , so 𝑃𝑂′ = 𝑂 ′𝑂
Construct circle with center 𝑂 ′ and radius 𝑂 ′𝑂
Join PA and PB
OR
Draw a circle of radius 6cm
Draw OA and Construct ∠ 𝐴𝑂𝐵 = 1200
Draw ∠ 𝑂𝐴𝑃 = ∠ 𝑂𝐵𝑃 = 900
PA and PB are required tangents
Join OP and apply tan∠𝐴𝑃𝑂 = tan 30° = 6 𝑃𝐴
⇒ Length of tangent = 6√3 cm
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 |
| ∑ 𝑓𝑖 = 300 | ∑ 𝑓𝑖𝑥𝑖 =12600 |
Mean age = 𝑥̅= ∑ 𝑓𝑖𝑥𝑖 ∑ 𝑓𝑖 = 12600 300
𝑥̅= 42
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 30 0 and 60 0 respectively.
i) Which of the two, boat or the ship is nearer to the light house. Find its distance from the lighthouse?
ii) Find the time taken by the boat to reach the light house 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 ∆ ACB, tan 600 = 𝐴𝐵 /𝐵𝐶
√3 = 40 𝐵𝐶
∴ BC = 40 √3
= 40√3 3 m
II)In ∆ ADB, tan 30 0 = 𝐴𝐵 𝐵𝐷
⇒ 1 √3 = 40 𝐷𝐵
∴ DB = 40√3𝑚
Time taken to cover this distance = ( 60 2000 × 40√3) minutes
= 60√3 100 = 2.076 minutes
Q.14 Krishnanagar is a small town in Nadia District of West Bengal. Krishnanagar clay dolls are unique in their realism and quality of their finish. They are created by modelling coils of clay over a metal frame. The figures are painted in natural colours and their hair is made either by sheep’s wool or jute. Artisans make models starting from fruits, animals, God, goddess, farmer, fisherman, 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, male doll of blue colour has 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 𝑟1𝑎𝑛𝑑 𝑟2 be respectively the radii of apples and oranges
∵ 2𝑟1: 2𝑟2 = 2: 3 ⇒ 𝑟1: 𝑟2 = 2: 3
4𝜋𝑟1 2 : 4𝜋𝑟2 2 = ( 𝑟1 𝑟2 ) 2 = ( 2 3 ) 2 = 4: 9
II)Let the height of the drum be h. Volume of the drum = volume of the cylinder + volume of the sphere π3 2h = (π3 2 × 8 + 4 3 π3 3 ) 𝑐𝑚3
⇒ ℎ = (8 + 4)𝑐𝑚
⇒ ℎ = 12𝑐𝑚
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CBSE Term 2 Class 10th Maths Sample Paper 2022: FAQs
Q. Where can I get CBSE Class 10th Maths Sample Paper 2022?
On this page, you will get CBSE Class 10th Maths Sample Paper 2022.
Q. When will CBSE announce the CBSE Term 1 Result 2021-22?
The CBSE is going to announce CBSE Term 1 Result 2021-22 soon. The CBSE Term 1 Result 2021-22 can be announced at any time in the 2nd or 3rd week of February 2022.
Q. What are the passing criteria of the CBSE Term 1 exam?
To pass the CBSE Term 1 Examination, students must have scored 33% of the total marks in each subject.