**Maths Basic Term 2 Sample Paper 2022** **(Solved)**

**Class 10 Maths Basic Term 2 Sample Paper 2022**,** (Maths) exams are Students are taught thru NCERT books in some of the state board and CBSE Schools. As the chapter involves an end, there is an exercise provided to assist students to prepare for evaluation. Students need to clear up those exercises very well because the questions inside the very last asked from those.**

**Sometimes, students get stuck inside the exercises and are not able to clear up all of the questions. To assist students, solve all of the questions, and maintain their studies without a doubt, we have provided a step-by-step NCERT Sample Question Papers for the students for all classes. These answers will similarly help students in scoring better marks with the assist of properly illustrated Notes as a way to similarly assist the students and answer the questions right.**

**Class 10 Maths Standard Term 2 Sample Paper 2022**

**General Instructions: **

1. The question paper consists of 14 questions divided into 3 sections A, B, C.

2. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.

3. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.

4. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.

**Section – A **

**[2 Marks each]**

1. Find the roots of the quadratic equation 6x^{2} – x – 2 = 0.

2. If the 4^{th} term of an A.P is zero. Prove that the 25^{th} term of the A.P is three times its 11^{th} term.

OR

Find the 13^{th} term from the end of the A.P. 3, 10, 17, ……..136.

3. A solid sphere of radius 15 cm is melted and recast into a solid right circular cones of radius 25 cm and height 8 cm. Calculate the number of cones recast.

4. Find the median for the data, using an empirical relation when it is given that mode = 18.6 and mean = 16.2 .

5. Find the median of the following data:

OR

Find the mode of the following data:

6. In the given figure, AB is a diameter. The tangent at C meets AB produced at Q and ∠BAC = 34°. Find ∠ABC.

**Section – B**

**[3 Marks Each] **

7. The angles of depression of the top and bottom of a 8 m tall building from the top of a multi- storied building are 30° and 45°, respectively. Find the height of the multi-storied building.

OR

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building.

8. Find the 16^{th} term of the A.P. 7, 11, 15, 19, …. Also, find the sum of 6 terms.

9. In △ABC, right angled at B. A circle with centre O has been inscribed in the triangle. If AB = 12 cm and BC = 12 cm. Find the radius of the circle.

10. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 77. Find the numbers.

**Section – C**

**[4 Marks Each]**

11. Water flows at the rate of 10 metres per minute from a cylindrical pipe 5 mm in diameter. How long it will take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm.

OR

In a medical laboratory a lab technician composed a capsule in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of entire capsule is 21 mm and the diameter of the cylindrical portion is 8 mm. What will be the volume of 10 such capsules?

12. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

**Case Study – 1**

13. The maximum bowling speeds, in km/hr of 45 players at a cricket playground are given as follows.

From the above data answer the following questions.

(i) What is the sum of lower limit of modal class and upper limit of median class? [ 2 ]

(ii) What is the mode of bowling speed? [ 2 ]

**Case Study – 2**

14. A plane was scheduled to fly from New Delhi to New York.

After it’s take off it was observed that angle of elevation of the plane was 60° from a point A on the ground. After a flight of 15 seconds, the angle of elevation changes to 30°.

Based on above situation answer the following questions.

(i) If the plane is flying at a constant height of 1200 √3 m. What was the distance of plane covered in those 15 seconds? [ 2 ]

(ii) What was the speed of the plane in km/h. [ 2 ]

**Solution of Sample Paper**

**Section – A**

1. Given, 6x² – x – 2 = 0

⇒ 6x² + 3x – 4x – 2 = 0 ½

⇒ 3x(2x + 1) -2(2x + 1) = 0 ½

⇒ (2x + 1)(3x – 2) = 0 ½

Either 3x – 2 = 0 or 2x + 1 = 0

x = 2/3 or x = – 1/2

Roots of equation are 2/3 and – 1/2. ½

2. Given, a_{4} = 0

⇒ a + 3d = 0

⇒ a = -3d …..(i) ½

25^{th} term, a_{25} = a + 24d

= -3d + 24d [from (i)]

a_{25} = 21d ……..(ii) ½

11^{th} term, a_{11}= a + 10d

= -3d + 10d [from (i)]

i.e., a_{11} = 7d ……(iii) ½

From (ii) and (iii),

a_{25}= 3(a_{11}) Hence proved. ½

OR

The formula to find any term from end

= l – (n – 1)d ½

Here, l = 136, n = 13 and d = 10 – 3 = 7 ½

13^{th} term from last = 136 – (13 – 1)7

= 136 – 84

=52 1

3. Radius of sphere, R = 15 cm

Radius of cone, r = 2.5 cm and height (h) = 8 cm

4. Since,

5.

So lower limit of median class (l) = 18, *f* = 5 and h = 3

(Approx) 1

6. Given, ∠BAC = 34°

∠ACB = 90°

(Angle in a semi-circle is 90°) ½

In △ABC,

∠BAC + ∠ACB + ∠ABC = 180° (Angle sum property of a A) ½

⇒ 34° + 90° + ∠ABC = 180°

⇒ 124° + ∠ABC = 180°

⇒ ∠ABC = 180° – 124°

⇒ ∠ABC = 56°. 1

**Section – B**

7.

1

Let AB and CD be the multi-storied building and the building respectively.

Let the height of the multi-storied building = h m and the distance between the two buildings = x m.

AE = CD = 8m [Given]

BE = AB – AE = (h – 8) m

and AC = DE = x m [Given]

Also, ∠FBD = ∠BDE = 30° (Alternate angles)

and ∠FBC = ∠BCA = 45° (Alternate angles) 1

Now, In △CAB, ∠A = 90°

⇒ tan 45° = AB/AC

⇒ l = x/h

⇒ x = h …..(i)

In △DEB, ∠E = 90°

Hence, height multi-storied building is (12 + 4√3) m ½

OR

Hence, the height of the building = 20 m. ½

8. Given, A.P. = 7, 11, 15, 19, …….

First term (a) = 7 and common difference (d) = 4 ½

For nth term, T_{n} = a + (n – 1)d

T_{16} = 7 + (16 – 1)4

= 67 1

Hence, 16^{th} term of A.P. is 67 and sum of 6 terms is 102.

9.

According to the property that length of tangents drawn from same external point are equal in length.

Let BP = MB = a,

PC = CQ = b

and AQ = AM = c

From figure, it is clear that,

b + c = 12

a + b = 5

a + c = 13 ……(i) 1

Add all above equations,

2(a + b + c) = 30

a + b + c = 15 …(ii)

From (i) & (ii), we get

b = 2 cm

Hence, radius of the circle is 2 cm. 1

10. Let the three consecutive natural numbers be x, x + 1 and x + 2, then according to question. ½

(x + 1)² = (x + 2)² – x² + 77 1

⇒ x² + 2x + 1 = x² + 4x + 4 = x² + 77

⇒ x² – 2x – 80 = 0 ½

⇒ x² – 10x + 8x – 80 = 0

⇒ x(x – 10) + 8(x – 10) = 0

⇒ (x – 10) (x + 8) = 0

Thus x = 10

or x = -8

Rejecting -8, we get

x = 10

Hence, three numbers are 10, 11 and 12. 1

**Section – C**

11. Diameter of cylindrical pipe = 5 mm

Length of cylindrical pipe, (h) = 10 m = 1000 cm

Radius of cylindrical pipe = 5/20 = 1/4 cm ½

Volume of water that flows out in one minute = Volume of cylindrical pipe

= πr^{2}h

Radius of cylindrical and hemispherical parts

(r) = 4 mm

Length of cylindrical part = 21 – 4 – 4 = 13 mm ½

Volume of 10 capsules = 10 x 921.9 = 9219 mm³ 1

12. **Steps of construction :**

1. Draw a circle of radius 4 cm with O as centre.

2. Draw two radii OA and OB inclined to each other at an angle of 120°.

In quadrilateral,

∵ ∠AOB = 360° – (90° + 90° + 60°) = 120°.

2

3. Draw AP ⊥ OA at A and BP ⊥ OB at B which meet at P.

4. PA and PB are the required tangents inclined to each other at an angle of 60°. 2

**Case Study – 1**

13. (i) Highest frequency class = 115-130 So, it is a modal class.

∴ Lower limit of modal class = 115 ½

Again, n/2 = 45/2 = 22.5,

which lies in class interval 100-115.

So, it is a median class.

∴ Upper limit of median class = 115 ½

Hence, sum = 115 + 115

= 230. 1

(ii) Here, modal class = 115-130

∴ *f*₁ =15, *f*₂ = 7, *f*_{0} = 12, l = 115 and h = 15

Hence, mode of bowling speed is 119.09 km/hr (Approx).

**Case Study – 2**

14. (i)

The plane is flying at a constant height of 1200 √3 m.

i.e., BE = CD = 1200 √3 m.

According to the first observation,

Let the plane is at point E. BE is the height at which the plane is flying.

In △ABE, ∠B = 90°,

AC = 3600 m

BC= 3600 – 1200 = 2400 m

Distance covered in those 15 seconds is 2400 m. 1

(ii) The plane covers a distance of 2400 m in 15 seconds.