Class 10 Maths Basic Term 2 Sample Paper 2022 (Solved)

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. The mean of the following distribution is 18. Find the frequency f of the class 19- 21.

2. Check weather the equation 5x2 – 6x – 2 = 0 has real roots and if it has, find them by the method of quadratic formula.

3. If 10 times of its 10th term and 15 times of its 15th term are equal. Find the 25th term.

OR

Find the sum of n even natural numbers.

4. In Fig., AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB.

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5. A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel ?

OR

Two cubes have their volumes in the ratio 1: 27. Find the ratio of their surface areas.

6. Consider the following distribution:

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(a) Calculate the frequency of class 30 – 40.

(b) Calculate the class mark of the class 10-20.

 

Section – B

[3 Marks each]

7. In a cricket match, Harbhajan took three wickets less than twice the number of wickets taken by Zahir. The product of the number of wickets taken by these two is 20. Find the number of wickets Harbhajan has taken.

8. Prove that tangent drawn at any point of a circle is perpendicular to the radius through the point of contact.

9. On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot of tower respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.

10. How many terms of the Arithmetic Progression 45, 39, 33, …… must be taken so that their sum is 180? Explain the double answer.

OR

The sum of first 20 terms of an A.P is 400 and sum of first 40 terms is 1600. Find the sum of its first 10 terms.

 

Section – C

[4 Marks each]

11. Draw a circle of radius 5 cm. Take two points P and Q on one of its extended diameter each at a distance of 10 cm from its centre. Draw tangents to the circle from these two points P and Q.

12. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode of this data.

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OR

The arithmetic mean of the following distribution is 53. Find the value of k.

 

Case Study – 1

13. A well of diameter 8 m is dug 28 m deep, The earth taken out is evenly spread all around the well to form a 80 cm high embankment.

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Based on above situation answer the following questions:

(i) Find the volume of earth taken out.        [ 2 ]

(ii) Find the width of the embankment.        [ 2 ]

 

Case Study – 2

14. A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60° to 30°.

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Based on above informations, answer the following questions:

(i) What is the distance between the boat and light house after two minutes?              [ 2 ]

(ii) How much distance the boat has covered in those two minutes and at what speed?             [ 2 ]

 

Solution of Sample Paper

 

Section – A

1.

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2. The given equation is 5x² – 6x – 2 = 0

Comparing with ax² + bx + c = 0, we get

Here a = 5, b = -6 and c = -2             1

D=b² – 4ac = (-6)² -4(5) (-2) = 36 + 40 = 76            1

Since, D > 0, hence it has real roots

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3. Let the first term and common difference of the A.P. be a and d respectively.

Using, an = a + (n – 1)d

a15 = a + 14d and a10 = a +9d          ½

According to the question,

10(a + 9d) = 15 (a + 14d)

or 10a + 90d = 15a + 210d

or 5a + 120d = 0

or 5 (a + 24d) = 0         1

or a + 24d = 0

⇒ a25 = a + (25 – 1)d = 0            ½

Or

Consider 2, 4, 6,………. is in A.P.          ½

a = 2, d = 2

4. PA = PB               ½

or, ∠PAB = ∠PBA = 60°              ½

∴ △PAB is an equilateral triangle.           ½

Hence, AB = PA = 5 cm.            ½

5. Diameter of sphere = 6 cm

∴ Radius (r) = 3 cm

Diameter of cylindrical vessel = 12 cm

∴ Radius (R) = 6 cm

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= 36π cm3             1

∴ Volume of sphere = Increase volume in cylinder

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∴ Level of water rise in vessel = 1 cm.         1

OR

Let volumes of two cubes be V₁ and V₂ respectively, then

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Hence, the ratio of their surface areas is 1 : 9.            1

6.

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(a) So, frequency of the class 30-40 is 3.            ½

(b) Class mark of the class 10-20 = (10 + 20) / 2

= 30/2 = 15          ½

 

Section – B

7. Let the number of wickets taken by Zahir be x.

Then, the number of wickets taken by Harbhajan = 2x – 3

According to question,

x (2x – 3) = 20

⇒ 2x² – 3x = 20

∴ Required quadratic equation is,

2x² – 3x – 20 = 0                1

2x² – 8x + 5x – 20 = 0

2x (x – 4) + 5 (x – 4) = 0

(2x + 5)(x – 4) = 0

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Hence, 4 wickets are taken be Zahir

The number of wickets taken by

Harbhajan = 2x – 3

= 2 × 4 – 3

= 5             1

8. Given – A circle ( 10, OP ) and tangent at P

To prove – OP ⊥ PQ

Construction – Extend OR to Q, at AB.

Proof- We have –

OP = OR (radius)

OQ = OR + OR

Clearly OQ = OR

OQ = OP

The shortest line joining a point to any point on given line is ⊥ to that line

⇒ OP ⊥ AB

⇒ OP ⊥ PQ

9.

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10. Given AP is:

45, 39, 33, ……

Here, a = 45, d = 39 – 45 = -6 and Sn = 180

Since,

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⇒ 360 = n(90 – 6n + 6)

⇒ 360 = n(96 – 6n)

⇒ 180 = n (48 – 3n)

⇒ 3n² – 48n + 180 = 0               1

⇒ 3n² – 18n – 30m + 180 = 0

⇒ 3n(n – 6) – 30 (n – 6) = 0

⇒ (3n – 30) (n – 6) = 0

⇒ n = 10, n = 6

Hence, 10 terms or 6 terms can be taken to get the sum of AP as 180.

Verification

Now, sum of 6 terms i.e.,

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Hence verified.             1

and sum of 10 terms, i.e.,

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Hence verified.          1

OR

Sol. Let the first term be a and common difference be d and sum of first 20 terms be S20

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or, 400 = 10 [2a + 19d]

or 2a + 19d = 40 …(i)                     ½

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= 5 [2 + 9 × 2]

= 5 [2 + 18]

= 5 × 20 = 100                 1

Section – C

11. The tangents can be drawn on the given circle as follows.

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Steps of construction :

Step 1: Taking any point O on the given plane as centre, draw a circle of 5 cm radius.

Step 2: Take one of its diameters, PQ, and extend it on both sides. Locate two points on this diameter such that OR = OS = 10 cm

Step 3: Bisect OR and OS. Let T and U be the mid-points of OR and OS, respectively.

Step 4: Taking T and U as its centre and with TO and UO as radius, draw two circles. These two circles will intersect the circle at point V, W, X, Y respectively. Join RV, RW, SX, and SY. These are the required tangents.                  2

12. For Mode:

Here, maximum frequency is 10 and it corresponds to the class interval 30 – 35.

Therefore, Modal class = 30-35

l = 30, f₁ = 10, f0 = 9, f₂ = 3 and h = 5             1

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OR

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Case Study – 1

13. (i) Depth of the well = 28 m and radius of the well = 4 m

Volume of earth taken out from cylindrical well

= πr2h                  1

= 22/7 × 4 × 4 × 28

= 1408 m³          1

(ii) Let r be the width of the embankment.

The radius of outer circle of the embankment = 4 + r

Area of upper surface of embankment = Area of outer circle – Area of inner circle

= π [(4 + r)² – 4²]

Volume of the embankment = Volume of earth             1

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or r = 20 m            1

Hence, the width of the embankment is 20 m.

Case Study – 2

14. (i) Height of the light house, h = 100 m

Let initial distance be x m and the angle is 60°

From △DBA, ∠B = 90°

So, tan 60° = h/x

⇒ √3 = 100/x

⇒ x = 100/√3 m           1

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Now, after time 2 min, let the new distance from lighthouse be y and angle is 30°

From △ABC, ∠B = 90°

So, tan 30° = 100/y

⇒ 1/√3 = 100/y

⇒ y = 100 √3 m           1

(ii) Therefore, distance (d) travelled in 2 min = y – x

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