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

Maths Standard Term 2 Sample Paper 2022 (Solved)

Class 10 Maths Standard 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. All questions are compulsory .

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

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

5. 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 value of a25a15 for the AP : 6 , 9 , 12 , 15 , ………..

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 .

2. Find the value of m so that the quadratic equation mx ( 5x – 6 ) + q = 0 has two equal roots .

3. From a point P, two tangents PA and PB are drawn to a circle C (0, r) .

If OP = 2r , then find APB . What type of triangle is APB ?

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4. The curved surface area of a right circular cone is 12320 cm2. If the radius of its base is 56 cm , then find its height .

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 .

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 : 9x² – 6px + ( p² − q² ) = 0

Section – B

[ 3 Marks Each ]

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

8. Construct a pair of tangents to a circle of radius 4 cm , which are inclined to each other at an angle of 60⁰ .

9. The distribution given below shows the runs scored by batsmen in one – day cricket matches . Find the mean number of runs .

10. Two vertical poles of different heights are standing 20 m 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 60 ° and angle of elevation of the top of the second pole from the foot of the first pole is 30 ° . Find the difference between the heights of two poles . ( Take √3 = 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 60 ° . Calculate the height of the building . ( Take √3 = 1.73 )

Section – C

[ 4 Marks Each ]

11. The internal and external radil of a spherical shell are 3 cm and 5 cm respectively . It is melted and recast into a solid cylinder of diameter 14 cm , find the height of the cylinder . Also find the total surface area of the cylinder ( Taken π = 22 / 7 )

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 PTQ = 2 OPQ

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Case Study – 1

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 240 m 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 30 ° .

( 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 .            [ 2 ]

( ii ) After 10 minutes , the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240 ( √3 – 1 ) m . He immediately raised the alarm . What was the new angle of depression of the boat from the top of the observation tower ?                  [ 2 ]

Case Study – 2

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 ?           [ 2 ]

( ii ) Find the total number of push – ups performed by Nitesh up to the day his goal is achieved .         [ 2 ]

Solution of Sample Paper

Section – A

1 . a = 6 , d = 3 ;

a25 = 6 + 24 ( 3 ) = 78

a15 = 6 + 14 ( 3 ) = 48

a25 – a15 = 78 – 48

a25 – a15 = 30

OR

7 ( a + 6d ) = 5 ( a + 4d )

⇒ 2a + 22d = 0

⇒ a + 11d = 0

⇒ t12 = 0

Detailed Solution :

6 , 9 , 12 , 15 , …… is the given A.P.

First term = a = 6

Common Difference = d = 9 – 6 = 3

nth term = an = a + ( n – 1 ) d

25th term = a25

= 6 + ( 25 – 1 ) x 3

= 6 + ( 24 ) x 3

= 6 +72

= 78

15th = a15

= 6 + ( 15-1 ) x 3

= 6 + ( 14 ) x 3

= 6 + 42

= 48

a25 – a15 = 78 – 48 = 30

OR

Let ‘ a ‘ and ‘ d ’ be the first term and common difference of AP

nth term = an = a + ( n – 1 ) d

7th term = a7 = a + ( 7-1 ) d = a + 6d

5th term = a5 = a + ( 5-1 ) d = a + 4d

According to the Questions ,

7a7 = 5a5

7 ( a + 6d ) = 5 ( a + 4d )

7a + 42d = 5a + 20d

7a – 5a + 42d – 20d = 0

2a + 22d = 0

2 ( a + 11d ) = 0

a + 11d = 0 ….(i)

12th term = a12

= a + 11d

= 0 ( From ( i )

Hence , 12th term is zero .

2. Given 5mx² – 6mx + 9 = 0

Since , b² – 4ac = 0

for equal roots

( -6 m ) ² – 4 ( 5 m ) ( 9 ) = 0

⇒ 36m ( m – 5 ) = 0

⇒ m = 0,5 ; rejecting m = 0 ,

we get m = 5

Detailed Solution :

Since , mx ( 5x – 6 ) + 9 = 0 has equal roots

∴ Discriminant = 0

⇒ b² – 4ac = 0 ….(i)

Quadratic equation can be written as

5mx² – 6mx + 9 = 0

Here , a = 5m , b = -6m , c = 9

Put in ( i ) ,

( 6 m ) ² – 4 ( 5 m ) ( 9 ) = 0 (m = 0)

36m² – 180m = 0 or m = 5

= 36 m ( m – 5 ) = 0

Either m = 0 or m = 5

Since , m = 0 is not possible

Hence , m = 5

3.

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Let ∠APO = θ

Sinθ = OA / OP = 1/2 ⇒ θ = 30°

⇒ ∠APB = 2 θ = 60 °

Also ∠PAB = ∠PBA = 60 ° ( ∴ PA = PB )

⇒ △APB is equilateral

Detailed Solution :

Radius of given circle OA = r units

Let , ∠APO = θ

Radius is always perpendicular to tangent

= OA ⊥ AP

So , ∠OAP = 90 °

∴ △OAP is a right angled triangle

In △OAP ,

sin θ = OA / OP

sin θ = r/2r = 1/2

sin θ = sin 30⁰

⇒ θ = 30 °

∠APO = 30 °

∠APB = 2 x ∠APO = 2 x 30 ° = 60 °

PA = PB [ Length of tangents from an external point are equal in length ]

Let , ∠ABP = ∠BAP = x

So , from △APB ,

∠APB + ∠ABP + ∠BAP = 180 ° [ Angle sum property of triangle ]

60 ° + x + x = 180 °

or x = 60 °

∴ △APB is equilateral triangle .

4. CSA ( cone ) = πrl = 12320 cm²

22/7 x 56 x l = 12320

l = 70 cm

= 42 cm

Detailed Solution :

Let Height of cone = h

Radius of cone = 56 cm

C.S.A of cone = 12320 cm²

Let slant height of cone = l cm

Curved surface area of cone = πrl

⇒ 12320 = 22/7 x 56 x l

⇒ 12320 × 7 / 22 × 56 = l

⇒ l = 70 cm

Now, l² = h² + r²

⇒ ( 70 ) ² = h² + ( 56 ) ²

⇒ h² = 4900 – 3136

⇒ h² = 1764

⇒ h² = 42²

⇒ h = 42 cm

5. Modal class is 40-60 , l = 40 ,

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Detailed Solution :

Mode = 45

⇒ Modal class = 40 – 60

Lower limit of modal class ( l ) = 40

Size of modal class ( h ) = 20

Frequency corresponding to modal class

( F₁ ) = x

Frequency preceding to modal class

( F0 ) = 10

Frequency preceding to modal class ( F₂ ) = 6

6. Let the present age of Ritu be x years

( x – 5 ) ² = 5x + 11

x² – 15x + 14 = 0

( x – 14 ) ( x – 1 ) = 0

⇒ x = 1 or 14

x = 14 years ( rejecting x = 1 as in that case Ritu’s age 5 years ago will be -ve )

OR

9x² – 6px + ( p² – q² ) = 0

a = 9 , b = -6p , c = p² – q²

D = b² – 4ac

= ( -6p ) ² – 4 ( 9 ) ( p² – q² )

= 36q²

Detailed Solution :

Let present age of Ritu = x years

As per question given the following equation can be formed :

⇒ ( x – 5 ) ² = 11 + 5x

⇒ x² – 10x + 25 = 11 + 5x

⇒ x² – 10x – 5x + 25 – 11 = 0

⇒ x² – 15x + 14 = 0

⇒ x² – 14x – 1x + 14 = 0

⇒ x ( x – 14 ) – 1 ( x – 14 ) = 0

⇒ ( x – 14 ) ( x – 1 ) = 0

⇒ x – 14 = 0 , x – 1 = 0

x = 14 , x = 1

Since , her age can’t be 1 , so her present age will be 14 years .

OR

9x² – 6px + ( p² – q² ) = 0

In comparing with

ax² + bx + c = 0

a = 9 , b = -6p , c = p² -q²

Solve by quadratic formula ,

D = b² – 4ac

= ( -6p ) ² – 4 ( 9 ) ( p² – q² )

= 36p² – 36p² + 36q²

= 36q²

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Section – B

7 .

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Detailed Solution :

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Take = N/2 = 250/2 = 125

∴ Median class = 2 – 3

Lower limit of median class ( l ) = 2

Size of median class ( h ) = 1

Frequency corresponding to median class

( f ) = 62

Total number of observations

Frequency ( N ) = 250

Cumulative frequency preceding

Median class ( c.f. ) = 120

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∴ 50 % of students jumped below 2.08 m and 50 % of students jumped above 2.08 m .

8. Draw circle of radius 4cm

Draw OA and construct ∠AOB = 120 °

Draw ∠OBP = ∠OBP = 90 °

PA and PB are required tangents

Detailed Solution :

Steps of construction :

1. Draw a circle of radius 4 cm .

2. Draw two radii having an angle of 120 ° .

3. Let the radii intersect circle at A and B.

4. Draw angle of 90 ° on both A and B.

5. The point where both rays of 90 ° intersect is P.

6. PA and PB are the required tangents .

9 .

10 .

Detailed Solution :

Let the heights of two pole be y and x .

Distance between the poles is QS = 20 m .

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OR

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Height of Boy = AB = 1.7 m

= QR = 1.7 m

Distance between Boy and building

BR = 50 m

⇒ AQ = 50 m

In △PQA , ∠Q = 90 °

tan 60 ° = PQ / AQ

√3 = PQ / 50

50√3 = PQ

Total height of the building = PQ + QR

= 86.5 + 1.7

= 88.2 m .

Section – C

11. Volume of shell = Volume of cylinder

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Detailed Solution :

Internal radius ( r ) = 3 cm

External radius ( R ) = 5 cm

Radius of cylinder ( R₁ ) = 14 / 2 = 7 cm

Let height of cylinder = h cm

According to the question ,

[ When one shape is reshaped into another shape the volumes are same of both shapes ]

or 425.33 cm²

12 .

∠OAP + ∠OBP + ∠APB + ∠AOB = 360

⇒ 90 ° + 90 ° + ∠APB + ∠AOB = 360 ° ( ∴ Tangent ⊥ radius )

⇒ ∠APB + ∠AOB = 180 °

OR

Let ∠PTQ = θ

TPQ is an isosceles triangle

∠OPT = 90°

Detailed Solution :

Given : PA and PB are two tangents to a circle with centre O.

To Prove :

∠APB + ∠AOB = 180 °

Proof : □ OAPB is a quadrilateral , OA is radius and PA & PB are tangents .

∴ ∠OAP = ∠OBP = 90 ° [ Radius is always perpendicular to tangent ]

By angle sum property of quadrilateral .

∴ ∠AOB + ∠OBP + ∠APB + ∠OAP

= 360 °

∠AOB + 90 ° + ∠APB + 90 ° = 360 °

∴ ∠APB + ∠AOB = 180 °

Hence Proved .

OR

Given : A circle with centre O two tangents

TP and TQ to the circle where P and Q are the point of contact .

To Prove : ∠PTQ = 2∠OPQ

Proof : TP = TQ [ length of tangents drawn from same external point to a circle are equal ]

∴ ∠TQP = ∠TPQ …( i ) [ Angles opposite to equal sides are equal ]

Now , PT is a tangent & OP is radius

∴ OP ⊥ TP [ Radius and Tangent are perpendicular to each other ]

⇒ ∠OPT = 90 °

∠OPQ + ∠TPQ = 90 °

∠TPQ = 90 ° – ∠OPQ …( ii )

In △PTQ ,

By angle sum property of triangle ,

∠TPQ + ∠TQP + ∠PTQ = 180 ° [ From ( i ) ]

∠TPQ + ∠TPQ + ∠PTQ = 180 °

⇒ 2 ∠TPQ + ∠PTQ = 180 °

⇒ 2 ( 90 ° – ∠OPQ ) + ∠PTQ = 180 ° [ From ( ii ) ]

⇒ 180 – 2∠OPQ + ∠PTQ = 180 °

∠PTQ = 2∠OPQ

Hence Proved .

Case Study – 1

13. 

Detailed Solution :

Distance of foot from the foot of observation tower = 240 √3 m .

( ii ) After 10 minutes ,

Distance between boat and light house is reduced by

= 24 ( √3 – 1 ) m

Distance between boat and light house

= 240 √3 – 240 ( √3 – 1 )

= 240 √3 – 240 √3 + 240

= 240 m

Let the point C is the new position of boat ,

tan θ = AB / BC = 240/240

⇒ θ = 45 °

Case Study – 2

14. ( i ) 3000 , 3005 , 3010 , … , 3900

an = a + ( n – 1 ) d

3900 = 3000 + ( n – 1 ) 5

⇒ 900 = 5n – 5 ⇒ 5n = 905

⇒ n = 181

Minimum number of days of practice = n – 1

= 180 days

( ii )

Detailed Solution :

( i ) By the given situation the A.P. to be formed is :

3000 , 3005 , 3010 , …… 3900

First term = a = 3000

Common Difference = d = 3005 – 3000

d = 5

According to the problem

nth term ( an ) = 3900

To Find n

nth term = an

= a + ( n – 1 ) d

3900 = 3000 + ( n – 1 ) 5

3900 – 3000 = ( n – 1 ) 5

900 = ( n – 1 ) 5

900/5 = n – 1

180 + 1 = n

or n = 181

Minimum number of days he needs to practice before his goal is accomplished

= 181 – 1 [ Excluding the last day ]

= 180

( ii ) Total Number of push – ups performed means sum of all push – ups he did in 181 days .

∴ Sn = n/2 [ a + 1 ] ….( i)

Sn = Sum of all push – ups in 181 days .

n = number of days = 181

a = first term of the A.P.

= 3000

l = last term = 3900

Put all values in ( i )

Total number of Push – ups performed in 181 days

= 624450

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