# 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. Solve the following quadratic equation for x :

4x² – 4a²x + ( a4 – b4 ) = 0

OR

Find the value ( s ) of k for which the quadratic equation x² + 2√2 kx + 18 = 0 has equal roots

2. Show that ( a – b ) ² , ( a² + b ) and ( a + b ) ² are in A.P .

OR

Which term of the A.P. 3 , 15 , 27 , 39 , … will be 120 more than its 21st term ?

3. In the figure , quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB . If radius of incircle is 10 cm , then find the value of x . 4. Solve for x :

x2 + 5x – ( a² + a – 6 ) = 0

5. The 3/4th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water . The water emptied into a cylindrical vessel with internal radius 10 cm . Find the height of water in cylindrical vessel .

6. The mean of the following frequency distribution is 25. Find the value of p . ### Section – B

[ 3 Marks Each ]

7. The marks obtained by 100 students in an examination are given below : Find the median marks of the students .

8. Draw a circle of radius 2 cm with centre O and take a point P outside the circle such that OP = 6.5 cm . From P , draw two tangents to the circle .

9. Find the mode of the following distribution of marks obtained by the students in an examination : Given the mean of the above distribution is 53 , using empirical relationship estimate the value of its median .

10. A man on the top of a vertical observation tower observes a car moving at uniform speed coming directly towards it . If it takes 12 minutes for the angle of depression to change from 30 ° to 45 ° , how long will the car take to reach the observation tower from this point ?

OR

The angle of elevation of the top of a hill from the foot of a tower is 60 ° and the angle of depression from the top of the tower of the foot of the hill is 30 ° . If tower is 50 metre high , find the height of the hill .

### Section – C

[ 4 Marks Each ]

11. Rampal decided to donate canvas for 10 tents conical in shape with base diameter 14 m and height 24 m to a centre for handicapped person’s welfare . If the cost of 2 m wide canvas is ₹ 40 per metre , find the amount by which Rampal helped the centre .

12. In the given figure , two tangents TP and TQ are drawn to circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ . OR

a , b and c are the sides of a right triangle , where c is the hypotenuse . A circle , of radius r , touches the sides of the triangle . Prove that r = a + b – c / 2

### Case Study – 1

13. A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka , them being Nanda Devi ( height 7 , 816 m ) and Mullayanagiri ( height 1,930 m ) . The angles of depression from the satellite , to the top of Nanda Devi and Mullayanagiri are 30 ° and 60 ° respectively . If the distance between the peaks of two mountains is 1937 km , and the satellite is vertically above the midpoint of the distance between the two mountains . ( i ) Find the distance of the satellite from the top of Nanda Devi .         [ 2 ]

( ii ) Find the total distance of the satellite from the top of Mullayanagiri .       [ 2 ]

### Case Study – 2

14. Your elder brother wants to buy a car and plans to take loan from a bank for his car . He repays his total loan of ₹ 1,18,000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month , answer the following : ( i ) Find the amount paid by him in 30th installment .             [ 2 ]

( ii ) The amount paid by him in the 30 installments .        [ 2 ]

## Solution of Sample Paper

### Section – A

1. Given , 4x² – 4a²x + ( a4 – b4 ) = 0

Comparing with Ax2 + Bx + C = 0 , we get

Here , A = 4 , B = -4a² and C = ( a4 – b4 ) OR

Since , x² + 2√2 kx + 18 = 0 has equal roots then discriminant , D = 0

Comparing x² + 2√2 kx + 18 = 0 with ax² + bx + c = 0 ,

Now , a = 1 , b = 2√2 k and c = 18

D = 0

⇒ b² = 4ac          1

( 2√2k ) ² = 4 × 1 × 18

4 × 2 × k² = 72

k² = 72 / 8

k² = 9 ⇒ k = √9

k = ± 3 .            1

2. Given , ( a – b ) ² , ( a² + b² ) and ( a + b ) ²

Common difference ,

d₁ = ( a² + b² ) – ( a – b ) ²

= a² + b² – ( a² + b² – 2ab )

= a² + b² – a² – b² + 2ab

= 2ab            ½

and d2 = ( a + b ) ² – ( a² + b² )

= a² + b² + 2ab – a² – b²

= 2ab              ½

Since , d₁ = d₂

Hence , ( a – b ) ² , ( a² + b² ) and ( a + b ) ² are in an A.P.

Hence Proved .           1

OR

an = a21 + 120

= ( 3 + 20 x 12 ) + 120

= 363            1

∴ 363 = 3 + ( n – 1 ) x 12

⇒ n = 31            1

or 31st term is 120 more than a21

Detailed Solution :

Given A.P. is : 3 , 15 , 27 , 39

Here , first term , a = 3 and common difference , d = 12

Now , 21st term of A.P. is

t21 = a + ( 21 – 1 ) d

∴ t21 = 3 + 20 x 12 = 243

Therefore , 21st term is 243

We need to calculate term which is 120 more than 21st term

i.e., it should be 243 + 120 = 363             1

Therefore , tn = 363

∴ tn = a + ( n – 1 ) 12

363 = 3 + ( n – 1 ) 12

360 = 12 ( n – 1 )

n – 1 = 30

n = 31

So , 31st term is 120 more than 21st term .             1

3. ∠A = ∠OPA = ∠OSA = 90 °          ½

Hence , ∠SOP = 90 °

Also , AP = AS

Hence , OSAP is a square .

AP = AS = 10 cm           ½

CR = CQ = 27 cm

BQ = BC – CQ

= 38 – 27 = 11 cm         ½

BP = BQ = 11 cm

x = AB = AP + BP

= 10 + 11 = 21 cm             ½

Detailed Solution :

With O as centre , draw a perpendicular OP on AB .

Now , in quadrilateral APOS , ∠SAP = 90 ° ( Given )

∠APO = 90 ° ( By construction )

and ∠ASO = 90 ° ( Angle between tangent and radius )

Finally ∠SOP = 360 ° – ( 90 ° + 90 ° + 90 ° )

= 90 °               ½

AP = AS ( Tangents from external point A )

∴ OSAP is a square .

AP = AS = SO = 10 cm             ½

∵ CR = CQ ( Tangents from external point C )

∴ CR = CQ = 27 cm

But BC = 38 cm ( Given )

∴ BQ = BC – CQ = ( 38 – 27 ) cm

BQ = 11 cm          ½

BP = BQ ( Tangent from external point B )

∴ BP = 11 cm

So, x = AB = AP + PB

= ( 10 + 11 ) cm = 21 cm

Hence , the value of x is 21 cm .             ½

4. Given ,

x2 + 5x – ( a² + a – 6 ) = 0 Detailed Solution :

x² + 5x – ( a² + a – 6 ) = 0

⇒ x² + 5x – ( a² + 3a – 2a – 6 ) = 0          ½

⇒ x² + 5x – [ a ( a + 3 ) -2 ( a + 3 ) ] = 0

⇒ x² + 5x – ( a + 3 ) ( a – 2 ) = 0

⇒ x² + [ ( a + 3 ) – ( a – 2 ) ] x – ( a + 3 ) ( a – 2 ) = 0

⇒ x² + ( a + 3 ) x – ( a – 2 ) x – ( a + 3 ) ( a – 2 ) = 0

⇒ x [ x + ( a + 3 ) ] – ( a – 2 ) [ x + ( a + 3 ) ] = 0            ½

⇒ [ x + ( a + 3 ) ] [ x – ( a – 2 ) ] = 0

⇒ x = – ( a + 3 ) or x = a – 2

Hence , roots of given equation are – ( a + 3 ) and a – 2 .            1

5. Radius of conical vessel = 5 cm

and its height = 24 cm

Volume of cone vessel = 1/3πR²H

= 1/3 x π x 5 × 5 × 24

= 200 cm³ .            1

Internal radius of cylindrical vessel = 10 cm

Let the height of emptied water be h .

∴ Volume of water in cylinder

= 3/4 x Volume of cone

⇒ πr2h = 3/4 x Volume of cone

⇒ π x 10 x 10 x h = 150π

⇒ h = 1.5 cm

Hence the height of water = 1.5 cm            1

6 . 1

### Section – B

7 .  1

8. Steps of construction :

( i ) Draw a line segment OP = 6.5 cm .

( ii ) Taking O as centre and radius 2 cm , draw a circle .

( iii ) Draw perpendicular bisector of OP , intersect at OP at S.

( iv ) Taking S as centre draw another circle which intersects the first circle at Q and R.

( v ) Join P to Q and P to R.

Hence PQ and PR are two tangents .               2 9 . Modal class = 60 – 80 Empirical relationship ,

Mode = 3 median – 2 mean         ½

Mode = 68 and mean = 53 ( given )

∴ 3 Median = Mode + 2 Mean

3 Median = 68 + 2 × 53

Median = 174/3 = 58                  1

Hence , Median = 58 .

10. Let the speed of car by x m / minute

In △ABC , ∠B = 90 ° ⇒ y = 6x ( √3 + 1 ) …..(ii)

Hence , time taken from C to B = 6 ( √3 + 1 ) minutes .

OR

Let AB = 50 m be the height of the tower and CD be the height of hill .

Now , in △ABC ,

∠ABC = 90 ° ### Section – C

11. Diameter of tent = 14 m and height = 24 m

∴ radius of tent = 7 m  = 550 m²            1

Surface area of 10 tents = 550 × 10

= 5500 m² Hence , the amount by which Rampal helped the centre = ₹ 1,10,000              1

12. Let ∠OPQ be θ , then

∠TQP = 90 ° – θ            1

Since , TP = TQ

∴ ∠TPQ = 90 ° – θ ( opposite angles of equal sides )             1 Now , ∠TPQ + ∠TQP + ∠PTQ = 180 ° ( Angle sum property of a Triangle )              1

⇒ 90 ° – θ + 90 ° – θ + ∠PTQ = 180 °

⇒ ∠PTQ = 180 ° – 180 ° + 2θ

⇒ ∠PTQ = 2θ

Hence , ∠PTQ = 2∠OPQ             1

Hence Proved .

OR ½

Let circle touches CB at M , CA at N and AB at P.

Now OM ⊥ CB and ON ⊥ AC ( radius ⊥ tangent )

OM = ON ( radii )

CM = CN ( Tangents )          ½

∴ OMCN is a square .

Let OM = r = CM = CN           1

AN = AP , CN = CM and BM = BP ( tangent from external point )

AN = AP

⇒ AC – CN = AB – BP           1

b – r = c – BM

b – r = c – ( a – r )

b – r = c – a + r

2r = a + b – c

r = a + b – c / 2            1

Hence Proved .

#### Case Study – 1

13. √3 / 2 = 1937 / 2AF

AF = 1937 / √3

AF = 1118.36 km            2

( ii ) For △FPH ,

cos θ = PH / FP

cos 60 ° = 1937 / 2FP

1 / 2 = 1937 / 2FP

FP = 1937 km .           2

### Case Study – 2

14. ( i ) First term a = 1000

Common difference d = 100

30th term , a30 = a + ( n – 1 ) d

= 1000 + ( 30 – 1 ) 100

= 1000 + 2900

= ₹ 3900            2

( ii ) Sum of 30 installments 2

Total amount paid in 30 installments = ₹ 73500