HBSE Class 10 Sample Paper 2024 Answers Maths

HBSE
MODEL PAPER
2024
Practice Papers
MODEL PAPERS
Marking Scheme
ANSWER KEY

1
Marking Scheme Class X, Maths , 2023-24(English Medium)

Q. Expected solutions marks
no.

Section-A
1 (c) LCM(p, q)=a b 3 2
1

2 (c) 2+ 9 1

3 (b) 2-√3 1

4 (d) 15th 1

5 (d) 0,8 1

6 (c) 6 1

7 (b) similar but not congruent 1

8 (b) 700 1

9 (c) 500 1

10 (b) 0 1

11 tanA=
3 1
4
12 (c)
1 1
2
13 (c) 132 cm 1

14 � 1
(d) × 2 ��2
720°
15 (d) 16:9 1

16 Mode = 3Median – 2Mean 1
(b) 8
17 (b) 25 1

18 (c)
1 1
0.1

19 (c) Assertion (A) is true but Reason (R) is false. 1

2

20 (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct 1
explanation of Assertion (A).

Q. solution marks
no.

Section-B
21 Solve the following pair of linear equations:
�−�=�
� �
+ =�
� �
Solution:

� − � = 3 ⇒ � − � = 3 .................(1)
� �
+ = 6 ⇒ 2x + 3y = 36................(2) ½
3 2
----------------------------------------------------------------------------------------

Eq (2) - 2× Eq (1) ⇒ 2x + 3y – (2x – 2y ) ½
----------------------------------------------------------------------------------
⇒ 5y = 30
⇒y=6 ½
---------------------------------------------------------------------------------------
Putting value of y= 6 in eq (1) , we get x = 9 ½

22 A vertical pole of length 6 m casts a shadow 4 m long on the ground and at
the same time a tower casts a shadow 28 m long. Find the height of the
tower.

Solution:

3

Let x be the height of the Tower ½
----------------------------------------------------------------------------------
Two Triangles are similar as at the same time ∠E = ∠B ½
------------------------------------------------------------------------------------
6 � ½
∴ =
4 28
-------------------------------------------------------------------------------------
Or x = 42 m ½

OR

�� ��
In the fig.,
��
= �� ��� ∠��� = ∠���. Prove that PQR is an
isosceles triangle.

Solution:

PS PT
SQ
= TR ( Given )

So, ST ∥ QR (Converse of BPT)
½
-----------------------------------------------------------------------------------

∴ ∠PST = ∠PQR .............(1) ( Corresponding Angles )
½
-----------------------------------------------------------------------------------
Also ∠PST = ∠PRQ ...........(2) (Given)

∴ ∠PRQ = ∠PQR [From (1) and (2) ] ½
---------------------------------------------------------------------------------------

4

So PQ = PR (sides opposite to equal angles)
Hence PQR is an isosceles Triangle ½

23 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the
chord of the larger circle which touches the smaller circle.

Solution:

1/2

OA=5cm,OP=3cm
-----------------------------------------------------------------------------------------
OT ⊥ AB 1/2

---------------------------------------------------------------------------------------
1/2
Therefore AP = 52 − 32 = 16 = 4

----------------------------------------------------------------------------------------
1/2
∴ AB= 2AP

24 Evaluate the following:

� ���� ��° + � ���� ��° − ���� ��°
���� ��° + ���� ��°

Solution:

5 cos2 60° + 4 sec2 30° − tan2 45°
sin2 30° + cos2 30°

5
2 2
5(12) + 4( 2 ) − (1)2
= 3 1
2 2
(12) + ( 23)
---------------------------------------------------------------------------------
5 16
4
+ 3
−1
= 1 3
½
4
+ 4

----------------------------------------------------------------------------------

67 ½
=
12

25.
A chord of a circle of radius 15 cm subtends an angle of 600 at the centre.
Find the areas of the corresponding minor and major segments of the
circle.
(Use � = �. �� ��� � =1.73)
Solution:

� 1
Area of minor segment= × ��2 - × �2 sin � ½
360° 2

---------------------------------------------------------------------------------------

60 1
= × 3.14× (15)2 - × (15)2 sin 60°
360 2

1 3
= 3.14× 225× 6 - × 225 × ½
2 2

----------------------------------------------------------------------------------------

=117.75 - 97.312

= 20.4375 cm2 ½
---------------------------------------------------------------------------------------
Area of major segment = Area of circle – Area of minor segment
= 3.14 x (15)2 – 20.4375

6

= 706.5- 20.4375

= 686.0625 cm2 ½

OR

Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:
Cicumference of circle = 2 � � = 22
½

------------------------------------------------------------------------------
7
⇒�= cm
2 ½
----------------------------------------------------------------------------------
1
∴ Area of quadrant= × πr2 ½
4
-------------------------------------------------------------------------------------
1 22 7 7
= × × ×
4 7 2 2
77 ½
= cm2
8

Section-C
26. Prove that � is irrational.

Solution:
Let, if possible, 3 be a rational no. ½
-----------------------------------------------------------------------------------------
�
∴ 3 = , where p and q are co-prime integers and q≠ 0. ½
�
---------------------------------------------------------------------------------------
�2
⇒3= 2
�
⇒ p = 3 q2 ................................(i)
2

⇒ 3 divides p2 ⇒ 3 divides p also. ½

7

----------------------------------------------------------------------------------------
Let p = 3m,................................(ii) where m is any integer.

⇒ p2 = 9m2................................(iii)
½
------------------------------------------------------------------------------------
From (i) and (iii)
3q2 = 9m2
⇒ q2= 3m2
⇒ 3 divides q2 ⇒ 3 divides q also. ½
⇒ q = 3n......................................(iv)

---------------------------------------------------------------------------------------
From (i) and (iv) , p and q have 3 as common factor.
∴ p and q are not co-prime.

Hence our supposition is wrong.
∴ 3 is an irrational number. ½

27 Find the zeroes of the quadratic polynomial 6x2 -3 -7x and verify the
relationship between the zeroes and the coefficients.
Solution:

Given polynomial is 6x2 – 7x – 3 = (2x – 3)(3x + 1)
1
-----------------------------------------------------------------------------------

For zeroes , 2x – 3= 0, 3x+ 1= 0
3 1
⇒x= , x=-
2 3
3 1 1
⇒ Zeroes of polynomial are , -
2 3
--------------------------------------------------------------------------------------
3 1 7 (−7) − ����������� �� �
Sum of zeroes = - = =− = ����������� �� �2 ½
2 3 6 6

----------------------------------------------------------------------------------------

8

Product of zeroes =
3
×
−1
=-
1 3
=- =
�������� ���� ½
2 3 2 6 ����������� �� �2

28 Meena went to a bank to withdraw Rs 2000. She asked the cashier to give
her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how
many note Rs 50 and Rs 100 she received.

Solution:

Let the number of Rs. 50 and Rs. 100 notes be ‘x’ and ‘y’ respectively. ½
-----------------------------------------------------------------------------------

⇒ 50x + 100y = 2000

⇒ x+ 2y = 40 …… (1) ½

------------------------------------------------------------------------------------

Also, Meena got 25 notes in all.
½
⇒ x + y = 25 ……(2)

------------------------------------------------------------------------------------

(1) – (2) ⇒ x+2y–(x+y) =40–25

⇒ x+2y–x–y=40–25 ½
-----------------------------------------------------------------------------------

⇒ y = 15 ½
------------------------------------------------------------------------------------

Putting y = 15 in eq (1) , we get x = 10 ½

OR

Five years hence, the age of jacob will be three times that of his son. Five
years ago, Jacob's age was seven times that of his son. What are their present
ages?

9

Solution :

Let Jacob’s age be x years and his son’s age be y years. ½

-------------------------------------------------------------------------------------
Five years hence(later),
x+ 5 = 3 (y + 5) ½
⇒ x+ 5 = 3 y + 15
⇒ x - 3 y = 10.......(1)
----------------------------------------------------------------------------------------
Also, five years ago(before),
x-5 = 7 (y – )5
⇒ x-5 = 7y – 35 ½
⇒ x-7y = -30........(2)
----------------------------------------------------------------------------------------
Subtracting equation (2) from (1),
x - 3 y = 10
-x +7y = 30 (∵ eq.(2) changes its sign)
4y = 40
∵ 4 y = 40
∴ y = 10 ½
----------------------------------------------------------------------------------------
Put y = 10 in eq. (1),
x - 3(10) = 10 ⇒ x - 30 = 10 ½
⇒ x = 40
----------------------------------------------------------------------------------------
Thus, present age of Jacob=x=40 years and
½
present age of Jacob's son=y=10 years.

29 Prove that a parallelogram circumscribing a circle is a rhombus.

10

Solution:

½

--------------------------------------------------------------------------------------
Given :- ABCD be a parallelogram circumscribing a circle with centre O.
To Prove :- ABCD is a rhombus. ½
---------------------------------------------------------------------------------------
Proof:- We know that the tangents drawn to a circle from an exterior point are
equal is length.
∴ AP = AS, BP = BQ, CR = CQ and DR = DS. ½
----------------------------------------------------------------------------------------
AP+BP+CR+DR = AS+BQ+CQ+DS
(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
½
∴ AB+CD=AD+BC
----------------------------------------------------------------------------------------
or 2AB = 2AD (since AB = DC and AD=BC of parallelogram ABCD)
---------------------------------------------------------------------------------------- ½
∴ AB = BC = DC = AD
½
Therefore, ABCD is a rhombus.
30
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1

Solution:

sin θ + cos θ = √3 squaring on both sides

⇒ (sin θ + cos θ)2 = 3
1/2

11
------------------------------------------------------------------------------
1/2
⇒ sin2 θ + cos2 θ + 2sin θ cos θ = 3

---------------------------------------------------------------------------

⇒ 1 + 2sin θ cos θ = 3 (∵ sin2 θ + cos2 θ = 1 )

2sin θ cos θ = 3 - 1
1/2
2sin θ cos θ = 2

Divide both sides by 2

--------------------------------------------------------------------------------------
1/2
sin θ cos θ = 1 = sin θ + cos θ
2 2

--------------------------------------------------------------------------------------
1/2
1 = (sin2 θ + cos2 θ)/ sin θ cos θ

--------------------------------------------------------------------------------------

= tan θ + cot θ = 1 1/2

OR

���� �
LHS= 1+ �+������

������ �−�
=1 + ​ [∵ ���� �= ������ � − �] 1
� + ������
-------------------------------------------------------------------------------------------------
(������+�)(������−�)​
= 1+
(�+������)
1
-------------------------------------------------------------------------------------------------

12

= 1 + (cosec θ - 1)

= cosec θ = RHS
1

Hence proved

31
All the jacks, queens and kings are removed from a deck of 52
playing cards. The remaining cards are well shuffled and then one
card is drawn at random. Giving ace a value 1 similar value for
other cards, find the probability that the card has a value. (i) 7 (ii)
greater than 7 (iii) less than

Solution:

In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed,
then total no. of remaining cards = 52 - 3×4 = 40

(i) no. of favourable outcomes to card value 7= 4 because

card value 7 may be of a spade, a diamond, a club or a heart

no.of favourable outcomes to the event
∴ P(card value7) = =
Total no.of possible outcomes
4 1 1
= =
40 10

-------------------------------------------------------------------------------------

(ii)Cards having value greater than 7are from 8, 9 or 10⇒

13
∴ no. of favourable outcomes = 3x4 =12

12 3
∴ P(card having value greater than 7) = = = 1
40 10

------------------------------------------------------------------------------------

(iii) Cards having value less than 7are from 1,2,3,4,5or 6

∴ no. of favourable outcomes = 6×4 = 24

24 3
∴ P(card having value less than 7) = =5 1
40

32 SECTION-D

A train travels at a certain average speed for a distance of 63km and then
travels a distance of 72km at an average speed of 6 km/h more than its
original speed. If it takes 3 hours to complete the total journey, what is its
original average speed?

Solution:
Let original speed of the train be x km/h.
1/2
Then, time taken to travel 63 km = 63/x hours
--------------------------------------------------------------------------------------
New speed = (x + 6) km/hr
1
Time taken to travel 72 km = 72/(x + 6) hours
--------------------------------------------------------------------------------------
ATQ
1

1

14
x2 - 39x - 126 = 0
------------------------------------------------------------------------------------- 1
(x - 42)(x + 3) = 0
x = -3 or x = 42
------------------------------------------------------------------------------------- ½
As the speed cannot be negative, x = 42
Thus, the average speed of the train is 42 km/hr.

OR

A motor boat whose speed is 18 km /h in still water takes 1 hour more to
go 24Km upstream than to return downstream to the same spot.Find the
speed of the stream.
Solution

Let the speed of the stream be x km/h.

∴ The speed of the boat upstream = (18 – x) km/h
112

And the speed of the boat downstream = (18 + x) km/h

------------------------------------------------------------------------------------

We know that time = distance/speed

24 1
⇒ Time taken to go upstream = hours
18−�
24
Also, time taken to go downstream = hours
18+�

--------------------------------------------------------------------------------------

ATQ 1
24 24
- =1
18−� 18+�

15
½
⇒ 24 (18 + x) – 24 (18 – x) = (18 – x) (18 + x)

⇒ x2 + 48x – 324 = 0
½
----------------------------------------------------------------------------------------

⇒ (x + 54)(x- 6) = 0

--------------------------------------------------------------------------------------- ½

⇒ x= 6 or -54

Since x is the speed of the stream, it cannot be negative.

∴ x = 6 gives the speed of the stream as 6 km/h.
33 Prove that if a line is drawn parallel to one side of a triangle intersecting
the other two sides in distinct points, then the other two sides are divided
in the same ratio.
Solution:

½
Given: In ΔABC, DE||BC

½

To prove:
��
=
�� ½
�� ��
------------------------------------------------------------------------------------
Construction : Draw EM⊥AB and DN⊥AC. Join B to E and C to D ½

16

----------------------------------------------------------------------------------------

Proof: In ΔADE and ΔBDE

���� ������
�
×��×�� �� ½
���� �� ����
= �
� = �� --------------(i)
�
����

------------------------------------------------------------------------------------
In ΔADE and ΔCDE

�
����
½
���� ������ ��
���� �� ����
= �
� = ��
-----------------(ii)
�
����

----------------------------------------------------------------------------------
Since, DE||BC [Given]

1
∴ ar(ΔBDE) = ar(ΔCDE) --------------------------------------- (iii)
[Δs on the same base and between the same parallel sides are equal in area]
--------------------------------------------------------------------------------------

From eq. (i), (ii) and (iii)
1
�� ��
: = Hence proved.
�� ��
34 A juice seller was serving his customer using glasses as shown in the
figure. The inner diameter of the cylindrical glass was 5 cm but bottom of
the glass had a hemispherical raised portion which reduced the capacity
of the glass . If the height of the glass was 10cm, find the apparent and
actual capacity of the glass.
[Use � = �. ��]

17

Solution:

5
The inner radius of the glass = cm = 2.5 cm ½
2
Height of the glass = 10 cm
-------------------------------------------------------------------------------------
The apparent capacity of the glass = ��2 ℎ

112
=3.14×2.5×2.5×10 cm3=196.25 cm3

-------------------------------------------------------------------------------------
2 2
Volume of hemisphere = 3πr3=3×3.14×2.5×2.5×2.5 cm3=32.71 cm3
112

----------------------------------------------------------------------------------------

The actual capacity of the glass = apparent capacity of glass - volume of the
hemisphere

=(196.25−32.71) cm3

=163.54 cm3 112

OR

18

A tent is in the shape of a cylinder surmounted by a conical top. If the
height and diameter of the cylindrical part are 2.1 m and 4 m
respectively,and the slant height of the top is 2.8 m, find the area of the
canvas used for making the tent. Also, find the cost of canvas of the tent at
the rate of Rs 500per m2.(Note that the base of the tent will not covered
with canvas.)

Solution:
Radius of base of cylindrical portion = 2 m,
Height of the cylindrical portion = 2.1 m
slant height of conical top = 2.8 m 1

--------------------------------------------------------------------------------------
Curved surface area of cylindrical portion = 2πrh
=2π×2×2.1
1
=8.4 π m 2

----------------------------------------------------------------------------------------
Curved surface area of conical portion = πrl

=π×2×2.8
1
=5.6πm2
----------------------------------------------------------------------------------------
Total curved surface area= Area of canvas used = 2πrh + πrl
= 8.4π + 5.6π
1
=14×22/7=44m2

19

------------------------------------------------------------------------------------
Cost of canvas = Rate × Surface area
1
=500×44 = Rs.22000

35 The median of the following data is 525.find the values of x and y, if total
frequency is 100.

Class Interval Frequency
वर् अंतराल बारं बारता
0-100 2
100-200 5
200-300 x
300-400 12
400-500 17
500-600 20
600-700 y
700-800 9
800-900 7
900-1000 4

Solution:

Class Interval Frequency Cummulative Frequency
वर् अंतराल बारं बारता
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19 + x
400-500 17 36 + x
500-600 20 56 + x
600-700 y 56 + x + y
700-800 9 65 + x + y 1
800-900 7 72 + x + y
900-1000 4 76 + x + y

�
n = 100 ⇒ = 50
2 1
So, 76 + x + y = 100 ⇒ x + y = 24--------------------------(1)
---------------------------------------------------------------------------------------

20

Median = 525 ∴ Median class = 500 - 600

So, l = 500, f = 20, c f = 36 + x, h = 100
1

--------------------------------------------------------------------------------------
n
−cf
Median = l + ( )×h
2
f

50−36−x
⇒525 = 500 + ( ) × 100 1
20
-------------------------------------------------------------------------------------
⇒525−500 = (14− x) × 5
⇒25 = 70−5x
⇒ 5x = 70−25 = 45 ⇒ x = 9
½

-----------------------------------------------------------------------------------
From (1), we get 9 + y = 24
y = 24 - 9= 15
½

SECTION-E

36 Rahul wants to buy a car and plans to take loan from a bank for his car.
He repays his total loan of Rs 1,18,000 by paying every month starting
with the first instalment of Rs 1000.If he increases the instalment by Rs
100 every month. Based on the above information ,answer the following
questions:
(i) Find the amount paid by him in 30th instalment .
(ii)Find the amount paid by him in 30 instalments.
(iii) What amount does he still have to pay after 30th instalment?

21
OR
If total instalments are 40 then amount paid in the last instalment
SOLUTION
(i) Monthly instalment paid by Rahul are 1000, 1100, 1200, … 30 terms

a = 1000, d = 100, an = ?, n = 30

a30 = a + (29)d = 1000 + (29) 100

= 3900 1
So, the amount paid by him in 30th instalment = ₹ 3900.

----------------------------------------------------------------------------------------

(ii)Total amount of all 30 instalments paid = 1000 + 1100 + 1200 + … + 3900

Here, a = 1000, d = 100, n = 30
� 30
∴Sn=2[2a+(n−1)d]⇒S30= 2 [2×1000+(30−1)100]

=15[2000+2900]
1
=₹73500

----------------------------------------------------------------------------------

(iii)So, the loan amount left after 30th instalment
1
= ₹ 118000 – 73500 =

= ₹ 44500 1
Hence, he still has to pay ₹44500 after 30th instalment.

OR

a40 = a+ 39 d

22
= 1000 + 39(100) 1

--------------------------------------------------------------------------------------

= 4900

Amount paid in last instalment = ₹ 4900 1

Resident welfare Association (RWA)of a society put up three electric poles
37
A,B and C in a society’s park. Despite these three poles, some parts of the
park are still in dark.So, RWA decides to have one more electric pole D
in the park.

Based on the above information ,answer the following questions:
(i) Find the position of the pole C.
(ii) Find the distance of the pole B from corner O of the park.
(iii) Find the position of the fourth pole D so that four points A,B,C and D
form a parallelogram.
OR
Find the distance between poles A and C.
SOLUTION
(i) Position of point C(7,4)
1

---------------------------------------------------------------------------

23
(ii) Distance of pole B(4,9) from corner O(0,0)
= (4 − 0)2 + (9 − 0)2 = 97 units 1

--------------------------------------------------------------------------
(iii) A(1,5),B(4,9) ,C(7,4) are three vertices of parallelogram ABCD
and let D(x,y) be the fourth vertex
Mid-point of diagonal AC = Mid-point of BD 1

-----------------------------------------------------------------------

7+1 5+4 �+4 9+�
( , ) = ( , )
2 2 2 2

⇒ x=4 ,y=0
1
∴ D(4,0)
-------------------------------------------------------------------------

OR

Distance between Pole A and C = (7 − 1)2 + (4 − 5)2 1

= 36 + 1 = 37 1

38. A group of students of class X visited India Gate on an educational trip.
The teacher and students had interest in history as well. The teacher
narrated that India Gate , official name Delhi Memorial, originally called
All-India War Memorial, monumental sandstone arch in New
Delhi,dedicated to the troops of British India who died in wars fought
between 1914 and 1919.The teacher also said that India Gate ,which is
located at the eastern end of the Rajpath(formely called the Kingsway),is
about 138 feet (42 metres) in height.

Based on the above information answer the following questions:
(i)What is the angle of elevation if they are standing at a distance of 42 m

24
away from the monument?
(ii) They want to see the tower at an angle of 600.So, they want to know
the distance where they should stand and hence find the distance.
(iii)If the altitude of the Sun is at 600, then find the height of the vertical
tower that will cast a shadow of length 20m.
OR
The ratio of the length of a rod and its shadow is 1:1. Find the angle of
elevation of the Sun .

SOLUTION:

42 1/2
(i) tanθ = 42 = 1

-------------------------------------------------------------------------------
⇒ � = 45° 1/2

----------------------------------------------------------------------------------------
(ii)

42
tan60° = � ½
------------------------------------------------------------------
42 42 ½
3= ⇒�= ⇒ � = 14 3 m
� 3

25
----------------------------------------------------------------------------------
(iii)

ℎ
1
= tan60°
20

---------------------------------------------------------------------------------
1
⇒ ℎ = 20 3 �

OR

� 1
tanθ = �
------------------------------------------------------------------------------

⇒ tanθ= 1⇒ � = 45° 1