HBSE Class 10 Sample Paper 2025 Answers Maths Basic

Board Of School Education Haryana

मॉडल पेपर
उत्तर
2025

MARKING SCHEME BSEH PRACTICE PAPER 2, 10TH MATHS(BASIC) ,
March2025
(ENGLISH MEDIUM)
Q. Expected solutions marks
no.

Section-A

1 (d)60 1

2 (d)more than 3 1

3 (c )(x+2)(x-1)=x2-2x-3 1

4 (c )3 units 1
5 (a) -12 1
6 (a) 50° 1
7 (d) 55° 1

8 (b)
b 1
a2 +b2
9 (a)600 1
10 (b) 10 2 1

11 (d) 3 1

12 (a )
1 1
5
13 Irrational number 1
14 119 cm 1
15 tanθ =a b 1
16 1 1
2
17 77
cm2 or 4 cm2
49� 1
2
18 False 1
19 (a)Both Assertion(A) and Reason (R) are true and Reason (R) is the 1
correct explanation of Assertion(A).

20 (b) Both Assertion(A) and Reason (R) are true but Reason (R) is the not 1
correct explanation of Assertion(A).

SECTION-B
21. x/2 + 2y/3 = -1 1/2
(a) 3x + 4y = -6 …………………………. (i)

….……………………………………………………………………………………………………………………………..
x-y/3 = 3
3x – y = 9 ……………………………. (ii) 1/2

….……………………………………………………………………………………………………………………………..

When the equation (ii) is subtracted from equation (i) we get,
5y = -15
y = -3 ………………………………….(iii) 1/2

….……………………………………………………………………………………………………………………………..

When the equation (iii) is substituted in (i) we get,
3x – 12 = -6
3x = 6
1/2
x=2
Hence, x = 2 , y = -3
21. Using the property of a rectangle,
(b) We know that, 1/2
Lengths are equal,
i.e., CD = AB
Hence, x + 3y = 13 …(i)
….……………………………………………………………………………………………………………………………..

Breadths are equal,
i.e., AD = BC 1/2
Hence, 3x + y = 7 …(ii)
….……………………………………………………………………………………………………………………………..

On multiplying Eq. (ii) by 3 and then subtracting Eq. (i),
We get, 1/2
8x = 8
So, x = 1
….……………………………………………………………………………………………………………………………..

On substituting x = 1 in Eq. (i),
We get, 1/2
y=4
Therefore, the required values of x and y are 1 and 4, respectively.
22. Let (– 4, 6)divide AB internally in the ratio k : 1.
Using the section formula, we get
3�−6 −8�+10
(– 4, 6) = (
�+1
, �+1 ) 1
….…………………………………………………………………………………………………………………………

3�−6
So, – 4 =
�+1 1/2
.….…………………………………………………………………………………………………………………………

⇒ – 4k – 4 = 3k – 6
⇒ 7k =2
⇒, k : 1 = 2 : 7 1/2
We can check for the y-coordinate also.
So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and
B(3, – 8) in the ratio 2 : 7.
23.

In ∆PBC and ∆PDE,

∠BPC = ∠EPD [vertically opposite angles]

PB/PD = 5/10 = ½ … (i) 1

PC/PE = 6/12 = ½ … (ii)

.….…………………………………………………………………………………………………………………………

From equation (i) and (ii),

We get,

PB/PD = PC/PE

Since, ∠BPC of ∆PBC = ∠EPD of ∆PDE and the sides including these. 1/2

.….…………………………………………………………………………………………………………………………

Then, by SAS similarity criteria
1/2
∆PBC ∼ ∆PDE
24. We know that,

(b)
cos 60° = 1/2

sec 30° = 2/√3
1/2
tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2
….…………………………………………………………………………………………………………………………

Now, substitute the values in the given problem, we get

(5cos260° + 4sec230° – tan245°)/(sin2 30° + cos2 30°) 1/2

= {5(1/2)2+4(2/√3)2-1}/(1/2)2+(√3/2)2
….…………………………………………………………………………………………………………………………

= (5/4+16/3-1)/(1/4+3/4) 1/2

={ (15+64-12)/12}/(4/4)
….…………………………………………………………………………………………………………………………

1/2
= 67/12
24.
1+sinA
(a) LHS =
1−sinA
=
1+���� 1+���� 1/2
= ×
1−���� 1+����
….…………………………………………………………………………………………………………………………

1+sinA
= 1/2
1−sin2 A
….…………………………………………………………………………………………………………………………

1+sinA
= 1/2
cos2 A
….…………………………………………………………………………………………………………………………

1+sinA
= cosA 1/2

= secA +tanA = RHS

25.

1/2
Area swept by the minute hand in 60 minutes = Area of the circle with
radius equal to the length of the minute hand = πr2
….…………………………………………………………………………………………………………………………

Area swept by minute hand in 1 minute = πr2/60 1/2
….…………………………………………………………………………………………………………………………

Thus, area swept by minute hand in 5 minutes = (πr2/60) × 5 = πr2/12
1/2

[ ∵Length of the minute hand (r) = 14 cm]

….…………………………………………………………………………………………………………………………

= 1/12 × 22/7 × 14 × 14 cm2
1/2
= 154/3 cm 2

SECTION-C

26. Prove that 2 is irrational.

Solution: 1/2
Let, if possible, 2 be a rational no.

-----------------------------------------------------------------------------------------
p
∴ 2 = , where p and q are co-prime integers and q≠ 0. 1/2
q
---------------------------------------------------------------------------------------
p2
⇒2=
q2
⇒ p2 = 2 q2 ................................(i) 1/2

⇒ 2 divides p2 ⇒ 2 divides p also.

----------------------------------------------------------------------------------------
Let p = 2m,................................(ii) where m is any integer.

⇒ p2 = 4m2................................(iii) 1/2

------------------------------------------------------------------------------------
From (ii) and (iii)
2q2 = 4m2
⇒ q2= 2m2
1/2
⇒ 2 divides q2 ⇒2 divides q also.
⇒ q = 2n......................................(iv)

---------------------------------------------------------------------------------------
From (i) and (iv) , p and q have 2 as common factor.
∴ p and q are not co-prime.
1/2
Hence our supposition is wrong.
∴ 2 is an irrational number.

27. 6x2−3−7x=6x2−7x−3=0
⇒6x2+2x−9x−3=0

⇒2x(3x+1)−3(3x+1)=0
1
⇒(2x−3)(3x+1)=0
Zeros = 3/2,−1/3
….…………………………………………………………………………………………………………………………

α+β=−b/a⇒(3/2)+ (−1/3)=7/6=-(−7)6=−b/a 1

….…………………………………………………………………………………………………………………………

αβ=c/a⇒(3/2)(−1/3)=−1/2=−3/6=c/a 1
Hence proved.

28. Let Rahul’s age be x years and his son’s age be y years.
(a) 1/2
….…………………………………………………………………………………………………………………………

Five years hence(later),

x+ 5 = 3 (y + 5)

⇒x+ 5 = 3 y + 15 1/2

⇒x - 3 y = 10.......(1)

….…………………………………………………………………………………………………………………………

Also, five years ago(before),

x-5 = 7 (y – 5) 1/2

⇒ x-5 = 7y – 35

⇒x-7y = -30........(2)
….…………………………………………………………………………………………………………………………

Subtracting equation (2) from (1),

x - 3 y -x +7y = 10+ 30 1/2
(∵eq.(2) changes its sign)

4y = 40

⇒y = 10
….…………………………………………………………………………………………………………………………

Put y = 10 in eq. (1),
1/2
x - 3(10) = 10
⇒x - 30 = 10

⇒x = 40
….…………………………………………………………………………………………………………………………

Thus, present age of Rahul=x=40 years and 1/2

present age of Rahul's son=y=10 years.
28. Let the larger angle = x
(b) Smaller angle = y
As both angles are supplementary, 1
x + y = 180
⇒ x = 180 - y .... (i)
….…………………………………………………………………………………………………………………………

Difference is 18 degrees.
So, x - y = 18 1/2
⇒ x = 18 + y .... (i)
….…………………………………………………………………………………………………………………………

Substituting the value of x in equation (i) we get,
⇒ 18 + y = 180 - y
⇒ - y - y = 18 - 180
⇒ - 2y = -162
⇒ y=−162/−2
⇒ y = 81 1/2
….…………………………………………………………………………………………………………………………

Substituting the value of y in equation (i), we get,
⇒ x = 180 - 81 = 99 1/2

….…………………………………………………………………………………………………………………………

1/2
Hence, the angles are 99° and 81°.
29. We know that the distance between the two points is given
by the Distance Formula = (x2 − x1 )2 + (y2 − y1 )2 By 1/2
substituting the values of points P (2, - 3) and Q (10, y) in the distance
formula, we get
….…………………………………………………………………………………………………………………………

PQ = (2 − 10)² + ( − 3 − �)² = 10

1/2
PQ = ( − 8)² + (3 + �)² = 10

….…………………………………………………………………………………………………………………………

Squaring on both sides, we get
64 + (y + 3)2 = 100 1/2
….…………………………………………………………………………………………………………………………

(y + 3)2 = 36
y + 3 = 36 1/2
y+3=±6
….…………………………………………………………………………………………………………………………

y + 3 = 6 or y + 3 = - 6
Therefore, y = 3 or - 9 are the possible values for y. 1

30. Given, cos A + cos² A = 1
(a)
⇒ cos A = 1 - cos² A 1
⇒cos A = sin² A [ ∵ sin² A = 1 - cos² A ]
………………………(i)

….…………………………………………………………………………………………………………………………

1/2
LHS=(sin² A + sin⁴ A) = (sin²A + (sin²A)²)
….…………………………………………………………………………………………………………………………

= (sin²A + (cos A)²) [using (i)] 1

….…………………………………………………………………………………………………………………………

= sin²A + cos²A 1/2
= 1 = RHS
30. LHS=(sinA+cosecA)²+(cosA+secA)²
(b)
1/2
=sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA

….…………………………………………………………………………………………………………………………

=sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA 1

[∵cosecA=1/sinA and secA=1/cosA ]

….…………………………………………………………………………………………………………………………

=1+cosec²A+sec²A+2+2
[∵sin²A+cos²A=1] 1/2

….…………………………………………………………………………………………………………………………

=5+(1+cot²A)+(1+tan²A)

[∵1+tan²A=sec²A and 1+cot²A=cosec²A ] 1

=7+tan²A+cot²A= RHS

….…………………………………………………………………………………………………………………………

31.

1/2

….…………………………………………………………………………………………………………………………

Given the height of the observer be DE = 1.5 m
That is AB = 1.5 m
Let BC = h is the height of the chimney
Hence AC = (h – 1.5) m 1/2
Given the distance between the observer and the chimney
is AD = BE = 28.5 m

….…………………………………………………………………………………………………………………………
o
In right △CAD,θ=45
1
tan 45o=AC/AD
⇒1=(h−1.5)/28.5
….…………………………………………………………………………………………………………………………

⇒28.5=h−1.5 1
⇒h=28.5+1.5= 30 m

Thus the height of the chimney is 30 m.

SECTION-D
32. Given, 1
(a) 2nd term, a₂ = 14
3rd term, a₃= 18

Common difference, d = a₃ - a₂ = 18 - 14 = 4
...............................................................................................................................

We know that nth term of an AP is, aₙ = a + (n - 1)d
a₂ = a + d
14 = a + 4
a = 10 1
...............................................................................................................................

Sum of n terms of AP is given by Sₙ = n/2 [2a + (n - 1) d] 1
...............................................................................................................................

S₅₁ = 51/2 [2 × 10 + (51 - 1) 4] 1
...............................................................................................................................

= 51/2 [20 + 50 × 4]
= 51/2 × 220
= 51 × 110
= 5610 1

32.
(b) nth term of an AP aₙ = a + (n - 1)d 1/2
Let a be the first term and d the common difference.
.............................................................................................................................
According to the question, a₃ = 16 and a₇ - a₅ = 12
a + (3 - 1)d = 16 1
a + 2d = 16 ........ (1)
.............................................................................................................................

Using a₇ - a₅ = 12
[a + (7 - 1) d] - [a + (5 - 1) d] = 12
1
[a + 6d] - [a + 4d] = 12 12
2d = 12
d=6
.............................................................................................................................

By substituting this in equation (1), we obtain
a + 2 × 6 = 16

a + 12 = 16
a=4 1
.............................................................................................................................

Therefore, A.P. will be 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, ... 1
Hence, the sequence will be 4, 10, 16, 22, ...

33.
(a)

Statement:Basic Proportionality Theorem
Prove that if a line is drawn parallel to one side of a triangle ,the other two
1
sides are divided in the same ratio.

….………………………………………………………………………..

1/2
Given: In ΔABC, DE||BC

..........................................................

1/2

..........................................................
.......................................
�� �� 1/2
To prove: =
�� ��

----------------------------------------------------------
-----------------
Construction : Draw EM⊥AB and DN⊥AC. Join B to E and C to D
1/2

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

Proof: In ΔADE and ΔBDE

�
���� ������ ���� ��
���� �� ����
= �
� = �� --------------(i) 1/2
�
����

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

�
���� ������ ���� ��
���� �� ����
= �
� = ��
--------------(ii) 1/2
�
����

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

1/2
∴ 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)

�� �� 1/2
: = Hence proved.
�� ��

33.
(b)

1/2

...........................................................................................................................................

Given, ΔABC ∼ ΔPQR
⇒ ∠ABC = ∠PQR (corresponding angles) --------- (1) 1
⇒ AB/PQ = BC/QR (corresponding sides)
......................................................................................................................................
⇒ AB/PQ = (BC/2) / (QR/2)
1
⇒ AB/PQ = BD/QM (D and M are mid-points of BC and QR) ------------ (2)
.......................................................................................................................................
In ΔABD and ΔPQM,
∠ABD = ∠PQM (from 1)
1
AB/PQ = BD/QM (from 2) 12
⇒ ΔABD ∼ ΔPQM (SAS criterion)

.......................................................................................................................................
. 1
⇒ AB/PQ = BD/QM = AD/PM (corresponding sides)
⇒ AB/PQ = AD/PM
Hence proved.

34.
(a)

Depth of each conical depression, h₁ = 1.4 cm
Radius of each conical depression, r = 0.5 cm 1
Dimensions of the cuboid are 15 cm × 10 cm × 3.5 cm

….………………………………………………………………………………………………………….
Volume of wood in the entire pen stand = volume of the wooden
cuboid - 4 × volume of the conical depression 1

….………………………………………………………………………………………………………….

= l × b × h - 4 × 1/3 πr2h₁ 1
….………………………………………………………………………………………………………….

= (15 cm × 10 cm × 3.5 cm) - (4 × 1/3 × 22/7 × 0.5 cm × 0.5 cm × 1.4 cm)
1

….………………………………………………………………………………………………………….

= 525 cm3 - 1.47 cm3
= 523.53 cm3 1

The volume of wood in the entire stand is 523.53 cm3.

34. The total surface area of the cube
(b)

=6×(edge)2=6×5×5 cm2=150 cm2. 1

….………………………………………………………………………………………………………….…………………………

The surface area of the block = Total Surface Area of cube - base area
of hemisphere + Curved Surface Area of hemisphere 1

….………………………………………………………………………………………………………….…………………………

=150−πr2+2πr2
1
2
=(150+πr ) cm , 2

….………………………………………………………………………………………………………….…………………………

=150 cm2+(22/7×4.2/2×4.2/2) cm2
1
2
=(150+13.86) cm
….………………………………………………………………………………………………………….…………………………

=163.86 cm2 1

35.
(a) class interval class-mark (xi ) Number of fi xi
children(fi )

11-13 12 7 84
13-15 14 6 84
15-17 16 9 144
17-19 18 13 234
1+1
19-21 20 f 20f
21-23 22 5 110
23-25 24 4 96
fi =44+f fi xi =752+20f

..................................................................................................……………………………………….

fi xi
Mean = x = 1/2
fi

...........................................................................................………………………………………...

752+20f 1/2
⇒ 18 =
44+f

...............................................................................................…………………………………...
⇒ 18(44+f) =752+20f
1/2

⇒ 792 +18f = 752+ 20f
.......................................................................................................................
⇒792-752 = 20f -18 f 1/2

................................................................................................................
⇒40 = 2f

⇒ f = 20 1
Hence, missing frequency f =20

35.
(b)
Number of Cars Frequency
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8

From the table, it can be observed that the maximum class frequency is 20,
1
belonging to class interval 40 − 50
Therefore, modal class = 40 − 50

….………………………………………………………………………………………………………….…………………………

Class size, h = 10
Lower limit of modal class, l = 40
Frequency of modal class, f₁ = 20 1
Frequency of class preceding modal class, f₀ = 12

Frequency of class succeeding the modal class, f₂ = 11

….………………………………………………………………………………………………………….…………………………

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂) ]× h
1

….………………………………………………………………………………………………………….…………………………

= 40 + [(20 - 12)/(2 × 20 - 12 - 11)] × 10 1/2

….………………………………………………………………………………………………………….…………………………

= 40 + [8/(40 - 23)] × 10
= 40 + (8/17) × 10 1
= 40 + 4.705

….………………………………………………………………………………………………………….…………………………

= 44.705
≈ 44.7
1/2
Hence, the mode is 44.7

Section - E
36. Distance 1
(i) Time =
Speed
(ii)Let the usual speed of plane be x km/h
New increased speed of plane= (x + 250)km/h
Total distance=1500 km
According to question
1500 1500 1
− = 1/2
� � + 250 2

….……………………………………………………………………………………………………………..

1500(� + 250) − 1500� 1
=
�(� + 250) 2

1500x + 375000 − 1500x 1
=
x(x + 250) 2
2
X +250 x = 750000

X2 +250 x - 750000=0 1/2

(iii)(a) X2 +250 x - 750000=0
X2 +1000x - 750x- 750000=0
X(x+1000)-750(x +1000)=0
(x+1000)(x-750)=0 1

….……………………………………………………………………………………………………………..

X=-1000 or x = 750
Reject x=-1000, because speed cannot be negative.
1
Hence,usual speed of plane is 750km/h.

(iii)(b) X2 +250 x - 750000=0
X2 +1000x - 750x- 750000=0
X(x+1000)-750(x +1000)=0
(x+1000)(x-750)=0 1
….……………………………………………………………………………………………………………..

X=-1000 or x = 750
Reject x=-1000, because speed cannot be negative.
1
Hence,new speed of plane is x+250= 750+250=1000km/h.

37. (i) Since,radius at a point of contact is perpendicular to tangent.
∴ By Pythagoras theorem, we have
PA = ��2 + ��2 = 122 + 52 = 169 = 13 cm 1

….……………………………………………………………………………………………………………..

(ii)one common tangent can be drawn when two circles touch externally. 1

….……………………………………………………………………………………………………………..

(iii)(a) By Pythagoras theorem, we have
BQ = TQ2 + TQ2 = 32 + 42 = 25 = 5 cm 1

….……………………………………………………………………………………………………………..

QY= BQ- BY = 5-4 =1 cm 1

….……………………………………………………………………………………………………………..

(iii) (b) PK= PA + AK = 13 + 5 = 18 cm
1

….……………………………………………………………………………………………………………..

XY = XK + KY= 10 +8 = 18 cm 1

38. (i) Total no. of fish in the aquarium = 13+18+12+11= 54
Number of male fish in the aquarium = 36
∴Number of female fish in the aquarium =54- 36 =18
no. of favourable outcomes
So,probability of selecting a female fish= =
total no. of possible outcomes
18 1
=
54 3
1

….……………………………………………………………………………………………………………..

(ii)The probability of selecting a flowerhorn fish=
no. of favourable outcomes 18 1
= 54 = 1
total no. of possible outcomes 3

….……………………………………………………………………………………………………………..

(iii) (a) The probability of selecting a koi fish
no. of favourable outcomes 12 2
= = =
total no. of possible outcomes 54 9 1

….……………………………………………………………………………………………………………..

no. of favourable outcomes 13
P(selecting a guppy fish) = =
total no. of possible outcomes 54 1

(iii) (b)Total no. of angel fish and flowerhorn fish= 18 +11 =29

29
P (selecting either angel fish or flowerhorn fish)= 1
54

….……………………………………………………………………………………………………………..

P (selecting neither angel fish nor flowerhorn fish) =
=1- P (selecting either angel fish or flowerhorn fish)
29 25
= 1- = 54 1
54

MARKING SCHEME BSEH PRACTICE PAPER 2, 10TH गणित (आधार ) ,
March2025
(हिंदी माध्म)
Q. Expected solutions marks
no.

खण्-क

1 (d)60 1

2 (d) 3 से अधिक
1

3 (c )(x+2)(x-1)=x2-2x-3 1

4 1
(c )3 इकाई

5 (a) -12 1
6 (a) 50° 1
7 (d) 55° 1

8 (b)
b 1
a2 +b2
9 (a)600 1
10 (b) 10 2 1

11 (d) 3 1

12 1 1
(a ) 5
13 अपरिमेय संखयय 1
14 119 cm 1
15 tanθ =a b 1
16 1 1
2
17 77 49� 1
2
cm2 or 4
cm2
18 असतय 1
19 (a) अभिकथन (A) औि तक् (R) दोनो सही है औि तक् (R), अभिकथन (A) 1

की सही वययखयय कितय है ।

20 (b) अभिकथन (A) औि तक् (R) दोनो सही है औि तक् (R), अभिकथन(A) 1

की सही वययखयय नहीं है ।

खण् –ख

21. x/2 + 2y/3 = -1 1/2
(a) 3x + 4y = -6 …………………………. (i)
….……………………………………………………………………………………………………………………………..
x-y/3 = 3
1/2
3x – y = 9 ……………………………. (ii)

….……………………………………………………………………………………………………………………………..

जब समीकरण (ii) को समीकरण (i) से घटाया जाता है तो हमे पार होता है,
5y = -15
1/2
y = -3 ………………………………….(iii)

….……………………………………………………………………………………………………………………………..

जब समीकरण (iii) को (i) मे पधततिाधथत ककया जाता है तो हमे पार होता है,
3x – 12 = -6
1/2
3x = 6
x=2
अत:, x = 2 , y = -3
21. एक आयत के गुण का उथयोग करते हए,
(b) 1/2
हम जानते है कक,
लंबाई समान है,
i.e., CD = AB
Hence, x + 3y = 13 …(i)
….……………………………………………………………………………………………………………………………..

चौडाई बराबर है, 1/2

i.e., AD = BC
अत:, 3x + y = 7 …(ii)
….……………………………………………………………………………………………………………………………..
समीकिण (ii) को 3 से गण
ु य किके औि फिि समीकिण (i) से घटयने पि हमे पयरत होतय है ,
1/2

8x = 8
So, x = 1
….……………………………………………………………………………………………………………………………..

समीकरण(i) मे x = 1 रखने थर ,
1/2
हमे धमलता है,
y=4
इसधलए, x और y के वांधित मान कमरम 1 और 4 है
22. मान लीधजए (- 4, 6) AB को आंतररक रथ से k: 1 के अनुथात मे धवभाधजत करता है।
विियजन सूत का उथयोग करते हए, हमे भमलतय है
3�−6 −8�+10 1
(– 4, 6) = (
�+1
, �+1 )
….…………………………………………………………………………………………………………………………

3�−6
So, – 4 =
�+1 1/2
.….…………………………………………………………………………………………………………………………

⇒ – 4k – 4 = 3k – 6
⇒ 7k =2
1/2
⇒, k : 1 = 2 : 7
हम y-धनर्रांक की भी जांच कर सकते है।
अत:, बबंरु (- 4, 6), बबंरु A(- 6, 10)और बबंरु B(3, – 8)
को धमलाने वाले रे खाखंड को 2:7 के अनुथात मे धवभाधजत करता है ।

23.

∆PBC और ∆PDE मे,

∠BPC = ∠EPD [भिरय्भिमुु कोण]
1
PB/PD = 5/10 = ½ … (i)

PC/PE = 6/12 = ½ … (ii)

.….…………………………………………………………………………………………………………………………

समीकरण (i) और (ii) से,

हमे धमलता है,

PB/PD = PC/PE कयोफक,∆PBC कय ∠BPC= ∆PDE कय ∠EPD तथय उनकी सम्मभलत
िुजयएं िी समयनुपयती है
1/2
.….…………………………………………………………………………………………………………………………

∴ SAS समरपतय कसौटी दियिय
∆PBC ∼ ∆PDE
1/2
24. हम जयनते है फक
(b) cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1 1/2

sin 30° = 1/2

cos 30° = √3/2
….…………………………………………………………………………………………………………………………

उपिोकत मयनो को ददए गए प्न मे िुने पि
(5cos260° + 4sec230° – tan245°)/(sin2 30° + cos2 30°)

= {5(1/2)2+4(2/√3)2-1}/(1/2)2+(√3/2)2 1/2
….…………………………………………………………………………………………………………………………

= (5/4+16/3-1)/(1/4+3/4)

={ (15+64-12)/12}/(4/4) 1/2
….…………………………………………………………………………………………………………………………

= 67/12 1/2
24.
1+sinA
(a) LHS =
1−sinA
=

1/2
1+���� 1+����
= 1−����
×
1+����
….…………………………………………………………………………………………………………………………

1+sinA 1/2
=
1−sin2 A
….…………………………………………………………………………………………………………………………

1+sinA
= 1/2
cos2 A
….…………………………………………………………………………………………………………………………

1+sinA
= cosA 1/2

= secA +tanA = RHS

25.
धमनट की सुई दारा 60 धमनट मे तय ककया गया केतफल =धमनट की सूई की लंबाई
के बराबर धतरया वाले वृत कय केतफल=πr2 1/2
….…………………………………………………………………………………………………………………………

धमनट की सुई दारा 1 धमनट मे तय ककया गया केतफल = πr2/60 1/2

….…………………………………………………………………………………………………………………………

अत:,धमनट की सुई दारा 5 धमनट मे तय ककया गया केतफल = (πr2/60) × 5 = 1/2
2
πr /12

[ ∵धमनट की सुईकी लमबाई (r) = 14 cm]

….…………………………………………………………………………………………………………………………

= 1/12 × 22/7 × 14 × 14 cm2 1/2

= 154/3 cm2

खण् –ग
26. मयन लीमजए,यदद संिि हो तो , 2 एक परिमेय संखयय है l
1/2

----------------------------------------------------------------- ------------------------
p
∴ 2 = , जहयँ p औि q सह अियजय पूणयणक है तथय q≠ 0. 1/2
q
---------------------------------------------------------------------------------------
p2
⇒2=
q2
1/2
⇒ p2 = 2 q2 ................................(i)

⇒ 2 , p2को विियमजत कितय है ⇒ 2 , p को िी विियमजत कितय है l

----------------------------------------------------------------------------------------
मयनय p = 2m,................................(ii) जहयँ m कोई पणू यणक है l
⇒ p2 = 4m2................................(iii)
1/2

------------------------------------------------------------------------------------
(i) औि(iii) से
2q2 = 4m2
⇒ q2= 2m2
⇒ 2 , q2को विियमजत कितय है ⇒ 2 , q को िी विियमजत कितय है l 1/2
⇒ q = 2n......................................(iv)

---------------------------------------------------------------------------------------
(ii) औि (iv) से, p औि q कय उियननष् गण ु नुंड 2 है ।
∴ p औि q सह-अियजय पूणयणक नहीं है
अतः हमयिी कलपनय गलत है । 1/2
∴ 2 एक अथररमेय संखया है l

27. 6x2−3−7x=6x2−7x−3=0
⇒6x2+2x−9x−3=0
⇒2x(3x+1)−3(3x+1)=0 1
⇒(2x−3)(3x+1)=0
िन
ू यकα,β= 3/2,−1/3
….…………………………………………………………………………………………………………………………

1
α+β=−b/a⇒(3/2)+ (−1/3)=7/6=-(−7)6=−b/a

….…………………………………………………………………………………………………………………………

1
αβ=c/a⇒(3/2)(−1/3)=−1/2=−3/6=c/a
अत: धसद हआ।
28.
माना कक राहल की आयु x वर् है और उसके थुत की आयु y वर् है। 1/2
(a)
….…………………………………………………………………………………………………………………………

थाँच वर् बार (बार मे),
x+ 5 = 3 (y + 5)

⇒x+ 5 = 3 y + 15
1/2
⇒x - 3 y = 10.......(1)

….…………………………………………………………………………………………………………………………

साि ही, थांच साल थूव्(थहले),
x-5 = 7 (y – 5)
1/2
⇒ x-5 = 7y – 35

⇒x-7y = -30........(2)
….…………………………………………………………………………………………………………………………

समीकरण (2) को (1) से घटाने थर,
x - 3 y -x +7y = 10+ 30

(∵समीकरण(2) का धचनह बरलता है) 1/2
4y = 40

⇒y = 10
….…………………………………………………………………………………………………………………………

समीकरण(1) मे y = 10 रखने थर।

x - 3(10) = 10 1/2
⇒x - 30 = 10

⇒x = 40
….…………………………………………………………………………………………………………………………
अत: राहल की वत्मान आयु=x=40 वर् और

1/2
राहल के थुत की वत्मान आयु=y=10 वर्.

28.
माना बडा कोण = x
(b)
िोटा कोण = y
1
चूँकक रोनो कोण संथूरक है,
∴ x + y = 180
⇒ x = 180 - y .... (i)
….…………………………………………………………………………………………………………………………

अंतर 18 धडगी है. 1/2
∴ x - y = 18
⇒ x = 18 + y .... (i)
….…………………………………………………………………………………………………………………………

समीकरण (i) मे x का मान रखने थर हमे धमलता है,
⇒ 18 + y = 180 - y
⇒ - y - y = 18 - 180
⇒ - 2y = -162
⇒ y= −162/−2 1/2

⇒ y = 81
….…………………………………………………………………………………………………………………………

समीकरण (i) मे y का मान रखने थर, हमे धमलता है, 1/2
x = 180 - 81 = 99
….…………………………………………………………………………………………………………………………

इसधलए, कोण 99° और 81° है। 1/2

29.
हम जानते है कक रो बबंरओ
ु के बीच की रूरी, रूरी सूत
= (x2 − x1 )2 + (y2 − y1 )2 दारा री जाती है
1/2
रूरी सूत मे बबंरु P (2, - 3) और Q (10, y) के मानो को पधततिाधथत करने थर,
हमे धमलता है
….…………………………………………………………………………………………………………………………

PQ = (2 − 10)² + ( − 3 − �)² = 10 1/2

PQ = ( − 8)² + (3 + �)² = 10

….…………………………………………………………………………………………………………………………

रोनो तरफ वग् करने थर, हमे धमलता है 1/2

64 + (y + 3)2 = 100
….…………………………………………………………………………………………………………………………

(y + 3)2 = 36
1/2
y + 3 = 36
y+3=±6
….…………………………………………………………………………………………………………………………

y + 3 = 6 यय y + 3 = - 6 1
इसधलए, y = 3 या - 9, y के धलए संभाधवत मान है।
30. ददयय है : cos A + cos² A = 1
(a)
⇒ cos A = 1 - cos² A 1
⇒cos A = sin² A [ ∵ sin² A = 1 - cos² A ]
………………………(i)

….…………………………………………………………………………………………………………………………

1/2
LHS= (sin² A + sin⁴ A) = (sin²A + (sin²A)²)
….…………………………………………………………………………………………………………………………

= (sin²A + (cos A)²) [(i)कय उपयोग किने पि] 1

….…………………………………………………………………………………………………………………………

1/2
= sin²A + cos²A
= 1 = RHS
30. LHS=(sinA+cosecA)²+(cosA+secA)²
(b)
1/2
=sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA

….…………………………………………………………………………………………………………………………

=sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA 1

[∵cosecA=1/sinA and secA=1/cosA ]

….…………………………………………………………………………………………………………………………

=1+cosec²A+sec²A+2+2
[∵sin²A+cos²A=1] 1/2

….…………………………………………………………………………………………………………………………

=5+(1+cot²A)+(1+tan²A)

[∵1+tan²A=sec²A and 1+cot²A=cosec²A ] 1

=7+tan²A+cot²A= RHS

31.

1/2

….…………………………………………………………………………………………………………………………
पेकक की ऊं चाई DE = 1.5 मीटर री गई है
AB = DE=1.5 m
माना BC = h धचमनी की ऊँ चाई है
अत: AC = (h – 1.5) m 1/2
पेकक और धचमनी के बीच की रूरी AD = BE = 28.5 मीटर है
….…………………………………………………………………………………………………………………………
o
समकोण △CAD मे ,θ=45
1
tan 45o=AC/AD
⇒1=(h−1.5)/28.5
….…………………………………………………………………………………………………………………………

⇒28.5=h−1.5 1
⇒h=28.5+1.5= 30 m

अतम धचमनी की ऊं चाई 30 मीटर है।

खणड-घ

32. ददयय है :
(a) दस
ू िय पद a₂ = 14 1
तीसिय पद a₃= 18
सयि् अंति d = a₃ - a₂ = 18 - 14 = 4
...............................................................................................................................

हम जयनते है फक AP कय nियँ पद होतय है : aₙ = a + (n - 1)d
a₂ = a + d
1
14 = a + 4
a = 10
...............................................................................................................................

1
AP के n पदो कय योगिल होतय है :Sₙ = n/2 [2a + (n - 1) d]
...............................................................................................................................

1
∴ S₅₁ = 51/2 [2 × 10 + (51 - 1) 4]
...............................................................................................................................

= 51/2 [20 + 50 × 4]
= 51/2 × 220
= 51 × 110 1
= 5610

32.
(b) AP कय nियँ पद होतय है : aₙ = a + (n - 1)d 1/2
जहयँ a पथम पद तथय d सयि् अंति होतय है

.............................................................................................................................
प्न के अनुसयि , a₃ = 16 औि a₇ - a₅ = 12 1
a + (3 - 1)d = 16
a + 2d = 16 ........ (1)
.............................................................................................................................

a₇ - a₅ = 12 कय उपयोग किते हुए
[a + (7 - 1) d] - [a + (5 - 1) d] = 12

[a + 6d] - [a + 4d] = 12 1
12
2d = 12
d=6
.............................................................................................................................

इसे समीकिण (1) मे िुने पि हमे पयरत होतय है : a + 2 × 6 = 16
a + 12 = 16
a=4 1
.............................................................................................................................

अत:अभिष् AP होगी : 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, ...
यय 4, 10, 16, 22, ... 1

33.
(a)

किन: आिारभूत समानुथाधतकता पमेय
धसद कीधजए कक यकर ककसी धतभुज की एक भुजा के समानांतर एक रे खा खीची जाए,
1
तो अनय रो भुजाएँ समान अनुथात मे धवभाधजत हो जाती है।

….………………………………………………………………………..

हद्ा िै : ΔABC मे, DE||BC 1/2

............................................................................................

1/2

.................................................................................................
�� ��
सिदध करना िै : = 1/2
�� ��
---------------------------------------------------------------------------
रचना : EM⊥AB तथय DN⊥AC ुींचिएl B को E से तथय C को D
1/2

से भमलयइये l

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

पमयण: ΔADE तथय ΔBDE मे

����का क्षेत �
���� ��
= �
� = --------------(i) 1/2
����का क्षेत �
���� ��

--------------------------------------------------------------------------
ΔADE तथय ΔCDE मे

�
����का केतफल ×��×�� �� 1/2
= �
� = --------------(ii)
����का केतफल �
���� ��

---------------------------------------------------------------------------
कयोफक DE||BC [ददयय है ]

1/2
∴ (ΔBDE) का केतफल = (ΔCDE)का केतफल ----------------------------- (iii)
[ ∵ एक ही आिार थर और एक ही समानांतर भुजाओ के बीच बनी धतभुजो का
केतफल बराबर होता है]

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

समीकरण (i), (ii) और (iii)से

�� ��
: = यही धसद करना िा l 1/2
�� ��

33.
(b)

1/2

...........................................................................................................................................

ददयय है : ΔABC ∼ ΔPQR
⇒ ∠ABC = ∠PQR (संगत कोण) --------- (1) 1
⇒ AB/PQ = BC/QR (संगत िज
ु यएं)
...........................................................................................................................................
⇒ AB/PQ = (BC/2) / (QR/2) 1
⇒ AB/PQ = BD/QM (∵D औि M ,BC तथयQR के मधय बिंद ु हैl) ------------ (2)
..........................................................................................................................................
ΔABD तथय ΔPQM मे ,
∠ABD = ∠PQM ((1)से ) 1
12
AB/PQ = BD/QM ((2)से )
⇒ ΔABD ∼ ΔPQM (SAS कसौटी दियिय )

........................................................................................................................................
⇒ AB/PQ = BD/QM = AD/PM (संगत िज
ु यएं ) 1
⇒ AB/PQ = AD/PM
यही भसद् किनय थय l

34.
(a)

प्येक रंकाकार गडे की गहराई, h₁=1.4 सेमी
1
प्येक रंकाकार गडे की धतरया, r = 0.5 सेमी
घनाभ का आयाम 15 सेमी × 10 सेमी × 3.5 सेमी है
….………………………………………………………………………………………………………….
थूरे थेन तटैड मे लकडी का आयतन = लकडी के घनाभ का आयतन - 4 × रंकाकार गडे
1
का आयतन
….………………………………………………………………………………………………………….

= l × b × h - 4 × 1/3 πr2h₁ 1

….………………………………………………………………………………………………………….

= (15 cm × 10 cm × 3.5 cm) - (4 × 1/3 × 22/7 × 0.5 cm × 0.5 cm × 1.4 cm) 1

….………………………………………………………………………………………………………….

= 525 cm3 - 1.47 cm3
= 523.53 cm3 1

थूरे तटैड मे लकडी का आयतन 523.53 cm3है।

34.
घन का कु ल थृषीय केतफल
(b)

=6×(िुजय)2=6×5×5 cm2=150 cm2. 1

….………………………………………………………………………………………………………….…………………………

बललक का थृषीय केतफल = घन का कु ल थृषीय केतफल - गोलाि् का आिार केतफल
1
+ गोलाि् का वक थृषीय केतफल
….………………………………………………………………………………………………………….…………………………

=150−πr2+2πr2

=(150+πr2) cm2, 1

….………………………………………………………………………………………………………….…………………………

=150 cm2+(22/7×4.2/2×4.2/2) cm2

1
=(150+13.86) cm2
….………………………………………………………………………………………………………….…………………………

=163.86 cm2 1

35.
(a) िग् -अंतियल िग् -चिनह(xi) बचो की संखया(fi ) fi xi

11-13 12 7 84
13-15 14 6 84
15-17 16 9 144
17-19 18 13 234
1+1
19-21 20 f 20f
21-23 22 5 110
23-25 24 4 96
fi =44+f fi xi =752+20f

..................................................................................................……………………………………….

fi xi
मयधय = x = 1/2
fi

...........................................................................................………………………………………...

752+20f 1/2
⇒ 18 =
44+f

...............................................................................................…………………………………...
⇒ 18(44+f) =752+20f
1/2

⇒ 792 +18f = 752+ 20f
.......................................................................................................................
⇒792-752 = 20f -18 f 1/2

................................................................................................................
⇒40 = 2f

⇒ f = 20 1
अत:लुरत ियिं ियितय f =20

35.
(b)
कयिो की संखयय ियिं ियितय
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8

ताधलका से, यह रेखा जा सकता है कक अधिकतम वग् ियिं ियितय 20 है, जो वग् अंतराल 40 - 50 से 1
संबंधित है।
इसधलए, बहलक वग् = 40 − 50
….………………………………………………………………………………………………………….…………………………

िग् आमयप h = 10
बहलक वग् की धनचली सीमा, l = 40 1
बहलक वग् की बारं बारता, f₁=20

बहलक वग् से थहले वाले वग् कीबारं बारता, f₀ = 12
बहलक वग् के बार वाले वग् की बारं बारता, f₂ = 11
….………………………………………………………………………………………………………….…………………………

िहुलक = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂) ]× h
1

….………………………………………………………………………………………………………….…………………………
1/2
= 40 + [(20 - 12)/(2 × 20 - 12 - 11)] × 10

….………………………………………………………………………………………………………….…………………………

= 40 + [8/(40 - 23)] × 10 1
= 40 + (8/17) × 10
= 40 + 4.705

….………………………………………………………………………………………………………….…………………………

= 44.705 1/2
≈ 44.7
अत: िहुलक 44.7

खणड-ङ

36. दिू ी
1
(i) समय =
गनत
(ii)माना धवमान की सामानय गधत x ककमी/घंटा है
धवमान की नई बढी हई गधत = (x + 250) ककमी/घंटा
कु ल रूरी=1500 ककमी
प् के अनुसार
1500 1500 1
− = 1/2
� � + 250 2

….……………………………………………………………………………………………………………..

1500(� + 250) − 1500� 1
=
�(� + 250) 2

1500x + 375000 − 1500x 1
=
x(x + 250) 2
2
X +250 x = 750000

1/2
X2 +250 x - 750000=0

(iii)(a) X2 +250 x - 750000=0
X2 +1000x - 750x- 750000=0
X(x+1000)-750(x +1000)=0 1
(x+1000)(x-750)=0

….……………………………………………………………………………………………………………..

X=-1000 अथिय x = 750
x=-1000 को अतवीकार करे , कयोकक गधत ऋणा्मक नही हो सकती। 1

अतम, धवमान की सामानय गधत 750 ककमी/घंटा है।
(iii)(b) X2 +250 x - 750000=0
X2 +1000x - 750x- 750000=0
X(x+1000)-750(x +1000)=0
(x+1000)(x-750)=0 1
….……………………………………………………………………………………………………………..

X=-1000 अथिय x = 750
x=-1000 को अतवीकार करे , कयोकक गधत ऋणा्मक नही हो सकती। 1
अतम, धवमान की नई गधत x+250= 750+250=1000 ककमी/घंटा है।
37.
(i) चूँकक, संथक् बबंरु थर धतरया तथर् रे खा के लंबवत होती है।
∴ थाइिागोरस पमेय से 1
PA = ��2 + ��2 = 122 + 52 = 169 = 13 cm

….……………………………………………………………………………………………………………..

(ii)जब रो वृत बाह रथ से तथर् करते है तो एक उभयधनष तथर्रेखा खीची जा 1

सकती है।
….……………………………………………………………………………………………………………..

(iii)(a) पयइथयगोिस पमेय से
1
BQ = TQ2 + TQ2 = 32 + 42 = 25 = 5 cm

….……………………………………………………………………………………………………………..

QY= BQ- BY = 5-4 =1 cm
1

….……………………………………………………………………………………………………………..

(iii) (b) PK= PA + AK = 13 + 5 = 18 cm
1

….……………………………………………………………………………………………………………..

XY = XK + KY= 10 +8 = 18 cm
1

38.
(i) एके ररयम मे कु ल मिधलयाँ = 13+18+12+11= 54

एके ररयम मे कु ल नर मिधलयाँ = 36
∴एके ररयम मे मारा मिधलयाँ =54- 36 =18
अनुकूल परिणयमोकी संखयय 18 1
∴मारा मिली के चयन की पाधयकता = = = 1
कुल संियवित परिणयमो की संखयय 54 3

….……………………………………………………………………………………………………………..

अनक
ु ू ल परिणयमोकी संखयय
(ii)फलावरहलन् मिली को चुनने की पाधयकता = =
कुल संियवित परिणयमो की संखयय 1
18 1
=
54 3

….……………………………………………………………………………………………………………..

अनुकूल थररणामोकी संखया
(iii) (a) ‘कोई’ मिली के चुनने की पाधयकता= = 1
कु ल संभाधवत थररणामो की संखया
12 2
=
54 9

….……………………………………………………………………………………………………………..

अनुकूल थररणामोकी संखया 13 1
गपथी मिली के चुनने की पाधयकता= = 54
कु ल संभाधवत थररणामो की संखया

(iii) (b)एंजल और फलावरहलन् मिधलयोकी कु ल संखया= 18 +11 =29
1
29
एंजल मिली या फलावरहलन् मिली को चुनने की पाधयकता=
54

….……………………………………………………………………………………………………………..

न ही एंजल मिली और न ही फलावरहलन् मिली को चुनने की पाधयकता=
1
=1- एंजल मिली या फलावरहलन् मिली को चुनने की कमरम पाधयकता
29 25
= 1- = 54
54

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