HBSE Class 9 Sample Paper 2025 Answers Maths

Board Of School Education Haryana

मॉडल पेपर
उत्तर
2025

MARKING SCHEME BSEH PRACTICE PAPER 1, 9TH MATHS , March2025
(HINDI MEDIUM)
Q. Expected solutions marks
no.

Section-A

1 (C) √9 1
2 (B) 40° 1
3 (C) SSA 1
4 (D) 3.4 cm 1
5 (A) 75º 1
6 T 1
7 (A) 24 cm2 1
8 C) आधा 1
9 (A) 137.5 1
10 (D) 2 1

11 (D) 7 1
12 (B) 3 1
13 (A) - 6 1
14 (B) 1

15 (A) 65o 1
16 3600 1
17 (B) 50o 1
18 90o 1
19 D) A असत्य है लेकिन R सत्य है 1
20 B) A और R दोनों सत्य हैं लेकिन R, A िी सही व्याख्या नहीीं है । 1

SECTION-B
21.
हम जानते हैं कि 5+1=6

= = और = =
1

….…………………………………………………………………………………………………….........
पाींच पररमेय सींख्याएँ =
1

22. (3√5 -5√2 )(4√5 +3√2)
3√5(4√5 +3√2) - 5√2(4√5 +3√2) 1
….……………………………………………………………………………………………………………..

60+9√10-20√10-30
1
30-11√10
OR
.....................................................................................................................
X 1

..............................................................................................
1
23.
X

1
=
......................................................................................................................

1

24. 104 × 97
104×97= (100+4) × (100-3)

यहाँ, x = 100 , a=4 , b = -3 1

सर्वसममिा , (x+a) (x+b) = x2 + (a+b) x + ab द्र्ारा

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

= (100)2+(4-3)100+(4×-3)

= 10000+100-12 1

= 10088

25.
यदद x – 2, p(x) िा एि गण
ु नखींड है तो P(2) = 0

गुणनखींड प्रमेय द्र्ारा
1
3 2
p(2) = 4(2) -3(2) - 4(2)+3k=0
.......................................................................................................................

= 32-12-8+3k=0
=12+3k=0
1
3k= -12
K= - 4
..............................................OR.................................................................

x-3 =0 लेने पर x = 3
1
x=3 बहुपद में रखने पर (3)3-4(3)2+3+6

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

= 27-36+3+6= 0

अतः गुणनखींड प्रमेय द्र्ारा x-3, बहुपद x3-4x2+x+6 िा एि गुणनखींड है I 1

SECTION-C
26.
y, x िे अनुक्रमानुपाती है
1
y x
y = kx
.......................................................................................................
y=12 रखने पर जब x=4 है
12=4k 1
k=3
.......................................................................................................
इसमलए y=3x
x=5 रखने पर y=3x5=15 1

27. िैप्सूल िा व्यास = 3.5 mm
िैप्सल mm
ू िी त्रिज्या = 1
……………………………………………………………………………...

िैप्सल
ू में दर्ा िा आयतन = πr3 1

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

.

22
= x x x x 1

= 22.46 मममी2

.........................................................OR............................................................................

माना चींद्रमा िी त्रिज्या = r
माना पथ् ृ र्ी िी त्रिज्या = 4r 1
..........................................................................................................................................
.
चींद्रमा िा पष्ृ ठीय क्षेिफल
=
π
पथ्ृ र्ी िा पष्ृ ठीय क्षेिफल π
1
..........................................................................................
=
π
π
1

= = 1:16
16

28.

The expression 64m3 – 343n3 , can be written as (4m)3-(7n)3 1

3 3
64m3 – 343n3= (4m) -(7n)

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

We know that, x3-y3 = (x-y)(x2+xy+y2)
1
= (4m-7n)[(4m)2+(4m)(7n)+(7n)2
............................................................................................................................. ............................................................

.
1
2 2
= (4m-7n)(16m +28mn+49n )

29.

2x + 3y = 12

X=0 रखने पर 2x0 + 3xY = 12
3Y = 12
Y=4 1

(i) पहला हल (0,4)
......................................................................................................................
X=1 रखने पर 2x1+ 3xY = 12
3Y=10
1
10
Y=
10
(ii) दस ू रा हल (1, )
.......................................................................................................................
X=2 रखने पर 2x2 + 3Y= 12
3Y= 8 1

Y=
(iii) तीसरा हल (2,

30.
माना x = = 0.2353535……………… (i)
1
(i) िो 10 से दोनों तरफ गण
ु ा िरने पर
10x = 2.353535............................ (ii)
.....................................................................................................................
(ii) िो दोनों तरफ 100 से गण
ु ा िरने पर 1
1000x = 235.353535......................... (iii)
.....................................................................................................................
(iii) – (ii) िरने पर
990x = 233.0000 1
233
x =
990

31. The expression, 8X3 + 27Y3 + 36X2Y + 54XY2

can be written as (2X)3 + (3Y)3 +3(2X)2(3Y) +3(2X)(3Y)2 1

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

=(2X)3 + (3Y)3 +3(2X)2(3Y) +3(2X)(3Y)2
1
सर्वसमिका द्र्ारा (x + y)3 = x3 + y3 + 3xy (x + y)

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

=(2X)3 + (3Y)3 +3(2X)(3Y)(2X +3Y) 1

=(2X+3Y)3

=(2X+3Y) (2X+3Y) (2X+3Y)
OR
31. 8X3 + Y3 + 27Z3 – 18XYZ िो
1
इस प्रिार मलख सिते है (2X) + Y +(3Z) -3(2X)(Y)(3Z)
3 3 3

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

सर्वसमिका द्र्ारा x3 + y3+z3–3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
1

=(2X+Y+3Z)((2X)2 +Y2 +(3Z)2 -2XY-Y(3Z)-3Z(2X))

......................................................................................................
(2X+Y+3Z)(4X2 +Y2 +9Z2 -2XY-3YZ-6ZX)
1

SECTION-D
32.
माना एि रे खाखण्ड AB िे दो मध्य त्रबन्द ु C तथा C’ हैं I 1
.......................................................................................................
यदद C, रे खाखण्ड AB िा मध्य त्रबन्द ु है I
1
तो , AC=BC
AC = AB ........................(i)
.......................................................................................................
माना C’ , रे खाखण्ड AB िा मध्य त्रबन्द ु है I
1
इसमलए , AC’=BC’
AC’ = AB .......................(ii)

......................................................................................................
समीिरण (i) र् (ii) से
1
AC=AC’
......................................................................................................
C=C’
1
इसमलए C र् C’ एि ही त्रबींद ु है अथावत सींपाती है I
अतः एि रे खाखण्ड िा एि ही मध्य त्रबींद ु होता है I

33

1
ददया हुआ है :- a : b = 2 : 3 तथा ∠POY = 90°
........................................................................................................................
ज्ञात िरना है :- a तथा b िा मान I
1
.......................................................................................................................

हल :- a : b = 2 : 3 मान लीजजए a =2x और b =3x है

हम जानते हैं कि रै खखि यग्ु मों िा योग हमेशा 180° िे बराबर होता है

1
इसमलए, POY +a +b = 180°

जैसा कि प्रश्न में ददया गया है POY = 90° िा मान रखने पर,

a+b = 90°

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

2x+3x = 90° इसे हल िरने पर हमें प्राप्त होता है

5x = 90°
1
So, x = 18°

a = 2×18° = 36°

b = 3×18° = 54°

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

आरे ख से, b+c भी एि सीधी रे खा पर िोण बनाता है ,

इसमलए, b+c = 180°

c+54° = 180°
1
c = 126°

.........................................................OR.....................................................................

1

ददया हुआ है :- AB II CD , APQ = 50° तथा APR = 127°
...................................................................................................................................
ज्ञात िरना है :- x तथा y िा मान 1
...................................................................................................................................
हल :- AB II CD तथा PQ एि ततयवि रे खा है ( ददया हुआ है )
इसमलए, APQ=PQR ( अींतः एिाींतर िोण )

APQ = 50° और PQR = x िा मान रखने पर 1

x = 50°

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

APR = PRD (अींतःएिाींतर िोण)
1
APR = 127° ( क्योकि PRD = 127° ददया हुआ है )

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

हम र्ह जानते हैं APR =APQ+QPR
1
अब, QPR = y और APR = 127° िा मान रखने पर,

हम पाते हैं 50°+ y = 127°

y = 77°

x = 50° और y = 77°

34.
1
माना त्रिभुज िी तीन भुजाएँ हैं 12x , 17x, 25x
540
अर्द्व पररमाप (s) = = 270 सेमी
.......................................................................................................................

s =

1
12 17 25
270 =

54x= 540
x=10
........................................................................................................................
.
पहली भुजा = 12x10= 120 सेमी 1
दस
ू री भुजा = 17x10 = 170 सेमी
तीसरी भुजा = 25x10 = 250 सेमी
........................................................................................................................
.
1
िा क्षे =

= 270 270 120 270 170 270 250

......................................................................................................
1
= 270 150 100 20

= 9000 र्गव सेमी

OR
34.

शींक्र्ािार टोपी िी त्रिज्या (r) = 7 सेमी 1

शींक्र्ािार टोपी िी ऊींचाई (h) = 24 सेमी

........................................................................................................................
शींक्र्ािार टोपी िी ततयवि ऊींचाई (L) =
= 24
= 49 57
= 625
1
= 25 सेमी
........................................................................................................................
10 टोपपयाँ बनाने िे मलए आर्श्यि शीट िा क्षेिफल= 10 πrL
1
.......................................................................................................................

22
= 10x x7x25
1
= 5500 र्गव सेमी

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

10 टोपपयों िा िुल मूल्य = 5500 x 0.35 = 1925 रु.

1

35.
1
ददया गया है : AC = AD और रे खाखींड AB , A िो समद्पर्भाजजत िरती है I

........................................................................................................................
1
मसर्द् िरना है : ΔABC  ΔABD
..........................................................................................................................................
प्रिाण:
1
त्रिभुजों ΔABC और ΔABD में

(i) AC = AD ( ददया गया है )

(ii) AB = AB (उभयतनष्ठ)

(iii) CAB = DAB (क्योंकि AB िोण A िा समद्पर्भाजि है )

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

इसमलए, ΔABC  ΔABD. (SAS सर्ाांगसमता िसौटी िे अनुसार)
1

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

प्रश्न िे दस
ू रे भाग िे मलए, BC =BD हैं। (C.P.C.T िे तनयम िे अनुसार 1

SECTION-E

36.
(i) त्रिभुज ABC िी भुजाएँ क्रमशः 122 मीटर, 22 मीटर और 120 मीटर हैं

1
अब, पररमाप (122+22+120) = 264 मीटर होगा

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

1
(ii) िा क्षे = जहाँ s = (a+b+c)/2

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

(i) अर्द्व पररमाप (s) = 264/2 = 132 मी.

हीरोन िे सूि िा प्रयोग िरने पर,

त्रिभज
ु िा क्षेिफल =

= 132 132 122 132 22 132 120 2

= 132 10 110 12

=1320 m2

OR

हम जानते हैं कि प्रतत र्र्व पर्ज्ञापन िा किराया = 5000 प्रतत र्गव मीटर

एि दीर्ार िा 3 महीने िा किराया = रु. (1680×5000×3)/12

= रु. 2100000 2

37. (i) राहुल तथा नीतू िे बीच िी दरू ी = 2- (-2) = 4 इिाई 1

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

(ii) अींकित (III) तीसरे चतथ
ु ाांश में बैठा है I 1

........................................................................................................................
(iii) चार ममिों िे स्थानों िो क्रम से ममलाने पर आयत िी आिृतत बनती है 2

OR

(iii) लम्बाई = 8 इिाई चौड़ाई = 4 इिाई 2

क्षेिफल = 8X4 = 32 र्गव इिाई
38. (i) राजनीतति दल A ने सबसे अधधि सीटे जीती I 1

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

(ii) राजनीतति दल E ने सबसे िम सीटें जीतीीं I 1

........................................................................................................................
(iii) दण्ड आलेख

2

(राजनीतति दल )

MARKING SCHEME BSEH PRACTICE PAPER 1, 9TH MATHS , March2025
(ENGLISH MEDIUM)
Q. Expected solutions marks
no.

Section-A

1 (C) √9 1
2 (B) 40° 1
3 (C) SSA 1
4 (D) 3.4 cm 1
5 (A) 75º 1
6 T 1
7 (A) 24 cm2 1
8 C) Halved 1
9 (A) 137.5 1
10 (D) 2 1

11 (C) 2 1
12 (B) 3 1
13 (A) - 6 1
14 (B) 1

15 (A) 65o 1
16 3600 1
17 (B) 50o 1
18 90o 1
19 D) A is false but R is true 1

20 B) Both A and R are true but R is not the correct explanation of A. 1

SECTION-B
21.
We know that 5+1=6

= = and = =
1
….…………………………………………………………………………………………………….........
Five Rational Numbers =
1

22. (3√5 -5√2 )(4√5 +3√2)
3√5(4√5 +3√2) - 5√2(4√5 +3√2) 1
….……………………………………………………………………………………………………………..
60+9√10-20√10-30
30-11√10 1
OR
.....................................................................................................................
X
1

..............................................................................................
1
23.
X Rationalizing the denominator

1
=
......................................................................................................................

1

24. 104 × 97
104×97= (100+4) × (100-3)

Here, x = 100 , a=4 , b = -3 1
By using identity , (x+a) (x+b) = x2 + (a+b) x + ab

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

= (100)2+(4-3)100+(4×-3)

= 10000+100-12
1
= 10088
25.
If x – 2 is a factor of p(x), then P(2) = 0

By factor theorem
1

p(2) = 4(2)3 -3(2)2- 4(2)+3k=0
.......................................................................................................................
= 32-12-8+3k=0
=12+3k=0 1
3k= -12
K= - 4
..............................................OR.................................................................
By putting x-3 =0 we get x = 3

By substituting x=3 in polynomial we get (3)3-4(3)2+3+6
1
.....................................................................................................................
= 27-36+3+6= 0

Hence by factor theorem x-3 is a factor of polynomial x3-4x2+x+6
1

SECTION-C
26.
Since y is proportional to x given
y x
1
y = kx .................(i)
.......................................................................................................
We put y=12 when x=4
12=4k
1
k=3
.......................................................................................................
By using k=3 in (i) we get y=3x
1
when we put x=5 we get y=3x5=15

27. Diameter of capsule = 3.5 mm
Radius of capsule = mm
1
.....................................................................................................................
Volume of medicine in capsule = πr3
1
................................................................................................................................................................

22
= x x x x

= 22.46 mm2
1

.........................................................OR............................................................................

Let radius of Moon = r
Let radius of earth = 4r
..........................................................................................................................................
.
π
=
π
1
..........................................................................................
=
π
π
1

= = 1:16
16

1
3 3 , 3 3
28. The expression 64m – 343n can be written as (4m) -(7n)
3 3 1
64m3 – 343n3= (4m) -(7n)

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

We know that, x3-y3 = (x-y)(x2+xy+y2)

= (4m-7n)[(4m)2+(4m)(7n)+(7n)2 1
............................................................................................................................. ............................................................

= (4m-7n)(16m2+28mn+49n2) 1

29.
2x + 3y = 12
By taking X=0 we get 2x0 + 3xY = 12
3Y = 12 1
Y=4
(i) First Solution (0,4)
......................................................................................................................
By taking X=1 we get 2x1+ 3xY = 12
1
3Y=10
10
Y=
10
(ii) Second solution (1, )
.......................................................................................................................

By taking X=2 we get 2x2 + 3Y= 12
1
3Y= 8
Y=
(iii) Third solution (2,

30.
Let x = = 0.2353535……………… (i)
1
Multiplying (i) by 10 on both sides
10x = 2.353535............................ (ii)
.....................................................................................................................
Multiplying (ii) by 100 on both sides
1
1000x = 235.353535......................... (iii)
.....................................................................................................................
By (iii) – (ii) we get
990x = 233.0000
233 1
x=
990

31. The expression, 8X3 + 27Y3 + 36X2Y + 54XY2
1
3 3 2 2
can be written as (2X) + (3Y) +3(2X) (3Y) +3(2X)(3Y)
......................................................................................................
=(2X)3 + (3Y)3 +3(2X)2(3Y) +3(2X)(3Y)2

By using identity (x + y)3 = x3 + y3 + 3xy (x + y) 1

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

=(2X)3 + (3Y)3 +3(2X)(3Y)(2X +3Y)
1
3
=(2X+3Y) =(2X+3Y) (2X+3Y) (2X+3Y)

OR
31. 3 3 3
The expression 8X + Y + 27Z – 18XYZ 1

Can be written as (2X)3 + Y3 +(3Z)3 -3(2X)(Y)(3Z)

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

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) 1

=(2X+Y+3Z)((2X)2 +Y2 +(3Z)2 -2XY-Y(3Z)-3Z(2X))
…………………………………………………………………………….

(2X+Y+3Z)(4X2 +Y2 +9Z2 -2XY-3YZ-6ZX) 1

SECTION-D
32.
Let there are two mid points C and C’ of line AB 1

.......................................................................................................
If C is mid point of line AB
then , AC=BC
1

AC = AB ........................(i)
.......................................................................................................
If C’ is mid point of line AB
then , AC’=BC’ 1

AC’ = AB .......................(ii)
......................................................................................................
From (i) and (ii)
AC=AC’ 1

......................................................................................................
C=C’
Therefore C and C’ are same point 1
Hence there is one and only one mid point of a line.

33

1

Given :- a : b = 2 : 3 and POY = 90°
.....................................................................................................................................
To Find :- The value of a and b.
1
Solution :- a : b = 2 : 3 Let a =2x and b =3x

We know that sum of angles of linear pair are 180°

therefore, POY +a +b = 180°

As given in question, by putting POY = 90° 1
a+b = 90°
..........................................................................................................................................

2x+3x = 90°

5x = 90°

So, x = 18° 1

a = 2×18° = 36°

b = 3×18° = 54°
..........................................................................................................................................
In fig. b+c is on Straight line

therefore, b+c = 180°

c+54° = 180° 1

c = 126°
...........................................................OR..........................................................................

1

Given :- AB II CD , APQ = 50° and APR = 127°
..........................................................................................................................................

To Find :- The value of x and y
....................................................................................................................... 1
Solution :- AB II CD and PQ is a transversal line. (Given)

1
APQ = PQR (Alternate Angles)

By putting APQ = 50° and PQR = x

x = 50°
..........................................................................................................................................

Also APR = PRD (Alternate Angles)

Or, APR = 127° ( Given PRD = 127°) 1

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

We know that APR =APQ+QPR
1

Now by putting values, QPR = y and APR = 127°

We get 50°+ y=127°

y = 77°

x = 50° and y = 77°

34.
1
Let three sides of triangle 12x , 17x, 25x
540
Semi perimeter (s) = = 270 cm
.......................................................................................................................
s =

12 17 25 1
270 =

54x= 540
x=10
........................................................................................................................
First Side = 12x10= 120 cm
Second Side = 17x10 = 170 cm 1

Third Side = 25x10 = 250 cm
........................................................................................................................
=
1
= 270 270 120 270 170 270 250

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

= 270 150 100 20 1

= 9000 cm2

OR

34. Radius of conical cap (r) = 7 cm
Height of conical cap (h) = 24 cm
........................................................................................................................ 1

Slant height of conical cap (L) =
= 24
1
= 49 57
= 625
= 25 cm
........................................................................................................................
Required area of sheet to make 10 caps = 10 πrL
1
.......................................................................................................................
22
= 10x x7x25
1
= 5500 cm2
........................................................................................................................
1
Total cost of 10 caps = 5500 x 0.35 = 1925 Rs.

35.

Given :- AC = AD line AB bisect A 1

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

To Prove :- ΔABC  ΔABD 1
..........................................................................................................................................
Proof:- In ΔABC and ΔABD

(i) AC = AD ( Given)
1
(ii) AB = AB (Common)

(iii) CAB = DAB (Because AB is bisector of A Given)
..........................................................................................................................................
1
Hence, ΔABC  ΔABD. ( By S.A.S)
..........................................................................................................................................
1
Hence BC =BD ( By C.P.C.T )

SECTION-E

36.
(i) Sides of Triangle ABC 122m, 22m and 120m

1
Perimeter of Triangle = (122+22+120) = 264m

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

1
(ii) = Where S=
.....................................................................................................................
(i) Semi Perimeter (s) = 264/2 = 132 मी.

=

= 132 132 122 132 22 132 120
2

= 132 10 110 12

=1320 m2

OR

We know Rent of Advertisement per year = 5000 m2
2

Rent of 3 months of the wall = Rs.(1680×5000×3)/12= Rs. 2100000

37. (i) Distance between Rahul and Neetu = 2 - (-2) = 4 Unit 1

........................................................................................................................
(ii) Ankit is sitting in (III) third quadrant 1

........................................................................................................................
(iii) By joining four places in order we get a rectangle 2

OR

(iii) Length = 8 Units Breadth = 4 Units 2
Area = 8X4 = 32 m 2

38. (i) Political Party A won maximum number of Seats. 1

........................................................................................................................
(ii) Political party E won minimum number of seats.
1
........................................................................................................................
(iii) Bar Graph

2

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