HBSE Class 9 Sample Paper 2027 Mathematics Marking Scheme

FOR HARYANA CLASS 9 PREPARATION

HBSE Class 9 Sample
Paper 2027
Mathematics Marking
Scheme
EXAM YEAR TYPE SUBJECT

Haryana Class 9 2027 Answer Key Maths

Notes · Sample Papers · Previous Year Papers · Mock Tests

m
m .co
c. o

m
MARKING SCHEME, HBSE,SAMPLE PAPER 1, 9TH MATHS,
s e
s em (2026-27)
g l a
(ENGLISH MEDIUM) a
Q. no. Expected solutions marks

Section-A

m
m c. o
co m
1. 5 1
. e
(d)

m la s
4

s e g
2.
l a a
1
27 15

ag
(d) 9 3
3. (d) not defined 1

4. 3�3/2 1
(c) �2 +
�

5. (d) 1 1

6. (c) infinitely many solutions 1
m
co
7. (c) 10 1
.
em
8. (b)2 1

l a s
g
9. equal(बराबर) 1
10. obtuse angle(अधिक कोण ) a 1

11. (a) 10° 1
12. (c) SSA 1
13. (a) DF = 7cm, ∠E = 60° 1
14. ∠OAB = 40°, 1
15. (a) 50° 1

m
co
16. 1
सत्
o m32 = 4 2 cm m .
c
18. . r = 7cm
17. 1

em (a)Both Assertion(A) and Reason (R) are true and Reason (R) is thegcorrect
l a explanation of se 1

las19.
a
1

g
a 20. (d) Assertion(A) is false but Reason(R) is true.
Assertion(A).

1

Section-B
21.(a) सरल कीजिए : 5 81/3 + 271/3
3 1/4

3 1/4
Simplify : 5 81/3 + 271/3

m .
c. o + 3
3 1/4

e m
s
3 1/4 1/3 3 1/3

em
5 81/3 + 271/3 = 5 23 =
a
1/2

s g l
la a
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 1 of 24

-----------------------------------------------------------------------------------------------------------------------
1
= 5 2+3 3 4
1/2

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

1
= 5 5 34

1/2
1
4 4
= 5

---------------------------------------------------------------------------------------------------------------------------------
= 5

1/2

OR 21.(b) 21.(b) 0.47 को � के रू मे व् कीजिए, िहाँ p और q ूूराणक है और q ≠ 0 है।
�
�
Express 0.47 in form where p and q are integers and q ≠ 0.
�

Let x = 0.47 = 0.4777……….. ........(1)
1/2
We multiply both sides of equation (1) by 10.
10x = 4.777 ........(2)

-----------------------------------------------------------------------------------------------------------------------------
We will subtract equation (1) from equation (2). 1/2
10x - x = 4.777…….... - 0.4777………….…

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

9x = 4.3 1/2

….…………………………………………………………………………………………………………………………………..
9x = 43/10
x = 43/90 1/2
Thus, 0.47 = 43/90

22. 22. ्दि बिंिु C, िो बिंिओ
ु A और B के िीच इस पकार ज्थि है दक AC = BC है, िो जसद कीजिए दक AC = 2
1

AB है। जचत िनाकर समझाइए।
1
If a point C lies between two points A and B such that AC = BC , prove that AC = 2 AB . Explain
by drawing the figure.

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 2 of 24

m
m .co
c. o

e m
em l a s
s a g 1/2

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

AC = BC (Given)
Adding AC on both sides, we get (According to Euclid's axioms, we know that when equals
1/2
are added to equals, the wholes are equal.)

m
m c. o
-------------------------------------------------------------------------------------------------------------------------

co m
⇒ AC + AC = BC + AC (BC + AC coincides with AB)
⇒ 2 AC = AB
m . e
s 1/2
a s e g la
l a
ag
-------------------------------------------------------------------------------------------------------------------------
⇒ AC = 1/2 AB (According to Euclid's axioms, things which are halves of the same thing are
1/2
equal to one another)

23. ABC एक समकोण त्रिभ है ,जभसमे AB=AC और ∠A = 900 है | ∠B और ∠C
जाञ कीजभए ।
ABC is a right angled triangle in which AB=AC and ∠� = 900 . Find ∠B and ∠C
m
co
.
s em
g l a
a

m
m .co
c. o se
m
s em g la
gla a
a
It is given that,

AB = AC
∴ ∠C = ∠B (Angles opposite to equal sides are also equal) 1/2

Let ∠B = ∠C = x
o m m .
. c s e
e m
-------------------------------------------------------------------------------------------------------
l a
l a s a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 3 of 24

In ΔABC,
∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)
1
90°+ x + x = 180°

-----------------------------------------------------------------------------------------------------
90°+ 2x = 180°
2x = 90° 1/2
x = 45°
∴ ∠B = ∠C = 45°

24.(a) ABC और DBC एक ही आधार BC ूर िने िो समज्िाब जतरुि है (जचत िेख)े । दराशाए दक ∠ABD =
∠ACD है।
ABC and DBC are two isosceles triangles on the same base BC(See fig.)Show that ∠ABD =
∠ACD.

Given: ABC and DBC are isosceles triangles
To Prove: ∠ABD = ∠ACD
Let's join point A and point B.

In △DAB and △DAC,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common side) 1/2

….………………………………………………………………………………………………………….
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 4 of 24

m
m .co
c. o

e m
em l a s
s g
1
∴ △ ABD ≅ △ ACD (By SSS congruence rule)
a
-------------------------------------------------------------------------------------------------------
∴ ∠ABD = ∠ACD (By CPCT)
1/2

m
OR 24.(b) AD एक समज्िाब जतरुि ABC का शीर्लंि है, जिसमे AB = AC है। दराशाए दक (i) AD, BC को

m c. o
co
समज्राजिि करिा है (ii) AD, ∠A को समज्राजिि करिा है।

. e m
m s
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i)AD bisects BC

s e
(ii)AD bisects ∠� .
a g la
l a
ag
Given: AB = AC
Let's construct an isosceles triangle ABC in which AB = AC as shown below.

m
. co
s em
g l a
a
(i) In ΔBAD and ΔCAD,
∠ADB = ∠ADC (Each 90° as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
∴ ΔBAD ≅ ΔCAD (By RHS Congruence rule) 1

m
m .co
-------------------------------------------------------------------------------------------------------
c. o ∴ BD = CD (By CPCT)
se
m
s em Hence, AD bisects BC.
g l a
la
1/2

g a
a ----------------------------------------------------------------------------------------------------
(ii) Since, ΔBAD ≅ ΔCAD
By CPCT, ∠BAD = ∠CAD
Hence, AD bisects ∠A. 1/2

25. अलग-अलग िनगरना वर् मे दिलली राज् की िनसंख्ा नीचे िी गई है:
m .
co m
1961 1971 1981 1991 2001
. e
िनगरना वर्

e m l as
िनसंख्ा (लाखो मे) 30
l a s 55 70 110 150
a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 5 of 24

ऊूर िी गई िानकारी को दं ड आलेख के रू मे निरूूञ कीजभए ।
The population of Delhi State in different census years is as given below:
Census year 1961 1971 1981 1991 2001

Population in Lakhs 30 55 70 110 150

Represent the above information with the help of a bar graph.
Let us consider the horizontal and vertical axes represent the years and population in lakhs
respectively. The heights of the rectangles are proportional to the population in lakhs. 1/2

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

1
12

SECTION-C
26.(a) ्दि 2x³ - ax² - bx - 14 का एक गुरनखंड x + 2 है और इसे x + 3 से राग िेने ूर शेरफल - 26 िचिा है, िो a और
b के मान जाि कीजिए।
If 2x3 - ax2 - bx - 14 has x + 2 as a factor and leaves remainder - 26 when divided by x + 3 , find
values of a and b .
x+2 is a factor of 2x3 - ax2 - bx - 14
1/2
∴ 2 −2 3 - a −2 2 - b −2 - 14 = 0 ⇒-16-4a +2b -14=0

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

⇒-4a+2b -30 = 0 1/2

⇒2a-b= -15----------(i)
---------------------------------------------------------------------------------------------------------------------------

2x3 - ax2 - bx - 14 leaves remainder - 26 when divided by x + 3 1/2
∴ 2 −3 3 - a −3 2 - b −3 - 14 = -26

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

⇒ -54-9a+3b -14 = -26 ⇒ -9a+3b = 42 ⇒3a-b= -14-------(ii) 1/2
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 6 of 24

m
m .co
c. o

e m
m a s
se a g l
---------------------------------------------------------------------------------------------------------------------------
Subtract Equation (i) from Equation (ii) to eliminate b

(3a-b)-(2a-b)=-14-(-15) 1/2
(3a-2a=-14+15
a=1

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

m
c. o
Now, substitute a = 1 back into Equation (i) to find b

m
2(1)-b=-15

.co m
2-b=-15
-b= -15-2
m e
s 1/2
s e la
-b=-17
b=17
l a a g
a g
OR 26.(b) गुरनखंड कीजिए : x³ + 13x² + 32x + 20
Factorize : x3 + 13x2 + 32x +20
Let p(x) = x3 + 13x2 + 32x + 20
We shall find a factor of p(x) by using some trial value of x, say x = -1. (Since all the terms are
positive.)

m
p(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20

c. o
= -1 + 13 - 32 + 20 = 0 1/2

em
Since the remainder of p(-1) = 0, by factor theorem we can say x + 1 is a factor of

s
p(x) = x3 + 13x2 + 32x + 20.

g l a
---------------------------------------------------------------------------------------------------------------------------
Now dividing p(x) by x + 1 using long division,
a

m
m .co
c. o se
m
s em g la
gla a
a
1/2

m .
.co s e m
e m l a
a s
---------------------------------------------------------------------------------------------------------------------------
l ag
ag
-
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 7 of 24

1/2
Therefore, x + 13x + 32x + 20 = (x + 1)(x + 12x + 20)
3 2 2

---------------------------------------------------------------------------------------------------------------------------
-
Now splitting the middle term of the given polynomial,
x2 + 12x + 20 = x2 + 10x + 2x + 20 1
= x(x + 10) + 2(x + 10)
= (x + 10) (x + 2)

---------------------------------------------------------------------------------------------------------------------------
Therefore, x3 + 13x2 + 32x + 20 = (x + 1) (x + 10) (x + 2) 1/2

27. यदद बिंिु (2,1) समीकरर 2x + ky = 7 का हल है, िो k का मान जाि कीजिए। k का मान पार करके और उसे दिए
गए समीकरर मे रखकर, समीकरर के चार और हल जाि कीजिए।
Find the value of k if point (2,1) is a solution of the equation 2x+ ky = 7 . After getting and
putting the value of k in given equation ,find four more solutions of the equation.
Since (2,1) is a solution to the equation 2x + ky = 7, substitute x = 2and y = 1 into it:
2(2)+k(1)=7
4+k =7
k=7-4 1/2
k= 3

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

Substituting k = 3 back into the original expression gives the complete linear equation: 1/2
2x+3y =7

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

To find more solutions, substitute arbitrary values for x and solve for y

 Solution 1: Let x = 5
2(5)+3y=7⇒10+3y=7⇒ 3y= -3⇒ y= -1 1/2
Point: (5, -1)
 -------------------------------------------------------------------------------------------------------------------
 Solution 2: Let x = -1
2(-1)+3y=7implies -2+3y=7⇒ 3y=9⇒ y=3
Point: (-1, 3) 1/2

 ------------------------------------------------------------------------------------------------------------------
 Solution 3: Let x = 8
2(8)+3y=7implies 16+3y=7implies 3y = -9⇒ y= -3 1/2
Point: (8, -3)
 ----------------------------------------------------------------------------------------------------------------------
 Solution 4: Let x = -4
2(-4)+3y=7implies -8+3y=7⇒ 3y=15⇒ y=5 1/2
Point: (-4, 5)

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 8 of 24

m
m .co
c. o कीजिए।

m
28. संलग आकृ जि मे, AB∥CD और CD∥EF है I साथ ही, EA⊥ AB. है। ्दि ∠BEF=55°है, िो x, y और z के मान जाि

m s e
se g l a
a
In the adjoining figure, AB∥CD and CD∥EF. Also,EA⊥ AB. If
∠BEF=55°,find the values of x,y and z.

m

m c. o
The required values are x = 125°, y = 125°, and z = 35°

.co s e m
m
Finding y:

s e
The lines CD ∥ EF are cut by the transversal line BE. The consecutive interior angles on the
a g la
l a
ag
same side of a transversal are supplementary (add up to 180°:
1/2
-------------------------------------------------------------------------------------------------------------------------
y+55°=180°
1/2
y=180-55=125°

--------------------------------------------------------------------------------------------------------------------------
Finding x:
1/2
Since AB ∥ CD, the angles x and y are equal because they are corresponding angles:
m
. co
--------------------------------------------------------------------------------------------------------------------

s em
x = y =125°

a
1/2
l
ag
------------------------------------------------------------------------------------------------------------------------

Finding z:
1/2
Since AB ∥ EF and EA ⊥ AB, the line EA is also perpendicular to EF ∠ AEF = 90°.
angle ∠ AEF is the sum of z and angle BEF:

------------------------------------------------------------------------------------------------------------------------
z+55°=90°
1/2
m
z=90-55=35°

m एक चञिरभिश है जभसमे P,Q,R और S रिभाओं AB,BC,CD और DA के मधयm.तबंदि है। AC c o
c. oABCD e
29.

m
e PQRS एक समांञर चञरभश है।
एक ूिकणश है . ददखाएँ कक (i) SR∥AC और SR= AC (ii)PQ=SR (iii)
l a s
s
1

la a g
2

g
ि ि
a ABCD is a quadrilateral in which P,Q,R and S are mid points of
the sides AB,BC,CD and DA. AC is a diagonal. Show that (i)
1
SR∥AC and SR= 2 AC (ii)PQ=SR (iii) PQRS is a parallelogram.

We will use the mid-point theorem here. It that states that the line segment joining the mid-

m .
co
points of any two sides of the triangle is parallel to the third side and is half of it.

.
(i) In ΔADC, S and R are the mid-points of sides AD and CD respectively. Thus, by using the
e m
m 1s
e l a
mid-point theorem

a s
∴ SR || AC and SR = 1/2AC ... (1)
l a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 9 of 24

---------------------------------------------------------------------------------------------------------------------------
(ii) In ΔABC, P and Q are mid-points of sides AB and BC. Therefore, by using the mid-point
theorem,
PQ || AC and PQ = 1/2 AC ... (2) 1/2

----------------------------------------------------------------------------------------------------------------------
Using Equations (1) and (2), we obtain PQ || SR and PQ = SR ... (3)
∴ PQ = SR 1/2

----------------------------------------------------------------------------------------------------------------------
(iii) From Equation (3), we obtained PQ || SR and PQ = SR
Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is 1
a parallelogram.

30.(a) 30(a).ABCD एक चकीय चञर
ि भ
िश है जभसके ूिकणश एक तबंद ि E ूर पनञतिे द करञे है । यदद
∠��� = 700 और ∠��� = 300 हो , ञो ∠��� जाञ कीजभए। ूि
ि ः यदद AB=BC ,ञब ∠���
जाञ कीजभए ।
ABCD is a cyclic quadrilateral whose diagonals intersect at point E. If ∠��� = 700, ∠��� =
300 then find ∠���. Further, if AB=BC, find ∠���.

In the triangles, ABD and BCD, ∠CAD = ∠CBD = 70°. (Angles in the same segment are
1/2
equal)

------------------------------------------------------------------------------------------------------------
Hence, ∠BAD = ∠CAB + ∠DAC
= 30° + 70° = 100°
Thus, ∠BAD = 100° 1/2
----------------------------------------------------------------------------------------------------------
Since ABCD is a cyclic quadrilateral, the sum of either pair of opposite angles of a cyclic
quadrilateral is 180º.
∠BAD + ∠BCD = 180°
∠BCD = 180° - 100°
= 80° 1
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 10 of 24

m
m .co
c. o

m
Thus, ∠BCD = 80°

s e
em
-----------------------------------------------------------------------------------------------------------

s
Also given AB = BC.
g l a
So, ∠BCA = ∠BAC = 30° (Base angles of isosceles triangle are equal)
a 1/2

----------------------------------------------------------------------------------------------------------------
∠ECD = ∠BCD - ∠BCA
= 80° - 30°
= 50° 1/2
Thus, ∠ECD = 50°

m
m c. o
co
OR 30.(b) 5 cm और 3 cm त्रया िाले दो ित
ृ दो तबनदओ
ि ं ूर पनञतिे द करञे है ञथा उिके केन्द के बीच की दरू ी
. e m
m
4 cm है । उभयनििष भीिा की लमबाा जाञ कीजभए ।
e la s
a s
Two circles of radii 5 cm and 3cm intersect at two points and the distance between their
l a g
ag
centres is 4 cm. Find the length of the common chord.

Solution:Let O and O' be the centres of the
circles with radii 5 cm and 3 cm
respectively.
And PQ be the common chord.
m
co
In triangle OMA (right triangle):
OP = 5 cm (radius of first circle),O'P = 3 cm
.
em
(radius of second circle)and O'O = 4 1/2
cm[Given]
l a s
a g
--------------------------------------------------------------------------------------------------------------
Since 52= 42+32 ⇒ OP 2 = O O' 2 + O'P 2
⇒ △ O O' P is a right angled triangle. 1
⇒ ∠O O' P = 90° ⇒ O ⇒ O O' ⊥ PQ

-------------------------------------------------------------------------------------------------------------------------
Thus O' lies on the common chord PQ and O O' ⊥ PQ
m
c. o
1/2
m
c. o m
⇒O O' bisects PQ

s e
---------------------------------------------------------------------------------------------------------------------------

s em ⇒O´P = O´Q = 3 cm
g l a
la a
1/2

ag ------------------------------------------------------------------------------------------------------------------------

⇒ PQ = 2 O´Q = 2 × 3 = 6 cm 1/2
31. रंकि के आकार का एक ञंबू 10 m ऊँचा है और उसके आिार की त्रया 24 m है । जाञ कीजभए :
(i )ञंबू की नञयशक ऊँचाा (ii)ञंबू मे लगे केििास की लागञ ,यदद 1 m2 केििास की लागञ
70 रूए है ।
A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the

m .
co
tent (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

. e m
e m l as
l a s a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 11 of 24

The total surface area of the cone is the sum of the curved surface area and area of the base
which is a circle.
Curved surface area of a right circular cone of base radius, 'r' and slant height, 'l' is πrl
Slant height, l = √r² + h² where h is the height of the cone and r is the radius of the base.
i) Radius, r = 24 m
Height, h = 10 m
Slant height, l = r² + h²
= (24)² + (10)²
1
= 576 + 100
= 676
= 26 m

-------------------------------------------------------------------------------------------------------------------------
ii) Curved surface area of the cone = πrl
= 22/7 × 24 m × 26 m
= 13728/7 m² 1

--------------------------------------------------------------------------------------------------------------------------
The cost of the canvas required to make the tent, at ₹ 70 per m² = 70 × Curved surface area
of the cone
= 13728/7 × 70
= ₹ 137280
Thus, slant height of the tent is 26 m and the cost of the canvas is ₹ 137280. 1

Section –D
32.(a) 11− 7
्दि a और b ूररमे् संख्ाएँ है और 11+ 7 = a − b 77 है, िो a और b के मान जाि कीजिए।
11− 7
If a and b are rational numbers and 11+ 7 = a − b 77 ,find the values of a and b .
11− 7 11− 7 11− 7
Solution: LHS= 11+ 7
= 11+ 7
× 11− 7
1

---------------------------------------------------------------------------------------------------------------------------
2
11+ 7
= 2 2
11 − 7 1

---------------------------------------------------------------------------------------------------------
11 + 7 − 2 77 1/2
=
11 − 7

---------------------------------------------------------------------------------------------------------------------------
18 − 2 77
= = 1/2
4

------------------------------------------------------------------------------------------------------------------------
18 2
= − 77
4 4

1
9 1
= 2 − 2 77 = a − b 77 = RHS (given)

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 12 of 24

m
m .co
c. o

m
---------------------------------------------------------------------------------------------------------------------------

s e
s em g l a
9 1
⇒ a = and b= 1

a
2 2

OR 32.(b) 1 1
संख्ा रे खा ूर 9.3 को जनरजूि करे । 3 और 2 के िीच चार ूररमे् संख्ाएँ री जाि कीजिए।
1
Represent 9.3 on the number line.Also,write down four rational numbers between 3 and
1
2
.
Let's look into the steps below to represent √9.3 on the number line.

m
Step I: Draw a line and take AB = 9.3 units on it.

m c. o
Step II: From B, measure a distance of 1 unit and mark C on the number line. Mark the

.co e m
midpoint of AC as O.

m s 1
gla
Step III: With ‘O’ as center and OC as radius, draw a semicircle.

se
1

la
Step IV: At B, draw a perpendicular to cut the semicircle at D.
a
2

g
Step V: With B as a center and BD as radius draw an arc to cut the number line at E. Thus,
a
taking B as the origin the distance BE = √9.3
Therefore, point E represents √9.3 on the number line.

m
. co
s em
g l a
a

1
12

m
m .co
c. o se
m
s em g la
gla a
a

---------------------------------------------------------------------------------------------------------------------------
1 1 1 2

o m 1 3

m .
c
Further ,3 and 2 = 3 × 2and 2 × 3.

m . ×. 1 se
a
2 3 2 5 3 5

e gl
= and = × and
s
6 6 6 5 6 5

g l a a
a For more Question Papers, Sample Papers, Notes & Syllabus visit Page 13 of 24

10 15
= 30 and 30 .
---------------------------------------------------------------------------------------------------------------------------
1 1 10 15 11 12 13 14
Four rational numbers between 3 and 2 or 30 and 30 are 30 , 30 , 30 and 30 1

33.(a) (i) गुरनखंड कीजिए : 4x² + y² + z² - 4xy - 2yz + 4xz (ii) ्दि a + b + c = 5 और ab + bc + ca = 10 है,
िो
दराशाए दक a³ + b³ + c³ - 3abc = -25
(i)Factorise : 4x2+ y2+z2-4xy-2yz+4xz(ii) If a+ b+ c = 5 and ab + b c + c a = 10, then
show that a3 + b3 + c3 - 3abc = - 25

The expression 4x² + y² + z² - 4xy - 2yz + 4xz
To find :
To factorise the expression
Formula:
Solution :
4x² + y² + z² - 4xy - 2yz + 4xz
1
=(2x)2+(−y)2+z2+2.2x.(−y)+2.(−y).z+2.2x.z

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

1/2
[ using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
---------------------------------------------------------------------------------------------------------------------------

=(2x−y+z)2
=(2x−y+z)(2x−y+z)
1/2

---------------------------------------------------------------------------------------------------------------------------
(ii)Solution:
Given, a + b + c = 5 and ab + bc + ca = 10
We have to prove that a³ + b³ + c³ - 3abc = - 25.
Using the algebraic identity,
x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)
a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - (ab + bc + ca))
1
= 5(a² + b² + c² - (10))

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

To find a² + b² + c², by using the algebraic identity,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(5)² = a² + b² + c² + 2(10)
25 = a² + b² + c² + 20
1
a² + b² + c² = 25 - 20
a² + b² + c² = 5

---------------------------------------------------------------------------------------------------------------------------
Now, a³ + b³ + c³ - 3abc = 5(a² + b² + c² - (10))
1
= 5(5 - 10)
= 5(-5)
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 14 of 24

m
m .co
c. o

m
= - 25

s e
em
Therefore, a³ + b³ + c³ - 3abc = - 25

s g l a
OR 33.(b) a
3x² - x - 4 का गुरनखंड कीजिए और उू्ु् सव्सजमका का उू्ोग करके (998)³ का मान जाि कीजिए।
Factorise 3x2- x - 4 and evaluate 998 3 using suitable identity.

1
3x² - x - 4 = 3x² + 3x - 4x - 4

om
m 1/2.
c
---------------------------------------------------------------------------------------------------------------------------

.co s e m
m
= 3x(x + 1) - 4(x + 1)

s e g l a
l a a
ag
---------------------------------------------------------------------------------------------------------------------------

1/2
= (x + 1) (3x - 4)

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

(998)3 = (1000 - 2)3 1/2

m
---------------------------------------------------------------------------------------------------------------------------

c. o
1

em
Identity: (x - y)3 = x3 - y3 - 3xy(x - y)

la s
---------------------------------------------------------------------------------------------------------------------------

Let us substitute x = 1000, y = 2 a g 1/2
(998)3 = (1000)3 - (2)3 - 3(1000) (2) (1000 - 2)
---------------------------------------------------------------------------------------------------------------------------

= 1000000000 - 8 - 6000 × 998 1/2

---------------------------------------------------------------------------------------------------------------------------
m
m .co
c. o
= 999999992 - 5988000

m
se
= 994011992
1/2

s em g la
gla a
a
34.(a) 34.(a) USA और कनाडा िैसे कु छ िेशो मे िाूमान को फारे नहाइइ मे माूा िािा है, ििदक रारि मे इसे
सेजलस्स मे माूा िािा है। सेजलस्स और फारे नहाइइ िाूमान के िीच संिंध इस पकार है:
9
F= 5
C+32.
(i) ्दि िाूमान 30°C है, िो फारे नहाइइ मे िाूमान क्ा होगा?

m .
(ii) ्दि िाूमान 95°F है, िो सेजलस्स मे िाूमान क्ा होगा?

.co
(iii) समुद िल ूर ूानी 100°C ूर उिलिा है। ्दि आू इसे फारे नहाइइ मे ििले, िो िाूमान क्ा होगा?
s e m
e m l a
a s
(iv) दकस िाूमान ूर सेजलस्स और फारे नहाइइ जडगी संख्ातमक रू से िरािर होिे है?
l a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 15 of 24

(v) दिए गए समीकरर को aC + bF + c = 0 के रू मे जलखे और a, b और c के मान जलखे।
34.(a)In some countries like USA and Canada,temperature is measured in Fahrenheit,whereas
in India,it is measured in Celsius.The relation between Celsius and Fahrenheit temperature is
given by :
9
F= 5 C+32.
(i)If the temperature is 30° C,then what will be the temperature in Fahrenheit?
(ii)If the temperature is 95° F,then what will be the temperature in Celsius?
(iii)Water boils at 100℃ at sea level. If you convert this to Fahrenheit,then what will be the
temperature?
(iv)At what temperature both Celsius and Fahrenheit degree are numerically equal?
(v)Express given equation as a C + b F +c = 0 and write down the values of a,b and c .

(i) Given: Temperature = 30°C
To find: F = ?
We know that, F = (9/5)C + 32
By Substituting the value of C = 30° C in the equation above,
F = (9/5)C + 32
= (9/5)30 + 32
= 54 + 32 1
= 86
Therefore, the temperature in Fahrenheit is 86° F .
….……………………………………………………………..
(ii) Given, Temperature = 95° F
To find, C = ?
We know that, F = (9/5)C + 32
By Substituting the value of temperature in the above equation,
95 = (9/5)C + 32
95 - 32 = (9/5)C
63 = (9/5)C 1
C = (63 × 5)/9
C = 35
Therefore, the temperature in Celsius is 35° C.
….……………………………………………………………..

(iii) We know that, F = (9/5)C + 32
If C = 100°, then by substituting this value in the above equation,
F = (9/5)100 + 32
F = 180 + 32 1
F = 212
Therefore, if C = 100°, then F = 212°
….…………………………………………………………….
(iv) We know that, F = (9/5)C + 32
Let us consider, F = C
By Substituting this value in the equation above,
F = (9/5)C + 32
(9/5 - 1)F + 32 = 0
(4/5)F = - 32
F = (- 32 × 5)/4
Hence, F = - 40
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 16 of 24

m
m .co
c. o

m
Yes, there is a temperature, -40°, which is numerically the same for both 1

s e
em
Fahrenheit and Celsius.
s g l a
---------------------------------------------------------------------------------------------
9
(v)F= 5 C+32. a
⇒ 5F = 9C + 160
⇒ -9C + 5F - 160 = 0
1
Comparing with a C + b F +c = 0, we get
a = -9 , b = 5 , c = -160
OR 34.(b) 34.(b) सोजन्ा ने अूने िनमदिन ूर एक अनाथाल् मे चॉकलेइ िांइी। उसने हर िचे को 5 चॉकलेइ और

m
c. o
िडो को कु ल 20 चॉकलेइ िी। मान लीजिए िचो की संख्ा x है और िांइी गई चॉकलेइ की संख्ा y है।

o m m
c
.c के मान ििाइए।
(i) िो चरो x और y मे रै जखक समीकरर जाि कीजिए। साथ ही, समीकरर को ax + by + c = 0 के रू मे
s e
m
eचॉकलेइ िांइी, िो अनाथाल् मे दकिने िचे थे? la
s
जलजखए और a, b िथा
(ii) ्दि उसनेa145
a g
glिचे थे, िो िांइी गई चॉकलेइ की संख्ा जाि कीजिए।
(iii) ्दिa20
(iv) ्दिy = 205 है, िोx का मान जाि कीजिए।
34.(b) On her birthday, Sonia distributed chocolates at an orphanage . She gave each child 5
chocolates and the adults a total of 20 chocolates.Let the number of children be x and the
number of chocolates distributed be y.
(i)Find linear equation in two variables x and y . Also,express the equation in the form ax +by
+ c = 0 and write down the values of a,b and c .

m
co
(ii)If she distributed 145 chocolates , how many children were there in the orphanage.

.
(iii)If there were 20 children , find the number of chocolates distributed.

em
(iv) If y = 205 ,find the value of x.

l a s
a g
Solution:(i)The number of chocolates given to children is 5x, and the number of chocolates
given to adults is 20. The total number of chocolates is the sum of these. Therefore, the linear
equation is y = 5x + 20 1

m
--------------------------------------------------------------------------------------------------------------

m
⇒.-5x+ y - 20 = 0
.co
c. o se
m
Comparing with ax+ by + c =0, we get a= -5, b= 1 , c = -20 1

s em --------------------------------------------------------------------------------------------------------------
g l a
gla a
a (ii)Substitute y = 145 into the equation y = 5x + 20 we get ,145 = 5x + 20
⇒ 125 = 5x ⇒ x = 25.
1
∴ 25 children were there in the orphanage.

--------------------------------------------------------------------------------------------------------------
(iii)Substitute x = 20 into the equation y = 5x + 20⇒ y = 5(20) + 20 = 100 + 20 = 120.
y= 120

m .
1

c. o e m
∴the number of chocolates distributed= 120

e m l as
s g
-----------------------------------------------------------------------------------------------------------------

g l a a
a For more Question Papers, Sample Papers, Notes & Syllabus visit Page 17 of 24

(iv)Substitute y = 205 into the equation y = 5x + 20 we get 205 = 5x + 20 ⇒ 185 = 5x
⇒ x = 37 1

35.(a) 35.(a) रामूाल के ूास एक जतरुिाकार खेि है जिसकी रुिाओ का अनुूाि 5 : 12 : 13 है और इसका
ूररमाू 300 मीइर है। वह इसमे िो अलग-अलग िरह की फसले उगाना चाहिा है। इसजलए, उसने सिसे
लंिी रुिा के सामने वाले शीर् से उस रुिा ूर एक लंि खीचकर खेि को िो रागो मे िांइने का फै सला दक्ा।
हीरोि के सूत का उू्ोग करके लंि की लंिाई जाि कीजिए।
Rampal has a triangular field whose sides are in the ratio 5 : 12 : 13 and its perimeter is 300
m . He wants to grow two different types of crops in it . So, he decided to divide the field into
two parts by drawing a perpendicular line from opposite vertex to the side whose length is
longest.Find the length of the perpendicular using Heron’s Formula.

Given sides of triangles are in the ratio 5:12:13
Let the sides of triangle are 5x, 12x and 13x
Perimeter of triangle = 300 m
⇒ 5x + 12x + 13x = 300
⇒ 30x = 300
1/2
⇒ x = 10

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

Therefore, sides of the given triangle are
⇒5x = 5 × 10 = 50 m,
⇒12x = 12 × 10 = 120 m 1/2
⇒13x = 13 × 10 = 130 m

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

Notice that 1302 = 502 + 1202
⇒ 16900 = 2500 + 14400 = 16900
⇒16900= 2500 + 14400
⇒ 16900 = 16900 1/2
Thus the given triangle is a right angled triangle with hypotenuse 130 m and the perpendicular
sides are 50 m and 120 m.

---------------------------------------------------------------------------------------------------------------
∴Area of triangle = (1/2) × base × height. Here, the base is 130 m.

= (1/2) × 130 × h m2 --------------(i) 1/2

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

Use the formula for the area of the triangle using the sides:
1/2
Calculate the semi-perimeter (s): s = (50 + 120 + 130) / 2 = 300 / 2 = 150 m.

-----------------------------------------------------------------------------------------------------------
Use Heron's formula to find the area:
Area = [s(s − a)(s − b)(s − c)] = √[150(150 - 50)(150 - 120)(150 - 130)]

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 18 of 24

m
m .co
c. o

m
= √[150 × 100 ×30 × 20].

s e
s em = [9000000] = 3000 m².
g l a 1

a
----------------------------------------------------------------------------------------------------------------
⇒3000 = (1/2) × 130 × h. [using (i)] 1/2

:
---------------------------------------------------------------------------------------------------------------
⇒3000 = 65h ⇒h = 3000 / 65 = 46.15 m (approximately).
1

m
o m c. o
.c m
OR 35.(b)
e
35(b).एक फलाईईवर की जतकोरी् साइड की
िीवारो का इ्िेमालm
e la s
s
जवजाूनो के जलए दक्ा

l a a g
g
ग्ा है। िीवारो की रुिाएँ 13 मीइर, 14 मीइर
और 15 a मीइर है। जवजाूनो से पजि वर् ₹2000
पजि वग् मीइर की कमाई होिी है। एक कं ूनी ने 6
महीने के जलए इसकी एक िीवार दकराए ूर ली।
उसने दकिना दकरा्ा दि्ा?
The triangular side walls of a flyover have been used for advertisements. The sides of the
walls are 13 m ,14 m and 15 m .The advertisements yield an earning of ₹ 2000 per m2 per
m
co
year. A company hired one of its walls for 6 months. How much rent did it pay?
.
em
Let a = 13 m
l a s
b = 14 m
c = 15 m a g
By Heron’s formula,
Area of triangle = s(s − a)(s − b)(s − c) 1

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

where s = semiperimeter

m
s = (a + b + c)/2

co
So, s = (13 + 14 + 15)/2
m .
c. o m
= 42/2

se
em -------------------------------------------------------------------------------------------------------------------------
s = 21 m
a
1

s g l
gla a
a
Area = √ 21(21 − 13)(21 − 14)(21 − 15)
= √21(8)(7)(6)
= √7 × 3 × 4 × 2 × 7 × 3 × 2
=7×3×2×2
= 21 × 4 1
Area = 84 m²

m .
c. o m
-------------------------------------------------------------------------------------------------------------------------

Cost of advertising per m s e
s eyear for 1 m² = Rs. 2000
l
g1 a
g la
Advertisement cost per year for 84 m² = 2000 × 84
a
a For more Question Papers, Sample Papers, Notes & Syllabus visit Page 19 of 24

= Rs.168000

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

Rent paid by the company for 6 months = 168000/2
= Rs. 84000 1
Therefore, the rent paid by the company is Rs. 84000/-

SECTION E
36. 36. जनमजलजखि आंकडा एक
जवशजववाल् की ्ोिना को िशा्िा है।
इस ्ोिना मे, बाभार मागश और
ूिसञकालय मागश एक िूसरे के लंिवि है।

कु छ ्थानो के संिर् बिंिु आंकडे मे
जचज्नि है। िी गई िानकारी के आधार
ूर, नीचे दिए गए सवालो के िवाि िे:
(i) पाकृनञक ूिजाि अकादमी , x-अक से
दकिनी िूरी ूर ज्थि है?
(ii) दिए गए ्थानो मे से कौन सा ्थान
मूल तबंद ू ूर ज्थि है?
(iii) ऐजिहाजसक सोसा्इी के ्थान के जनि्शांक जलखे?
(iv)चुंिकी् कं ूास दकस चिुथाणश मे ज्थि है?

[नोट :- कृपया अंग्रे माधयम वाल् चित को जयााा सही माने ]

36.The following figure indicates a
University ’s plan.In this plan , the Market
Street and Library Street are perpendicular
to each other.The reference points of
certain places are marked in the figure.
Based on the given information, answer
the following:
(i)At what distance is Academy of Natural
Sciences located from x-axis.
(ii)Which of the given places is located at
origin?
(iii)Write the coordinates of the location of
Historical Society?
(iv)In which quadrant does the magnetic compass lie?

(i) Distance of Academy of Natural Sciences from x-axis = 4 units. 1

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

1
(ii) Place located at the origin is City Hall.

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 20 of 24

m
m .co
c. o (iii) Coordinates of Historical Society are (2,-4)

m
---------------------------------------------------------------------------------------------------------------

s e
s em g l a
a 1

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

1
(iv)The Magnetic Compass lies in 3rd quadrant.

m
37. 37. आरव को उसके िनमदिन ूर उसकी िािी से उूहार मे एक जखलौना जमला। ्ह जखलौना एक सम-वृती्

m c. o
co
िेलन (right circular cylinder) के आकार का है, जिसके जनचले जसरे ूर एक अध्गोला और ऊूरी जसरे ूर

. e m
m s
एक शंकु लगा है। िेलनाकार राग की ऊँचाई और जतज्ा कमशम 13 cm और 5 cm है। अध्गोलाकार और

a s e g la
a
शंकाकार रागो की जतज्ाएँ िेलनाकार राग की जतज्ा के समान ही है और शंकाकार राग की ऊँचाई 12 cm

g l
है।
a
ऊूर िी गई िानकारी के आधार ूर जनमजलजखि प्ो के उतर िीजिए:
(i) जखलौने के शंकाकार राग की जि््क ऊँचाई (slant height) जाि कीजिए।
(ii)(a) िेलन का वक ूृषी् केतफल जाि कीजिए।
्ा (ii)(b) अध्गोले का वक ूृषी् केतफल जाि कीजिए।
(iii) जखलौने का कु ल ूृषी् केतफल जाि कीजिए।
m
co
36.Aarav got a toy as a gift on his birthday from his grandmother. The
.
em
toy is in the shape of a right circular cylinder with a hemisphere on the

s
lower end and a cone on the upper end. The height and radius of the

l a
cylindrical part are 13cm and 5 cm respectively. The radii of the
g
a
hemispherical and conical parts are the same as that of the cylindrical
part and height of the conical part is 12 cm.
Based on the above information answer the following questions:
(i) Find the slant height of the conical part of the toy.
(ii) (a)Find the curved surface area of the cylinder.
OR (ii)(b)Find the curved surface area of the hemisphere .
(iii) Find the total surface area of the toy .

m
m .co
(i)Slant height of cone, l = ​ �2 + ℎ2 1/2

c. o se
m
---------------------------------------------------------------------------------------------------------------

s em g l a
gla a 1/2
a
l= 5 + 12 = 25 + 144= 169=13 cm
2 2

-----------------------------------------------------------------------------------------------------------------
(ii)(a) Curved surface area of cylinder = 2πrH= 2× 22/7​ ×5 × 13
1/2
---------------------------------------------------------------------------------------------------------------

2860
= = 408.57 cm2 1/2
m .
c. o
7

e m
e m l as
s g
---------------------------------------------------------------------------------------------------------------

g l a a
a For more Question Papers, Sample Papers, Notes & Syllabus visit Page 21 of 24

OR (ii)(b)Curved surface area of hemisphere = 2πr2 = 2× 22/7​ ×5 × 5 1/2

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

1/2
1100
= 7
= 157.14 cm 2

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

(iii)Surface area of toy(S) = Curved surface area of cylinder + Curved surface area of
hemisphere + Curved surface area of cone.
S=2πrH+2πr2+πrl= πr(2H+2r+l) 1

=22/7​ ×5×(2×13+2×5+13)
---------------------------------------------------------------------------------------------------------------

=7110​ ×(26+10+13)=110/7​ ×49=110×7=770 cm2.
Hence, the surface area of the toy is 770 cm2. 1

38. 38. COVID-19 महामारी, जिसे कोरोना वा्रस महामारी के रू मे री िाना िािा है, गंरीर िीव शसन
बसंडोम कोरोना वा्रस 2 (SARS-CoV-2) के कारर बई थी। इसे ूहली िार दिसंिर 2019 मे चीन के वुहान
मे ूहचाना ग्ा था। सव् के िौरान, शहर के एक अ्ूिाल मे रि् COVID से संकजमि 80 रोजग्ो की आ्ु
िि् की गई और एकत दकए गए डेइा को आवृजत जविरर िाजलका मे िशा््ा ग्ा है।
आ्ु (वर् मे) 5-15 15-25 25-35 35-45 45-55 55-65

रोजग्ोकी संख्ा 6 11 21 23 14 5

ऊूर िी गई िानकारी के आधार ूर, नीचे दिए गए सवालो के िवाि िे:
(i) दकस वग् अंिराल की िारं िारिा सिसे अजधक है?
(ii) कौन सा आ्ु वग् सिसे कम पराजवि बआ? [1]
(iii)(a)दिए गए आँकडो के जलए आयञधच् िनाइए। [2]
अथवा
(iii) (b) दिए गए आँकडो के जलए िारं िारिा िबरुि िनाइए। [2]
38.The COVID-19 pandemic, also known as the corona virus pandemic was caused by severe
acute respiratory syndrome corona virus 2 (SARS-CoV-2).It was first identified in December
2019 in Wuhan , China.During survey , the ages of 80 patients infected by COVID and
admitted in the one of the city hospital were recorded and the collected data is represented
in the frequency distribution table .
Age(in years) 5-15 15-25 25-35 35-45 45-55 55-65
No. of patients 6 11 21 23 14 5
Based on the above the above information , answer the following questions:
(i) Which class interval is of highest frequency? [1]
(ii)Which age group was affected the least ? [1]
(iii)(a)Draw histogram for the given data.
OR (iii) (b) Draw frequency polygon for the given data. [2]

(i) The class interval with the highest frequency is 35-45 years with 23 patients.
1
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 22 of 24

o m
m .c
c. o ----------------------------------------------------------------------------------------------------------------

e m
m a s
se The class interval with the lowest frequency is 55-65 years withl 5 patients.
(ii)

ag 1

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

(iii) Draw histogram for the above data. (2 marks)
To draw a histogram:
The x-axis represents the age intervals (5-15, 15-25, etc.).
m
c. o
The y-axis represents the number of patients (frequency).
m
co
1/2
m
Each bar's height corresponds to the frequency of the respective age group.

m . s e
a s e g la
l a
ag

m
. co
em
1
12

l a s
a g

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

m
m .co
c. o
(iv) Draw frequency polygon for the above data. (2 marks)
m
se
To draw a frequency polygon:

em la
Find the class marks (mid-values) for each interval:

las 5-15: (5+15)/2 = 10
a g
ag 15-25: (15+25)/2 = 20
25-35: (25+35)/2 = 30
35-45: (35+45)/2 = 40
45-55: (45+55)/2 = 50
55-65: (55+65)/2 = 60
Plot the points (class mark, frequency): (10,6), (20,11), (30,21), (40,23), (50,14), (60,5)
Join the points with straight lines.
Optionally, add points at the beginning and end with frequency 0 for closure: (0,0) and (70,0). 1/2

m .
co
----------------------------------------------------------------------------------------------------------------

. e m
e m l as
l a s a g
ag For more Question Papers, Sample Papers, Notes & Syllabus visit Page 23 of 24

1
12

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 24 of 24