HBSE Class 10 Sample Paper 2025 Answers Maths Standard

Board Of School Education Haryana

मॉडल पेपर
उत्तर
2025

MARKING SCHEME BSEH PRACTICE PAPER 1, 10TH MATHS(Standard) ,
March2025
(ENGLISH MEDIUM)
Q. Expected solutions mar
no. ks

Section-A

1 (b)500 1
2 (a) both positive 1
3 (a)x2 -4x +3 2 =0 1
4 ((b)1 1
5 (c )± 4 1
6 (c )7 1
7 (a)30° 1
8 5 1
(a)2
9 (c) r2 sq. units 1
10 (d) 6 : π 1

11 (b) 8 [ use mode= 3median-2mean ] 1
12 (b) 14 1
13 a =3 1
14 diameter 1
15 1 1
16 A+B= 90° 1
17 r2 sq. units 1
18 10+15 = 25 1
19 (a)Both Assertion(A) and Reason (R) are true and Reason (R) is the 1
correct explanation of Assertion(A).

20 (d) Assertion(A) is false but Reason(R) is true. 1

SECTION-B
21. From the above question, we have the linear equations as,
(a) 2x + y = 23---------------(i)
4x - y = 19----------------(ii)
Adding the equation (i) and (ii),
6x = 42 1/2
x = 7.
….………………………………………………………………………………………………………...
Substituting x value in (i), we get,

2(7) + y = 23
14 + y = 23
y = 23 - 14 1/2
y=9

….…………………………………………………………………………………………………………
Substituting the values of x and y in 5y - 2x and y/ x - 2, we get,
5y - 2x = 5 × 9 - 2 × 7
= 45 - 14 1/2

= 31
….…………………………………………………………………………………………………………….
y/x - 2 = 9/7 -2
1/2
= -5/7.

21. In the above equation a1= 4, a2 = 2, b1 = p and b2 = 2.
(b) 1/2

….……………………………………………………………………………………………………………..
If the solution of a pair of linear equations is unique, then
1/2
a1/a2 ≠ b1/b2
….…………………………………………………………………. ……………………………………………………………………………………………
4/2 ≠ p/ 2
1/2
….…………………………………………………………………………………………………………….
4≠p

Thus, the pair of linear equations has a unique solution for all values of p 1/2

except 4

22. Given, DE||AB
We have to find the value of x.
From the figure,
CD = x+ 3
AD = 3x + 19
CE = x
BE = 3x + 4
∴ By Basic Proportionality Theorem 1/2
CD/DA = CE/EB
….……………………………………………………………………………………………………………..
⇒ (x+3)/(3x+19) = x/(3x+4) 1/2
On cross multiplication,

(x+3)(3x+4) = x(3x+19)

….………………………………………………………………………………………………………….
By multiplicative and distributive property,
3x² + 4x + 9x + 12 = 3x² + 19x
1/2
Cancelling out common terms,
13x + 12 = 19x
….………………………………………………………………………………………………………….
By grouping,
13x - 19x = -12
-6x = -12
x = 12/6 1/2
x=2
Therefore, the value of x is 2.

23. The chord of the larger circle is a tangent to the smaller circle as shown in
the figure below.

1/2

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

PQ is a chord of a larger circle and a tangent of a smaller circle.
Tangent PQ is perpendicular to the radius at the point of contact S.
Therefore, ∠OSP = 90°
In ΔOSP (Right-angled triangle)
By the Pythagoras Theorem,
OP2 = OS2 + SP2
52 = 32 + SP2
SP2 = 25 - 9
SP2 = 16 1
SP = ± 4
SP is the length of the tangent and cannot be negative
Hence, SP = 4 cm.
….………………………………………………………………………………………………………….
QS = SP (Perpendicular from center bisects the chord considering QP to
be the larger circle's chord)
Therefore, QS = SP = 4cm
Length of the chord PQ = QS + SP = 4 + 4 1/2
PQ = 8 cm
Therefore, the length of the chord of the larger circle is 8 cm.

24. sin(A - B) = 1/2 ⇒ Sin(A-B) = sin (30°) ⇒A - B = 30o …(1)
(a) 1/2

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

1/2
cos (A + B) = 1/2 ⇒ cos(A + B) = cos (60o) ⇒ A + B = 60o …(2)

.....................................................................................................................
1/2
On Adding Eq. (1) and (2), we get 2A = 90o ⇒ A = 45o

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

Now, Putting the value of A in Eq.(2), we get 45o + B =60o ⇒ B = 15o 1/2

Hence, A = 45o and B = 150
24. We have ,
(b) a2/x2−b2/y2

=a2/a2sin2θ−b2/b2tan2θ [∵x=asinθ,y=btanθ] 1/2

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

1/2
=1/sin θ−1/tan θ
2 2

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

=cosec2θ−cot2θ 1/2

….…………………………………………………………………………………………………………
[∵1+cot2θ=cosec2θ∴cosec2θ−cot2θ=1]
1/2
=1

25. Given, rectangular field of dimension 20m × 16m
A cow is tied with a rope of length 14 m at the corner of the rectangular
field.
We have to find the area of the field in which the cow can graze.

1/2

….……………………………………………………………………………………………………………
Let ABCD be the rectangular field.
From the figure,
We observe that the area that the cow can gaze is in the form of a sector
of a circle.
So, AGEF is a sector of a circle with radius 14 m. 1/2
Area of sector = πr²θ/360°
Here, θ = 90°
….……………………………………………………………………………………………………………
Area of sector = (22/7)(14)²(90°/360°)
= (22)(2)(14)(1/4)
1/2
= (22)(14)(1/2)
= 11(14)

….……………………………………………………………………………………
= 154 m² 1/2
Therefore, the area in which the cow can gaze is 154 m².

SECTION-C
26. Let us assume that
1/2
3−2 5 is rational.
.............................................................................………………………………...
Hence it can be written in the form

a
​ where a and b are co-prime and b≠ 0
b
a 1/2
Hence 3−2 5 =
b
.......................................................................……………………………………….
​

a 3b−a 1/2
​ ⇒2 5= 3− =
b b
..................................................................................………………………….
3b−a
⇒ 5= 1/2
2b
..............................................................................………………………….
3b−a
where 5 is irrational and 2b is rational.

because irrational number≠ rational number 1/2
.....................................................................…………………………………...

Therefore the above is a contradiction.
1/2
So our assumption is wrong.
Hence 3−2 5 is irrational.

27. Since α and β are the zeroes of the polynomial
f(x)=2x2 -7x +3
−7 7 3
∴ α + β =− = and αβ = 1
2 2 2
...........................................................................
2
Now α2 + β2 = α + β − 2αβ 1

..........................................................................
7 2 3
= −2× =
2 2 1
49 3 49−12 37
4
−
1
= 4
= 4
28. Let ₹ x is the fixed charge for the first two days and
(a) 1/2
₹ y is the additional charge for each.

….…………………………………………………………………………………………………………
From the first condition,
Latika paid ₹ 22 for a book kept for six days 1/2
x + 4y = 22---------(1)

….…………………………………………………………………………………………………………
According to the second condition,
Anand paid ₹16 for a book kept for four days 1/2
x + 2y = 16------------(2)
….…………………………………………………………………………………………………………

Now solve equations (1) and (2)
Subtracting (2) from (1), we get,
1/2
2y = 6
y = 3.
….…………………………………………………………………………………………………………
Substitute the value of y in (2), we get,
x + 2 x 3 = 16
x = 16 - 6 = 10 1/2
x = 10.
x = 10.
….…………………………………………………………………………………………………………

Therefore, the fixed charge = ₹ 10 and the charge for each extra day = ₹ 1/2
3.

28. Let the digits at tens place and units place in the first number be x and y
(b) respectively.
A number can be expressed in the expanded form as 10(x) + y.
On reversing the digits, x is the units digit and y is the tens digit. The
expanded notation for the second number be 10(y) + x 1/2

….…………………………………………………………………………………………………………
As per the question,
(10x + y) + (10y + x) = 66
⇒ 11(x + y) = 66 1/2

⇒ x + y = 6 ................. (1)
….…………………………………………………………………………………………………………
Also, it is given that the difference between the two digits is 2.
1/2
x - y = 2 ................... (2)
….…………………………………………………………………………………………………………
or y - x = 2 ................... (3)
1/2
When x - y = 2
….…………………………………………………………………………………………………………
Subtracting equation (2) from equation (1).
(x + y ) - (x - y )= = 6-2
2y = 4 1/2
y = 2 and x = 4.
∴ The two digit number is 10x + y = 40 + 2 = 42.
….…………………………………………………………………………………………………………
When y - x = 2
Subtracting equation (3) from equation (1).
(x + y ) - (y - x) =6- 2
2x = 4
x = 2 and y = 4 1/2
∴ The two digit number is 10y + x = 20 + 4 = 24

Thus, the two digits are 42 and 24

29.

1/2

......................................................................................................................
Given: A circle touching the side BC of ΔABC at P and AB, AC produced at
Q and R respectively.
1/2
To Prove: AQ=1/2(Perimeter of ΔABC)

......................................................................................................................
.
Proof: Lengths of tangents drawn from an external point to a circle are 1/2
equal.
⇒AQ = AR, BQ = BP, CP = CR.
......................................................................................................................
.
Perimeter of Δ ABC = AB + BC + CA
1/2
= AB + (BP + PC) + (AR - CR)
......................................................................................................................
.
= (AB + BQ) + (PC) + (AQ - PC) [ ∵AQ = AR, BQ = BP, CP = CR] 1/2

......................................................................................................................
.
= AQ + AQ
1/2
= 2AQ
⇒AQ=1/2 (Perimeter of ΔABC)
∴AQ is the half of the perimeter of ΔABC.

30. sinθ+cosθ= 3

(a) ​

⇒(sinθ+cosθ) 2 =3
1
⇒sin 2 θ+cos θ+2sinθcosθ=3
2

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

⇒1+2sinθcosθ=3

⇒2sinθcosθ=2
1/2

⇒sinθcosθ=1
................................................................................................

⇒sinθcosθ=sin 2θ+cos 2θ

sin2 �+cos2 � 1
⇒1=
sin � cos �
............................................................................................
​
1/2
⇒tanθ+cotθ=1

30. cotA −cosA
(b) LHS=
cotA+cosA
cosA
−cosA 1
sinA
= cosA
+cosA
sinA

….………………………………………………………………………….
1 1
cosA(sinA − 1)
= 1
cosA(sinA + 1)

….…………………………………………………………………………..
cosecA−1 1
= cosecA+1 = RHS

31. Given, a bag contains 24 balls of which x are red, 2x are white and 3x are
blue.
A ball is selected at random.
Given, x + 2x + 3x = 24
6x = 24
x = 24/6 1/2
x=4
….……………………………………………………………………………………………………………
Number of red balls = x
=4
Number of white balls = 2x
= 2(4)
=8
Number of blue balls = 3x
= 3(4) 1/2
= 12
….…………………………………………………………………………………………………………..
(i)The probability of selecting a ball that is not red is given by
Favourable outcomes = balls other than red
= white balls + blue balls
Number of favourable outcomes = 8 + 12 = 20
Number of possible outcomes = 24
The probability of selecting a ball that is not red = number of favourable
outcomes / number of possible outcomes 1
Probability = 20/24= 10/12= 5/6
….……………………………………………………………………………………………………………..
(ii)The probability of selecting a white ball = number of favourable
outcomes / number of possible outcomes 1
Probability = 8/24= 1/3

SECTION-D
32. We have to find the marks Sunita scored in the test.
(a) Let the actual marks be x 1/2
Total marks = 30
….………………………………………………………………………………………………………
Given, 9(x + 10) = x² 1

….……………………………………………………………………………………………………..
9x + 90 = x²
x² - 9x - 90 = 0 1

….………………………………………………………………………………………………………..
On factoring,
x² - 15x + 6x - 90 = 0
x(x - 15) + 6(x - 15) = 0
1+1
(x + 6)(x - 15) = 0
Now, x + 6 = 0
x = -6
Also, x - 15 = 0
x = 15
….……………………………………………………………………………………………………….. 1/2
Negative term x = -6 is neglected.
So, x = 15
Therefore, Sunita scored 15 marks in the examination.

32. Let the first integer be x.
(b) The next consecutive positive integer will be x + 1. 1/2
According to the given question, the sum of squares of x and x + 1 is 365.
….……………………………………………………………………………………………………………

x2 + ( x + 1)2 = 365
x2 + (x2 + 2x + 1) = 365 [ ∵ (a + b)2 = a2 + 2ab + b2] 1
2x2 + 2x + 1 = 365
2x2 + 2x + 1- 365 = 0
….………………………………………………………………………………………………………
2x2 + 2x - 364 = 0
2(x2 + x - 182) = 0 1

x2 + x - 182 = 0
….…………………………………………………………………………………………………….
x2 + 14x - 13x - 182 = 0
x (x + 14) - 13 (x + 14) = 0 1+1
(x - 13) (x + 14) = 0
x - 13 = 0 and x + 14 = 0
x = 13 and x = - 14
….………………………………………………………………………………………………………..
The value of x cannot be negative (because it is given that the integers
are positive). Thus, we ignore x = -14. 1/2

∴ x = 13 and x + 1 = 14

33.
(a)

1/2

Given ∆NSQ ≅ ∆MTR and ∠1 = ∠2
To prove : ∆PTS ~ ∆PRQ

….…………………………………………………………………………………………………………..
Proof : Since, ∆NSQ ≅ ∆MTR
So, SQ = TR ………….. (i) 1
12
Also, ∠1 = ∠2 ⇒ PT = PS ………… (ii)
[since, sides opposite to equal angles are also equal]
….…………………………………………………………………………………………………………….
From Eqs.(i) and (ii), PS/SQ=PT/TR
⇒ ST || QR [by converse of basic proportionality theorem] 1
12
∴ ∠1 = ∠PQR [Corresponding angles]
and ∠2 = ∠PRQ
….……………………………………………………………………………………………………….
In ∆PTS and ∆PRQ,
∠P = ∠P [common angles]
∠1 = ∠PQR 1
12
∠2 = ∠PRQ
∴ ∆PTS ~ ∆PRQ [by AAA similarity criterion]
33. Consider the trapezium ABCD as shown below.
(b)

1

….………………………………………………………………………………………………………..
In trapezium ABCD,
AB || CD 1
Also, AC and BD intersect at point O.
Construct XY parallel to AB and CD (XY || AB, XY || CD) through point O
….…………………………………………………………………………………………………………
In ΔABC
OY || AB (construction)
According to Basic Proportionality Theorem
1
BY/CY = AO/OC................. (1)
….……………………………………………………………………………………………………………..

In ΔBCD
OY || CD (construction)
According to Basic Proportionality Theorem 1
BY/CY = OB/OD................. (2)

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

From equations (1) and (2) 1/2
OA/OC = OB/OD

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

⇒ OA/OB = OC/OD
1/2
Hence proved.

34 The height and diameter of the base of the cone is 4 cm and 8 cm.
(a) We have to find the volume of the toy.
We have to find the difference of the volumes of the cube and the toy
and total surface area of the toy.

1/2

1/2

Diameter of base of cone = diameter of hemisphere = 8 cm
Radius = 8/2 = 4 cm

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

1/2
Volume of hemisphere = (2/3)πr³
= (2/3)(22/7)(4)³
= 134.095 cm³
….……………………………………………………………………………………………………………..

Volume of the cone = (1/3)πr²h
1/2
= (1/3)(22/7)(4)²(4)
= 67.047 cm³

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

Volume of the toy = volume of hemisphere + volume of cone 1/2
= 134.095 + 67.047
= 201.142 cm³
Therefore, the volume of the toy is 201.42 cm³
….……………………………………………………………………………………………………………..

Given, a cube circumscribes the toy
1/2
So, the edge of the cube = diameter of the hemisphere = 8 cm
Volume of cube = a³

= (8)³
= 512 cm³
….……………………………………………………………………………………………………………..

1/2
Difference in volume of cube and toy = volume of cube - volume of toy
= 512 - 201.142
= 310.858 cm³
1/2

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

Curved surface area of the cone = πrl 1/2
Slant height, l = ℎ² + �² = 4² + 4²= 32= 5.657 cm
….………………………………………………………………………………………………………….
Curved surface area of the cone = (22/7)(4)(5.657)
= 71.117 cm²
….………………………………………………………………………………………………………….

Curved surface area of hemisphere = 2πr² 1/2
= 2(22/7)(4)²
= (44/7)(16)
= 100.571 cm²
….………………………………………………………………………………………………………….
Total surface area of the toy = curved surface area of the cone + curved
surface area of hemisphere.
= 71.117 + 100.571
= 171.688 cm²
Therefore, the total surface area of the toy is 171.688 cm².

34.
(b)

1

Radius of base of cylinder, r= 3.5 cm Height, h = 10 cm.

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

Total area of article = Curved Surface area of Cylinder + 2 × Curved 1
Surface area of Hemisphere=

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

1
= 2πrh + 2 × 2πr2

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

1
22 22
= 2× 7 × 3.5 × 10 + 2 × 2 × 7 × (3.5)2

….……………………………………………………………………………………………………
22
= 2× 7 × 3.5 ×(10 +7)

1/2

….……………………………………………………………………………………………………
22
= 7× 7 × (17)

= 374 cm2 1/2

35. Class Intervals Frequency Cumulative Frequency
(a)
0-100 2 2

100-200 5 7

200-300 x 7+x

300-400 12 19+x

400-500 17 36+x

500-600 20 56+x
1
600-700 y 56+x+y

700-800 9 65+x+y

800-900 7 72+x+y

900-1000 4 76+x+y

….…………………………………………………………………………………………………………..
It is given that n=100 1
So , 76 +x+y = 100 or x+y = 24…………..(1)
….…………………………………………………………………………………………………………..

The median is 525 which lies in the class 500-600
So, l = 500, f= 20, cf =36 +x, h= 100 1

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

� 1/2
−��
Using the formula: Median = l +
2
×ℎ
�

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

50−36−� 1/2
525 = 500 +( 20
)× 100

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

525-500= (14-x)× 5 1/2

25=70- 5x

5x= 70-25=45
x=9
….……………………………………………………………………………………………………

Therefore, from (1),we get 9+y =24 1/2
y=15

35.
(b) Daily Wages(C.I.) Class Mark (�� ) No. of workers (�� ) �� ��
100-120 110 12 1320
120-140 130 14 1820
140-160 150 8 1200
1
160-180 170 6 1020 22
180-200 190 10 1900
�� = 50 �� �� =7260

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

�� ��
Mean daily wages =
�� 1

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

7260
= 1
50

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

1/2
= ₹ 145.2

SECTION-E
36. (i)A.P.= 4,7,10,13,…………..
d=7-4=3
an= a +(n-1)d 1

a15=4 +(15-1)3 = 4 +42= 46

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

(ii)A.P.= 4,7,10,13,…………..
d = 7-4=3
an= a +(n-1)d
136= 4 +(n-1)3
136= 4 + 3n - 3
136 - 1 = 3n 1
135
3
=n
n = 45
….……………………………………………………………………………………………………

(iii)(a)A.P.= 4,7,10,13,…………..
d = 7-4=3 1
a=4
n=30
….……………………………………………………………………………………………………

30
S30= 2 [2 × 4 + (30 − 1)3]

S30= 15 [8 + 29 × 3] 1

S30= 15 × 95=1425
….……………………………………………………………………………………………………

(iii)(b)A.P.= 4,7,10,13,………….. 1
d = 7-4=3
a=4
….……………………………………………………………………………………………………

1
an= a +(n-1)d
a20= 4+(20-1)3=4 +19× 3= 4 +57=61

37. (i)The coordinates of the vertices of ∆PQR are P(4,6 ), Q(3,2) and R(6,5)

1

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

(ii)(a) PQ= (3 − 4)2 + (2 − 6)2 = ( − 1)2 + ( − 4)2 = 17 m
2

QR= (6 − 3)2 + (5 − 2)2 = (3)2 + (3)2 = 18 m = 3 2 m

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

(ii) (b) Let S(x,y) be the point which divides the line segment joining 1
points P(4,6) and R(6,5)
In the ratio 2 :1 internally.
2×6+1×4 2×5+1×6
By section formula S(x,y)= S ( , )
2+1 2+1 1
….………………………………………………………………………………………………….
12+4 10+6
= S( ,
3
)=
3

16 16
= S( 3 , 3 )

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

(iii)PQ= (3 − 4)2 + (2 − 6)2 = ( − 1)2 + ( − 4)2 = 17 m

QR= (6 − 3)2 + (5 − 2)2 = (3)2 + (3)2 = 18 m = 3 2 m

1
PR= (6 − 4)2 + (5 − 6)2 = (2)2 + ( − 1)2 = 5 m
PQ≠ QR ≠ PR
∴ ∆PQR is not an isosceles triangle but a scalene triangle.

38. (i)

2

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

(ii)In right Δ ADB, tan 45° = BD/AD
∴ AD = BD/tan 45° 1/2
AD= BD = OB - OD = (10 - h) m

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

In right ΔADC 1/2
tan 60° = CD/AD = ( 10 + h)/( 10 − h)

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

⇒ ( 10 + h)/( 10 − h ) = √3
1/2
⇒ 10 + h = 10√3 - √3h
⇒ (√3 + 1)h = 10(√3 - 1)

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

∴ h = 10 (√ 3 − 1)/( √ 3 + 1) 1/2
h = 10( √ 3 − 1)(√ 3 - 1)/( √ 3 + 1)(√ 3 - 1)
h = 10(√ 3 - 1)2/ 2
⇒ h = 2.67 m using 3 = 1.73

MARKING SCHEME BSEH PRACTICE PAPER 1, 10TH गणित(मानक)
March2025
(हिंदी माध्म )
Q. Expected solutions mar
no. ks

खण्-क

1 (b)500 1
2 (a) दोनो धनातमक
1

3 (a)x2 -4x +3 2 =0 1
4 (b)1 1
5 (c )± 4 1
6 (c )7 1
7 (a)30° 1
8 5 1
(a)2
9 (c) r2 वर् इकाई 1
10 (d) 6 : π 1

11 (b) 8 [ बहुलक = 3माध्क-2माध् का उप्ोर करके ] 1
12 (b) 14 1
13 a =3 1
14 व्ाा 1
15 1 1
16 A+B= 90° 1
17 r2 वर् इकाई 1
18 10+15 = 25 1
19 (a) अभिकथन (A) और तक् (R) दोनो ाही है और तक् (R), अभिकथन (A) 1

की ाही व्ाा्ा करता है ।

20 (d) अभिकथन (A) गलत है , परनतु तक् (R) ाही हैI 1

खण् –ख

21.
उपरोक प् से, हमारे पास रै णखक समीकरि इस पकार है,
(a)

2x + y = 23---------------(i)
4x - y = 19----------------(ii)
समीकरि (i) तथा (ii) को जोडने पर, 1/2
6x = 42
x = 7.
….………………………………………………………………………………………………………...
(i) मे x के मान को पणततथाणपत करने पर, हमे णमलता है,
2(7) + y = 23
14 + y = 23
y = 23 - 14 1/2
y=9

….…………………………………………………………………………………………………………
5y - 2x और y/ x - 2 मे x और y के मानो को पणततथाणपत करने पर, हमे णमलता है,
5y - 2x = 5 × 9 - 2 × 7 1/2
= 45 - 14
= 31
….…………………………………………………………………………………………………………….
y/x - 2 = 9/7 -2 1/2
= -5/7.

21.
उपरोक प् मे , a1= 4, a2 = 2, b1 = p और b2 = 2. 1/2
(b)

….……………………………………………………………………………………………………………..
यदद रै णखक समीकरिो के एक युगम का हल अण्तीय है, तब
a1/a2 ≠ b1/b2 1/2

….…………………………………………………………………. ……………………………………………………………………………………………
4/2 ≠ p/ 2
1/2
….…………………………………………………………………………………………………………….
4≠p

1/2
इस पकार, रै णखक समीकरिो के युगम मे 4 को छोडकर p के सभी मानो के णलए एक
अण्तीय हल है
22. दद्ा है : DE||AB

हमे x का मान जात करना है।
आकृ णत से

CD = x+ 3
AD = 3x + 19
CE = x
BE = 3x + 4 1/2
∴आधारिूत ामानुपाततकता पमे् दवारा
CD/DA = CE/EB
….……………………………………………………………………………………………………………..
⇒ (x+3)/(3x+19) = x/(3x+4)
1/2
आर पार रण
ु ा करने पर
(x+3)(3x+4) = x(3x+19)

….………………………………………………………………………………………………………….
गुिनातमक और णितरिातमक गुि ्ारा,
3x² + 4x + 9x + 12 = 3x² + 19x
उि्तनष् पदो को काटने पर 1/2
13x + 12 = 19x
….………………………………………………………………………………………………………….
13x - 19x = -12
-6x = -12
x = 12/6
x=2 1/2
इसणलए, x का मान 2 है।
23.
बडे िृत की जीिा छोटे िृत की तपपररेखा है जैसा दक नीचे ददए गए णचत मे ददखाया
गया है।

1/2

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

PQ एक बडे िृत की जीिा और एक छोटे िृत की तपपररेखा है।
तपपर रे खा PQ तपपर बबंदु S पर णतरया के लंबित है।
इसणलए, ∠OSP = 90°
समकोि ΔOSP मे
पाइथागोरस पमेय ्ारा
OP2 = OS2 + SP2 1
52 = 32 + SP2
SP2 = 25 - 9
SP2 = 16
SP = ± 4
SP तपपर रे खा की लंबाई है और ऋिातमक नही हो सकती
अतः,SP=4सेमी
.….………………………………………………………………………………………………………….
QS = SP (QP को बडे िृत की जीिा मानते हए के ं से लंब जीिा को समण्भाणजत
करता है)
इसणलए, QS = SP = 4 सेमी
जीिा की लंबाई PQ = QS + SP = 4 + 4 1/2

PQ = 8 सेमी
अत: बडे िृत की जीिा की लंबाई 8 सेमी है।
24. sin(A - B) = 1/2 ⇒ Sin(A-B) = sin (30°) ⇒A - B = 30o …(1)
(a) 1/2

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

1/2
cos (A + B) = 1/2 ⇒ cos(A + B) = cos (60o) ⇒ A + B = 60o …(2)

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

समीकरि (1) और (2)को जोडने पर. , हमे पार होता है 2A = 90o ⇒ A = 45o 1/2

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

अब, A का मान समीकरि(2) मे रखने पर, हमे पार होता है 45o + B =60o 1/2
⇒ B = 15o

अत:, A = 45o और B = 150
24.
(b) a2/x2−b2/y2
1/2
=a /a sin θ−b /b tan θ [∵x=asinθ,y=btanθ]
2 2 2 2 2 2

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

=1/sin2θ−1/tan2θ 1/2
….……………………………………………………………………………………………………………

=cosec2θ−cot2θ 1/2

….…………………………………………………………………………………………………………
[∵1+cot2θ=cosec2θ∴cosec2θ−cot2θ=1]

1/2
=1

25. दद्ा है : 20m × 16m आ्ाम का एक आ्ताकार मैदान
एक रा् को आ्ताकार मैदान के कोने पर 14 मीटर लंबी रसाी ाे बांधा र्ा है ।
हमे खेत का िह केतेल जात करना है णजसमे गाय चर सकती है।

1/2

….……………………………………………………………………………………………………………
मान लीणजए दक ABCD एक आयताकार मैदान है।
आकृ णत मे ,हम देखते है दक गाय णजस केत को चर सकती है िह एक िृत के एक

णतरयखं् के रप मे है।
अतः, AGEF 14 मीटर णतरया िाले िृत का एक णतरयखं् है।
णतरयखं् का केतेल = πr²θ/360° 1/2

यहाँ,θ = 90°
….……………………………………………………………………………………………………………
णतरयखं् का केतेल = (22/7)(14)²(90°/360°)
= (22)(2)(14)(1/4) 1/2
= (22)(14)(1/2)
= 11(14)

….…………………………………………………………………………………… 1/2
= 154 m²
अतः, गाय णजस केत मे चर सकती है िह 154 िगर मीटर है।

खण् –ग

26. मान लीजिए कक 3−2 5 पररमे् ांा्ा है
1/2

.............................................................................………………………………...
a
अतः इसे रप मे णलखा जा सकता है
b

​ जहाँ a और b सह-अभारय है और b≠ 0 1/2
a
​ अत: 3−2 5 =
b
.......................................................................……………………………………….
​

a 3b−a 1/2
​ ⇒2 5= 3−b=
b
..................................................................................………………………….
3b−a
⇒ 5= 2b 1/2
..............................................................................………………………….
3b−a
िहाँ 5 अपररमे् है तथा
पररमे् है l 1/2
2b
क्ोकक अपररमे् ांा्ा≠ पररमे् ांा्ा

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

अतः उपरोकत एक ववरोधािाा है ।
इाभलए हमारी कलपना गलत है . 1/2

अतः 3−2 5 अपररमेय है।

27.
चूँदक α और β बहपद f(x)= 2x2 -7x +3 के पूनयक है

−7 7 3
∴ α + β =− = और αβ = 1
2 2 2
...........................................................................
2
अब α2 + β2 = α + β − 2αβ 1

..........................................................................
7 2 3
= −2× =
2 2 1
49 3 49−12 37
4
−1 = 4
= 4
28.
मान लीणजए दक ₹ x पहले दो ददनो के णलए णनधारररत पुलक है और 1/2
(a)
पतयेक अणतररक ददन के णलए पुलक ₹ y है।
….…………………………………………………………………………………………………………
पहली णतथणत से,
लणतका ने छह ददनो तक रखी एक पुततक के णलए ₹ 22 का भुगतान दकया 1/2

x + 4y = 22---------(1)

….…………………………………………………………………………………………………………
दूसरी पतर के अनुसार,
आनंद ने चार ददनो तक रखी एक दकताब के णलए ₹16 का भुगतान दकया 1/2
x + 2y = 16------------(2)
….…………………………………………………………………………………………………………

ामीकरण (2) को (1)ाे घटाने पर, हमे भमलता है
2y = 6

y = 3. 1/2
….…………………………………………………………………………………………………………
y का मान ामीकरण (2) मे रखने पर, हमे भमलता है
x + 2 x 3 = 16
x = 16 - 6 = 10
x = 10. 1/2
x = 10.
….…………………………………………………………………………………………………………

इसणलए, णनधारररत पुलक = ₹ 10 और पतयेक अणतररक ददन का पुलक = ₹ 3. 1/2
28.
मान लीणजए दक पहली संखया मे दहाई के तथान पर और इकाई के तथान पर अंक
(b)
कमपः x और y है।
एक संखया को णितताररत रप मे 10(x) + y के रप मे वक दकया जा सकता है।
अंको को उलटने पर, x इकाई का अंक है और y दहाई का अंक है। दूसरी संखया के णलए 1/2

णितताररत अंकन 10(y) + x है
….…………………………………………………………………………………………………………
प् के अनुसार,
(10x + y) + (10y + x) = 66
⇒ 11(x + y) = 66 1/2
⇒ x + y = 6 .............. (1)
….…………………………………………………………………………………………………………
साथ ही, यह भी ददया गया है दक दोनो अंको के बीच का अंतर 2 है।
1/2
∴ x - y = 2 ...............(2)
….…………………………………………………………………………………………………………
्ा y - x = 2............. (3)
1/2
िब x - y = 2
….…………………………………………………………………………………………………………
समीकरि (2) को (1)से घटाने पर, हमे णमलता है
(x + y ) - (x - y )= = 6-2
2y = 4 1/2
y = 2 और x = 4.
∴ दो अंको की संखया है= 10x + y = 40 + 2 = 42.
….…………………………………………………………………………………………………………
िब y - x = 2

समीकरि (3) को (1)से घटाने पर, हमे णमलता है

(x + y ) - (y - x) =6- 2
1/2
2x = 4
x = 2 और y = 4
∴ दो अंको की संखया है=10y + x = 20 + 4 = 24

अत: दो अंक 42 और 24 है
29.

1/2

......................................................................................................................
ददया गया है: एक िृत ΔABC की भुजा BC को P पर और भुजा AB तथा AC को
आगे बढ़ाने पर कमपः Q और R पर पर तपपर करता है।
णसद करना है : AQ=1/2(ΔABC का पररमाप) 1/2

......................................................................................................................
.
पमाि: दकसी बाह बबंदु से िृत पर खीची गई तपपर रे खाओ की लंबाई बराबर होती है। 1/2
⇒AQ = AR, BQ = BP, CP = CR.
......................................................................................................................

Δ ABCका पररमाप = AB + BC + CA
1/2
= AB + (BP + PC) + (AR - CR)
......................................................................................................................
.
1/2
= (AB + BQ) + (PC) + (AQ - PC) [ ∵AQ = AR, BQ = BP, CP = CR]

......................................................................................................................
= AQ + AQ
= 2AQ
1/2
⇒AQ=1/2 (ΔABCका पररमाप )
∴AQ, ΔABC के पररमाप का आधा भाग है।

30.
ददया है : sinθ+cosθ= 3
(a)
​

⇒(sinθ+cosθ) 2 =3 1

⇒sin 2θ+cos 2θ+2sinθcosθ=3
..................................................................................................

⇒1+2sinθcosθ=3

⇒2sinθcosθ=2 1/2

⇒sinθcosθ=1
................................................................................................

⇒sinθcosθ=sin 2θ+cos 2θ
1
sin2 θ+cos2 θ
⇒1=
sin θ cos θ
............................................................................................
​
1/2
⇒tanθ+cotθ=1 ्ही भादध करना था l
30. cotA −cosA
(b) LHS=
cotA+cosA
cosA
−cosA 1
sinA
= cosA
+cosA
sinA

….………………………………………………………………………….

1 1
cosA(sinA − 1)
= 1
cosA( + 1)
sinA

….…………………………………………………………………………..
cosecA−1 1
= cosecA+1 = RHS

31.
ददया गया है, एक थैले मे 24 गेदे है णजनमे से x लाल, 2x सेे द और 3x नीली है।
एक गेद यादृण्छक रप से चुनी जाती है।
⇒x + 2x + 3x = 24
6x = 24
x = 24/6 1/2

x=4
….……………………………………………………………………………………………………………
लाल गेदो की संखया = x= 4

ाफेद गेदो की संखया = 2x = 2(4)
=8 1/2
नीली गेदो की संखया = 3x= 3(4)
= 12
….…………………………………………………………………………………………………………..
(i)ऐसी गेद को चुनने की पाणयकता जो लाल नही है, णनम ्ारा दी गई है
अनुकूल पररिाम = लाल के अलािा अनय गेदे
= सफे द गेद + नीली गेद
अनुकूल पररिामो की संखया = 8 + 12 = 20 1
संभाणित पररिामो की संखया = 24
ऐसी गेद चुनने की पाणयकता जो लाल नही है = अनुकूल पररिामो की संखया/संभाणित
पररिामो की संखया
पाणयकता = 20/24=10/12=5/6
….……………………………………………………………………………………………………………..

(ii) सेे द गेद चुनने की पाणयकता = अनुकूल पररिामो की संखया/संभाणित पररिामो
1

की संखया
पाणयकता = 8/24= 1/3
खण्-घ

32.
हमे सुनीता ्ारा परीका मे पार अंक जात करने है।
(a)
माना दक िाततणिक अंक x है
1/2
कु ल अंक = 30
….………………………………………………………………………………………………………
प्ानुसार 9(x + 10) = x² 1

….……………………………………………………………………………………………………..
9x + 90 = x²
x² - 9x - 90 = 0 1

….………………………………………………………………………………………………………..
x² - 15x + 6x - 90 = 0
x(x - 15) + 6(x - 15) = 0
(x + 6)(x - 15) = 0
्ा x + 6 = 0
1+1
x = -6
अथिा x - 15 = 0
x = 15
….………………………………………………………………………………………………………..
ऋिातमक पद x = -6 को छोडने पर
1/2
x = 15
इसणलए, सुनीता ने परीका मे 15 अंक पार दकये।
32.
माना पहला पूिाणक x है।
(b)
अगला कमागत धनातमक पूिाणक x + 1 होगा। 1/2

ददए गए प् के अनुसार, x और x + 1 के िग् का योग 365 है।
….……………………………………………………………………………………………………………

x2 + ( x + 1)2 = 365 1

x2 + (x2 + 2x + 1) = 365 [ ∵ (a + b)2 = a2 + 2ab + b2]

2x2 + 2x + 1 = 365
2x2 + 2x + 1- 365 = 0
….………………………………………………………………………………………………………
2x2 + 2x - 364 = 0
2(x2 + x - 182) = 0
1
x2 + x - 182 = 0
….…………………………………………………………………………………………………….
x2 + 14x - 13x - 182 = 0
x (x + 14) - 13 (x + 14) = 0
(x - 13) (x + 14) = 0 1+1
x - 13 = 0 और x + 14 = 0
x = 13 और x = - 14
….………………………………………………………………………………………………………..
x का मान ऋिातमक नही हो सकता (कयोदक यह ददया गया है दक पूिाणक धनातमक है)।
इस पकार, हम x = -14 को अनदेखा करते है। 1/2

∴ x = 13 और x + 1 = 14
33.
(a)

1/2

दद्ा है : ∆NSQ ≅ ∆MTR और ∠1 = ∠2
भादध करना है : ∆PTS ~ ∆PRQ

….…………………………………………………………………………………………………………..
पमाण : क्ोकक ∆NSQ ≅ ∆MTR
इाभलए SQ = TR ………….. (i) 1
12
ााथ ही ∠1 = ∠2 ⇒ PT = PS ………… (ii)

[∵ समान कोिो की सममुख भुजाएँ भी बराबर होती है]
.…………………………………………………………………………………………………………….

समीकरिो (i) और (ii) से, PS/SQ=PT/TR

⇒ ST || QR [आधारभूत समानुपाणतकता पमेय के णिलोम ्ारा ] 1
12
∴ ∠1 = ∠PQR [ांरत कोण ]
और ∠2 = ∠PRQ
….……………………………………………………………………………………………………….
∆PTS तथा ∆PRQ मे ,
∠P = ∠P [उि्तनष् कोण ]
1
∠1 = ∠PQR 12

∠2 = ∠PRQ
∴ ∆PTS ~ ∆PRQ [AAA ामरपता काौटी दवारा ]
33.
नीचे ददखाए गए समलंब ABCD पर णिचार करे ।
(b)

1

….………………………………………………………………………………………………………..
समलंब ABCD मे,
AB || CD
साथ ही, AC और BD बबंदु O पर पणत्छेद करते है। 1
बबंदु O से होकर AB और CD (XY || AB, XY || CD) के समानांतर XY की रचना करे
….…………………………………………………………………………………………………………
ΔABC मे
OY || AB (रचना )
आधारभूत समानुपाणतकता पमेय के अनुसार 1

BY/CY = AO/OC................. (1)
….……………………………………………………………………………………………………………..

ΔBCD मे
OY || CD (रचना )
आधारभूत समानुपाणतकता पमेय के अनुसार
1

BY/CY = OB/OD................. (2)

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

1/2
ामीकरण (1) और (2) ाे
OA/OC = OB/OD

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

1/2
⇒ OA/OB = OC/OD
यही णसद करना था l
34
पंकु के आधार की ऊँ चाई और वास 4 सेमी और 8 सेमी है।
(a)
हमे णखलौने का आयतन जात करना है।
हमे घन और णखलौने के आयतन का अंतर और णखलौने के कु ल पृषीय केतेल जात
करना है।

1/2

पंकु के आधार का वास = अधरगोले का वास = 8 सेमी
णतरया = 8/2 = 4 सेमी
….……………………………………………………………………………………………………………..

अधरगोले का आयतन = (2/3)πr³ 1/2
= (2/3)(22/7)(4)³
= 134.095 cm³
….……………………………………………………………………………………………………………..

पंकु का आयतन = (1/3)πr²h
= (1/3)(22/7)(4)²(4) 1/2
= 67.047 cm³

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

णखलौने का आयतन = अधरगोले का आयतन + पंकु का आयतन
= 134.095 + 67.047
= 201.142 cm³ 1/2
इाभलए, खखलौने का आ्तन 201.42 cm³ है
….……………………………………………………………………………………………………………..

ददया गया है, णखलौने के चारो ओर एक घन है
∴ घन की भुजा = अधरगोले का वास = 8 सेमी
1/2
घन का आयतन = a³= (8)³
= 512 cm³
….……………………………………………………………………………………………………………..

घन और णखलौने के आयतन मे अंतर = घन का आयतन - णखलौने का आयतन
= 512 - 201.142 1/2
= 310.858 cm³

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

पंकु का िक पृषीय केतेल= πrl
णतयरक ऊं चाई, l = ℎ² + �² = 4² + 4²= 32= 5.657 cm 1/2
….………………………………………………………………………………………………………….
पंकु का िक पृषीय केतेल= (22/7)(4)(5.657)

= 71.117 cm² 1/2
….………………………………………………………………………………………………………….

अधरगोले का िक पृषीय केतेल= 2πr²
= 2(22/7)(4)² 1/2
= (44/7)(16)
= 100.571 cm²
….………………………………………………………………………………………………………….
णखलौने का कु ल पृषीय केतेल = पंकु का िक पृषीय केतेल + अधरगोले का िक पृषीय
केतेल
= 71.117 + 100.571 1/2
= 171.688 cm²
इसणलए, णखलौने का कु ल पृषीय केतेल 171.688 सेमी² है।
34.
(b)

1

बेलन के आधार की त्र्ा, r= 3.5 cm , ऊँचाई, h = 10 cm.

….……………………………………………………………………………………………………………..
1
वसतु का कुल के्फल = बेलन का वक पषृ ्ठ् के्फल + 2 × अध्रोले का वक पषृ ् के्फल=

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

= 2πrh + 2 × 2πr2 1

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

22 22 1
= 2× 7 × 3.5 × 10 + 2 × 2 × 7 × (3.5)2

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

22
= 2× 7 × 3.5 ×(10 +7)
1/2

….……………………………………………………………………………………………………
22
= 7× × (17)
7
1/2
= 374 cm2

35. िगर अंतराल बारं बारता संचयी बारं बारता
(a)
0-100 2 2

100-200 5 7

200-300 x 7+x

300-400 12 19+x

400-500 17 36+x

500-600 20 56+x
1
600-700 y 56+x+y

700-800 9 65+x+y

800-900 7 72+x+y

900-1000 4 76+x+y

….…………………………………………………………………………………………………………..
्ह दद्ा र्ा है कक n=100
1
इाभलए , 76 +x+y = 100 ्ा x+y = 24…………..(1)
….…………………………………………………………………………………………………………..

माधयक 525 है जो िगर 500-600 मे णतथत है
1
इाभलए , l = 500, f= 20, cf =36 +x, h= 100

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

�
−��
ाू् : माध्क = l + × ℎ का उप्ोर करने पर
2
1/2
�

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

50−36−� 1/2
525 = 500 +( 20
)× 100

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

525-500= (14-x)× 5

25=70- 5x
1/2

5x= 70-25=45
x=9
….……………………………………………………………………………………………………

अत:, (1) से, हमे पार होता है
1/2
9+y =24
y=15

35.
(b) दैणनक मजदूरी (वर् वर् चचनह (�� ) मजदूरो की संखया (�� ) �� ��
अनतराल )
100-120 110 12 1320
120-140 130 14 1820 1
22
140-160 150 8 1200
160-180 170 6 1020
180-200 190 10 1900
�� = 50 �� �� =7260

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

�� ��
औसत दैणनक मजदूरी =
�� 1

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

7260 1
=
50

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

1/2
= ₹ 145.2
36. (i)A.P.= 4,7,10,13,…………..
d=7-4=3
an= a +(n-1)d 1

a15=4 +(15-1)3 = 4 +42= 46

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

(ii)A.P.= 4,7,10,13,…………..
d = 7-4=3
an= a +(n-1)d
136= 4 +(n-1)3
136= 4 + 3n - 3
136 - 1 = 3n 1
135
3
=n
n = 45
….……………………………………………………………………………………………………

(iii)(a)A.P.= 4,7,10,13,…………..
d = 7-4=3 1
a=4
n=30
….……………………………………………………………………………………………………

30
S30= 2 [2 × 4 + (30 − 1)3]

S30= 15 [8 + 29 × 3] 1

S30= 15 × 95=1425
….……………………………………………………………………………………………………

1
(iii)(b)A.P.= 4,7,10,13,…………..
d = 7-4=3
a=4
….……………………………………………………………………………………………………
1
an= a +(n-1)d
a20= 4+(20-1)3=4 +19× 3= 4 +57=61

37.
(i)∆PQR के पीर् के णनद्पांक P(4,6 ), Q(3,2) और R(6,5) है l 1

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

(ii)(a) PQ= (3 − 4)2 + (2 − 6)2 = ( − 1)2 + ( − 4)2 = 17 m
2

QR= (6 − 3)2 + (5 − 2)2 = (3)2 + (3)2 = 18 m = 3 2 m

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

(ii) (b) मान लीणजए S(x,y) िह बबंदु है जो बबंदओ
ु P(4,6) और R(6,5) को णमलाने
िाले रे खाखं् को आंतररक रप से 2:1 के अनुपात मे णिभाणजत करता है l

2×6+1×4 2×5+1×6 1
वविािन ाू् दवारा S(x,y)= S ( , )
2+1 2+1
….………………………………………………………………………………………………….
12+4 10+6
= S( 3 , 3 )=

16 16
= S( 3 , 3 ) 1

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

(iii)PQ= (3 − 4)2 + (2 − 6)2 = ( − 1)2 + ( − 4)2 = 17 m

QR= (6 − 3)2 + (5 − 2)2 = (3)2 + (3)2 = 18 m = 3 2 m

PR= (6 − 4)2 + (5 − 6)2 = (2)2 + ( − 1)2 = 5 m
1
PQ≠ QR ≠ PR
∴ ∆PQR एक समण्बाह णतभुज नही है बणलक एक णिरमबाह णतभुज है।

38. (i)

2

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

(ii)ामकोण Δ ADB मे , tan 45° = BD/AD

∴ AD = BD/tan 45° 1/2
AD= BD = OB - OD = (10 - h) m

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

1/2
ामकोण ΔADC मे
tan 60° = CD/AD = ( 10 + h)/( 10 − h)

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

⇒ ( 10 + h)/( 10 − h ) = √3 1/2
⇒ 10 + h = 10√3 - √3h
⇒ (√3 + 1)h = 10(√3 - 1)

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

1/2
∴ h = 10 (√ 3 − 1)/( √ 3 + 1)
h = 10( √ 3 − 1)(√ 3 - 1)/( √ 3 + 1)(√ 3 - 1)
h = 10(√ 3 - 1)2/ 2
⇒ h = 2.67 m [ 3 = 1.73 का उप्ोर करते हुए]

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