Solution

Section A
1.
(c) an irrational number
Explanation: π = 3.14159265359…….., which is non-terminating non-recurring.
Hence, it is an irrational number.
2.
(b) infinitely many solutions
Explanation:
5y+15
Given linear equation: 3x - 5y = 15 Or, x = 3

When y = 0, x = 15

3
=5
When y = 3, x = 30

3
= 10
When y = -3, x = 0

3
=0
xx 5 10 0

yy 0 3 -3
Plot the points A(5,0) , B(10,3) and C(0,−3). Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation.
Hence, the linear equation has infinitely many solutions.

3.
(c) a line parallel to y-axis
Explanation: Two points having same abscissa but different ordinate always make a line which is parallel to the y-axis as
abscissa is fixed and the only ordinate keeps changing.
4.
(c) 2
frequency of the class
Explanation: Adjusted frequency = ( width of the class
) × 5

8
Therefore, Adjusted frequency of 25 - 45 = 20
× 5 = 2

5.
(b) (3,0)
Explanation: 2x + 3y = 6 meets the X-axis.

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 Put y = 0,
2x + 3(0) = 6
x=3
Therefore, graph of the given line meets X-axis at (3, 0).
6. (a) A postulate
Explanation: Eucid's fourth postulate states that all right angles are equal to one another.
7.
(c) 140o
Explanation: 140o
From the figure it follows that
(3x + 7) + (x + 5) + 40 = 180
⇒ 4x + 52 = 180

⇒ 4x = 180 - 52 = 128

⇒ x = 32

Now,
∠AOD = ∠C OP + ∠P OB

⇒ ∠AOD = (3x + 7) + (x + 5)

⇒ ∠AOD = 4x + 12

⇒ ∠AOD = 4 × 32 + 12

⇒ ∠AOD = 128 + 12

⇒ ∠AOD = 140

8.
(b) 80o
Explanation:
Given,
ABCD is a rectangle

Diagonals AC & BD intersect each other at P

∠ ABD = 50o
∵ diagonals of rectangle bisect each other and are equal in length
⇒ ∠ ABD = ∠ PDC [alternate angles]

⇒ ∠ PDC= ∠ PCD = 50°

In △DPC
⇒ ∠ DPC + ∠ PCD + ∠ PDC = 180o
⇒ ∠ DPC + 50o + 50o = 180o
⇒ ∠ DPC = 180o - 100o = 80o

9. (a) every real number

Explanation: Zero of the zero polynomial is any real number.
e.g., Let us consider zero polynomial be 0(x - k), where k is a real number.
For determining the zero, put x - k = 0 ⇒ x = k Hence, zero of the zero polynomial be any real number.
5x−7
10. (a) y = 2

Explanation: 5x - 2y = 7
- 2y = 7 - 5x
2y = 5x - 7
y= 5x−7

11.
(d) 50 ∘

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 Explanation: In Rhombus, digonals bisect each other right angle. By using angle sum property in any of the four triangles
formed by intersection of diagonals, we get ∠ CBD = 50 and ∠ CBD = ∠ ADC ( alternate angles).
So, ∠ ADC = 50
12.
(d) 135o
Explanation:
Given,

ABCD is a quadrilateral
∠A = 45o,
∵ diagonals of quadrilateral bisects each other hence ABCD is a parallelogram,

⇒ ∠ A + ∠ B = 180o
⇒ 45o + ∠ B = 180o
⇒ ∠ B = 180o - 45o = 135o

13.
(d) 70 o

Explanation:

In, △ABD
∠ D = 180o - ∠ A - ∠ B
=180o - 110o = 70o
Since angles made by same chord at any point of circumference are equal so, ∠AC B = ∠ADB = 70 0

14. (a) 26

45
¯
¯¯ 57−5
Explanation: 0.57 = 90

= 52

90
= 26

45

15.
(d) (3, 7)
Explanation: Let us put x = 3 in the give equation,
Then, y = 2(3) + 3
y=6+3=9
So, the point will be (3, 9)
For x = 3, y = 9. But in the given option, y = 7
So, the given point (3, 7) will not lie on the line y = 2x + 3.
16.
(b) SAS
Explanation: In △PQR and △PQS
PR = PS = 8 cm
∠ RPQ = ∠ SPQ (Given)

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 PQ = PQ (Common)
∴ △PQR ≅ △PQS (By SAS congruency)

17.
(b) Length of the rectangle
Explanation: In, Histogram each rectangle is drawn, where width equivalent to class interval and height equivalent to the
frequency of the class.
18.
(d) 2 : 1
Explanation: CSA of cone = CSA of cylinder
π rl = 2π rh

l = 2h
l:h=2:1
19.
(c) A is true but R is false.
a+b+c
Explanation: s = 2
3+4+5
s= 2
= 6 cm
−−−−−−−−−−−−−−−−−
Area = √s(s − a)(s − b)(s − c)
−−−−−−−−−−−−−−−−−−−
= √(6)(6 − 3)(6 − 4)(6 − 5)
= √(6)(3)(2)(1) = 6 cm2
−−−−−−−−−−

20. (a) Both A and R are true and R is the correct explanation of A.
Explanation: Putting (1, 1) in the given equation, we have
L.H.S = 1 + 1 = 2 = R.H.S
L.H.S = R.H.S
Hence (1, 1) satisfy the x + y = 2. So it is the solution of x + y = 2.
Section B
21. Let △ABC be an isosceles triangle and let AL ⊥ BC
1 2
∴ × BC × AL = 192cm
2
1 2
⇒ × 24cm × h = 192cm
2

192
⇒ h = ( ) cm = 16cm
12

Now, BL = 1

2
(BC ) = (
1

2
× 24) cm = 12 cm and AL = 16 cm.
In △ABL AB2 = BL2 + AL2
⇒ a2 = BL2 + AL2
−−−−−−−−− −−−−−−−−− − −−−−−− −−
2 2 2 2
∴ a = √BL + AL = √(12) + (16) cm = √144 + 256cm
−−−
⇒ a = √400cm = 20cm

Hence, perimeter = (20 + 20 + 24) cm = 64 cm.

22.

Join BC,
Then, ∠ ACB = 90° (Angle in the semicircle)
Since DCBE is a cyclic quadrilateral.
∠ BCD + ∠ BED = 180°

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 Adding ∠ ACB both the sides, we get
∠ BCD + ∠ BED + ∠ ACB = ∠ ACB +180°

(∠ BCD + ∠ ACB) + ∠ BED = 90° + 180°

∠ ACD + ∠ BED = 270°
23. Outer diameter = 10 cm
∴ Outer radius (R) = cm = 5 cm 10

As Inner diameter = 9 cm
∴ Inner radius (r) = cm
9

2
4 4
Volume of the metal contained in the shell = 3
πR
3
−
3
3
πr

4 3 3 4 22 3 9
π(R − r )= × × [(5) − ( )]
3 3 7 3
4 22 729
= × × (125 − )
3 7 8
4 22 271 2981 3
= × × = cm
3 7 8 21

24. In the given diagram join AB. Also ∠ ABD = 90° (because angle in a semicircle is always 90°)
Similarly, we have ∠ ABC = 90°
So, ∠ ABD + ∠ ABC = 90° + 90° = 180°
Therefore, DBC is a line i.e., B lies on the line segment DC.
OR
∠ODB = ∠OAC = 50
∘
[angles in the same segment]
OB = OD ⇒ ∠OBD = ∠ODB = 50 ∘
= x
∘

=> xo = 50o
25. For x = 2, y = 1
x+y+4=0
L.H.S. = x + y + 4
=2+1+4=7
≠ R.H.S

∴ x = 2, y = 1 is not a solution of x + y + 4 = 0.

OR
x-2y=4
– –
Put x = √2 , y = 4√2 in given equation, we get
– – – – –
√2 − 2(4√2) = √2 − 8√2 = −7√2

which is not 4.
– –
∴ (√2, 4√2) is not a solution of given equation.
Section C
1 1
26. Here a = 3
,b= 2
,n=3
1 1 3−2 1
−
b−a 2 3 6 6 1
∴ = = = =
n+1 3+1 4 4 24

∴ Three rational numbers between 1

3
and 1

2
are
1 1 1 1 1 1
+ , + 2( ), + 3( )
3 24 3 24 3 24

3
+
1

24
, 1

3
+
12
1
, 1

3
+
1

8
3 5 11
, ,
8 12 24
–
27. p(x) = 2x + kx + √2 2

We know that according to the factor theorem

p (a) = 0, if x − a is a factor of p (x)
–
We conclude that if (x − 1) isaf actorof p(x) = 2x 2
+ kx + √2 then p (1) = 0
2 –
p (1) = 2(1) + k (1) + √2 = 0, or
–
2 + k + √2 = 0
–
k = − (2 + √2) .
–
Therefore, we can conclude that the value of k is − (2 + √2)
28. Let ABC be an equilateral triangle, O be the interior point and OQ, OR and OC are the perpendicular drawn from points O. Let
the sides of an equilateral triangle be a m.

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 Area of ΔAOB = 1

2
× AB × OP

[∵ Area of a triangle = 1

2
× (base × height) ]
=
1

2
× a × 14 = 7a cm
2
…(1)
Area of ΔOBC = 1

2
× BC × OQ =
1

2
× a × 10

= 5a cm2 …(2)
Area of ΔOAC = 1

2
× AC × OR =
1

2
× a× 6

= 3a cm2…(3)
∴ Area of an equilateral ΔABC

= Area of (ΔOAB + ΔOBC + ΔOAC )

= (7a + 5a + 3a) cm2
= 15a cm2…(4)
a+a+a
We have, semi-perimeter s = 2
3a
⇒ s= cm
2
−−−−−−−−−−−−−−−−−
∴ Area of an equilateral ΔABC = √s(s − a)(s − b)(s − c) [By Heron’s formula]
−−−−−−−−−−−−−−−−−−−−−−−−−
3a 3a 3a 3a
= √ ( − a) ( − a) ( − a)
2 2 2 2

−−−−−−−−−−−−−
3a a a a
= √ × × ×
2 2 2 2

√3
=
4
a
2
…(5)
From equations (4) and (5), we get
√3
2
a = 15a
4
15×4 60
⇒ a= =
√3 √3

60 √3 –
⇒ a= = = 20√3cm
√3 √3

–
On putting a = 20√3 in equation (5), we get
√3 – 2 √3 –
Area of ΔABC = 4
(20√3) =
4
× 400 × 3 = 300√3cm
2

–
Hence, the area of an equilateral triangle is 300√3cm . 2

OR
The sides of triangular side walls of flyover which have been used for advertisements are 13 m, 14 m, 15 m.
13+14+15 42
s= = = 21m
2 2
−−−−−−−−−−−−−−−−−−−−−−− −
= √21(21 − 13)(21 − 14)(21 − 15)
−−−−−−−−−− −
= √21 × 8 × 7 × 6
−−−−−−−−−−−−−−−−−−−−−− −
= √7 × 3 × 2 × 2 × 2 × 7 × 3 × 2

2
= 7 × 3 × 2 × 2 = 84m

It is given that the advertisement yield an earning of Rs. 2,000 per m2 a year.
∴ Rent for 1 m2 for 1 year = Rs. 2000
So, rent for 1 m2 for 6 months or 1

2
year = Rs(
1

2
× 2000) = Rs. 1,000.
∴ Rent for 84 m2 for 6 months = Rs. (1000 × 84) = Rs. 84,000.
29. 3x + 2y = 6
Put y = 0, we get
3x + 2(0) = 6
⇒ 3x = 6
⇒x=
6
= 2
3

∴ (2, 0) is a solution.
3x + 2y = 6

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 put x = 0, we get
3(0) + 2y = 6
⇒ 2y = 6
⇒y=
6
= 3
2

∴ (0, 3) is a solution.
5x+ 2y = 10
Put y = 0, we get
5x + 2(0) = 10
⇒ 5x = 10
⇒x= 10

5
= 2

∴ (2, 0) is a solution.
5x + 2y = 10
Put x = 0, we get
5(0) + 2y = 10
⇒ 2y = 10
⇒y= 10

2
= 5

∴ (0, 5) is a solution.
The given equations have a common solution (2, 0).
30. Let ABCD be a rhombus and P, Q, R and S be the mid-points of sides AB, BC, CD and DA, respectively (Fig.). Join AC and BD.

From triangle ABD, we have SP = 1

2
BD and
SP || BD (Because S and P are mid-points)
Similarly RQ = BD and RQ || BD
1

Therefore, SP = RQ and SP || RQ
So, PQRS is a parallelogram ...(1)
Also, AC ⊥ BD (Diagonals of a rhombus are perpendicular)
Further PQ || AC (From △BAC)
As SP || BD, PQ || AC and AC ⊥ BD,
therefore, we have BD ⊥ PQ, i.e. ∠SPQ = 90º. ..(2)
Therefore, PQRS is a rectangle. [From (1) and (2)]
OR
In △ABP
D is mid points of AB and DE∥ BP
∴ E is midpoint of AP

∴ AE = EP

Therefore,AP=2AE
1
Also PC = 2
AP
2PC = AP
2PC = 2AE
⇒ PC = AE
∴ AE = PE = PC
∴ AC = AE + EP + PC

AC = AE + AE + AE
⇒ AE = AC 1

Hence Proved.
31. Given LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units.

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 i. Coordinate of point P = (3,2)
Coordinate of point Q = (3,-1)
Coordinate of point R = (3, 0) [since its lies on X-axis, so its y coordinate is zero].
ii. Abscissa of point L = 3, abscissa of point M=3
∴ Difference between the abscissa of the points L and M = 3 – 3 = 0

Section D
32. LHS
7+3√5 7−3√5
= −
3+ √5 3− √5

7+3√5 3− √5 7−3√5 3+ √5
= × − ×
3+ √5 3− √5 3− √5 3+ √5

7×3−7√5+3√5×3−3√5× √5 7×3+7√5−3√5×3−3√5× √5
= −
2 2 2 2
3 − √5 3 − √5

21−7√5+9√5−15 21+7√5−9√5−15
= −
9−5 9−5

6+2√5 6−2√5
= −
4 4
6+2√5−6+2√5
=
4
0+4√5
=
4
–
= 0 + √5

We know that,
7+3√5 7−3√5 –
− = a + b√5
3+ √5 3− √5
– –
0 + √5 = a + b√5

a = 0 and b = 1
OR
3− √5
p =
3+ √5

3− √5 3− √5
= ×
3+ √5 3− √5

2
(3− √5)
=
2 2
3 − √5

9+5−6√5
=
9−5

14−6√5
=
4

7−3√5
=
2

3+ √5
q =
3− √5

3+ √5 3+ √5
= ×
3− √5 3+ √5
2
(3+ √5)
=
2 2
3 − √5

9+5+6√5
=
9−5

14+6√5
=
4

7+3√5
=
2

p2 + q2
2 2
7−3√5 7+3√5
= ( ) + ( )
2 2

49+45−42√5 49+45+42√5
= +
4 4

94−42√5 94+42√5
= +
4 4

47−21√5 47+21√5
= +
2 2

47−21√5+47+21√5
=
2
94
=
2

= 47
33. Six points: A,B,C,D,E,F
¯
¯¯¯¯
¯¯¯ ¯
¯¯¯¯¯
¯ ¯
¯¯¯¯¯
¯ ¯
¯¯¯¯¯¯
¯ ¯¯¯¯¯¯¯
¯
Five line segments: EG, FH, EF, GH, MN
−→ −
−→ −
−→ −
−→
Four rays: EP , GR, GB, HD

Page 15 of 18
 ←→ ←→ ←→ ←→
Four lines: = AB , C D , P Q , RS
Four collinear points: M,E,G,B
34. To Prove: ∠ROS = (∠QOS − ∠P OS)
1

Given: OR is perpendicular to PQ, or ∠ QOR = 90°

From the given figure, we can conclude that ∠ POR and ∠ QOR form a linear pair.
We know that sum of the angles of a linear pair is 180°.
∴ ∠ POR + ∠ QOR = 180°

or ∠ POR = 90°
From the figure, we can conclude that
∠ POR = ∠ POS + ∠ ROS

⇒ ∠ POS + ∠ ROS = 90°

⇒ ∠ ROS = 90° - ∠ POS...(i)

Again,
∠ QOS + ∠ POS = 180°
⇒
1

2
(∠QOS + ∠P OS) = 90
∘
.(ii)
Substitute (ii) in (i), to get
1
∠ROS = (∠QOS + ∠P OS) − ∠P OS
2
1
= (∠QOS − ∠P OS) .
2

Therefore, the desired result is proved.

OR
Since corresponding angles are equal.
∴x = y ... (i)
We know that the interior angles on the same side of the transversal are supplementary.
∴ y + 55o = 180o
⇒ y = 180o - 55o = 125o
So, x = y = 125o
Since AB || CD and CD || EF.
∴ AB || EF

⇒ ∠ EAB + ∠ FEA = 180o [∵ Interior angles on the same side of the transversal EA are supplementary]
⇒ 90o + z + 55o = 180o
⇒ z = 35o
35. The given polynomial is,
f(x) = x4 + ax3 - 7x2 - 8x + b
Now, x + 2 = 0 ⇒ x = -2
By the factor theorem, we can say: f(x) will be exactly divisible by (x + 2) if f(-2) = 0
Therefore, we have:
f(-2) = [(-2)4 + a × (-2)3 - 7 × (-2)2 - 8 × (-2) + b]
= (16 - 8a - 28 + 16 + b)
= (4 - 8a + b)
∴ f(-2) = 0 ⇒ 8a - b = 4 ...(i)

Also, x + 3 = 0 ⇒ x = -3
By the factor theorem, we can say: f(x) will be exactly divisible by (x + 3) if f(-3) = 0
Therefore, we have:
f(-3) = [(-3)4 + a × (-3)3 - 7 × (-3)2 - 8 × (-3) + b]
= (81 - 27a - 63 + 24 + b)
= (42 - 27a + b)
∴ f(-3) = 0 ⇒ 27a - b = 42 ...(ii)

Subtracting (i) from (ii), we have:

⇒ 19a = 38

⇒ a = 2

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 Putting the value of a, we get the value of b, i.e., 12
∴ a = 2 and b = 12

Section E
36. i. Expenses in 2009-10 = 9160 Million
Expenses in 2010-11 = 10300 Million
Total expenses from 2009 to 2011
= 9160 + 10300
= 19460 Million
ii. Expenses in 2009-10 = 9160 Million
Expenses in 2010-11 = 10300 Million
Thus percentage of no of expenses in 2009-10 over the expenses in 2010-11
9160
= 10300
× 100
= 88.93%
iii. The minimum expenses (in 2007-08) = 5.4 Million
The maximum expenses (in 2010-11) = 10300 Million
Thus percentage of no of minimum expenses over the maximum expenses
= × 100
5.4

10300

= 0.052%
OR
The expenses in 2010-11 = 10300 Million
The expenses in 2006-09 = 9060 Million
The difference = 10300 - 9060 Million
= 1240 Million
37. i. Diameter of golf ball = 4.2 cm
Radius of golf ball, R = 2.1 cm
Radius of dimple, r = 2mm = 0.2 cm
Surface area of each dimple = 2πr 2

2× 22

7
× (0.2)2 = 0.08 cm2
ii. Diameter of golf ball = 4.2 cm
Radius of golf ball, R = 2.1 cm
Radius of dimple ,r = 2mm = 0.2cm
Volume of the material dug out to make one dimple
= Volume of 1 dimple
2 3
= πr
3

cm3
0.016π
=
3

iii. Diameter of golf ball = 4.2 cm

Radius of golf ball, R = 2.1 cm
Radius of dimple, r = 2mm = 0.2cm
The total surface area exposed to the surroundings
= surface area of golf ball − surface area of 315 dimples
= 4π R2 - 315 × 0.08π
= 70.56π - 25.2π cm2
= 45.36π cm2
OR
Diameter of golf ball = 4.2 cm
Radius of golf ball, R = 2.1 cm
Radius of dimple, r = 2mm = 0.2cm
volume of the golf ball = volume of sphere − volume of 315 dimples
= 4

3
π R3 - 315 × 2

3
π r3
= 4

3
π (74.088 - 10.08)
= 97.344 π cm3

Page 17 of 18
38. i. △ADE and △CFE
DE = EF (By construction)
∠ AED = ∠ CEF (Vertically opposite angles)

AE = EC(By construction)
By SAS criteria △ADE ≅△CFE
ii. △ADE ≅ △CFE
Corresponding part of congruent triangle are equal
∠ EFC = ∠ EDA

alternate interior angles are equal

⇒ AD ∥ FC

⇒ CF ∥ AB

iii. △ADE ≅ △CFE

Corresponding part of congruent triangle are equal.
CF = AD
We know that D is mid point AB
⇒ AD = BD

⇒ CF = BD

OR
BC
DE = 2
{line drawn from mid points of 2 sides of △ is parallel and half of third side}
DE ∥ BC and DF ∥ BC
DF = DE + EF
⇒ DF = 2DE(BE = EF)

⇒ DF = BC

Page 18 of 18