Model Question paper-I

SECTION-A

1.C 2.D 3.C 4. C 5. A

6.D 7.B 8.C 9.B 10.

11.B 12.B 13.B 14.D 15.A

SECTION-B

16. 𝑃 = 𝑎, 𝑏, 𝑐 , 𝑄 = 𝑔, 𝑕, 𝑥, 𝑦 , 𝑅 = {𝑎, 𝑒, 𝑓, 𝑠}

𝑃 ∩ 𝑄 = { }

𝑅\ 𝑃 ∩ 𝑄 = {𝑎, 𝑒, 𝑓, 𝑠}

17. In given arrow diagram , every element has

a unique image. Hence it is a function.

18. 𝑎 = 16; 𝑟 =−48

16 = −3 ≠ 1

𝑆𝑛 =𝑎 1 − 𝑟𝑛

1 − 𝑟, 𝑟 ≠ 1

𝑆25 =16(1 − (−3)25)

1 − −3

= 4 1 + 325

19. Let P(x) be the required rational expression

Given that

𝑥3 − 1

𝑥2 + 2+ 𝑃 𝑥 =

2𝑥3 − 𝑥2 + 3

𝑥2 + 2

∴ 𝑃 𝑥 =2𝑥3 − 𝑥2 + 3

𝑥2 + 2−

𝑥3 − 1

𝑥2 + 2

=2𝑥3 − 𝑥2 + 3 − 𝑥3 + 1

𝑥2 + 2=

𝑥3 − 𝑥2 + 4

𝑥2 + 2

20. Sum of the roots =4+ 7

2+

4− 7

2

=4+ 7+4− 7

2 =

8

2= 4

Product of the roots= 4+ 7

2 ×

4− 7

2

=42− 7

2

4 =

16−7

4=

9

4

The required equation is

𝑥2 − 𝑆𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 𝑥 + 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡 = 0

𝑥2 − 4𝑥 +9

4= 0

4𝑥2 − 16𝑥 + 9

4= 0

4𝑥2 − 16𝑥 + 9 = 0

21. In general a 2 × 3 matrix is given by

𝐴 = 𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

Now 𝑎𝑖𝑗 = 2𝑖 − 3𝑗 . where 𝑖 = 1,2 and j=1, 2, 3

𝑎11 = 2 1 − 3 1 = −1 = 1

𝑎12 = 2 1 − 3 2 = −4 = 4

𝑎13 = 2 1 − 3 3 = −7 = 7

𝑎21 = 2 2 − 3 1 = 1

𝑎22 = 2 2 − 3 2 = 2

𝑎23 = 2 2 − 3 2 = 5

The required matrix = 1 4 71 2 5

22. 𝐴 = 4 −25 −9

, 𝐵 = 8 2−1 −3

6𝐴 = 6 × 4 −25 −9

= 24 −1230 −54

3𝐵 = 3 × 8 2−1 −3

= 24 6−3 −9

6𝐴 − 3𝐵 = 24 −1230 −54

23. If (7,3), 6,1 , (8,2) and (P,4) are the vertices of

a parallelogram taken in order, then find the value of P

We know that the diagonals of a parallelogram

bisect each other.

∴ midpoint of AC = midpoint of BD

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7 + 8

2,3 + 2

2 =

6 + 𝑃

2,1 + 4

2

⟹ 15

2,5

2 =

6 + 𝑃

2,5

2

Equating the X- co ordinates , We get

6 + 𝑃

2=

15

2

𝑃 = 15 − 6 =9

24. 𝐼𝑛 ∆ 𝐴𝐵𝐶 AD is the internal bisector of ∠𝐴.

𝐵𝐷

𝐷𝐶=

𝐴𝐵

𝐴𝐶 (Angle bisector theorem)

𝐷𝐶 = 𝐵𝐷×𝐴𝐶

𝐴𝐵

𝐷𝐶 = 2.5×4.2

5= 2.1 𝑐𝑚

𝐷𝐶 = 2.1 𝑐𝑚

25. Let GD be the girl’s height

∴ 𝐺𝐷 = 150 𝑐𝑚

Let GC be the length of the shadow

∴ 𝐺𝐶 = 150 3 𝑐𝑚

In ∆𝐶𝐺𝐷, 𝑡𝑎𝑛𝜃 =𝐷𝐺

𝐺𝐶=

150

150 3

tan 𝜃 =1

3

∴ 𝜃 = 30°

Angle of elevation of the top of the lamp post 𝜃 =

30°

26. 1−𝑠𝑖𝑛𝜃

1+𝑠𝑖𝑛𝜃

= 1 − 𝑠𝑖𝑛𝜃

1 + 𝑠𝑖𝑛𝜃×

1 − 𝑠𝑖𝑛𝜃

1 − 𝑠𝑖𝑛𝜃 =

(1 − 𝑠𝑖𝑛𝜃)2

12 − 𝑠𝑖𝑛2𝜃

= (1 − 𝑠𝑖𝑛𝜃)2

𝑐𝑜𝑠2𝜃

=1 − 𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃=

1

𝑐𝑜𝑠𝜃−

sin𝜃

cos𝜃= sec 𝜃 − tan 𝜃

27. Ratio of radii of two cylinders 𝑟1: 𝑟2 = 2: 3

∴ 𝑟1 = 2, 𝑟2 = 3

Ratio of height of two cylinders 𝑕1: 𝑕2 = 5: 3

∴ 𝑕1 = 5, 𝑕2 = 3

Ratio of their volumes 𝑉1:𝑉2 = 𝜋𝑟1

2𝑕1

𝜋𝑟22𝑕2

=22×5

32×5=

4×5

9×3=

20

27

Ratio of the volumes of cylinders 𝑉1: 𝑉2 = 20: 27

28. Smallest value S = 12

Range R = 59

Largest value L=?

We know that, 𝑅 = 𝐿 − 𝑆

∴ 𝐿 = 𝑅 + 𝑆

= 59 + 12 = 71

Largest Value =71

29. Total number of tickets available in bag =100

∴ 𝑛 𝑆 = 100

Let A be the event of getting a ticket with a number

divisible by 10.

∴ 𝐴 = 10,20,30,40,50,60,70,80,90,100

∴ 𝑛 𝐴 = 10

𝑃 𝐴 =𝑛(𝐴)

𝑛(𝑆)=

10

100=

1

10

30. a) Given 10

𝑦+

9

𝑥= −

8

𝑥𝑦

Multiply both sides by −8

𝑥𝑦

10

𝑦×

𝑥𝑦

−8+

9

𝑥×

𝑥𝑦

−8=

−8

𝑥𝑦×

𝑥𝑦

−8

10𝑥

−8+

9𝑦

−8= 1

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𝑥

−810

+𝑦

−89

= 1

𝑥 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑝𝑡𝑠 (𝑎) =−8

10

𝑦 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑝𝑡𝑠 𝑏 =−8

9

(OR)

Total surface area of cylinder = 3×CSA of cylinder

2𝜋𝑟 𝑕 + 𝑟 = 3 × 2𝜋𝑟𝑕

𝑕 + 𝑟 = 3𝑕

𝑕 − 3𝑕 = −𝑟

−2𝑕 = −𝑟

𝑕 =𝑟

2

SECTION-C

31. 𝐴\ 𝐵 ∪ 𝐶 = 𝐴\𝐵 ∩ (𝐴\𝐶)

From (I) and (II) 𝐴\ 𝐵 ∪ 𝐶 = 𝐴\𝐵 ∩ 𝐴\𝐶 is true.

32. f: A B defined by 𝑓 𝑥 = 2𝑥 + 1

Given: A= 0,1,2,3 B={1,3,5,7,9}

𝑓 0 = 2 0 + 1 = 1 similarly

𝑓 1 = 3; 𝑓 2 = 5 ; 𝑓 3 = 7

(i) A set of ordered pairs

The given function f can be represented as a set of

ordered pairs as 𝑓 = 0,1 , 1,3 , 2,5 , 3,7

(ii) Table form:

Let us represent 𝑓 using a table as shown below.

x 0 1 2 3

f(x) 1 3 5 7

(iii) Arrow Diagram:

Let us represent 𝑓 by an arrow diagram. We draw two

closed curves to represent the sets A and B

Then each element of A and its unique image element

in B are related with an arrow.

(iv) Graph:

𝑓 = {(𝑥, 𝑓 𝑥 )/𝑥 𝜖𝐴} ={ 0, 1 , 1, 3 , 2, 5 , 3, 7 }

Now, the points (0, 1), (1, 3), (2, 5) and (3, 7) areplotted

on the plane as shown below. The totality of all points

represent the graph of the function.

33. 𝑆𝑛 = 7 + 77 + 777 +……….to n terms.

= 7(1 + 11 + 111 +……. to n terms)

=7

9 9 + 99 + 999 + …… to n terms]

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=7

9[ 10 − 1 + 100 − 1 + (1000 − 1)+… to n terms

=7

9[10 + 102 + 103 +…. to n terms – (1 + 1 +…..+to

n terms)]

=7

9 10(10𝑛 − 1)

10 − 1− 𝑛

=70

81 10(10𝑛 − 1) −

7𝑛

9

34. 34. 𝑥3 − 2𝑥2 − 5𝑥 + 6

1 1 -2 -5 6

0 1 -1 -6

1 -1 -6 0

∴ 𝑥 − 1 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟.

The other factor is 𝑥2 − 𝑥 − 6 = 𝑥 − 3 𝑥 + 2

i.e., 𝑥3 − 2𝑥2 − 5𝑥 + 6 = 𝑥 − 1 𝑥 − 3 𝑥 + 2

35. 3𝑥2 − 𝑥 + 1

3𝑥2 9𝑥4 − 6𝑥3 + 7𝑥2 − 2𝑥 + 1

9𝑥4

6𝑥2 − 𝑥 -6𝑥3 + 7𝑥2

-6𝑥3 + 𝑥2

6𝑥2 − 2𝑥 + 1 6𝑥2 − 2𝑥 + 1

6 𝑥2 − 2𝑥 + 1

0

9𝑥4 − 6𝑥3 + 7𝑥2 − 2𝑥 + 1 = |3𝑥2 − 𝑥 + 1|

36. 𝐺𝑖𝑣𝑒𝑛:

𝐴 = −245

𝐵 = 1 3 −6

𝐴𝐵 = −245

1 3 −6 = −2 −6 124 12 −245 15 −30

(𝐴𝐵)𝑇 = −2 4 5−6 12 1512 −24 −30

--------------( I )

𝐵𝑇𝐴𝑇 = 13−6

−2 4 5

= −2 4 5−6 12 1512 −24 −30

--------------( 2 )

Thus (1) and (2) We get (AB)𝑇 = 𝐵𝑇𝐴𝑇

37. Area of the quadrilateral ABCD

Area of a quadrilateral

= 1

2 𝑥1𝑦2 + 𝑥2𝑦3 + 𝑥3𝑦4 + 𝑥4𝑦1 −

(𝑥2𝑦1 + 𝑥3𝑦2 + 𝑥4𝑦3 + 𝑥1𝑦4)

=1

2{

3 4 77 2 4

6 9

3 7

}

= {6 + 16 + 63 + 42) - (28+14+24+27) }

= 1

2{127 − 93}=

1

2× 34 = 17 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠

38. Let AD, BE , CF be the altitudes of ∆𝐴𝐵𝐶

Slope of BC= 𝑦2−𝑦1

𝑥2−𝑥1=

1−5

0+4= −1

Since the altitude AD is perpendicular to BC

∴ 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐷 = 1 [ ∵ 𝑚1𝑚2 = −1]

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Slope of AC =1−2

0−1= 1

Thus, slope of BE = −1 [∵ 𝐵𝐸 ⊥ 𝐴𝐶]

Also Slope of AB =5−2

−4−1=

−3

5

Slope of 𝐶𝐹 = 5

3 [∵ 𝐶𝐹 ⊥ 𝐴𝐵]

39. Let ABCD be a parallelogram such that its sides

touch a circle with centre O.

We know that the length of tangents from an

external point are of equal length.

∴ 𝐴𝑃 = 𝐴𝑆, 𝐵𝑃 = 𝐵𝑄, 𝐶𝑅 = 𝐶𝑄, 𝐷𝑅 = 𝐷𝑆

𝐴𝑃 + 𝐵𝑃 + 𝐶𝑅 + 𝐷𝑅 = 𝐴𝑆 + 𝐵𝑄 + 𝐶𝑄 + 𝐷𝑆

𝐴𝑃 + 𝐵𝑃 + 𝐶𝑅 + 𝐷𝑅

= 𝐴𝑆 + 𝐷𝑆 + (𝐵𝑄 + 𝐶𝑄)

𝐴𝐵 + 𝐶𝐷 = 𝐴𝐷 + 𝐵𝐶

2𝐴𝐵 = 2𝐵𝐶 [∵ 𝐴𝐵𝐶𝐷 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚]

𝐴𝐵 = 𝐵𝐶 [∵ 𝐴𝐵 = 𝐶𝐷 𝑎𝑛𝑑 𝐵𝐶 = 𝐴𝐷]

∴ 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐴𝐷.

Hence ABCD is a Rhombus.

40. Refer your text books (example No: 7.18, Page:209)

41. Radius of spherical solid material R = 18 cm

Let 𝑟1 , 𝑟2 , 𝑟3 be the radii of small spheres formed

𝑟1 = 2𝑐𝑚 𝑟2 = 12 𝑐𝑚 𝑟3 =?

Volume of the solid sphere = volume of 3 small spheres

4

3πR3 =

4

3πr1

3 +4

3πr2

3 +4

3πr3

3

4

3𝜋(18)3 =

4

3𝜋[(2)3 + (12)3 + (r3)3]

r33 = (18)3 − [8 + 1728]

= 5832 − 1736 = 4096

r33 = 163

𝑟 = 16 𝑐𝑚

raidus of the third sphere = 16 cm

42. Hemispherical portion :

Radius, r = 3.5 cm

Conical portion :

Radius, r = 3.5 cm

Height, h =17.5 – 3.5=14 cm

Volume of the wood = Volume of the hemisphere +

Volume of the cone

=2

3𝜋𝑟3 +

1

3𝜋𝑟2𝑕

=𝜋𝑟2

3 2𝑟 + 𝑕

=22

7×

3.5×.35

3 2 × 3.5 + 14

= 269.5

volume of the wood used in the toy = 269.5 cu.cm.

43. Given that 𝑥 = 35 and n = 5.

∴ 𝑥 = 𝑥

𝑛=

35

5= 7 , 1 = 5

Let us find 𝑥2

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(𝑥 − 9)2 = 82

(𝑥2 − 18𝑥 + 81) = 82

𝑥2 − 18 𝑥 + 81 1 = 82

𝑥2 − 18 × 35 + 81 × 5 = 82

𝑥2 − 630 + 405 = 82

𝑥2 = 307

To find (𝑥 − 𝑥 )2, Let us consider

(𝑥 − 9)2 = 82

(𝑥 − 7 − 2)2 = 82

[(𝑥 − 7) − 2)2] = 82

(𝑥 − 𝑥 )2 − 4 0 + (4 × 5)=8 [∴ (𝑥− 𝑥 ) = 0

(𝑥 − 𝑥 )2 = 62

𝑥2 = 307 and (𝑥 − 𝑥 )2 = 62

44. Let A denotes the event that A solves the given

problem

Let B denotes the event that B solves the given problem

Let C denote the event that C solves the given problem

𝑃 𝐴 =4

5; 𝑃 𝐴 ∩ 𝐵 =

8

15

𝑃 𝐵 =2

3 ; 𝑃 𝐵 ∩ 𝐶 =

2

7

𝑃 𝐶 =3

7 ; 𝑃 𝐶 ∩ 𝐴 =

12

35

𝑃 𝐴 ∩ 𝐵 ∩ 𝐶 =8

35

𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) denotes the event that the problem is

solved by atleast one of them.

To find 𝑃 𝐴 ∪ 𝐵 ∪ 𝐶 :

𝑃 𝐴 ∪ 𝐵 ∪ 𝐶 = 𝑃 𝐴 + 𝑃 𝐵 + 𝑃 𝐶 − 𝑃 𝐴 ∩ 𝐵 − 𝑃 𝐵 ∩ 𝐶

− 𝑃 𝐴 ∩ 𝐶 + 𝑃 𝐴 ∩ 𝐵 ∩ 𝐶

=4

2+

2

3+

3

7−

8

15−

2

7−

12

35+

8

35

𝑃 𝐴 ∪ 𝐵 ∪ 𝐶 =101

105

45. a) All the natural numbers between 400 and 600

which are divisible by 11

407,418,………594

that is 𝑆𝑛 = 407 + 418 +……..+594

𝑎 = 407, 𝑑 = 11, 𝑙 = 594

𝑛 =𝑙 − 𝑎

𝑑+ 1

=594 − 407

11+ 1

=187

11+ 1 = 18

𝑛 = 18

Hence 𝑆𝑛 =𝑛

2 𝑎 + 𝑙

𝑆18 =18

2 407 + 594 = 9 × 1001

𝑆18 = 9009

45. b L.H.S= 1−𝑥4

1−𝑥+

1−𝑥3

1+𝑥

= 14 − 𝑥4 1 + 𝑥 + 1 − 𝑥3 1 − 𝑥

1 − 𝑥 1 + 𝑥

=((12)2 − (𝑥2)2) 1 + 𝑥 + (1 − 𝑥3)(1 − 𝑥)

1 − 𝑥 1 + 𝑥

= 1 − 𝑥2 1 + 𝑥2 1 + 𝑥 + 1 − 𝑥3 1 − 𝑥

1 − 𝑥 1 + 𝑥

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=(1 + 𝑥)2 1 − 𝑥 1 + 𝑥2 + 1 − 𝑥3 1 − 𝑥

1 − 𝑥 1 + 𝑥

= (1 − 𝑥) (1 + 𝑥)2 1 + 𝑥2 + 1 − 𝑥3

1 − 𝑥 1 + 𝑥

= 𝑥2 + 2𝑥 + 1 1 + 𝑥2 + 1 − 𝑥3

1 + 𝑥

=𝑥2 + 2𝑥 + 1 + 𝑥4 + 2𝑥3 + 𝑥2 + 1 − 𝑥3

1 + 𝑥

=2𝑥2 + 2𝑥 + 2 + 𝑥3 + 𝑥4

1 + 𝑥

=2 𝑥2 + 𝑥 + 1 + 𝑥3 + 𝑥4

1 + 𝑥

= 𝑅. 𝐻. 𝑆

L.H.S=R.H.S

Section-D

46. (a) Refer your text book (example no 9.2)

Construction

(i) With O as the centre, draw a circle of radius 3.2 cm. (ii) Take a point P on the circle. (iii) Through P, draw any chord PQ.

(iv) Mark a point R distinct from P and Q on the circle

so that P, Q and R are in counter clockwise direction.

46. (b)

47. (a)

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47.(b)

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