Test No. 2

04-10-2020

A, B & C

Olympiads & Class IX-2021for

Test-2 (Answers) All India Aakash Test Series-2021 (Class IX)

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All India Aakash Test Series-2021 (Class IX) TEST - 2

Test Date : 04-10-2020 ANSWERS SECTION-I (Code-A)

1. (1) 2. (2) 3. (2) 4. (2) 5. (4) 6. (2) 7. (3) 8. (4) 9. (3) 10. (3) 11. (1) 12. (4) 13. (2) 14. (1) 15. (2) 16. (3) 17. (2) 18. (2) 19. (4) 20. (2)

21. (4) 22. (2) 23. (2) 24. (3) 25. (2) 26. (3) 27. (4) 28. (2) 29. (2) 30. (4) 31. (1) 32. (3) 33. (1) 34. (4) 35. (2) 36. (3) 37. (4) 38. (2) 39. (2) 40. (4)

41. (3) 42. (2) 43. (4) 44. (2) 45. (3) 46. (3) 47. (1) 48. (3) 49. (4) 50. (1) 51. (1) 52. (3) 53. (2) 54. (4) 55. (3) 56. (2) 57. (2) 58. (4) 59. (4) 60. (2)

61. (4) 62. (3) 63. (4) 64. (3) 65. (4) 66. (3) 67. (1) 68. (2) 69. (2) 70. (2) 71. (3) 72. (1) 73. (2) 74. (3) 75. (4) 76. (3) 77. (4) 78. (2) 79. (2) 80. (2)

81. (1) 82. (4) 83. (2) 84. (3) 85. (2) 86. (4) 87. (3) 88. (4) 89. (1) 90. (2) 91. (1) 92. (3) 93. (2) 94. (1) 95. (4) 96. (2) 97. (1) 98. (3) 99. (2) 100. (1)

SECTION-II (Code-B)

1. (4) 2. (3) 3. (2) 4. (1) 5. (2) 6. (3)

7. (3) 8. (4) 9. (3) 10. (3) 11. (2) 12. (4)

13. (1) 14. (2) 15. (3) 16. (2) 17. (1) 18. (2)

19. (3) 20. (2) 21. (4) 22. (3) 23. (2) 24. (4)

25. (2) 26. (3) 27. (2) 28. (4) 29. (3) 30. (1)

SECTION-III (Code-C)

1. (2) 2. (3) 3. (1)

4. (3) 5. (3) 6. (3)

7. (1) 8. (2) 9. (1)

10. (1) 11. (1) 12. (2)

13. (1) 14. (3) 15. (2)

All India Aakash Test Series-2021 (Class IX) Test-2 (Answers & Hints)

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Answers & Hints SECTION-I (Code-A)

1. Answer (1)

svt

=

35 10

12 60v ×=

×

125 m/s18

v =

2. Answer (2) Displacement = 2R = 14 m 3. Answer (2) 4. Answer (2) 5. Answer (4) 6. Answer (2) 7. Answer (3) 8. Answer (4) 9. Answer (3) 10. Answer (3) 11. Answer (1) Net displacement

( )1 1 15 10 5 2 5 2 10 2 52 2 2

= × × + + × + × + × ×

3525 20 52

= + + +

135 m2

=

Average velocity 135 2.25 m/s2 30

= =×

12. Answer (4) 13. Answer (2) Time taken by ball to return to the juggler

2 40 8 s10×

= =

So, he is throwing the ball after each 2 s. Let at some instant he is throwing ball number 4. Before 2 s of it he throws ball-3, so height of

ball-3:

( )23140 2 10 2 60 m2

h = × − × × =

Before 4 s he throws ball-2, so height of ball-2

( )22140 4 10 4 80 m2

h = × − × × =

For ball-1

( )21140 6 10 6 60 m2

h = × − × × =

14. Answer (1)

avg

3 3 3 3 5

xvx x xv v v

=+ +

× ×

avg90

1 1 1 463 9 15

v vv = =+ +

avg4523

vv =

15. Answer (2) Let a be the acceleration of system from equation of motion 2g – T = 2a ...(i) T– 3gsinθ = 3a ...(ii) From equation (i) and (ii) 2g – 3gsin30° = 5a

220 15 1 m/s5

a −= =

16. Answer (3) Rate of flow of water

3 6 310 cm / s 10 10 m /svt

−= = = ×

310 kgρ =

Area of pipe ( )2222 107

−= × = 4 222 10 m7

−×

Force mdv mv V v v vdt t t t At

ρ ρ= = = = ×

2v

t Aρ =

( )326

4

1010 10 22 107

−

−= × ×

×

37 10 N22

−= ×

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Test-2 (Answers & Hints) All India Aakash Test Series-2021 (Class IX)

17. Answer (2)

1 1d g dg g gR n R

′ = − = = −

11dR n

= −

⇒ 1nd Rn− = ×

18. Answer (2) 19. Answer (4)

∴ m vft

∆=

∆

m∆v = f∆t

( )1 14 8 2 2 2 22 2

m v∆ = × × − × − ×

m∆v = 16 – 2 – 4 = 10

10 5 m/s2

v = =

20. Answer (2)

∴ 2GMgR

=

3

2

4433

G Rg GR

R

× π × ρ= = π ρ

34

gGR

ρ =ρ

21. Answer (4) 22. Answer (2) 23. Answer (2) 24. Answer (3) 25. Answer (2) 26. Answer (3) 27. Answer (4) 28. Answer (2) 29. Answer (2) 30. Answer (4) 31. Answer (1)

Elements → copper, mercury, oxygen

Homogeneous mixture → Sugar in water, water in alcohol

32. Answer (3) 33. Answer (1) 34. Answer (4) 35. Answer (2) More is the melting point, more is the strength of

force of attraction between their constituent particles and hence more is the rigidity

36. Answer (3) 37. Answer (4) 38. Answer (2)

Concentration of solution ( )Mass of solute sugar

100Mass of solution

= ×

Mass of sugar30 100250

= ×

Mass of sugar = 75 g Let x g of sugar should be added

75 x50 100250 x

+ = × +

50 75 x100 250 x

+=

+

12500 + 50x = 7500 + 100x 5000 = 50x x = 100 g

39. Answer (2) 40. Answer (4)

41. Answer (3) Penicillium is a fungus which is multicellular. 42. Answer (2) 43. Answer (4) 44. Answer (2) 45. Answer (3)

All India Aakash Test Series-2021 (Class IX) Test-2 (Answers & Hints)

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46. Answer (3) 47. Answer (1) Ribosomes do not have membrane. 48. Answer (3) 49. Answer (4) 50. Answer (1) 51. Answer (1) 52. Answer (3) 53. Answer (2) 54. Answer (4) 55. Answer (3) 56. Answer (2) The given figure is of cuboidal epithelium. 57. Answer (2) 58. Answer (4) 59. Answer (4) 60. Answer (2) 61. Answer (4)

x2 + y2 + z2 ≠ 0

If x6 + y6 + z6 = 3x2y2z2

⇒ x2 = y2 = z2 62 Answer (3)

Area of triangle = 1 5 72

× ×

= 35 17.5 sq. units2

=

63. Answer (4)

x y y z z xxy yz zxa a a− − −

× ×

= ( ) ( ) ( )z x y x y z y z x

xyza− + − + −

= zx zy xy xz yz yx

xyza− + − + −

= a0 = 1 64. Answer (3)

AD is extended to E such that AD = DE ⇒ ABEC is a parallelogram ⇒ BE = AC So, In ∆ABE, AB + BE > AE ⇒ AB + AC > 2AD

⇒ 2AB ACAD+

>

65. Answer (4) D is mid-point of BC

⇒ 4 1 1 6 5 7, ,

2 2 2 2D + + = =

66. Answer (3)

Let ∠BAP = ∠DAP = θ ⇒ ∠APB = ∠DAP = θ [Alternate interior angles] ⇒ AB = BP = CP = CD

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Test-2 (Answers & Hints) All India Aakash Test Series-2021 (Class IX)

∠DCP = ∠DAB = 2θ

⇒ ∠CDP = ∠CPD = 90° – θ

⇒ ∠APD = 180° – θ – (90° – θ) = 90°

⇒ ∠ADP = 90° – θ

∴ ∠ADP = ∠CDP 67. Answer (1) 68. Answer (2)

22

1 44 4 2x xxx

+ − − + +

= 22

1 44 2 4x xxx

+ + + − −

= 21 2x

x + −

69. Answer (2)

AD = BD

In ∆CAD and ∆DBO,

∠CAD = ∠DBO [ ∠AOE = ∠BOD and ∠AEO

= ∠ADC = BDO = 90°]

and ∠ACD = ∠BOD

and OB = AC

∴ ∆CAD ≅ ∆OBD [By AAS]

⇒ AD = BD [By CPCT]

∴ AB = 10 2 cm

70. Answer (2)

x = 2 (180° – x)

⇒ 3x = 2 × 180°

⇒ x = 120°

∴ Its linear pair = 60°

71. Answer (3)

72. Answer (1) ∠AOD = ∠BOC = 40° ∴ ∠DOE = 90° – 40° = 50° 73. Answer (2) 74. Answer (3) Let the angles be 4x, 2x and 3x, ⇒ 4x + 2x + 3x = 9x = 180° ⇒ x = 20° ∴ Angles are 80°, 40° and 60° ∴ 80° – 40° = 40° 75. Answer (4)

p(x) = ax3 – 7x + b p(1) = a – 7 + b = 0 and p(2) = 8a – 14 + b = 0 ⇒ a + b = 7 …(i) And 8a + b = 14 …(ii) Solving (i) and (ii), we get a = 1 and b = 6 and –b = αβγ ⇒ –6 = 2γ ∴ γ = –3 76. Answer (3)

Let a = 4.23

⇒ 100a = 423.23

⇒ 99a = 423.23 4.23− = 419

⇒ a = 41999

⇒ p = 419, q = 99

All India Aakash Test Series-2021 (Class IX) Test-2 (Answers & Hints)

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Similarly,

670.6799

=

⇒ x = 67 and y = 99 77. Answer (4)

Let ∠BAD = ∠DAE = ∠CAE = θ ⇒ ∠ADE = θ + ∠B and ∠AED = θ + ∠C

⇒ ∠ADE > ∠AED [ ∠B > ∠C]

∴ AE > AD 78. Answer (2) Let C = (x, y)

⇒ ( ) 5 61, 1 ,2 2

x yO + + − =

∴ x = 2 – 5 = –3 and y = –2 – 6 = –8 ∴ C = (–3, –8) 79. Answer (2)

A is mid-point of DG [ E is mid-point of

CD and AE||CG]

⇒ F is mid-point of CG [ A is mid-point of DG and AF||CD]

∴ AD = AG and GF = CF 80. Answer (2) a + b = negative and a – b = positive

81. Answer (1)

( )

11302

1130 7 342

210 187

23

H M°

°

°

°

θ = −

θ = × − ×

θ = −

θ =

so reflex angle is 337°. 82. Answer (4) 83. Answer (2) 84. Answer (3) 85. Answer (2) 86. Answer (4) 87. Answer (3) 88. Answer (4) 89. Answer (1) 90. Answer (2) 91. Answer (1) 92. Answer (3) The code of letter Q is 34. 93. Answer (2) 94. Answer (1) 95. Answer (4) 96. Answer (2) Sohan is brother of Shashi, who is wife of Rakesh. 97. Answer (1) Sunita is mother of Shashi, who is mother of

Chinku. 98. Answer (3) 99. Answer (2) 100. Answer (1)

SECTION-II (Code-B) 1. Answer (4)

2. Answer (3)

3. Answer (2)

4. Answer (1)

5. Answer (2)

6. Answer (3)

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Test-2 (Answers & Hints) All India Aakash Test Series-2021 (Class IX)

7. Answer (3) 8. Answer (4) 9. Answer (3) 10. Answer (3) 11. Answer (2) 12. Answer (4) 13. Answer (1) 14. Answer (2) The given figure is of aerenchyma tissue, which is

found in aquatic plants. 15. Answer (3) 16. Answer (2) 17. Answer (1) 18. Answer (2) 19. Answer (3) 20. Answer (2) The given figure is of basophil. 21. Answer (4) 22. Answer (3) 23. Answer (2) 24. Answer (4) 25. Answer (2) 26. Answer (3) Krebs cycle occurs in the matrix of mitochondria. 27. Answer (2) 28. Answer (4) 29. Answer (3) 30. Answer (1) The given characteristics belong to phylum

Echinodermata.

SECTION-III (Code-C) 1. Answer (2)

( )( )

( )2 2 6 2 3

2 3 2 33 2 3

+ ++ + + −

+

= ( )( )( )

( )( )2 6 8 2 12 2 3

2 3 2 33 2 3 2 3

+ + −+ + + −

+ −

= ( )( )2 6 2 3 6 2 2 6 2

2 3 2 33

+ − + + ×+ + + −

= ( ) ( )22 2 2 6 6 3 2 6 2

2 3 2 33

+ − − ++ + + −

= ( )( )6 2 6 2 4 2 3 4 2 3

3 2

− + + + −+

= ( ) ( )2 2

3 1 3 16 23 2

+ + −−+

= ( )4 2 3 1 3 13 2

+ + + −

= 4 63

+

⇒ 4 , 13

a b= =

∴ 13

a b− =

2. Answer (3)

∆ABC is a right triangle.

⇒ Hypotenuse AC is diameter of its circum circle

∴ Its circum radius = ( ) ( )2 21 4 2 3 22

+ + +

= 1 36 252

+

= 61 units2

3. Answer (1)

Let p(x) = (x – 51) (x – 31) + ax + b

⇒ p(51) = 51a + b = 31 …(i)

and p(31) = 31a + b = 51 …(ii)

Solving (i) and (ii), we get

a = –1 and b = 82

∴ Polynomial p(x) = x2 –83x + 1663

All India Aakash Test Series-2021 (Class IX) Test-2 (Answers & Hints)

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4. Answer (3)

∆ADC ≅ ∆BDC [By SSS congruency]

⇒ ∠ACD = ∠BCD = 302

ACB∠= °

Now,

In ∆BDE and ∆BDC,

BE = BC [ BE = AB and AB = BC]

BD = BD [Common]

and ∠DBE = ∠DBC

∴ ∆BDE ≅ ∆BDC [By SAS congruency]

⇒ ∠BED = ∠BCD = 30°

5. Answer (3)

4(a + 1) + 3 (a – 1) = 1

⇒ 4a + 4 + 3a – 3 = 1

⇒ a = 0

and 2(b + 2) – 5 (b – 2) = 6

⇒ 2b + 4 – 5b + 10 = 6

⇒ 83

b =

∴ ba

is undefined

6. Answer (3)

Here,

GBY CDY∆ ≅ ∆ [By ASA]

⇒ Y is mid-point of CG.

⇒ ( )12

XY AB CD= −

and ( )12

EF AB CD= +

⇒ ( )2 214

EF XY AB CD⋅ = −

7. Answer (1)

Let, ( ) ( )3 15 17 255a b c k= = =

⇒ 3 2 1

15 ,17 and 255a b ck k k= = =

⇒ 1 3 2c a bk k k= × [ 255 = 15 × 17]

⇒ 1 3 2 3 2b ac a b ab

+= + =

⇒ ab = 3bc + 2ac

∴ ab – 3bc – 2ac = 0 8. Answer (2)

1 1 1 1.....9 10 10 11 11 12 80 81

+ + + ++ + + +

= 9 10 10 11 11 12 80 81......1 1 1 1

− − − −+ + + +

− − − −

= 10 9 11 10 12 11 ..... 81 80− + − + − + + −

= 81 9 9 3 6− = − =

9. Answer (1) For, y = 0 x = 3 10. Answer (1)

Let and a x b y= =

⇒ x3 + y3 = 341 and x2y + xy2 = 330

⇒ xy (x + y) = 330 …(i)

and (x + y)3 = x3 + y3 + 3xy (x + y)

⇒ (x + y)3 = 341 + 3 × 330 = 1331

⇒ x + y = 11 …(ii)

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Test-2 (Answers & Hints) All India Aakash Test Series-2021 (Class IX)

From (i),

330 330 3011

xyx y

= = =+

Also,

( )2 4x y x y xy− = ± + −

121 120= ± − x – y = ± 1 …(iii) From (ii) and (iii), x = 6, y = 5 or x = 5, y = 6 Now,

a ÷ b = x2 ÷ y2

= 2536

11. Answer (1)

In ∆ABC, 2θ = 2α + ∠B …(i) and In, ∆BCG,

θ = ∠BGC + 2B∠

⇒ 2θ = 2∠BGC + ∠B …(ii) Subtracting (ii) from (i), 0 = 2α – 2∠BGC

⇒ ∠BGC = α = 2A∠

Similarly, ∠AHC = 2B∠

and ∠AIB = 2C∠

∴ ∠BGC + ∠AHC + ∠AIB = 2

A B C∠ + ∠ + ∠

= 90°

12. Answer (2)

In ∆ACD and ∆ABE,

AC = AE

AD = AB

∠CAD = 60° + ∠A = ∠BAE

∴ ∆ACD ≅ ∆AEB [By SAS congruency]

⇒ CD = BE [By CPCT]

13. Answer (1)

14. Answer (3)

For ∆CEF,

∠AFE = ∠ACE + ∠BEC …(i)

and For ∆BDG,

∠AGB = ∠ADB + ∠DBE …(ii)

Adding (i) and (ii),

∠AFE + ∠AGB = ∠B + ∠C + ∠D + ∠E

⇒ ∠A + ∠AFE + ∠AGB = ∠A + ∠B + ∠C + ∠D + ∠E

∴ ∠A + ∠B + ∠C + ∠D + ∠E = 180°

15. Answer (2)

AB = 13 cm

and 2

ABOP = [By mid-point theorem]

Edition: 2020-21