test no. 1 · test-1 (answers & hints) all india aakash test series (junior) - 2020 (class ix)...

Test No. 1

Olympiads & Class IX-2020for

14-07-2019

A, B & C

1/6

Test-1 (Answers) All India Aakash Test Series (Junior) - 2020 (Class IX)

1. (2)

2. (3)

3. (3)

4. (3)

5. (2)

6. (4)

7. (2)

8. (2)

9. (2)

10. (2)

11. (4)

12. (2)

13. (4)

14. (1)

15. (3)

16. (1)

17. (3)

18. (2)

19. (4)

20. (1)

21. (1)

22. (3)

23. (2)

24. (3)

25. (2)

26. (4)

27. (2)

28. (2)

29. (4)

30. (1)

31. (2)

32. (1)

33. (1)

34. (3)

35. (1)

36. (4)

37. (1)

38. (2)

39. (3)

40. (2)

41. (3)

42. (2)

43. (1)

44. (3)

45. (3)

46. (2)

47. (1)

48. (2)

49. (4)

50. (4)

51. (4)

52. (3)

53. (4)

54. (4)

55. (3)

56. (2)

57. (2)

58. (1)

59. (4)

60. (3)

61. (1)

62. (4)

63. (1)

64. (1)

65. (3)

66. (2)

67. (1)

68. (3)

69. (4)

70. (3)

71. (3)

72. (2)

73. (4)

74. (2)

75. (2)

76. (3)

77. (1)

78. (2)

79. (4)

80. (4)

ANSWERS

TEST - 1

All India Aakash Test Series (Junior) - 2020 (Class IX)

Test Date : 14-07-2019

SECTION-I (Code-A)

81. (4)

82. (2)

83. (1)

84. (2)

85. (4)

86. (3)

87. (2)

88. (4)

89. (2)

90. (4)

91. (3)

92. (4)

93. (1)

94. (2)

95. (3)

96. (1)

97. (4)

98. (2)

99. (3)

100. (4)

SECTION-II (Code-B)

1. (3)

2. (4)

3. (2)

4. (4)

5. (2)

6. (2)

7. (2)

8. (3)

9. (1)

10. (2)

11. (2)

12. (3)

13. (2)

14. (3)

15. (2)

16. (3)

17. (2)

18. (1)

19. (3)

20. (4)

21. (2)

22. (2)

23. (3)

24. (3)

25. (4)

26. (2)

27. (2)

28. (4)

29. (3)

30. (2)

SECTION-III (Code-C)

1. (2)

2. (2)

3. (4)

4. (2)

5. (3)

6. (2)

7. (1)

8. (4)

9. (4)

10. (4)

11. (3)

12. (3)

13. (4)

14. (4)

15. (4)

All India Aakash Test Series (Junior) - 2020 (Class IX) Test-1 (Answers & Hints)

2/6

SECTION-I (Code-A)

1. Answer (2)

2. Answer (3)

3. Answer (3)

4. Answer (3)

5. Answer (2)

21

2S ut at

210 10 4

2

= 80 cm

6. Answer (4)

12 1

2X u a n

15 1 2 4 1

2

= 1.5 m

7. Answer (2)

8. Answer (2)

v = u + at

v = 40% of u = 0.4 u

Hence 0.4u = u – at

So 0.6u

ta

3

5

u

a

9. Answer (2)

10. Answer (2)

11. Answer (4)

v = v0 + at

= 0 + 10 × 3

= 30 ms–1

12. Answer (2)

Distance covered in 2 seconds = 2110 2

2

= 20 m

Total distance covered in next 3 seconds

2110 3

2

= 45 m

H = 100 – 20 – 45 = 35 m

13. Answer (4)

14. Answer (1)

15. Answer (3)

73.5 rad/s

2

v

r

16. Answer (1)

Here

AOB and BCD

C

AO CD

OB B

604

8CD

= 30 m

Net displacement

= Area under the curve in v - t graph

= AOB – BED + EFG

1 1 160 8 30 6 60 4

2 2 2

= 240 – 90 + 120

= 270 m

17. Answer (3)

v2 = u2 + 2as

2

4

v D

v D

2

vv

Hints to Selected Questions

All India Aakash Test Series (Junior) - 2020 (Class IX)

TEST - 1

Test-1 (Answers & Hints) All India Aakash Test Series (Junior) - 2020 (Class IX)

3/6

18. Answer (2)

19. Answer (4)

21

2S ut at

140 0 16

2a

a = 5

v = u + at

= 0 + 5 × 4

= 20 m/s

Again

v = u + at

v = 20 + 5 ×16

= 100 m/s

20. Answer (1)

Vr = 120 – 100 = 20 m/s

L = 80 + 120 = 200 m

LV

t

20010 s

20t

21. Answer (1)

22. Answer (3)

23. Answer (2)

24. Answer (3)

25. Answer (2)

26. Answer (4)

27. Answer (2)

28. Answer (2)

29. Answer (4)

30. Answer (1)

31. Answer (2)

32. Answer (1)

33. Answer (1)

34. Answer (3)

35. Answer (1)

36. Answer (4)

37. Answer (1)

38. Answer (2)

39. Answer (3)

40. Answer (2)

41. Answer (3)

42. Answer (2)

Aspergillus is an example of multicellular fungi.

43. Answer (1)

44. Answer (3)

45. Answer (3)

46. Answer (2)

47. Answer (1)

48. Answer (2)

49. Answer (4)

50. Answer (4)

Golgi apparatus is not a semi-autonomous cell

organelle.

51. Answer (4)

52. Answer (3)

53. Answer (4)

Prokaryotic cells do not contain nucleus and plasma

membrane is composed of phospholipids and protein.

54. Answer (4)

55. Answer (3)

56. Answer (2)

57. Answer (2)

58. Answer (1)

59. Answer (4)

60. Answer (3)

61. Answer (1)

62. Answer (4)

63. Answer (1)

64. Answer (1)

65. Answer (3)

66. Answer (2)

67. Answer (1)

68. Answer (3)

69. Answer (4)

70. Answer (3)


4/6

71. Answer (3)

72. Answer (2)

73. Answer (4)

74. Answer (2)

Given, x – y = 5

(x – y)3 = 125

x3 – y3 – 3xy (x – y) = 125

x3 – y3 – 15xy = 125 x3 – y3 – 15xy – 100 = 25

75. Answer (2)

76. Answer (3)

(x3 + 6x2 + 11x + 6) = (x + 1) (x + 2) (x + 3)

Since, (x + 1), (x + 2) and (x + 3) are three

consecutive terms and x is an integer.

So, the largest number by which the given

expression is divisible is 6.

77. Answer (1)

78. Answer (2)

79. Answer (4)

80. Answer (4)

27y6 – 28y3 + 8

27y6 + 8y3 + 8 – 36y3

(3y2)3 + (2y)3 + (2)3 – 3 × 3y2 × 2y × 2

(3y2 + 2y + 2) (9y4 + 4y2 + 4 – 6y3 – 4y – 6y2)

(3y2 + 2y + 2) (9y4 – 6y3 – 2y2 + 4)

Hence, (3y2 + 2y + 2) is the factor of the polynomial.

81. Answer (4)

Difference of 156.

82. Answer (2)

Pattern is n2 0 (n + 1)3

83. Answer (1)

84. Answer (2)

85. Answer (4)

n : n3 – n

86. Answer (3)

ab : (a + b)3 – 1

87. Answer (2)

88. Answer (4)

89. Answer (2)

90. Answer (4)

91. Answer (3)

92. Answer (4)

93. Answer (1)

a b c | a b c d | a b c d e | a b c d e f

94. Answer (2)

x y z a b c | x y z a b c | x y z a b c | x y z a b c

95. Answer (3)

a p b q c r | a p b q c r | a p b q c r

96. Answer (1)

First series : 3, 5, 7, 9

Second series : 5, 19, 41, ?

Difference of second series are 14, 22, 30 etc

Next term is 41 + 30 i.e, equal to 71

97. Answer (4)

The given sequence is a combination of three series:

(i) 1st, 4th, 7th terms i.e, 2, 4, 7

(ii) 2nd, 5th, 8th terms i.e, 3, 6, 12

(iii) 3rd, 6th, 9th terms i.e, 4, 8, 16

In each one of (i), (ii) and (iii), each term is twice the

preceding term.

So, 7 is wrong and must be replaced by (4*2) = 8

98. Answer (2)

99. Answer (3)

100. Answer (4)

SECTION-II (Code-B)

1. Answer (3)

2. Answer (4)

3. Answer (2)

White blood cells and Amoeba can change their

shape.

4. Answer (4)

5. Answer (2)

6. Answer (2)

7. Answer (2)

8. Answer (3)

9. Answer (1)

Test-1 (Answers & Hints) All India Aakash Test Series (Junior) - 2020 (Class IX)

5/6

10. Answer (2)

Camillo Golgi carried out a revolutionary method of

staining individual nerve and cell structures.

11. Answer (2)

12. Answer (3)

13. Answer (2)

14. Answer (3)

Intake of liquid food material through cell membrane

is known as pinocytosis.

15. Answer (2)

16. Answer (3)

17. Answer (2)

18. Answer (1)

19. Answer (3)

20. Answer (4)

Chloroplast and mitochondria contain their own

circular DNA and 70S ribosomes.

21. Answer (2)

22. Answer (2)

23. Answer (3)

24. Answer (3)

25. Answer (4)

26. Answer (2)

27. Answer (2)

Krebs cycle occurs in the matrix while oxysomes

contain metabolic enzymes for ATP synthesis.

28. Answer (4)

29. Answer (3)

30. Answer (2)

SECTION-III (Code-C)

1. Answer (2)

2. Answer (2)

3. Answer (4)

4. Answer (2)

5. Answer (3)

44 24 2 2 11 6 2

2

2 3 2

2 3 2

6 2 2

11 6 2 3 2

2 3 2 3 2

2 1

k

6 2 2 3 2

2 1

k

3 3 2

2 1k

k = 3

13

k

6. Answer (2)

7. Answer (1)

8. Answer (4)

Centroid of triangle 1 5 9 1 3 2

,3 3

(5, 2)

Hence, (5, 2) is the centroid.

9. Answer (4)

10. Answer (4)

Given, 53 4

x y

4x + 3y = 60

4x = 60 – 3y

x = 3

154y

y = 0, 4, 8, 12, 16, 20

x = 15, 12, 9, 6, 3, 0

So, the ordered pairs are (0, 15), (4, 12), (8, 9),

(12, 6), (16, 3) and (20, 0)

Hence, equation has six ordered pairs.


6/6

��

11. Answer (3)

12. Answer (3)

Given a + b + c 4

3

x

4x – (3a + 3b + 3c) = 0

On adding 4 4

3 , 33 3

x xa b

and 33

xc

,

we get,

4 4 43 3 3 4 3 3 3

3 3 3

x x xa b c x a b c

4 4 43 3 3 0

3 3 3

x x xa b c

3 3 3

4 4 43 3 3

3 3 3

x x xa b c

4 4 43 3 3 3

3 3 3

x x xa b c

4 4 43 3 3 3

3 3 3

x x xa b c

4 4 43 3 3 3

3 3 3

x x xa b c

[∵ a3 + b3 + c3 = 3abc]

0

13. Answer (4)

14. Answer (4)

Mid-point of (–3, 2) and (5, 4) = 3 5 2 4

,2 2

= (1, 3)

Since, (1, 3) is the mid-point of (– 3, 2) and (5, 4).

(– 3, 2), (1, 3) and (5, 4) are collinear points.

15. Answer (4)

Given, 3 35 5 5 5 5 25

x

3

3 325 5 5 5 25

x

3

3225 5 5 5x

19

2245 5x

192

24x

19

48x

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