Test No. 1...3/7 Test-1 (Code A) - (Answers & Hints) All India Aakash Test Series (Junior) - 2019 (Class IX) 12. Answer (1) a = v t Distance travelled in t seconds l 1 = 1 . 2 2 at

Test No. 1

Olympiads & Class IX-2019for

15-07-2018

A

1/7

Test-1 (Code A) - (Answers & Hints) All India Aakash Test Series (Junior) - 2019 (Class IX)

1. (4)

2. (4)

3. (2)

4. (2)

5. (3)

6. (4)

7. (4)

8. (2)

9. (1)

10. (1)

11. (3)

12. (1)

13. (4)

14. (3)

15. (1)

16. (4)

17. (4)

18. (3)

19. (2)

20. (3)

21. (1)

22. (4)

23. (4)

24. (4)

25. (2)

26. (3)

27. (3)

28. (4)

29. (3)

30. (4)

31. (1)

32. (1)

33. (3)

34. (3)

35. (3)

36. (2)

37. (2)

38. (4)

39. (3)

40. (3)

41. (1)

42. (4)

43. (2)

44. (4)

45. (1)

46. (1)

47. (4)

48. (2)

49. (1)

50. (3)

51. (1)

52. (4)

53. (3)

54. (2)

55. (4)

56. (1)

57. (3)

58. (1)

59. (3)

60. (1)

61. (2)

62. (1)

63. (1)

64. (1)

65. (2)

66. (4)

67. (2)

68. (1)

69. (1)

70. (1)

71. (3)

72. (2)

73. (3)

74. (3)

75. (1)

76. (1)

77. (1)

78. (3)

79. (1)

80. (4)

81. (3)

82. (2)

83. (3)

84. (2)

85. (2)

86. (2)

87. (2)

88. (2)

89. (2)

90. (2)

91. (3)

92. (3)

93. (3)

94. (4)

95. (3)

96. (1)

97. (2)

98. (2)

99. (4)

100. (2)

101. (4)

102. (1)

103. (4)

104. (4)

105. (2)

106. (2)

107. (2)

108. (2)

109. (4)

110. (2)

111. (2)

112. (4)

113. (4)

114. (4)

115. (2)

116. (3)

117. (1)

118. (2)

119. (2)

120. (4)

121. (3)

122. (1)

123. (2)

124. (4)

125. (3)

126. (1)

127. (2)

128. (3)

129. (4)

130. (2)

131. (2)

132. (3)

133. (2)

134. (1)

135. (3)

136. (2)

137. (4)

138. (3)

139. (2)

140. (4)

141. (2)

142. (1)

143. (3)

144. (2)

145. (2)

146. (1)

147. (3)

148. (4)

149. (3)

150. (4)

ANSWERS

TEST - 1 (Code-A)

All India Aakash Test Series (Junior) - 2019 (Class IX)

Test Date : 15-07-2018

2/7

All India Aakash Test Series (Junior) - 2019 (Class IX) Test-1 (Code A) - (Answers & Hints)

Hints to Selected Questions

[ SCIENCE]

1. Answer (4)

2. Answer (4)

aavg

(Cheetah) = 25 0

2

aavg

(Tiger) = 25 0

4.5

(Cheetah) 4.5 9

(Tiger) 2 4

avg

avg

a

a

3. Answer (2)

a = 2v

t

a = 40

t,

t = R

v

t = 22 14

7 20

t = 44

20

a = 40 20

44

,

a = 2800 200

ms44 11

4. Answer (2)

The distance between them will increase till velocity

of Q becomes equal to velocity of P.

at t = 1 second

Distance travelled by P = 4 × 1 = 4 m

Distance travelled by Q = 1

4 1 2 m2

Maximum distance = 2 m

5. Answer (3)

h = 21

(2 )2g t (distance travelled by first drop)

h1 =

21( )

2g t (distance travelled by second drop)

2 21 1 15(2 ) ( )

2 2 4g t g t

23 15

2 4g t

21 15( )

2 12g t

h = 21 15

(2 ) 42 12g t

15

3

= 5 m

6. Answer (4)

7. Answer (4)

8. Answer (2)

Average velocity = Net displacement

time

5 380 40

2 2

4

200 60

4

140 70

4 2 = 35 kmh–1

9. Answer (1)

P Q

PQ = l = 21

2at

l = 21

( 2)2a t

2 21 1( 2) 40

2 2at a t

2 24 4 40

2

at t t⎡ ⎤ ⎣ ⎦

5[4t – 4] = 40

4t = 8 + 4

t = 12

3 s4

t = 21

10 (3)2

l = 45 m

10. Answer (1)

11. Answer (3)

3/7


12. Answer (1)

a = v

t

Distance travelled in t seconds

l1 =

21.

2a t

Distance travelled in (t – 2) seconds

l2 =

21( 2)

2a t

Distance travelled in last 2 seconds in

l1 –

2

1

2l a [t2 – t2 – 4 + 4t]

= 1

[4 4]2a t

= 2a[t – 1]

= 2a

v a

a

⎡ ⎤⎢ ⎥⎣ ⎦

= 2(v – a)

13. Answer (4)

14. Answer (3)

Velocity-time graph

2010 30 (s)

100

(ms )–1

Distance = area of velocity – time graph

= 1

(30 10) 1002

= 40 × 50

= 2000 m

15. Answer (1)

vA

= 20 ms–1

v B

= 10 ms–1

200 m

Velocity of A with respect to B

vAB

= 20 – 10 = 10 m/s

0 = (10)2 + 2(a) × 200

a = 2100 1

m/s400 4

16. Answer (4)

17. Answer (4)

18. Answer (3)

19. Answer (2)

Sn = u + (2 1)

2

an

S5 = 0 + (2 5 1)

2

a

S5 =

9

2

a

S4

=

7

2

a

5

4

9

7

S

S

20. Answer (3)

vavg

=

Total distance

Total time

=

2

2 4

3 3(2 )

R

R R

v v

2

2 2

3 3

R

R R

v v

= 3

2v

21. Answer (1)

22. Answer (4)

23. Answer (4)

24. Answer (4)

25. Answer (2)

26. Answer (3)

27. Answer (3)

T(K) = 273 + t(°C)

28. Answer (4)

29. Answer (3)

30. Answer (4)

31. Answer (1)

32. Answer (1)

33. Answer (3)

34. Answer (3)

35. Answer (3)

36. Answer (2)

4/7


37. Answer (2)

38. Answer (4)

Temperature below 0°C water exist in solid state

only.

39. Answer (3)

40. Answer (3)

41. Answer (1)

42. Answer (4)

According to fluid mosaic model, two types of

proteins are present in a sandwich of phospholipid

bilayer.

43. Answer (2)

44. Answer (4)

a - Cytoplasm

b - Cell wall

c - Cell membrane

d - Ribosome

45. Answer (1)

46. Answer (1)

47. Answer (4)

48. Answer (2)

49. Answer (1)

50. Answer (3)

51. Answer (1)

52. Answer (4)

53. Answer (3)

54. Answer (2)

Number of chromosomes in Dog = 78, Cat = 38,

Human = 46, Chimpanzee = 48 and fruit fly = 8.

55. Answer (4)

56. Answer (1)

57. Answer (3)

58. Answer (1)

59. Answer (3)

Smooth endoplasmic reticulum is involved in the

detoxification of drugs and poisons and also in

formation of poisons.

60. Answer (1)

61. Answer (2)

62. Answer (1)

63. Answer (1)

Process 'X' is dif fusion and process 'Y' is

osmosis.

64. Answer (1)

65. Answer (2)

H2O

2 toxicity in a cell is controlled by catalase

enzyme present in peroxisomes.

66. Answer (4)

67. Answer (2)

Knoll and Ruska invented electron microscope and

porter discovered endoplasmic reticulum. Fluid

mosaic model was given by Singer and Nicolson.

68. Answer (1)

69. Answer (1)

70. Answer (1)

71. Answer (3)

72. Answer (2)

Lysosomes are the suicidal bags formed by golgi

bodies. Hydrolytic enzyme present in them are

formed by rough endoplasmic reticulum.

73. Answer (3)

74. Answer (3)

Only trans/concave or maturing face of golgi

apparatus is involved in packaging of cellular

contents in vesicles.

75. Answer (1)

76. Answer (1)

Leucoplast are colorless plastids that store fats

whereas chromoplast and chloroplast are colored

plastids, chloroplast being green in color and

chromoplast stores carotenoids.

77. Answer (1)

Process 'A' is endosmosis which makes the plant

cell turgid and process 'B' is exosmosis which

plasmolysed the plant cell.

78. Answer (3)

79. Answer (1)

Depicted process is endocytosis through which

Amoeba acquires its food by forming pseudopodia

through folding of membrane.

80. Answer (4)

A is Mitochondria

B is Golgi bodies

C is Centrioles

5/7


[ MATHEMATICS

]

81. Answer (3)

82. Answer (2)

83. Answer (3)

84. Answer (2)

85. Answer (2)

86. Answer (2)

87. Answer (2)

88. Answer (2)

89. Answer (2)

90. Answer (2)

91. Answer (3)

92. Answer (3)

93. Answer (3)

94. Answer (4)

95. Answer (3)

96. Answer (1)

4

4

147x

x

2

2 2

2 2

1 1– 2 47x x

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2

17x

x

[∵ x > 0]

2

1 1– 2 7x x

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

13x

x

3 2

3 2

1 1 1 1x x x x

x xx x

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

3

3

118x

x

6

3

118

x

x

97. Answer (2)

98. Answer (2)

99. Answer (4)

100. Answer (2)

x5 + x + 1 = x5 – x2 + (x2 + x + 1)

= x2 (x3 – 1) + (x2 + x + 1)

= x2(x – 1) (x2 + x + 1) + (x2 + x + 1)

= [x2(x – 1) + 1] (x2 + x + 1)

= (x3 – x2 + 1) (x2 + x + 1)

101. Answer (4)

32222 = (32)1111 = (9)1111

25555 = (25)1111 = (32)1111

54444 = (54)1111 = (625)1111

63333 = (63)1111 = (216)1111

9 < 32 < 216 < 625

32222 < 25555 < 63333 < 54444

102. Answer (1)

103. Answer (4)

104. Answer (4)

105. Answer (2)

106. Answer (2)

107. Answer (2)

a3(b – c) + b3(c – a) + c3(a – b)

= a3(b – c) + b3(c – a) – c3(b – a)

= a3(b – c) + b3(c – a) – c3 [(b – c) + (c – a)]

= [a3(b – c) – c3(b – c)] + [b3(c – a) – c3(c – a)]

= (b – c) (a3 – c3) + (c – a) (b3 – c3)

= (b – c) (a – c) (a2 + ac + c2) + (c – a) (b – c)

(b2 + bc + c2)

= (b – c) (c – a) [–(a2 + ac + c2) + (b2 + bc + c2)]

= (b – c) (c – a) [b2 – a2 + bc – ac]

= (b – c) (c – a) [(b – a) (b + a) + c(b – a)]

= (b – c) (c – a) (a – b) [–(b + a) – c]

= –(a + b + c) (a – b) (b – c) (c – a)

108. Answer (2)

109. Answer (4)

1

2011 2010 2012

x y

=

2

4022 2 2010 2012

=

2

2012 2010 2 2012 2010

= 2

2012 2010

6/7


[ MENTAL ABILITY

]

121. Answer (3)

23 × 3, 23 × 5, 23 × 7, 23 × 9 and 23 × 11.

122. Answer (1)

General term is (17n – 2), put n = 1, 2, 3... .

123. Answer (2)

Put n = 1, 2, 3, 4... in n2 + 3

124. Answer (4)

Each term is multiply by 15.

125. Answer (3)

Each term is divided by 0.03

126. Answer (1)

127. Answer (2)

128. Answer (3)

129. Answer (4)

a b c d | a b c d | a b c d

130. Answer (2)

a b c d d | a b c c d | a b b c d | a a b c d

131. Answer (2)

abcd : a × b × c × d

132. Answer (3)

abc : (a + b) – c

133. Answer (2)

a : a4 + 1

134. Answer (1)

135. Answer (3)

136. Answer (2)

137. Answer (4)

138. Answer (3)

139. Answer (2)

140. Answer (4)

=

2( 2012 2010)

( 2012 2010)( 2012 2010)

= 2( 2012 2010)

2012 2010

= 1

( 2012 2010)2

= 1006 1005 x = 1006 and y = 1005

So, x + y = 1006 + 1005 = 2011

110. Answer (2)

111. Answer (2)

112. Answer (4)

113. Answer (4)

114. Answer (4)

115. Answer (2)

116. Answer (3)

a – b = 3 and b – c = 1

c – a = (c – b) + (b – a)

= – (b – c) – (a – b)

= –1 – 3

= –4

So, a2 + b2 + c2 – ab – bc – ca

=1

2

2 2 2( ) ( ) ( )a b b c c a⎡ ⎤ ⎣ ⎦

2 2 21(3) (1) ( 4)

2⎡ ⎤ ⎣ ⎦

19 1 16

2

=26

132

117. Answer (1)

118. Answer (2)

119. Answer (2)

5 5a and 5 5b 2

5 5a and 2

5 5b [∵ (x2)3 + (y2)3 = (x2 + y2)3 – 3x

2y

2 (x2 + y2)]

a6 + b6 =

3

(5 5) (5 5)⎡ ⎤ ⎣ ⎦ – 3

(5 5)(5 5) (5 5) (5 5)⎡ ⎤ ⎣ ⎦

a6 + b6 = (10)3 – 3(52 – 5) × 10

a6 + b6 = 1000 – 600

a6 + b6 = 400

120. Answer (4)

7/7


� � �

141. Answer (2)

142. Answer (1)

143. Answer (3)

1

5

9

13

2

6

10

14

3

7

11

15

4

8

12

16

Dot is placed at the position of prime numbers.

144. Answer (2)

145. Answer (2)

146. Answer (1)

Pattern is n2 – 1, where n = 13, 14, 15, .....

147. Answer (3)

Pattern of prime numbers.

148. Answer (4)

Pattern is written in the form (x |0| x3).

149. Answer (3)

150. Answer (4)

768

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Documents

Test No. 1...3/7 Test-1 (Code A) - (Answers & Hints) All India Aakash Test Series (Junior) - 2019 (Class IX) 12. Answer (1) a = v t Distance travelled in t seconds l 1 = 1 . 2 2 at