test no. 1 · test-1 (answers & hints) all india aakash test series (junior) - 2020 (class ix)...
TRANSCRIPT
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)
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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)
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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)
All India Aakash Test Series (Junior) - 2020 (Class IX) Test-1 (Answers & Hints)
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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)
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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.
All India Aakash Test Series (Junior) - 2020 (Class IX) Test-1 (Answers & Hints)
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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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