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Test No. 2
Olympiads & Class IX-2020for
13-10-2019
A, B & C
1/8
Test-2 (Answers) All India Aakash Test Series for Class IX (2020)
1. (2)
2. (1)
3. (2)
4. (2)
5. (3)
6. (1)
7. (1)
8. (3)
9. (3)
10. (3)
11. (3)
12. (4)
13. (2)
14. (3)
15. (2)
16. (1)
17. (3)
18. (1)
19. (3)
20. (3)
21. (3)
22. (4)
23. (3)
24. (4)
25. (3)
26. (4)
27. (1)
28. (4)
29. (3)
30. (4)
31. (4)
32. (3)
33. (3)
34. (4)
35. (4)
36. (4)
37. (3)
38. (2)
39. (3)
40. (3)
41. (3)
42. (1)
43. (4)
44. (2)
45. (3)
46. (3)
47. (4)
48. (4)
49. (3)
50. (2)
51. (4)
52. (4)
53. (2)
54. (4)
55. (3)
56. (1)
57. (2)
58. (3)
59. (4)
60. (3)
61. (2)
62. (3)
63. (4)
64. (1)
65. (3)
66. (2)
67. (4)
68. (3)
69. (4)
70. (1)
71. (1)
72. (3)
73. (4)
74. (3)
75. (2)
76. (2)
77. (1)
78. (2)
79. (2)
80. (3)
ANSWERS
TEST - 2
All India Aakash Test Series for Class IX (2020)
Test Date : 13-10-2019
SECTION-I (Code-A)
81. (2)
82. (3)
83. (4)
84. (3)
85. (1)
86. (2)
87. (4)
88. (3)
89. (2)
90. (1)
91. (3)
92. (4)
93. (4)
94. (2)
95. (3)
96. (4)
97. (1)
98. (3)
99. (3)
100. (4)
SECTION-II (Code-B)
1. (3)
2. (2)
3. (1)
4. (4)
5. (2)
6. (1)
7. (3)
8. (2)
9. (3)
10. (2)
11. (4)
12. (2)
13. (1)
14. (3)
15. (4)
16. (2)
17. (3)
18. (1)
19. (2)
20. (1)
21. (2)
22. (4)
23. (4)
24. (2)
25. (3)
26. (3)
27. (4)
28. (2)
29. (2)
30. (4)
SECTION-III (Code-C)
1. (2)
2. (2)
3. (3)
4. (3)
5. (1)
6. (1)
7. (3)
8. (4)
9. (2)
10. (4)
11. (3)
12. (2)
13. (2)
14. (3)
15. (4)
All India Aakash Test Series for Class IX (2020) Test-2 (Answers & Hints)
2/8
SECTION-I (Code-A)
1. Answer (2)
2. Answer (1)
3. Answer (2)
4. Answer (2)
F = ma
F = 12 × 5 = 60 N
5. Answer (3)
6. Answer (1)
7. Answer (1)
35002 g/cm
250
M
v
8. Answer (3)
9. Answer (3)
10. Answer (3)
11. Answer (3)
Relative speed of car moving in opposite direction
= 5
75 km/hr4/60
Speed of third car = 75 – 45 = 30 km/hr
12. Answer (4)
21
2S at
22 1
3 2
Sa t
2St
a
4 2
3 3
St t
a
t – t = 2
3t t
21
3t t
Hints to Selected Questions
All India Aakash Test Series for Class IX (2020)
TEST - 2
13. Answer (2)
22 62 m/s
4a
S1 = 6 × 4 +
1
2× (–2) × (4)2
S1 = 8 m
S2 = – 2 × 2 = – 4 m
S3 = 4 × 2 = 8 m
S = 8 + 8 – 4 = 12 m
Vavg
= 12
1.5 m/s8
14. Answer (3)
From law of conservation of momentum,
mu = 2mv
2
uv
Retardation 2
f
m
v2 = u2 + 2as
2
0 22 2
u fx
m
2
4
u fx
m
4fxu
m
15. Answer (2)
260 12 484 m/s
7 3 2 12a
For the block of 2 kg,
T2 – 12 = 2 × 4
T2 = 8 + 12 = 20 N
For the block of 3 kg,
T1 – T
2 = 3 × 4
T1 = 12 + 20 = 32 N
Test-2 (Answers & Hints) All India Aakash Test Series for Class IX (2020)
3/8
16. Answer (1)
1 2
1 2
m m ga
m m
Since both the blocks travel equal distance in opposite
directions
21 2
1 2
1
2 2
m mhgt
m m
1 2
1 2
m mht
g m m
17. Answer (3)
2hT
g
2 1255 s
10T
5
100 2 5 1
2S
S5 = 45 m
18. Answer (1)
1d
dg g
R
641
100
dg g
R
64 361
100 100
d
R
366400
100d
d = 2304 km
19. Answer (3)
Mg = Bnet
v × 2.4 × g = 0.82 2
v vg x g
2.4 = 0.4 + 2
x
x = 4 g/cm3
20. Answer (3)
When system is at rest,
W + T = B
When system is moving upward with acceleration 2
g,
02 2
b l
g gv g T v g
0
3
2l b
gT v
When system is moving downward with acceleration
4
g,
4 4b l
g gv g T v g
3
4l b
gT v p p
0
2
TT
21. Answer (3)
22. Answer (4)
23. Answer (3)
24. Answer (4)
25. Answer (3)
26. Answer (4)
27. Answer (1)
28. Answer (4)
29. Answer (3)
30. Answer (4)
31. Answer (4)
32. Answer (3)
33. Answer (3)
34. Answer (4)
35. Answer (4)
36. Answer (4)
37. Answer (3)
38. Answer (2)
39. Answer (3)
40. Answer (3)
Concentration of solution made by Rama (w/w)
25 25100 100 20%
100 25 125
Concentration of solution made by Kanta (w/w)
50 50100 100 25%
150 50 200
Concentration of solution made by Shyam (w/w)
50100 25%
200
All India Aakash Test Series for Class IX (2020) Test-2 (Answers & Hints)
4/8
41. Answer (3)
42. Answer (1)
Squamous epithelium is also called pavement
epithelium.
43. Answer (4)
44. Answer (2)
Wuchereria is a nematode.
45. Answer (3)
46. Answer (3)
47. Answer (4)
48. Answer (4)
49. Answer (3)
50. Answer (2)
51. Answer (4)
52. Answer (4)
53. Answer (2)
54. Answer (4)
Smooth muscles are found in iris of eye.
55. Answer (3)
56. Answer (1)
57. Answer (2)
58. Answer (3)
59. Answer (4)
60. Answer (3)
The given features are of phylum Mollusca.
61. Answer (2)
62. Answer (3)
x6 – 729 = (x3)2 – (33)2 = (x3 + 33)(x3 – 33)
= (x + 3)(x – 3)(x2 + 9 + 3x)(x2 + 9 – 3x)
63. Answer (4)
64. Answer (1)
65. Answer (3)
66. Answer (2)
67. Answer (4)
2 22 29 3 3 9 3 3
3 0 6 02 2 2 2
2 26 3 0 = 3 units
Triangle is equilateral with side = 3 units
Perimeter = 9 units
68. Answer (3)
69. Answer (4)
70. Answer (1)
71. Answer (1)
72. Answer (3)
x2 + x – 6 is factor of x3 + ax2 + bx + 6,
(x + 3), (x – 2) are factors of x3 + ax2 + bx + 6
(–3)3 + a(–3)2 + (–3)b + 6 = 0
9a – 3b = 21
3a – b = 7 ...(i)
and (2)3 + a(2)2 + 2b + 6 = 0
4a + 2b = –14
2a + b = –7 ...(ii)
Solving (i) and (ii), we get
5a = 0 a = 0 b = –7
73. Answer (4)
210 8 5 8 5 2 40 13 4 10y x
y = 13, x = –4
|y| – |x| = 13 – 4 = 9
74. Answer (3)
In BCE and BAG,
BC = BA [Sides of a square]
BE = BG [Sides of a square]
and CBE = 90° – ABE = ABG
[∵ ABC = EBG = 90°]
BCE BAG [By SAS]
CE = AG = 4 cm [By CPCT]
75. Answer (2)
2 22 5 2 2 7 units,AB
2 24 2 4 2 2 10 units,BC
2 24 3 4 4 7 units,CD
and 2 25 3 4 2 2 10 units,AD
ABCD is a parallelogram
[∵ Opposite sides are equal]
Test-2 (Answers & Hints) All India Aakash Test Series for Class IX (2020)
5/8
76. Answer (2)
77. Answer (1)
78. Answer (2)
A
B C 15° 75°
D
60°120°
45°45°
BAC = 90°
BAD = CAD = 45°
ADC = 45° + 15° = 60° and
ADB = 45° + 75° = 120°
In ABD and ACD,
AB > BD > AD ...(i)
and AD > AC > CD ...(ii)
CD < AD < BD [From (i) and (ii)]
79. Answer (2)
2 2 2 3 3 31 1 3
13 3 3
a b c a b c abc
bc ac ab abc abc
[∵ a + c = –b a + b + c = 0. So, a3 + b3 +
c3 = 3abc]
80. Answer (3)
3 3
3
5 5– 600
55
x x
3 3
5 5600
125 5
x x
3 1 255 600
125
x
3
5 60025
125 24
x
53x = 125 × 25 = 55
3x = 5
5
3x
81. Answer (2)
82. Answer (3)
83. Answer (4)
84. Answer (3)
85. Answer (1)
86. Answer (2)
87. Answer (4)
The opposite letter of G is t.
88. Answer (3)
The opposite letter of Z is a.
89. Answer (2)
90. Answer (1)
91. Answer (3)
A is father of N, who is daughter of M. Therefore A
is husband of M.
92. Answer (4)
K is wife of C, who is brother of B and B is father
of Q.
93. Answer (4)
94. Answer (2)
1130H M
2
1130 9 24
2
270 132
138
95. Answer (3)
96. Answer (4)
97. Answer (1)
98. Answer (3)
99. Answer (3)
100. Answer (4)
SECTION-II (Code-B)
1. Answer (3)
2. Answer (2)
3. Answer (1)
4. Answer (4)
5. Answer (2)
6. Answer (1)
Study of fungi is called mycology.
7. Answer (3)
8. Answer (2)
9. Answer (3)
10. Answer (2)
All India Aakash Test Series for Class IX (2020) Test-2 (Answers & Hints)
6/8
11. Answer (4)
12. Answer (2)
13. Answer (1)
The given features are of division Pteridophyta.
14. Answer (3)
Reticulate venation is found in dicots.
15. Answer (4)
16. Answer (2)
17. Answer (3)
18. Answer (1)
19. Answer (2)
20. Answer (1)
21. Answer (2)
Mesosome is formed by folding of cell membrane.
22. Answer (4)
Here, ‘a’ is Porifera, ‘b’ is Coelenterata, ‘c’ is
Aschelminthes and ‘d’ are Annelida to Chordata.
23. Answer (4)
Here, ‘P’ is chloroplast, ‘Q’ is nucleus, ‘R’ is golgi
apparatus and ‘S’ is smooth endoplasmic reticulum.
24. Answer (2)
25. Answer (3)
26. Answer (3)
27. Answer (4)
28. Answer (2)
Here, ‘A’ is endoplasmic reticulum, ‘B’ is nucleus and
‘C’ is vacuole.
29. Answer (2)
30. Answer (4)
SECTION-III (Code-C)
1. Answer (2)
Let A = (–3, 2), B = (–5, –5), C = (2, –3) and D = (4, 4)
Sides of quadrilateral
2 23 5 2 5 53 unitsAB
2 25 2 5 3 53 units,BC
2 22 4 3 4 53 units,CD
and 2 24 3 4 2 53 unitsAD
ABCD is a rhombus.
Now, 2 23 2 2 3 5 2 unitsAC
and 2 25 4 5 4 9 2 unitsBD
Area of 1 5 2 9 2
2 2ABCD AC BD
= 45 sq. units
2. Answer (2)
Draw bisector of B intersecting AC at P and Join
PD.
A
B D
P
C
y y y
x x
Let C = y
B = 2y and let, A = 2x
BP = CP [∵ CBP = BCP = y]
ABP DCP
[By SAS as AB = CD, BP = CP and ABP = DCP = y]
BAP = CDP = 2x [By CPCT]
and DAP = ADP [∵ AP = DP]
In ABD,
ADC = ABD + BAD
x + 2x = 2y + x
x = y
CAB = CBA
AC = BC = 10 cm
[Sides opposite to equal angles]
3. Answer (3)
In ABD,
B + BAD = 90°
B + BAE – DAE = 90° ...(i)
and In ACD,
C + CAD = 90°
C + CAE + DAE = 90°
C + BAE + DAE = 90° ...(ii)
[∵ BAE = CAE]
Test-2 (Answers & Hints) All India Aakash Test Series for Class IX (2020)
7/8
Subtracting (ii) from (i),
B – C – 2DAE = 0
70 40
152 2
B CDAE
AED = 90° – DAE = 90° – 15° = 75°
4. Answer (3)
2 2
7 4 3 4 3 2 4 3x
24 3 2 3
1 1 2 3
2 34 32 3x
1
4x
x
Now,
33/2
3/2 3
1 1x x
xx
1 13x x
x x
[∵ (x + y)3 = x3 + y3 + 3xy (x + y)]
3/2 3
3/2
14 3 4 52x
x
5. Answer (1)
Let p(x) = a0x100 + a
1x99 + a
2x98 + a
3x97........
p(1) = a0 + a
1 + a
2 + a
3 .....
and p(–1) = a0 – a
1 + a
2 – a
3.....
p(1) + p(–1) = 2(a0 + a
2 + a
4 +....)
a0 + a
2 + a
4 ..... 1 1
2
p p
6. Answer (1)
7. Answer (3)
n4 – 2n3 –n2 + 2n = n2(n2 – 1) – 2n(n2 – 1)
= (n2 – 1) (n2 – 2n)
= (n – 2)(n – 1)n (n + 1)
= Product of 4 consecutive integers = divisible by 2, 4
and 3 i.e., 24
8. Answer (4)
In ABD and ACE,
AB = AC an AD = AE [Given]
and BAD = BAC + CAD = DAE + CAD = CAE
[∵ BAC = DAE]
ABD ACE [By SAS]
AEC = ADB = 30° [By CPCT]
9. Answer (2)
x x2
– – 12 x x3
– 9 – 7
– – – 12x x x3 2
x x2
+ 3 – 7
– – – 12x x2
4 + 5x
x + 1
+ +
+ +
Remainder = 4x + 5 = ax + 5
a = 4
–2a 10. Answer (4)
Let 4a = 14b = 8c = 7d = k
4 = k1/a, 14 = k1/b, 8 = k1/c, 7 = k1/d
k1/a × k1/b = k1/c × k1/d [∵ 4 × 14 = 8 × 7]
1 1 1 1
a b c dk k
1 1 1 1
a b c d
a b c d
ab cd
ab a b
cd c d
11. Answer (3)
12. Answer (2)
13. Answer (2)
A
B
C
D E F
G
In BDE,
BE + DE > BD
BE + DE > BC + AC ...(i) [∵ CD = AC]
and In ABC,
BC + AC > AB ...(ii)
BE + DE > AB ...(iii) [From (i) and (ii)]
All India Aakash Test Series for Class IX (2020) Test-2 (Answers & Hints)
8/8
Similarly, EG + GF > BE ...(iv)
Adding (iii) and (iv),
DE + EG + GF > AB
EG + GF > AB – DE
14. Answer (3)
2 2 2 2 2 2
4 5 5 6 6 7....
4 5 5 6 6 7
2 2
143 144
143 144
5 4 6 5 7 6 ............
144 143 144 4 = 12 – 2 = 10
4000
400 2010
15. Answer (4)
f(3) = 0 [By Factor Theorem]
34 – 33 –7(3)2 + 3a + b = 0
3a + b = 9 ...(i)
and f(2) = –12 [By Remainder Theorem]
24 –23 –7(2)2 + 2a + b = –12
2a + b = 8
b = 8 – 2a ...(ii)
Substituting (ii) in (i),
3a + 8 – 2a = 9
a = 9 – 8 = 1
b = 8 – 2 = 6
6 16 : 1
6a b
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