Test No. 3 29-11-2020 D, E & F Olympiads, NTSE & Class X-2021 for
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Test No. 3
29-11-2020
D, E & F
Olympiads, NTSE & Class X-2021for
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Test-3 (Answers) All India Aakash Test Series-2021 (Class X)
1/6
All India Aakash Test Series - 2021 (Class X)
TEST - 3
Test Date : 29-11-2020 ANSWERS SECTION-I (Code-D)
1. (2)
2. (2)
3. (4)
4. (4)
5. (3)
6. (2)
7. (2)
8. (2)
9. (1)
10. (2)
11. (1)
12. (3)
13. (2)
14. (4)
15. (4)
16. (4)
17. (2)
18. (2)
19. (3)
20. (4)
21. (2)
22. (4)
23. (3)
24. (2)
25. (3)
26. (3)
27. (4)
28. (2)
29. (3)
30. (4)
31. (3)
32. (2)
33. (2)
34. (3)
35. (4)
36. (3)
37. (2)
38. (4)
39. (3)
40. (4)
41. (3)
42. (1)
43. (1)
44. (3)
45. (1)
46. (3)
47. (1)
48. (2)
49. (4)
50. (4)
51. (3)
52. (3)
53. (4)
54. (1)
55. (1)
56. (2)
57. (3)
58. (4)
59. (2)
60. (3)
61. (2)
62. (2)
63. (3)
64. (2)
65. (2)
66. (3)
67. (3)
68. (1)
69. (1)
70. (4)
71. (4)
72. (1)
73. (1)
74. (3)
75. (3)
76. (4)
77. (1)
78. (4)
79. (1)
80. (2)
81. (1)
82. (2)
83. (3)
84. (4)
85. (3)
86. (2)
87. (1)
88. (4)
89. (2)
90. (3)
91. (4)
92. (1)
93. (3)
94. (1)
95. (4)
96. (1)
97. (2)
98. (3)
99. (4)
100. (2)
SECTION-II (Code-E)
1. (4) 2. (1) 3. (4) 4. (3) 5. (2) 6. (3)
7. (4) 8. (3) 9. (2) 10. (2) 11. (2) 12. (1)
13. (3) 14. (2) 15. (2) 16. (3) 17. (2) 18. (2)
19. (1) 20. (3) 21. (4) 22. (2) 23. (3) 24. (Deleted)
25. (1) 26. (4) 27. (3) 28. (3) 29. (4) 30. (3)
SECTION-III (Code-F)
1. (1) 2. (4) 3. (3)
4. (4) 5. (4) 6. (3)
7. (2) 8. (1) 9. (1)
10. (2) 11. (4) 12. (2)
13. (2) 14. (3) 15. (3)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series-2021 (Class X) Test-3 (Answers & Hints)
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Answers & Hints SECTION-I (Code-D)
1. Answer (2)
2. Answer (2)
3. Answer (4)
4. Answer (4)
5. Answer (3)
6. Answer (2)
7. Answer (2)
F = qvBsinθ
= 1.6 × 10–19 × 105 × 5 × sin90°
= 8 × 10–14 N
8. Answer (2)
qvB = mg
mgvqB
=
–3
–510 1010 1
×=
×
= 103 m/s
9. Answer (1)
10. Answer (2)
11. Answer (1)
12. Answer (3)
13. Answer (2)
14. Answer (4)
15. Answer (4)
34 2a
IB
aµ
=
34 2b
IB
bµ
=
Bnet = Ba – Bb
3 1 1–8
Ia b
= µ
3 ( – )8
b aIab
= µ
16. Answer (4)
17. Answer (2)
( )sin sin4
IB
rµ
= α + βπ
–74 10 4(sin30 sin45 )
4 4π × × ° + °
=π ×
–7 1 1102 2
= +
–71 2 10 T2
+= ×
18. Answer (2)
19. Answer (3)
20. Answer (4)
BA = BB
2
2 2 ( – )I I
x x aµ µ
=π π
⇒ 2 1–x x a
=
⇒ x = 2a
21. Answer (2)
22. Answer (4)
23. Answer (3)
24. Answer (2)
25. Answer (3)
26. Answer (3)
27. Answer (4)
28. Answer (2)
29. Answer (3)
30. Answer (4)
31. Answer (3)
32. Answer (2)
33. Answer (2)
Sunlight4 2 3
' X'CH Cl CH Cl HCl+ → +
34. Answer (3)
Number of covalent bonds in benzene and methanol are 15 and 5 respectively.
35. Answer (4)
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Test-3 (Answers & Hints) All India Aakash Test Series-2021 (Class X)
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36. Answer (3)
37. Answer (2)
38. Answer (4)
39. Answer (3)
40. Answer (4)
41. Answer (3)
42. Answer (1)
43. Answer (1)
44. Answer (3)
45. Answer (1)
46. Answer (3)
47. Answer (1)
48. Answer (2)
49. Answer (4)
50. Answer (4)
51. Answer (3)
Planaria reproduces through regeneration.
52. Answer (3)
Plasmodium causes malaria.
53. Answer (4) 54. Answer (1)
55. Answer (1)
56. Answer (2)
57. Answer (3)
58. Answer (4)
59. Answer (2)
60. Answer (3)
61. Answer (2)
62. Answer (2)
63. Answer (3)
64. Answer (2)
65. Answer (2)
66. Answer (3)
67. Answer (3)
68. Answer (1)
69. Answer (1)
70. Answer (4)
71. Answer (4)
72. Answer (1)
Since, ∠DAC = 30°
∴ ∠OAC = 15°
Now, in ∆OAC
sin15 OCOA
° =
⇒ ( )sin 45 30 rOA
° − ° =
⇒ sin45 cos30 cos45 sin30rOA
= ° ° − ° °
⇒ ( )3 1
2 2r
OA
−=
⇒ ( )
( )( )
( )3 1 2 2 3 12 223 1 3 1
rrOA+ +
= × =− +
⇒ ( )2 3 1OA r= + …(i)
Now, in ∆AOB
sin60 OBOA
° =
⇒ 32
OBOA
=
⇒ ( )6 3 1
2
rOB
+=
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series-2021 (Class X) Test-3 (Answers & Hints)
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73. Answer (1)
74. Answer (3)
Given ∆PQR is an isosceles triangle with
PQ = PR = K cm.
Let PS be perpendicular bisector on QR.
Now, in ∆PQS,
sin sin PSPQSPQ
∠ = θ =
⇒ 35
PSK
= 4cos5
θ =
⇒ 35KPS =
Now, cos QSPQ
θ =
⇒ 45
QSK
=
⇒ 45KQS =
Now, area of triangle = 12
PS QR× ×
1 3 4 22 5 5
K K= × × × [ QR = 2QS]
212
25K
=
⇒ 212
25KM =
Clearly, 2 2
4 2K KM< <
75. Answer (3)
76. Answer (4)
77. Answer (1)
In the given figure, ∠BDC = ∠BEC = 90°.
[Angle in semi-circle]
In ∆ADC,
∠ADC = 90°
∴ AC2 = AD2 + DC2 [Using Pythagoras Theorem]
⇒ (25)2 = AD2 + (20)2
⇒ AD2 = 625 – 400
⇒ AD = 15 cm
⇒ BD = 15 cm
∴ ∆ABC is an isosceles triangle.
∴ AD = BD and CD ⊥ AB.
∴ AC = BC = 25 cm
Area of ∆ABC = 1 12 2
CD AB BE AC× × = × ×
⇒ 20 × 30 = BE × 25
⇒ BE = 24 cm
Also, AT is tangent to the circle.
∴ AT2 = AD × AB
⇒ AT2 = 15 × 30
⇒ 15 2 cmAT =
78. Answer (4)
79. Answer (1)
80. Answer (2)
81. Answer (1)
Difference of 112, 212, 312, 412, and so on.
82. Answer (2)
83. Answer (3)
I S R O | I S R O | I S R O | I S R O
84. Answer (4)
85. Answer (3)
86. Answer (2)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Test-3 (Answers & Hints) All India Aakash Test Series-2021 (Class X)
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87. Answer (1)
88. Answer (4)
89. Answer (2)
90. Answer (3)
91. Answer (4)
92. Answer (1)
Position of opposite letter written in reverse order.
93. Answer (3)
94. Answer (1)
95. Answer (4)
abc : a × c + b
96. Answer (1)
Except F all letters are placed at prime number position.
97. Answer (2)
98. Answer (3)
Letters at prime number position.
99. Answer (4)
100. Answer (2)
SECTION-II (Code-E)
1. Answer (4)
2. Answer (1)
3. Answer (4)
4. Answer (3)
5. Answer (2)
6. Answer (3)
7. Answer (4)
8. Answer (3)
9. Answer (2)
10. Answer (2)
The process shown here is regeneration.
11. Answer (2)
Self-pollination does not produce variations.
12. Answer (1)
13. Answer (3)
14. Answer (2)
15. Answer (2)
16. Answer (3)
17. Answer (2)
18. Answer (2)
19. Answer (1)
20. Answer (3)
Cervical cap is a barrier method.
21. Answer (4)
22. Answer (2)
23. Answer (3)
24. Deleted
25. Answer (1)
26. Answer (4)
27. Answer (3)
28. Answer (3)
Oxygen was not used in Miler and Urey experiment.
29. Answer (4)
30. Answer (3)
SECTION-III (Code-F)
1. Answer (1)
2. Answer (4)
Let ABC be the required triangle since, its angle
are in the ratio 1 : 2 : 3.
∴ Let ∠C : ∠B : ∠A = 1 : 2 : 3.
⇒ ∠C = 30°, ∠B = 60° and ∠A = 90°
⇒ BC = 32 cm
Now, in ∆ABC,
sin60 ACBC
° =
⇒ 3 16 32
BCAC ×= = [ BC = 32 cm]
cos60 ABBC
° =
⇒ 32 16 cm2 2
BCAB = = = [ BC = 32 cm]
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series-2021 (Class X) Test-3 (Answers & Hints)
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Now perimeter = AB + BC + AC
16 32 16 3= + +
16 3 48= +
( )16 3 3 1 cm= +
3. Answer (3)
Join CD.
Since, AD = BC and ∠A = ∠B = 90°.
∴ ABCD is a rectangle.
Now, 20 2 cmCD AB= =
Now, in ∆ODC,
OD2 + OC2 = DC2
⇒ 2OD2 = DC2
[ OD = OC = Radius of circle]
⇒ 2 400 22
OD ×=
⇒ OD = 20 cm
Now, Area of sector ( )2
4OCπ
=
( )2204
π=
= π × 100
= 100π cm2
Area of 12
COD OD OC∆ = × ×
1 20 202
= × ×
= 200 cm2
∴ Area of segment (DEC) = 100π – 200
= 100(π – 2) cm2…(i)
Area of 12
COD CD OE∆ = × ×
⇒ 1200 20 22
OE= × ×
⇒ 10 2OE =
∴ 25 2 10 2AD = −
∴ 15 2 cmAD =
Now, area of shaded region = Area of rectangle ABCD – area of
segment DEC
= AB × AD – 100(π – 2) cm2
( ) 220 2 15 2 100 200 cm= × − π +
= (600 + 200 – 100π) cm2
= 100(8 – π) cm2 4. Answer (4) 5. Answer (4) 6. Answer (3) 7. Answer (2) 8. Answer (1) Largest triangle must be an isosceles triangle
inscribed in semi-circle. 9. Answer (1) 10. Answer (2) 11. Answer (4) 12. Answer (2) 13. Answer (2) 14. Answer (3) PQ = PR and PS = SR
∴ PQ = 2PS Since, PS is a tangent.
∴ PS2 = PT × PQ
⇒ 2
2PQ PT PQ = ×
⇒ PQ = 4PT
⇒ PQ = 4 × 4.5
⇒ PQ = 18 cm 15. Answer (3)
Edition: 2020-21
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