Test No. 1

02-08-2020

D, E & F

Olympiads, NTSE & Class X-2021for

Test-1 (Answers) All India Aakash Test Series-2021 (Class X)

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All India Aakash Test Series - 2021 (Class X) TEST - 1

Test Date : 02-08-2020 ANSWERS SECTION-I (Code-D)

1. (4) 2. (2) 3. (2) 4. (2) 5. (2) 6. (3) 7. (2) 8. (2) 9. (3) 10. (3) 11. (2) 12. (4) 13. (1) 14. (1) 15. (2) 16. (4) 17. (4) 18. (4) 19. (3) 20. (4)

21. (3) 22. (2) 23. (4) 24. (1) 25. (4) 26. (2) 27. (1) 28. (2) 29. (3) 30. (1) 31. (4) 32. (2) 33. (4) 34. (3) 35. (2) 36. (4) 37. (1) 38. (3) 39. (3) 40. (2)

41. (3) 42. (2) 43. (2) 44. (4) 45. (3) 46. (1) 47. (4) 48. (2) 49. (3) 50. (2) 51. (1) 52. (3) 53. (3) 54. (1) 55. (1) 56. (2) 57. (2) 58. (3) 59. (1) 60. (1)

61. (2) 62. (1) 63. (3) 64. (4) 65. (1) 66. (4) 67. (2) 68. (2) 69. (3) 70. (3) 71. (2) 72. (4) 73. (1) 74. (2) 75. (4) 76. (3) 77. (4) 78. (1) 79. (2) 80. (2)

81. (1) 82. (3) 83. (4) 84. (2) 85. (4) 86. (3) 87. (1) 88. (4) 89. (3) 90. (2) 91. (3) 92. (1) 93. (1) 94. (4) 95. (3) 96. (4) 97. (1) 98. (4) 99. (1) 100. (2)

SECTION-II (Code-E)

1. (4) 2. (1) 3. (2) 4. (1) 5. (3) 6. (2)

7. (4) 8. (2) 9. (3) 10. (4) 11. (2) 12. (2)

13. (2) 14. (3) 15. (1) 16. (4) 17. (2) 18. (3)

19. (2) 20. (1) 21. (3) 22. (3) 23. (2) 24. (1)

25. (3) 26. (2) 27. (4) 28. (1) 29. (2) 30. (4)

SECTION-III (Code-F)

1. (4) 2. (Deleted) 3. (3)

4. (1) 5. (3) 6. (Deleted)

7. (3) 8. (3) 9. (1)

10. (2) 11. (4) 12. (1)

13. (4) 14. (2) 15. (2)

All India Aakash Test Series-2021 (Class X) Test-1 (Answers & Hints)

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Answers & Hints SECTION-I (Code-D)

1. Answer (4) 2. Answer (2) 3. Answer (2) 4. Answer (2)

∴ 1Pf

=

100 20 cm5

f = =

5. Answer (2)

Height of mirror required 4.8 2.4 ft.2 2H

= = =

6. Answer (3) δ = 360° – 2θ δ = 360° – 2 × 60° δ = 240° 7. Answer (2) 8. Answer (2) Let h be the height of bird above water surface then

2 8aw

h+ =

µ

42 83

h+ × =

9 4.5 m2

h = =

Actual distance between bird and fish = 4.5 + 2 = 6.5 m 9. Answer (3) 10. Answer (3)

From Snell’s law sin60° × 1 = sinθ × µ …(i)

and sinθ × µ = sin2θ × 1 …(ii) From equation (i) and (ii) Sin60° = sin2θ 2θ = 60° θ = 30° From Snell’s law sin60 1 sin30° × = ° × µ

3µ =

11. Answer (2) 12. Answer (4)

∴ 1

xct

=

2

5xvt

=

∴ 2 2

1 15 5t tc x

v t x tµ = = × =

13. Answer (1) r = 2i sini × µ = sinr × 1 sini × µ = sin2i

3 sin 2sin cosi i i× =

3cos cos302

i = = °

i = 30° 14. Answer (1)

From Snell’s law sin30 sin60d r° × µ = ° × µ

sin30 1 2 1sin60 2 3 3

r

d

µ °= = × =

µ °

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Test-1 (Answers & Hints) All India Aakash Test Series-2021 (Class X)

15. Answer (2) Let the vessel is filled upto height x, then

3–2

x

H x =

2 53 3x xH x= + =

35Hx =

16. Answer (4)

∴ –f vmf

=

1–12 –2–12

v+ =

24 = 12 + v1

v1 = + 12 cm

2–12 ––2–12

v=

–24 = 12 + v2

v2 = –36 cm

∆v = 36 + 12 = 48 cm 17. Answer (4) For end closer to mirror u = –40 cm

∴ 1 1 1f v u

= +

1

1 1 1–20 –40v

= +

1

1 1 1 1– –40 20 40v

= =

v1 = –40 cm For other end u = –44 cm

2

1 1 1–20 –44v

= +

22

1 1 1 5 – 11 –6 220– –20 44 220 220 6

vv

= + = = ⇒ =

Length of image of rod 220 2040 – cm6 6

= =

18. Answer (4) 19. Answer (3) 20. Answer (4)

∴ sin( – )cos

t i rdr

=

From Snell’s law

sin60 1 sin 3r° × = ×

1sin sin30 302

r r= = ° ⇒ = °

5 3 sin(60 – 30 ) 5 3 tan30 5 cmcos30

d ° °= = ° =

°

21. Answer (3) 22. Answer (2) 23. Answer (4) 24. Answer (1)

4Na(s) + O2(g) → 2Na2O(s) 25. Answer (4)

( ) ( ) ( )Δ3 2ZnCO s ZnO s CO g→ +

26. Answer (2) 27. Answer (1) ( ) ( )

( )( ) ( )2 2 4 4

Whitecolour

BaCl aq Na SO aq BaSO s 2NaCl aq+ → +

28. Answer (2) 29. Answer (3)

2 2Fe 2HCl FeCl H+ → + ↑

30. Answer (1) 31. Answer (4) 32. Answer (2) 33. Answer (4) 34. Answer (3) 35. Answer (2) 36. Answer (4)

( )( )

( ) ( ) ( )2 2Quick 'X'lime

CaO s H O l Ca OH aq+ →

( ) ( )( )

( ) ( ) ( )2 3 22'Y'' X'

Ca OH aq CO g CaCO s H O l+ → +

All India Aakash Test Series-2021 (Class X) Test-1 (Answers & Hints)

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37. Answer (1) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) 38. Answer (3) 39. Answer (3)

( ) ( ) ( ) ( ) ( )Δ3 2 222Pb NO s 2PbO s 4NO g +O g→ +

40. Answer (2)

( )

( )( )

Δ4 2 4

Green crystal ColourlessFeSO ·7H O FeSO s→

( ) ( ) ( ) ( )Δ4 2 3 2 3

'P' 'Q' 'R ' 'S'FeSO s Fe O s SO g SO g→ + +

41. Answer (3) 42. Answer (2) 43. Answer (2) 44. Answer (4) 45. Answer (3) 46. Answer (1) 47. Answer (4) 48. Answer (2) 49. Answer (3) Monophyodont teeth grows only once in life. . 50. Answer (2) 51. Answer (1) 52. Answer (3) 53. Answer (3) The labelled part R is lacteal. 54. Answer (1) The amount of energy produced by the hydrolysis

of one molecule of ATP is 30.5 kJ/mol. 55. Answer (1) 56. Answer (2) 57. Answer (2) 58. Answer (3) 59. Answer (1) The given reaction represents the splitting of water

molecule. 60. Answer (1) 61. Answer (2) Let the numbers be a – 3d, a –d, a + d and a + 3d. ⇒ 4a = 36 a = 9

Now,

2 2

2 2– 9 9

10–a da d

=

⇒ (81 – 9d2)10 = 9(81 – d2)

d = ±1

∴ The numbers are 6.8, 10, 12

62. Answer (1)

αβ – 3 (α + β) + 9

–7 –2– 3 93 3

= +

7 2611–3 3

= =

63. Answer (3)

24 6 8 42 3

2n n n

n n+ +

= + +

⇒ All possible values of n are ±1, ±2, ±4

∴ Sum of –1, –2, –4, 1, 2, 4 = 0

64. Answer (4)

4 74 9 10

k kk k

+= ≠

+

⇒ k2 – k – 2 = 0

⇒ k = 2, –1

And 28 40,3 53

k k≠ ≠

65. Answer (1)

a5 + a12 + a19 = a12 – 5d + a12 + a12 + 5d

= 3 × 30 = 90

66. Answer (4)

(4 3)(4 – 3) 13ba

= + =

– 4 3 4 – 3 8ca

= + + =

5 6 5 – 6 10d = + + =

And (5 6)(5 – 6) 19e = + =

∴ 138

b cade

+=

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Test-1 (Answers & Hints) All India Aakash Test Series-2021 (Class X)

67. Answer (2)

am – bl = bn – cm = 0

⇒ am = bl and bn = cm

⇒ anda b b cl m m n

= =

a b cl m n

= =

68. Answer (2)

18m + 5 = 9(2m) + 5 = 9q + 5

18m + 5 = 3(6m + 1) + 2 = 3q + 2

18m + 5 = 6(3m) + 5 = 6q + 5

18m + 5 = 2(9m + 2) + 1 = 2q + 1

69. Answer (3)

p(x) = ax2– bx + c

⇒ p(0) = c > 0

Vertex of parabola – –,2 4

b Da a

⇒ 02ba

<

⇒ b < 0 [a > 0 as parabola open upwards]

70. Answer (3)

k α=

β

⇒ –2–21k

k α β + = + β α

–22 2 α + β

= αβ

2

22

( )

( ) – 2

αβ=

α + β αβ

2

22

–314442401

5 –3– 24 4

= =

71. Answer (2)

∴ 5x – a + 3 = 5x + b

⇒ a + b = 3

72. Answer (4)

Here 18 – 16 = 2, 12 – 10 = 2, 20 – 18 = 2 and 38 – 36 = 2 and LCM of 18, 12, 20 and 38 = 3420

∴ the required number = 3420 – 2 = 3418

73. Answer (1)

Let the speed of stream be x,

⇒ 40 40 64–20 – 20 60x x

=+

⇒ 40 × 60 (20 + x – 20 + x) = 64(400 – x2)

⇒ x2 + 75x – 400 = 0

⇒ x = 5, –80

∴ Speed of stream = 5 km/hr

74. Answer (2)

2 3 7 3 2 43and24 144m b m b

+ = + = , where m and b

represents number of days taken by a man and a boy respectively to do the work.

Let 1 1and ,x ym b

= =

⇒ 72 324

x y+ = ...(i)

And 433 2144

x y+ = ...(ii)

Solving (i) and (ii), we get

1 1and16 18

x y= =

⇒ b = 18

⇒ 14 days4 2b

=

All India Aakash Test Series-2021 (Class X) Test-1 (Answers & Hints)

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75. Answer (4)

1 1 α + β+ =

α β αβ

122

3 34

= 1 32 4

and α + β = αβ =

And 1 1 1 43

× = =α β αβ

∴ Required polynomial

2 2 4–3 3xk x = +

2(3 – 2 4)3k x x= +

76. Answer (3) A – x = B – A = y – B ⇒ 2A = B + x and 2B = A + y

⇒ 22

A yA x+= +

⇒ 222

A y xA + +=

⇒ 3A = 2x + y And 3B = x + 2y

∴ 22

B x yA x y

+=

+

77. Answer (4) x = 2n + 1 for n being any integer

⇒ x2 – 1 = 4n2 + 4n = 4n(n + 1) ∴ It is always divisible by 2 × 4 = 8 78. Answer (1)

2 2 2 3 2.........+ +

( )10 2 2 9 22

= +

55 2= 79. Answer (2) 80. Answer (2) Let the number of rows be x and number of

columns be y, xy = (x – 5)(y + 10) = (x + 5)(y – 5) ⇒ 0 = 10x – 5y – 50 = 5y – 5x – 25

⇒ 2x – y = 10 …(i)

And x – y = –5 …(ii) Solving (i) and (ii), we get x = 15 and y = 20

∴ Total number of students = 15 × 20

= 300 81. Answer (1) m n o L | m n o L | m n o L | m n o L 82. Answer (3) a a b b c c | b b c c d d | c c d d e e 83. Answer (4) Letter is codded to square of (its position +1) 84. Answer (2) 85. Answer (4)

Pattern is n2 , where n = 81, 71, 61, 51, 41

86. Answer (3) Pattern obtained by difference of the prime

numbers starting from 41. 87. Answer (1)

n3 – (n – 1)2, where n = 1, 2, 3, 4, 5 and so on.. .

88. Answer (4) 89. Answer (3) Letters at prime positions. 90. Answer (2) First two letter and their opposite letter. 91. Answer (3) 92. Answer (1) 93. Answer (1)

n : (n + 1)2 – n

94. Answer (4)

abc : 2 × (a + b)c

95. Answer (3) abc : (a + b) × c 96. Answer (4) 97. Answer (1) 98. Answer (4) 99. Answer (1) 100. Answer (2)

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Test-1 (Answers & Hints) All India Aakash Test Series-2021 (Class X)

SECTION-II (Code-E) 1. Answer (4)

2. Answer (1)

3. Answer (2)

4. Answer (1)

5. Answer (3)

6. Answer (2)

7. Answer (4)

8. Answer (2)

9. Answer (3)

10. Answer (4)

11. Answer (2)

12. Answer (2)

13. Answer (2)

14. Answer (3)

15. Answer (1)

16. Answer (4)

17. Answer (2)

18. Answer (3)

19. Answer (2)

20. Answer (1)

21. Answer (3)

22. Answer (3)

The labelled parts P is Bowman’s capsule and R is proximal convoluted tubule.

23. Answer (2)

24. Answer (1)

25. Answer (3)

26. Answer (2)

27. Answer (4)

The labelled parts a, b, c and d are vena cava, aorta, pulmonary vein and left ventricle respectively.

28. Answer (1)

29. Answer (2)

30. Answer (4)

SECTION-III (Code-F) 1. Answer (4) 2. Deleted 3. Answer (3)

36000 = 25 × 32 × 53 Total number of factors = (5 + 1)(2 + 1)(3 + 1) = 72 And number of odd factors = (2 + 1)(3 + 1) [Combination of all powers of 3 and 5] = 12

⇒ Number of even factors = 72 – 12 = 60

∴ Product = 60 × 12 = 720 4. Answer (1)

Let 1 1, ,– 1 – 2

a bx y

= =

⇒ 2a + 3b = –2 …(i) And 4a – 9b = 1 …(ii) Solving (i) and (ii), we get

–1 1 –1 1and2 1 3 – 2

a bx y

= = = =+

⇒ x = –1 and y = –1

∴ x + y = –2 5. Answer (3)

x2 = 42 + x and y2 = 132 + y

⇒ x2 – x – 42 = 0 and y2 – y – 132 = 0

⇒ (x – 7)(x + 6) = 0 and (y – 12)(y + 11) = 0

⇒ x = 7 and y = 12 [ x, y cannot be negative] 6. Deleted 7. Answer (3) Let speed of train and car be x and y respectively,

⇒ 450 350 554x y

+ =

And 300 500 956x y

+ =

Let 1 1, ,u vx y

= =

All India Aakash Test Series-2021 (Class X) Test-1 (Answers & Hints)

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⇒ 55450 3504

u v+ = …(i)

And 95300 5006

u v+ = … (ii)

Solving (i) and (ii) we get,

1 1and 90 40

u v= =

∴ Difference of speeds = 90 – 40 = 50 km/hr 8. Answer (3)

2002 2

2002( 4 ) 4 2449.25x x x

x+

= ×

⇒ x2 + 4x – 9797 = 0

⇒ –4 16 4 97972

x ± + ×=

–4 1982±

=

= 97, –101 9. Answer (1) 10. Answer (2)

7805 = (74)201 × 7

= (2401)201 × 7

∴ When 2401201 is divided by 24, the remainder is 1. [2401 = 24 × 100 + 1]

⇒ Remainder when 7805 is divided by 24 = 7. 11. Answer (4) As AM > GM

13( )

3a b c abc+ +

> …(i)

And 13( )

3ab bc ca ab bc ca+ +

> ⋅ ⋅ …(ii)

Multiplying (i) and (ii), we get

1 23 3( )( ) ( ) ( )

9a b c ab bc ca abc abc+ + + +

> ×

⇒ ( )( )9

a b c ab bc ca abc+ + + +>

⇒ ( )( ) 9a b c ab bc caabc

+ + + +>

12. Answer (1)

Let , ,5

xy xx

= = α β+

⇒ 51–

yxy

=

⇒ 25 510 – 1 0

1– 1–y yy y

+ =

⇒ 26y2 – 52y + 1 = 0 ∴ Required equation is 26x2 – 52x + 1 = 0 13. Answer (4)

1 1 1 – 1 – 1αβ + βγ + αγ+ + =

α β γ αβγ

–13 1– 1–12 12

= =

14. Answer (2) Constant term is zero if line passes through origin. ⇒ c = n = f = 0 15. Answer (2) mam = nan

⇒ ma + m(m – 1)d = na + n (n – 1)d ⇒ (m – n)a = (n2 – n – m2 + m)d

⇒ ( ) ( )( )( – ) – – –m n a m n m n m n d = +

⇒ a = (1 – m – n)d ⇒ a + (m + n – 1)d = 0 ⇒ am + n = 0 ⇒ x = 0 ∴ 2496780 = 1

Edition: 2020-21