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Page 1
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
ANSWER KEYS
1. 4 2. 4 3. 1 4. 3
5 2 6. 4 7. 3 8. 2
9. 2 10. 2 11. 4 12. 1
13. 1 14. 3 15. 3 16. 4
17. 3 18. 3 19. 2 20. 3
21. 4 22. 4 23. 2 24. 1
25. 4 26. 1 27. 1 28. 2
29. 4 30. 1 31. 4 32. 4
33. 3 34. 2 35. 4 36. 4
37. 3 38. 2 39. 4 40. 2
41. 3 42. 2 43. 1 44. 2
45. 2 46. 4 47. 2 48. 3
49. 1 50. 2 51. 4 52. 3
53. 3 54. 3 55. 3 56. 2
57. 4 58. 1 59. 1 60. 4
61. 4 62. 2 63. 4 64. 2
65. 3 66. 1 67. 2 68. 3
69. 2 70. 1 71. 3 72. 1
73. 2 74. 4 75. 4 76. 2
77. 2 78. 1 79. 3 80. 3
81. 4 82. 3 83. 3 84. 1
85. 2 86. 3 87. 3 88. 3
89. 2 90. 2 91. 4 92. 3
93. 3 94. 1 95. 4 96. 3
97. 3 98. 4 99. 3 100. 1
Page 2
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
ANSWER KEYS
1. 4 2. 4 3. 1 4. 3
5 2 6. 4 7. 3 8. 2
9. 2 10. 2 11. 4 12. 1
13. 1 14. 3 15. 3 16. 4
17. 3 18. 3 19. 2 20. 3
21. 4 22. 4 23. 2 24. 1
25. 4 26. 1 27. 1 28. 2
29. 4 30. 1 31. 4 32. 4
33. 3 34. 2 35. 4 36. 4
37. 3 38. 2 39. 4 40. 2
41. 3 42. 2 43. 1 44. 2
45. 2 46. 4 47. 2 48. 3
49. 1 50. 2 51. 4 52. 3
53. 3 54. 3 55. 3 56. 2
57. 4 58. 1 59. 1 60. 4
61. 4 62. 2 63. 4 64. 2
65. 3 66. 1 67. 2 68. 3
69. 2 70. 1 71. 3 72. 1
73. 2 74. 4 75. 4 76. 2
77. 2 78. 1 79. 3 80. 3
81. 4 82. 3 83. 3 84. 1
85. 2 86. 3 87. 3 88. 3
89. 2 90. 2 91. 4 92. 3
93. 3 94. 1 95. 4 96. 3
97. 3 98. 4 99. 3 100. 1
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
HINTS & SOLUTIONS
1. 4
Sol. A
B
C
S K R
Q
P
M
N
x
2x
x
AB = 10
AC = 10
P, Q are mid points of AB & AC
? AP = PB = 5
AQ = QC = 5
BC = 12
? x + 2x + x = 12 ? x = 3
? BS = 3, SR = 6, RC = 3
?ABC is isosceles
AK ? BC and BK = KC
? BS = SK = KR = RC = 3
PQ || BC
? ABK ?
BS BP
SP || AK PSB AKB
SK PA
= ? ? ? ?
? PS ? BC and
PS 1
AK 2
=
1
PS AK
2
=
Similarly, QR ? BC
22
ACK AK 10 6 8 ? ? = - =
1
PS AK PS 4
2
= ? =
and similarly QR = 4
Page 3
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
ANSWER KEYS
1. 4 2. 4 3. 1 4. 3
5 2 6. 4 7. 3 8. 2
9. 2 10. 2 11. 4 12. 1
13. 1 14. 3 15. 3 16. 4
17. 3 18. 3 19. 2 20. 3
21. 4 22. 4 23. 2 24. 1
25. 4 26. 1 27. 1 28. 2
29. 4 30. 1 31. 4 32. 4
33. 3 34. 2 35. 4 36. 4
37. 3 38. 2 39. 4 40. 2
41. 3 42. 2 43. 1 44. 2
45. 2 46. 4 47. 2 48. 3
49. 1 50. 2 51. 4 52. 3
53. 3 54. 3 55. 3 56. 2
57. 4 58. 1 59. 1 60. 4
61. 4 62. 2 63. 4 64. 2
65. 3 66. 1 67. 2 68. 3
69. 2 70. 1 71. 3 72. 1
73. 2 74. 4 75. 4 76. 2
77. 2 78. 1 79. 3 80. 3
81. 4 82. 3 83. 3 84. 1
85. 2 86. 3 87. 3 88. 3
89. 2 90. 2 91. 4 92. 3
93. 3 94. 1 95. 4 96. 3
97. 3 98. 4 99. 3 100. 1
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
HINTS & SOLUTIONS
1. 4
Sol. A
B
C
S K R
Q
P
M
N
x
2x
x
AB = 10
AC = 10
P, Q are mid points of AB & AC
? AP = PB = 5
AQ = QC = 5
BC = 12
? x + 2x + x = 12 ? x = 3
? BS = 3, SR = 6, RC = 3
?ABC is isosceles
AK ? BC and BK = KC
? BS = SK = KR = RC = 3
PQ || BC
? ABK ?
BS BP
SP || AK PSB AKB
SK PA
= ? ? ? ?
? PS ? BC and
PS 1
AK 2
=
1
PS AK
2
=
Similarly, QR ? BC
22
ACK AK 10 6 8 ? ? = - =
1
PS AK PS 4
2
= ? =
and similarly QR = 4
Now in ?QRS ? ?QRS = 90°
2 2 2 2
SQ SR RQ 6 4 2 13 ? = + = + =
? PQRS will be a rectangle
[PQRS] = 6 × 4 = 24
? [PQS] = [SRQ] = 12
Now [PSQ]
1
SQ PM 12
2
= ? ? =
1
2 13 PM 12
2
? ? =
12
PM
13
=
Similarly,
12
RN
13
=
Now, In right angle triangle ?PSM
2
2 2 2
12 8
SM PS PM 4
13 13
??
= - = - =
??
??
Similarly,
8
QN
13
=
Now, MN = SQ – SM – QN
8 8 26 16 10
2 13
13 13 13 13
-
= - - = =
4. 3
Sol. Thyroxin regulates carbohydrate, protein and fat metabolism in the body so as to provide the
best balance for growth.
6. 4
Sol. Fertilisation takes place in fallopian tube.
8. 2
Sol. Hibiscus and mustard are bisexual flowers.
9. 2
Sol.
A
E
D
B
F
x 15
9 15
C
O
22
AB 15 9 12 cm BD AC CE = - = = = =
Let OF = x cm then
2
DE 2 225 x =-
Using
abc
R
4
=
?
( )
( )
2
2
24 24 2 225 x
15
1
4 15 x 2 225 x
2
? ? -
=
? ? + -
Page 4
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
ANSWER KEYS
1. 4 2. 4 3. 1 4. 3
5 2 6. 4 7. 3 8. 2
9. 2 10. 2 11. 4 12. 1
13. 1 14. 3 15. 3 16. 4
17. 3 18. 3 19. 2 20. 3
21. 4 22. 4 23. 2 24. 1
25. 4 26. 1 27. 1 28. 2
29. 4 30. 1 31. 4 32. 4
33. 3 34. 2 35. 4 36. 4
37. 3 38. 2 39. 4 40. 2
41. 3 42. 2 43. 1 44. 2
45. 2 46. 4 47. 2 48. 3
49. 1 50. 2 51. 4 52. 3
53. 3 54. 3 55. 3 56. 2
57. 4 58. 1 59. 1 60. 4
61. 4 62. 2 63. 4 64. 2
65. 3 66. 1 67. 2 68. 3
69. 2 70. 1 71. 3 72. 1
73. 2 74. 4 75. 4 76. 2
77. 2 78. 1 79. 3 80. 3
81. 4 82. 3 83. 3 84. 1
85. 2 86. 3 87. 3 88. 3
89. 2 90. 2 91. 4 92. 3
93. 3 94. 1 95. 4 96. 3
97. 3 98. 4 99. 3 100. 1
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
HINTS & SOLUTIONS
1. 4
Sol. A
B
C
S K R
Q
P
M
N
x
2x
x
AB = 10
AC = 10
P, Q are mid points of AB & AC
? AP = PB = 5
AQ = QC = 5
BC = 12
? x + 2x + x = 12 ? x = 3
? BS = 3, SR = 6, RC = 3
?ABC is isosceles
AK ? BC and BK = KC
? BS = SK = KR = RC = 3
PQ || BC
? ABK ?
BS BP
SP || AK PSB AKB
SK PA
= ? ? ? ?
? PS ? BC and
PS 1
AK 2
=
1
PS AK
2
=
Similarly, QR ? BC
22
ACK AK 10 6 8 ? ? = - =
1
PS AK PS 4
2
= ? =
and similarly QR = 4
Now in ?QRS ? ?QRS = 90°
2 2 2 2
SQ SR RQ 6 4 2 13 ? = + = + =
? PQRS will be a rectangle
[PQRS] = 6 × 4 = 24
? [PQS] = [SRQ] = 12
Now [PSQ]
1
SQ PM 12
2
= ? ? =
1
2 13 PM 12
2
? ? =
12
PM
13
=
Similarly,
12
RN
13
=
Now, In right angle triangle ?PSM
2
2 2 2
12 8
SM PS PM 4
13 13
??
= - = - =
??
??
Similarly,
8
QN
13
=
Now, MN = SQ – SM – QN
8 8 26 16 10
2 13
13 13 13 13
-
= - - = =
4. 3
Sol. Thyroxin regulates carbohydrate, protein and fat metabolism in the body so as to provide the
best balance for growth.
6. 4
Sol. Fertilisation takes place in fallopian tube.
8. 2
Sol. Hibiscus and mustard are bisexual flowers.
9. 2
Sol.
A
E
D
B
F
x 15
9 15
C
O
22
AB 15 9 12 cm BD AC CE = - = = = =
Let OF = x cm then
2
DE 2 225 x =-
Using
abc
R
4
=
?
( )
( )
2
2
24 24 2 225 x
15
1
4 15 x 2 225 x
2
? ? -
=
? ? + -
x 4.2 cm ?=
12. 1
Sol.
C
B
A
X
Z
Y
Q
R
P
By using Menalau’s theorem
CR : RQ : QX = BP : PR : RZ = AQ : QP : PY = 3 : 3 : 1
Now, AX : XB = 1 : 2
? ? ? ? ?=
2
BXC ABC
3
CR : RX = 3 : 4
? ? ? ? ? ? ? = =
48
BRX BXC ABC
7 21
Join P to X, Since RP : PB = 1:1
? ? ? ? ? ?
14
XPR BRX ABC
2 21
? = =
RQ : QX = 3 : 1
? ? ? ? ? ?
31
PQR XPR ABC
47
? = =
( )
2
13
ar PQR 14 14 7 3 cm
74
? = ? ? ? =
15. 3
Sol. Arteries always carry blood away from heart, while veins always carry blood towards the
heart.
16. 4
Sol. 0.001 M NaOH ? [OH
–
] = 10
–3
M
? pOH = 3, so pH = 11
17. 3
Sol. Draw AQ ? ED
Then EQ = QD = 1 cm
Also, BP = PC = 1 cm
Join A to D and E
Now AP 3 cm, PQ 2cm ==
( )
AQ 3 2 cm ? = +
A
E D
C
2
2
B
1 1
1 1
P
2 2
Q
By Pythagoras theorem
( )
2
AD AE 3 2 1 = = + +
Now, given circle is circumcircle of ?AED
( ) ( ) ( )
( )
AE ED AD
Radius
4 ar AED
?=
?
( )
( )
( )
2
3 2 1 2
2 cm
1
4 2 3 2
2
??
++
??
??
==
? ? ? +
Page 5
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
ANSWER KEYS
1. 4 2. 4 3. 1 4. 3
5 2 6. 4 7. 3 8. 2
9. 2 10. 2 11. 4 12. 1
13. 1 14. 3 15. 3 16. 4
17. 3 18. 3 19. 2 20. 3
21. 4 22. 4 23. 2 24. 1
25. 4 26. 1 27. 1 28. 2
29. 4 30. 1 31. 4 32. 4
33. 3 34. 2 35. 4 36. 4
37. 3 38. 2 39. 4 40. 2
41. 3 42. 2 43. 1 44. 2
45. 2 46. 4 47. 2 48. 3
49. 1 50. 2 51. 4 52. 3
53. 3 54. 3 55. 3 56. 2
57. 4 58. 1 59. 1 60. 4
61. 4 62. 2 63. 4 64. 2
65. 3 66. 1 67. 2 68. 3
69. 2 70. 1 71. 3 72. 1
73. 2 74. 4 75. 4 76. 2
77. 2 78. 1 79. 3 80. 3
81. 4 82. 3 83. 3 84. 1
85. 2 86. 3 87. 3 88. 3
89. 2 90. 2 91. 4 92. 3
93. 3 94. 1 95. 4 96. 3
97. 3 98. 4 99. 3 100. 1
NTSE STAGE – I (HARYANA STATE)
(2020 – 21)
(For Class – X)
SCHOLASTIC APTITUDE TEST
HINTS & SOLUTIONS
1. 4
Sol. A
B
C
S K R
Q
P
M
N
x
2x
x
AB = 10
AC = 10
P, Q are mid points of AB & AC
? AP = PB = 5
AQ = QC = 5
BC = 12
? x + 2x + x = 12 ? x = 3
? BS = 3, SR = 6, RC = 3
?ABC is isosceles
AK ? BC and BK = KC
? BS = SK = KR = RC = 3
PQ || BC
? ABK ?
BS BP
SP || AK PSB AKB
SK PA
= ? ? ? ?
? PS ? BC and
PS 1
AK 2
=
1
PS AK
2
=
Similarly, QR ? BC
22
ACK AK 10 6 8 ? ? = - =
1
PS AK PS 4
2
= ? =
and similarly QR = 4
Now in ?QRS ? ?QRS = 90°
2 2 2 2
SQ SR RQ 6 4 2 13 ? = + = + =
? PQRS will be a rectangle
[PQRS] = 6 × 4 = 24
? [PQS] = [SRQ] = 12
Now [PSQ]
1
SQ PM 12
2
= ? ? =
1
2 13 PM 12
2
? ? =
12
PM
13
=
Similarly,
12
RN
13
=
Now, In right angle triangle ?PSM
2
2 2 2
12 8
SM PS PM 4
13 13
??
= - = - =
??
??
Similarly,
8
QN
13
=
Now, MN = SQ – SM – QN
8 8 26 16 10
2 13
13 13 13 13
-
= - - = =
4. 3
Sol. Thyroxin regulates carbohydrate, protein and fat metabolism in the body so as to provide the
best balance for growth.
6. 4
Sol. Fertilisation takes place in fallopian tube.
8. 2
Sol. Hibiscus and mustard are bisexual flowers.
9. 2
Sol.
A
E
D
B
F
x 15
9 15
C
O
22
AB 15 9 12 cm BD AC CE = - = = = =
Let OF = x cm then
2
DE 2 225 x =-
Using
abc
R
4
=
?
( )
( )
2
2
24 24 2 225 x
15
1
4 15 x 2 225 x
2
? ? -
=
? ? + -
x 4.2 cm ?=
12. 1
Sol.
C
B
A
X
Z
Y
Q
R
P
By using Menalau’s theorem
CR : RQ : QX = BP : PR : RZ = AQ : QP : PY = 3 : 3 : 1
Now, AX : XB = 1 : 2
? ? ? ? ?=
2
BXC ABC
3
CR : RX = 3 : 4
? ? ? ? ? ? ? = =
48
BRX BXC ABC
7 21
Join P to X, Since RP : PB = 1:1
? ? ? ? ? ?
14
XPR BRX ABC
2 21
? = =
RQ : QX = 3 : 1
? ? ? ? ? ?
31
PQR XPR ABC
47
? = =
( )
2
13
ar PQR 14 14 7 3 cm
74
? = ? ? ? =
15. 3
Sol. Arteries always carry blood away from heart, while veins always carry blood towards the
heart.
16. 4
Sol. 0.001 M NaOH ? [OH
–
] = 10
–3
M
? pOH = 3, so pH = 11
17. 3
Sol. Draw AQ ? ED
Then EQ = QD = 1 cm
Also, BP = PC = 1 cm
Join A to D and E
Now AP 3 cm, PQ 2cm ==
( )
AQ 3 2 cm ? = +
A
E D
C
2
2
B
1 1
1 1
P
2 2
Q
By Pythagoras theorem
( )
2
AD AE 3 2 1 = = + +
Now, given circle is circumcircle of ?AED
( ) ( ) ( )
( )
AE ED AD
Radius
4 ar AED
?=
?
( )
( )
( )
2
3 2 1 2
2 cm
1
4 2 3 2
2
??
++
??
??
==
? ? ? +
Circumference = 4 ? cm
19. 2
Sol. Tall dwarf
TT tt
?
Tt ? Tt F1 generation
?
TT
Tt
Tt
tt
T t
T
t
Homozygous tall – 1
Heterozygous tall – 2
Homozygous dwarf – 1
20. 3
Sol. One mole SO 2 means 64 g of it = 6.022 ? 10
23
molecules.
21. 4
Sol. Growing two or more crops but indefinite row pattern is known as mixed cropping.
23. 2
Sol.
A
P
B
D C
Q
x
6-x
r
O
r
6cm 6cm
4cm 4cm
Let radius of circle = r cm
and OQ = x cm then OP = (6 – x) cm
By Pythagoras theorem,
r
2
= x
2
+ 36 and r
2
= 16 + (6 – x)
2
on solving both equations we get,
2
4 340
x and r
39
==
Distance of chord EF from centre = - =
45
3 cm
33
So, length of chord EF
2
2
5
2r
3
??
=-
??
??
340 25
2 2 35
99
= - =
140 K 140 = ? =
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