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 Page 1


   
 
CHANDIGARH NTSE STAGE 1 2019-20  
(SCHOLASTIC APTITUDE TEST) 
 
 ANSWER KEYS 
 
 
 1. A 2. B 3. A 4. A 5. C 
 6. A 7. A 8. C 9. D 10. D 
 11. D 12. A 13. B 14. B 15. C 
 16. D 17. C 18. A 19. C 20. A 
 21. A 22. C 23. A 24. D 25. C 
 26. C 27. D 28. D 29. A 30. C 
 31. C 32. D 33. D 34. C 35. B 
 36. B 37. D 38. D 39. B 40. D 
 41. C 42. D 43. C 44. D 45. C 
 46. C 47. C 48. C 49. B 50. D  
 51. A 52. B 53. A 54. A 55. D 
 56. C 57. ** 58. A 59. B 60. B 
 61. B 62. B 63. D 64. B 65. B 
 66. B 67. D 68. D 69. A 70. B 
 71. C 72. C 73. B 74. B 75. C 
 76. C 77. B 78. C 79. B 80. B 
 81. B 82. ** 83. B 84. C 85. D 
 86. D 87. A 88. C 89. B 90. B 
 91. D 92. C 93. C 94. A 95. A 
 96. D 97. D 98. A 99. A 100. B  
 
** None of the option is correct  
Page 2


   
 
CHANDIGARH NTSE STAGE 1 2019-20  
(SCHOLASTIC APTITUDE TEST) 
 
 ANSWER KEYS 
 
 
 1. A 2. B 3. A 4. A 5. C 
 6. A 7. A 8. C 9. D 10. D 
 11. D 12. A 13. B 14. B 15. C 
 16. D 17. C 18. A 19. C 20. A 
 21. A 22. C 23. A 24. D 25. C 
 26. C 27. D 28. D 29. A 30. C 
 31. C 32. D 33. D 34. C 35. B 
 36. B 37. D 38. D 39. B 40. D 
 41. C 42. D 43. C 44. D 45. C 
 46. C 47. C 48. C 49. B 50. D  
 51. A 52. B 53. A 54. A 55. D 
 56. C 57. ** 58. A 59. B 60. B 
 61. B 62. B 63. D 64. B 65. B 
 66. B 67. D 68. D 69. A 70. B 
 71. C 72. C 73. B 74. B 75. C 
 76. C 77. B 78. C 79. B 80. B 
 81. B 82. ** 83. B 84. C 85. D 
 86. D 87. A 88. C 89. B 90. B 
 91. D 92. C 93. C 94. A 95. A 
 96. D 97. D 98. A 99. A 100. B  
 
** None of the option is correct  
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-2 
   
 
HINTS & SOLUTION 
 
BIOLOGY 
 
41. C.  
Sol: Apomixis is the asexual formation of seed from the maternal tissues of ovule, avoiding the process of fertilization. 
42. D 
Sol: Because intercalary meristem is present at nodes & internodes. 
43. C 
Sol: Because pteridophytes and gymnosperms have xylem and phloem but absent in bryophytes and thallophytes. 
44. D 
Sol: Pepsin helps in the digestion of proteins 
45. C 
Sol: Because colour blindness is a recessive X-linked disease. 
46. C 
Sol: Because ‘B’ tube has 5% salt solution and no change noticed as the solution in cylinder is isotonic with respect to 
B. 
47. C 
Sol: Because mitochondria and chloroplasts contain DNA and ribosomes (protein synthesizing machinery) 
48. C 
Sol: Haptopropism is the movement in which plant moves or grow in response to touch or contact stimuli. 
49. B 
Sol: Because chemical signals are released from the axonal end of the neuron. 
50. D 
Sol: Because these organs have common organization and different functions. 
51. A 
Sol: Due to biomagnification the non-biodegradable pollutants retained in the body tissues. 
52. B 
Sol: Because unisexual flowers are those in which sexes are separate 
53. B 
Sol: Because of anaerobic respiration in muscles. 
54. A 
Sol: Urine formation occurs in kidney which enters into ureters and then into urinary bladder and comes out through 
urethra.   
MATHEMATICS 
55. D 
 Sol: BD = 1 cm 
 BM = 
1
2
CM 
 AM = 
2
2
1
5
2
??
-
??
??
 
 = 
99 3 11
22
= 
 AC = 2 AM = 3 11 
 Area of Rhombus = ( )( )
1
AC BD
2
= 
( )
1
3 11 .1
2
= 
( ) 3 3.32
2
= 
9.92
2
= 4.96 cm
2
  
5 cm 5 cm 
B 
A C 
M 
D 
5 cm 5 cm 
1
cm
2
Page 3


   
 
CHANDIGARH NTSE STAGE 1 2019-20  
(SCHOLASTIC APTITUDE TEST) 
 
 ANSWER KEYS 
 
 
 1. A 2. B 3. A 4. A 5. C 
 6. A 7. A 8. C 9. D 10. D 
 11. D 12. A 13. B 14. B 15. C 
 16. D 17. C 18. A 19. C 20. A 
 21. A 22. C 23. A 24. D 25. C 
 26. C 27. D 28. D 29. A 30. C 
 31. C 32. D 33. D 34. C 35. B 
 36. B 37. D 38. D 39. B 40. D 
 41. C 42. D 43. C 44. D 45. C 
 46. C 47. C 48. C 49. B 50. D  
 51. A 52. B 53. A 54. A 55. D 
 56. C 57. ** 58. A 59. B 60. B 
 61. B 62. B 63. D 64. B 65. B 
 66. B 67. D 68. D 69. A 70. B 
 71. C 72. C 73. B 74. B 75. C 
 76. C 77. B 78. C 79. B 80. B 
 81. B 82. ** 83. B 84. C 85. D 
 86. D 87. A 88. C 89. B 90. B 
 91. D 92. C 93. C 94. A 95. A 
 96. D 97. D 98. A 99. A 100. B  
 
** None of the option is correct  
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-2 
   
 
HINTS & SOLUTION 
 
BIOLOGY 
 
41. C.  
Sol: Apomixis is the asexual formation of seed from the maternal tissues of ovule, avoiding the process of fertilization. 
42. D 
Sol: Because intercalary meristem is present at nodes & internodes. 
43. C 
Sol: Because pteridophytes and gymnosperms have xylem and phloem but absent in bryophytes and thallophytes. 
44. D 
Sol: Pepsin helps in the digestion of proteins 
45. C 
Sol: Because colour blindness is a recessive X-linked disease. 
46. C 
Sol: Because ‘B’ tube has 5% salt solution and no change noticed as the solution in cylinder is isotonic with respect to 
B. 
47. C 
Sol: Because mitochondria and chloroplasts contain DNA and ribosomes (protein synthesizing machinery) 
48. C 
Sol: Haptopropism is the movement in which plant moves or grow in response to touch or contact stimuli. 
49. B 
Sol: Because chemical signals are released from the axonal end of the neuron. 
50. D 
Sol: Because these organs have common organization and different functions. 
51. A 
Sol: Due to biomagnification the non-biodegradable pollutants retained in the body tissues. 
52. B 
Sol: Because unisexual flowers are those in which sexes are separate 
53. B 
Sol: Because of anaerobic respiration in muscles. 
54. A 
Sol: Urine formation occurs in kidney which enters into ureters and then into urinary bladder and comes out through 
urethra.   
MATHEMATICS 
55. D 
 Sol: BD = 1 cm 
 BM = 
1
2
CM 
 AM = 
2
2
1
5
2
??
-
??
??
 
 = 
99 3 11
22
= 
 AC = 2 AM = 3 11 
 Area of Rhombus = ( )( )
1
AC BD
2
= 
( )
1
3 11 .1
2
= 
( ) 3 3.32
2
= 
9.92
2
= 4.96 cm
2
  
5 cm 5 cm 
B 
A C 
M 
D 
5 cm 5 cm 
1
cm
2
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-3 
   
 
56. C  (Rs. should be omitted from options) 
 
Sol: 12 oranges sales price= 1Rs 
 1 orange sales price= 
1
12
Rs 
 S. P of one orange  = 
1
12
Rs 
 Loss = 20% 
 Cp =  
 If SP is 80 Rs. then CP =100 Rs. 
 If SP is Rs. 1 then CP =
100
80
Rs. 
 If SP is Rs. 1/12  then CP 
5
48
Rs. = 
 For 20% gain S.P of one orange = 
5 120 1
48 100 8
x Rs. = 
 In Rs.1, He should sell 8 oranges  
 
57.  
 
Sol:  (Given Options are incorrect) 
 Tn  = Sn  – Sn–1 
 = 
22
3 5 3 1 1
5
2 2 2 2
n n (n ) n ?? -- ??
+ - +
?? ??
??
??
 
 Tn = 3n + 1 
 T25 = 25 (3) + 1 = 76 
 
58. A 
 
Sol: Total days 366 
 n (E)= 2  
 n (S)= 7 
 SM, MT, TW, WTh, ThF, F S, S S  
P(E) = 
2
7
n(E)
n(S)
= 
 
59. B 
 
Sol: A + B = 90° 
 
0 0 2 0
02
90 90 90
90
tanA tan( A) tanAcot( A) sin ( A
sinA sec( A) cos A
- + - -
-
-
 
 ?
2
2
tanA cot A tanA tanA cos A
sinA.cosecA cos A
+
- 
 ? 1 + tan
2
A –1 
 = tan
2
A = tan
2
 (90
0
–B) = cot
2
B 
 
60. B 
Sol: Area of unshaded portion = 
2
3
2
4
x
??
??
??
??
 
 Area of ?ABC = 
2
3
3
4
( x) 
 
Area of Shaded portion
area of ABC
=
?
22
2
3
92
7
4
9 3
9
4
( x x )
( x )
-
=  
  
B D E C
G F
A
2 AF AG x ==
BF BD FD EC CG EG DE x = = = = = = =
Page 4


   
 
CHANDIGARH NTSE STAGE 1 2019-20  
(SCHOLASTIC APTITUDE TEST) 
 
 ANSWER KEYS 
 
 
 1. A 2. B 3. A 4. A 5. C 
 6. A 7. A 8. C 9. D 10. D 
 11. D 12. A 13. B 14. B 15. C 
 16. D 17. C 18. A 19. C 20. A 
 21. A 22. C 23. A 24. D 25. C 
 26. C 27. D 28. D 29. A 30. C 
 31. C 32. D 33. D 34. C 35. B 
 36. B 37. D 38. D 39. B 40. D 
 41. C 42. D 43. C 44. D 45. C 
 46. C 47. C 48. C 49. B 50. D  
 51. A 52. B 53. A 54. A 55. D 
 56. C 57. ** 58. A 59. B 60. B 
 61. B 62. B 63. D 64. B 65. B 
 66. B 67. D 68. D 69. A 70. B 
 71. C 72. C 73. B 74. B 75. C 
 76. C 77. B 78. C 79. B 80. B 
 81. B 82. ** 83. B 84. C 85. D 
 86. D 87. A 88. C 89. B 90. B 
 91. D 92. C 93. C 94. A 95. A 
 96. D 97. D 98. A 99. A 100. B  
 
** None of the option is correct  
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-2 
   
 
HINTS & SOLUTION 
 
BIOLOGY 
 
41. C.  
Sol: Apomixis is the asexual formation of seed from the maternal tissues of ovule, avoiding the process of fertilization. 
42. D 
Sol: Because intercalary meristem is present at nodes & internodes. 
43. C 
Sol: Because pteridophytes and gymnosperms have xylem and phloem but absent in bryophytes and thallophytes. 
44. D 
Sol: Pepsin helps in the digestion of proteins 
45. C 
Sol: Because colour blindness is a recessive X-linked disease. 
46. C 
Sol: Because ‘B’ tube has 5% salt solution and no change noticed as the solution in cylinder is isotonic with respect to 
B. 
47. C 
Sol: Because mitochondria and chloroplasts contain DNA and ribosomes (protein synthesizing machinery) 
48. C 
Sol: Haptopropism is the movement in which plant moves or grow in response to touch or contact stimuli. 
49. B 
Sol: Because chemical signals are released from the axonal end of the neuron. 
50. D 
Sol: Because these organs have common organization and different functions. 
51. A 
Sol: Due to biomagnification the non-biodegradable pollutants retained in the body tissues. 
52. B 
Sol: Because unisexual flowers are those in which sexes are separate 
53. B 
Sol: Because of anaerobic respiration in muscles. 
54. A 
Sol: Urine formation occurs in kidney which enters into ureters and then into urinary bladder and comes out through 
urethra.   
MATHEMATICS 
55. D 
 Sol: BD = 1 cm 
 BM = 
1
2
CM 
 AM = 
2
2
1
5
2
??
-
??
??
 
 = 
99 3 11
22
= 
 AC = 2 AM = 3 11 
 Area of Rhombus = ( )( )
1
AC BD
2
= 
( )
1
3 11 .1
2
= 
( ) 3 3.32
2
= 
9.92
2
= 4.96 cm
2
  
5 cm 5 cm 
B 
A C 
M 
D 
5 cm 5 cm 
1
cm
2
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-3 
   
 
56. C  (Rs. should be omitted from options) 
 
Sol: 12 oranges sales price= 1Rs 
 1 orange sales price= 
1
12
Rs 
 S. P of one orange  = 
1
12
Rs 
 Loss = 20% 
 Cp =  
 If SP is 80 Rs. then CP =100 Rs. 
 If SP is Rs. 1 then CP =
100
80
Rs. 
 If SP is Rs. 1/12  then CP 
5
48
Rs. = 
 For 20% gain S.P of one orange = 
5 120 1
48 100 8
x Rs. = 
 In Rs.1, He should sell 8 oranges  
 
57.  
 
Sol:  (Given Options are incorrect) 
 Tn  = Sn  – Sn–1 
 = 
22
3 5 3 1 1
5
2 2 2 2
n n (n ) n ?? -- ??
+ - +
?? ??
??
??
 
 Tn = 3n + 1 
 T25 = 25 (3) + 1 = 76 
 
58. A 
 
Sol: Total days 366 
 n (E)= 2  
 n (S)= 7 
 SM, MT, TW, WTh, ThF, F S, S S  
P(E) = 
2
7
n(E)
n(S)
= 
 
59. B 
 
Sol: A + B = 90° 
 
0 0 2 0
02
90 90 90
90
tanA tan( A) tanAcot( A) sin ( A
sinA sec( A) cos A
- + - -
-
-
 
 ?
2
2
tanA cot A tanA tanA cos A
sinA.cosecA cos A
+
- 
 ? 1 + tan
2
A –1 
 = tan
2
A = tan
2
 (90
0
–B) = cot
2
B 
 
60. B 
Sol: Area of unshaded portion = 
2
3
2
4
x
??
??
??
??
 
 Area of ?ABC = 
2
3
3
4
( x) 
 
Area of Shaded portion
area of ABC
=
?
22
2
3
92
7
4
9 3
9
4
( x x )
( x )
-
=  
  
B D E C
G F
A
2 AF AG x ==
BF BD FD EC CG EG DE x = = = = = = =
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-4 
   
 
61. B 
 
Sol:  
 Volume of solid = 3 volume of cone 
 ?r
2
H + 
1
3
?
2
 h = 
2
1
3
3
rh
??
?
??
??
 
 H = 
2
3
h
  
 
62. B 
Sol: 
0
3150
60 3 tan
x
== …….. 1 
 
0
3150 1
30
3
h
tan
x
-
== …….. 2 
 ? 
3150
3
3150 h
=
-
? 3150 = 9450-3h 
   3h = 6300 
     H = 2100 m  
 
63. D 
 
Sol: SI = 
82
100
P x x
 
 CI = 
2
8
1
100
PP
??
+-
??
??
 
 = 
27 27
1
25 25
??
-
??
??
P 
 CI – SI = 6.40 Rs.  
 
729 625 4 32
625 25 5
PP
- ? ? ? ?
-=
? ? ? ?
? ? ? ?
 
 
104 100 32
625 5
P
- ??
=
??
??
P 
 ? 
4 32
625 5
Px = 
 P = 1000 Rs.  
 
64. B 
 
Sol: a, b, c, d, e are in continuous proportion  
 
a b c d
x
b c d e
= = = = (say) 
 d = ex  
  c = dx = ex
2
  
 b = cx = ex
3 
a = bx = ex
4  
? 
4
a
x
e
= 
 
4
4
4
a
x
b
= 
  
r 
H 
h 
x
C D
A
h
B
3150m
0
60
3150 h -
0
30
Page 5


   
 
CHANDIGARH NTSE STAGE 1 2019-20  
(SCHOLASTIC APTITUDE TEST) 
 
 ANSWER KEYS 
 
 
 1. A 2. B 3. A 4. A 5. C 
 6. A 7. A 8. C 9. D 10. D 
 11. D 12. A 13. B 14. B 15. C 
 16. D 17. C 18. A 19. C 20. A 
 21. A 22. C 23. A 24. D 25. C 
 26. C 27. D 28. D 29. A 30. C 
 31. C 32. D 33. D 34. C 35. B 
 36. B 37. D 38. D 39. B 40. D 
 41. C 42. D 43. C 44. D 45. C 
 46. C 47. C 48. C 49. B 50. D  
 51. A 52. B 53. A 54. A 55. D 
 56. C 57. ** 58. A 59. B 60. B 
 61. B 62. B 63. D 64. B 65. B 
 66. B 67. D 68. D 69. A 70. B 
 71. C 72. C 73. B 74. B 75. C 
 76. C 77. B 78. C 79. B 80. B 
 81. B 82. ** 83. B 84. C 85. D 
 86. D 87. A 88. C 89. B 90. B 
 91. D 92. C 93. C 94. A 95. A 
 96. D 97. D 98. A 99. A 100. B  
 
** None of the option is correct  
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-2 
   
 
HINTS & SOLUTION 
 
BIOLOGY 
 
41. C.  
Sol: Apomixis is the asexual formation of seed from the maternal tissues of ovule, avoiding the process of fertilization. 
42. D 
Sol: Because intercalary meristem is present at nodes & internodes. 
43. C 
Sol: Because pteridophytes and gymnosperms have xylem and phloem but absent in bryophytes and thallophytes. 
44. D 
Sol: Pepsin helps in the digestion of proteins 
45. C 
Sol: Because colour blindness is a recessive X-linked disease. 
46. C 
Sol: Because ‘B’ tube has 5% salt solution and no change noticed as the solution in cylinder is isotonic with respect to 
B. 
47. C 
Sol: Because mitochondria and chloroplasts contain DNA and ribosomes (protein synthesizing machinery) 
48. C 
Sol: Haptopropism is the movement in which plant moves or grow in response to touch or contact stimuli. 
49. B 
Sol: Because chemical signals are released from the axonal end of the neuron. 
50. D 
Sol: Because these organs have common organization and different functions. 
51. A 
Sol: Due to biomagnification the non-biodegradable pollutants retained in the body tissues. 
52. B 
Sol: Because unisexual flowers are those in which sexes are separate 
53. B 
Sol: Because of anaerobic respiration in muscles. 
54. A 
Sol: Urine formation occurs in kidney which enters into ureters and then into urinary bladder and comes out through 
urethra.   
MATHEMATICS 
55. D 
 Sol: BD = 1 cm 
 BM = 
1
2
CM 
 AM = 
2
2
1
5
2
??
-
??
??
 
 = 
99 3 11
22
= 
 AC = 2 AM = 3 11 
 Area of Rhombus = ( )( )
1
AC BD
2
= 
( )
1
3 11 .1
2
= 
( ) 3 3.32
2
= 
9.92
2
= 4.96 cm
2
  
5 cm 5 cm 
B 
A C 
M 
D 
5 cm 5 cm 
1
cm
2
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-3 
   
 
56. C  (Rs. should be omitted from options) 
 
Sol: 12 oranges sales price= 1Rs 
 1 orange sales price= 
1
12
Rs 
 S. P of one orange  = 
1
12
Rs 
 Loss = 20% 
 Cp =  
 If SP is 80 Rs. then CP =100 Rs. 
 If SP is Rs. 1 then CP =
100
80
Rs. 
 If SP is Rs. 1/12  then CP 
5
48
Rs. = 
 For 20% gain S.P of one orange = 
5 120 1
48 100 8
x Rs. = 
 In Rs.1, He should sell 8 oranges  
 
57.  
 
Sol:  (Given Options are incorrect) 
 Tn  = Sn  – Sn–1 
 = 
22
3 5 3 1 1
5
2 2 2 2
n n (n ) n ?? -- ??
+ - +
?? ??
??
??
 
 Tn = 3n + 1 
 T25 = 25 (3) + 1 = 76 
 
58. A 
 
Sol: Total days 366 
 n (E)= 2  
 n (S)= 7 
 SM, MT, TW, WTh, ThF, F S, S S  
P(E) = 
2
7
n(E)
n(S)
= 
 
59. B 
 
Sol: A + B = 90° 
 
0 0 2 0
02
90 90 90
90
tanA tan( A) tanAcot( A) sin ( A
sinA sec( A) cos A
- + - -
-
-
 
 ?
2
2
tanA cot A tanA tanA cos A
sinA.cosecA cos A
+
- 
 ? 1 + tan
2
A –1 
 = tan
2
A = tan
2
 (90
0
–B) = cot
2
B 
 
60. B 
Sol: Area of unshaded portion = 
2
3
2
4
x
??
??
??
??
 
 Area of ?ABC = 
2
3
3
4
( x) 
 
Area of Shaded portion
area of ABC
=
?
22
2
3
92
7
4
9 3
9
4
( x x )
( x )
-
=  
  
B D E C
G F
A
2 AF AG x ==
BF BD FD EC CG EG DE x = = = = = = =
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-4 
   
 
61. B 
 
Sol:  
 Volume of solid = 3 volume of cone 
 ?r
2
H + 
1
3
?
2
 h = 
2
1
3
3
rh
??
?
??
??
 
 H = 
2
3
h
  
 
62. B 
Sol: 
0
3150
60 3 tan
x
== …….. 1 
 
0
3150 1
30
3
h
tan
x
-
== …….. 2 
 ? 
3150
3
3150 h
=
-
? 3150 = 9450-3h 
   3h = 6300 
     H = 2100 m  
 
63. D 
 
Sol: SI = 
82
100
P x x
 
 CI = 
2
8
1
100
PP
??
+-
??
??
 
 = 
27 27
1
25 25
??
-
??
??
P 
 CI – SI = 6.40 Rs.  
 
729 625 4 32
625 25 5
PP
- ? ? ? ?
-=
? ? ? ?
? ? ? ?
 
 
104 100 32
625 5
P
- ??
=
??
??
P 
 ? 
4 32
625 5
Px = 
 P = 1000 Rs.  
 
64. B 
 
Sol: a, b, c, d, e are in continuous proportion  
 
a b c d
x
b c d e
= = = = (say) 
 d = ex  
  c = dx = ex
2
  
 b = cx = ex
3 
a = bx = ex
4  
? 
4
a
x
e
= 
 
4
4
4
a
x
b
= 
  
r 
H 
h 
x
C D
A
h
B
3150m
0
60
3150 h -
0
30
ANSWER KEYS_CHANDIGARH_NTSE STAGE 1 2019-20_SAT.DOCX-5 
   
 
A
P Q B
2 1
12 ( , ) 2 (p, ) - 34 ( , ) -
A
Q B
1
12 ( , ) 34 ( , ) - (x,y)
1
2r
2
65. B 
    
Sol:   
   
 
 
 
 
 
11 2 3 7
33
( ) ( )
pp
+
= ? = 
 
1 2 2 4
2
3
( ) ( ) +-
=- 
 
 
 
 
  
 
 
Q
5
3
,q
??
??
??
 
 Then q = 0 
 x = 
21 13 5
33
( ) ( ) +
= 
 y = 
2 2 1 4
0
3
( ) ( ) +-
= 
 p = 
7
0
3
,q = 
66. B 
 
Sol: For maximum area of ? ABC = 
11
1
2
2
( r ).r 
 r1 = 
16
r
 
 maximum area of   ?ABC =  
2
256
r
 
 
67. D 
 
Sol: x + y + z = 0 x ? 0, y ? 0, z ? 0 
 
2 2 2 3 3 3
x y z x y z
yz xz xy xyz
++
+ + = 
 of x + y + z = 0 then x
3
 + y
3
 + z
3
 = 3xyz  
 ? 
3 3 3
3
x y z
xyz
++
= 
68. D 
 
Sol: x
2
 + y
2
 + z
2
 = r
2
  
 x = r cos a cos b 
 y = r cos a sin b  
 ? r
2
 cos
2
a cos
2
b + r
2
 cos
2
a sin
2
b + z
2
 = r
2
 
 r
2
cos
2
a(cos
2
b + sin
2
b) + z
2
 =r
2
 
 r
2
 cos a + z
2
 =r
2 
 
z
2
= r
2
( 1- cos
2
a) 
 z
2
 = r
2
sin
2
a 
 ? z = r sin a 
  
C B
A
1
r

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Document Description: Chandigarh: State Level Answer Key (SAT) - 2019-20 for Class 10 2024 is part of NTSE for Class 10 preparation. The notes and questions for Chandigarh: State Level Answer Key (SAT) - 2019-20 have been prepared according to the Class 10 exam syllabus. Information about Chandigarh: State Level Answer Key (SAT) - 2019-20 covers topics like and Chandigarh: State Level Answer Key (SAT) - 2019-20 Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Chandigarh: State Level Answer Key (SAT) - 2019-20.

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