Page 1
Answers & Solutions
for
JEE (Main)-2023 (Online) Phase-2
(Mathematics, Physics and Chemistry)
08/04/2023
Morning
Time : 3 hrs. M.M. : 300
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) The Test Booklet consists of 90 questions. The maximum marks are 300.
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry
having 30 questions in each part of equal weightage. Each part (subject) has two sections.
(i) Section-A: This section contains 20 multiple choice questions which have only one correct
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer.
(ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be
rounded off to the nearest integer.
Page 2
Answers & Solutions
for
JEE (Main)-2023 (Online) Phase-2
(Mathematics, Physics and Chemistry)
08/04/2023
Morning
Time : 3 hrs. M.M. : 300
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) The Test Booklet consists of 90 questions. The maximum marks are 300.
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry
having 30 questions in each part of equal weightage. Each part (subject) has two sections.
(i) Section-A: This section contains 20 multiple choice questions which have only one correct
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer.
(ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be
rounded off to the nearest integer.
MATHEMATICS
SECTION - A
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4 choices
(1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer:
1. Let
2 1 0
1 2 1 .
0 1 2
A
??
??
=-
??
?? -
??
If ( ) ( ) ( ) 2 16 ,
n
adj adj adj A =
then n is equal to
(1) 8 (2) 10
(3) 9 (4) 12
Answer (2)
Sol. |A| = 2(3) – 1(2) = 4
? Now |adj(adj(adj(2A)))|
?
3
( 1)
2
n
A
-
8
24 8
2 2 4 A = = ?
= 2
4
= 16
10
? 10 n =
2. If the points with position vectors
ˆ ˆ ˆ
10 13 , i j k ? + +
9
ˆ ˆ ˆ ˆ ˆ ˆ
6 11 11 , 8
2
i j k i j k + + + ? - are collinear, then
(19? – 6?)
2
is equal to
(1) 36 (2) 25
(3) 49 (4) 16
Answer (1)
Sol. • A(?, 10, 13)
• B(6, 11, 11)
•
9
, , 8
2
C
??
?-
??
??
Since, A, B, C are collinear
8 13
11
1
k
k
-+
=
+
? 11k + 11 = –8k + 13
? 19k = 2
?
2
19
k =
? Ratio = 2 : 19
19 9
6
21
?+
=
? 19? = 117
?
117
19
?=
2 190
11
21
?+
=
?
41
2
?=
? (19? – 6?)
2
= (117 – 123)
2
= 36
3. Let R be the focus of the parabola y
2
= 20x and the
line y = mx + c intersect the parabola at two points
P and Q. Let the points G(10, 10) be the centroid of
the triangle PQR. If c – m = 6, then (PQ)
2
is
(1) 296 (2) 325
(3) 317 (4) 346
Answer (2)
Sol.
( )
22
12
51
10
3
tt ++
=
?
22
12
5 tt += ...(i)
( )
12
10
10
3
tt +
=
?
12
3 tt += ...(ii)
Page 3
Answers & Solutions
for
JEE (Main)-2023 (Online) Phase-2
(Mathematics, Physics and Chemistry)
08/04/2023
Morning
Time : 3 hrs. M.M. : 300
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) The Test Booklet consists of 90 questions. The maximum marks are 300.
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry
having 30 questions in each part of equal weightage. Each part (subject) has two sections.
(i) Section-A: This section contains 20 multiple choice questions which have only one correct
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer.
(ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be
rounded off to the nearest integer.
MATHEMATICS
SECTION - A
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4 choices
(1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer:
1. Let
2 1 0
1 2 1 .
0 1 2
A
??
??
=-
??
?? -
??
If ( ) ( ) ( ) 2 16 ,
n
adj adj adj A =
then n is equal to
(1) 8 (2) 10
(3) 9 (4) 12
Answer (2)
Sol. |A| = 2(3) – 1(2) = 4
? Now |adj(adj(adj(2A)))|
?
3
( 1)
2
n
A
-
8
24 8
2 2 4 A = = ?
= 2
4
= 16
10
? 10 n =
2. If the points with position vectors
ˆ ˆ ˆ
10 13 , i j k ? + +
9
ˆ ˆ ˆ ˆ ˆ ˆ
6 11 11 , 8
2
i j k i j k + + + ? - are collinear, then
(19? – 6?)
2
is equal to
(1) 36 (2) 25
(3) 49 (4) 16
Answer (1)
Sol. • A(?, 10, 13)
• B(6, 11, 11)
•
9
, , 8
2
C
??
?-
??
??
Since, A, B, C are collinear
8 13
11
1
k
k
-+
=
+
? 11k + 11 = –8k + 13
? 19k = 2
?
2
19
k =
? Ratio = 2 : 19
19 9
6
21
?+
=
? 19? = 117
?
117
19
?=
2 190
11
21
?+
=
?
41
2
?=
? (19? – 6?)
2
= (117 – 123)
2
= 36
3. Let R be the focus of the parabola y
2
= 20x and the
line y = mx + c intersect the parabola at two points
P and Q. Let the points G(10, 10) be the centroid of
the triangle PQR. If c – m = 6, then (PQ)
2
is
(1) 296 (2) 325
(3) 317 (4) 346
Answer (2)
Sol.
( )
22
12
51
10
3
tt ++
=
?
22
12
5 tt += ...(i)
( )
12
10
10
3
tt +
=
?
12
3 tt += ...(ii)
? t
1
= 1, t
2
= 2
? P ? (5, 10) Q ? (20, 20)
? Equation of
10
10 ( 5)
15
PQ y x = - = -
3y – 30 = 2x – 10
2 20
33
yx =+
? PQ
2
= 225 + 100 = 325
4. The number of arrangements of the letters of the
word “INDEPENDENCE” in which all the vowels
always occur together is
(1) 16800 (2) 33600
(3) 18000 (4) 14800
Answer (1)
Sol. Vowels : I, E, E, E, E
Consonants : N N N D D P C
3 , 2 , , I E E E E N D P C
Number of required words
8! 5!
3! 2! 4!
=?
= 16800
5. The shortest distance between the lines
4 2 3
4 5 3
x y z - + +
== and
1 3 4
3 4 2
x y z - - -
== is
(1) 63 (2) 26
(3) 62 (4) 36
Answer (4)
Sol.
12
ˆ ˆ ˆ
4 5 3
3 4 2
i j k
ll ?=
ˆ ˆ ˆ
2i j k = - + +
( ) ( )
12
12
a b l l
d
ll
- ? ?
=
?
( ) ( )
ˆ ˆ ˆ ˆ ˆ ˆ
3 5 7 2
6
i j k i j k - - ? - + +
=
657
6
- - -
=
36 =
6. Let ? ?
sin cos 2
( ) , 0, .
sin cos 4
x
f x x
xx
+ - ? ??
= ? ? -
??
-
??
Then
77
12 12
ff
?? ? ? ? ?
??
? ? ? ?
? ? ? ?
is equal to
(1)
2
9
(2)
2
3
-
(3)
1
33
-
(4)
2
33
Answer (1)
Sol.
sin cos 2
()
sin cos
xx
fx
xx
+-
=
-
2 sin 2
4
()
2 sin
4
x
fx
x
? ??
+-
??
??
=
?? ? ??
-
?? ??
?? ??
sin 1
4
()
sin
4
x
fx
x
? ??
+-
??
??
=
? ??
-
??
??
cos 1
4 sin
x
fx
x
?- ??
+=
??
??
= tan
2
x
-
? ( ) tan
28
x
fx
? ??
= - -
??
??
( )
2
1
sec
2 2 8
x
fx
-? ??
? = -
??
??
2
1
( ) sec tan
2 2 8 2 8
xx
fx
?? ?? ? ? ? ?
?? = - -
?? ? ? ? ?
? ? ? ? ??
77
tan
12 24 8
f
? ? ? ? ? ? ?
= - -
? ? ? ?
? ? ? ?
4
tan
24
? ??
=-
??
??
1
tan
6
3
? ??
= - = -
??
??
Page 4
Answers & Solutions
for
JEE (Main)-2023 (Online) Phase-2
(Mathematics, Physics and Chemistry)
08/04/2023
Morning
Time : 3 hrs. M.M. : 300
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) The Test Booklet consists of 90 questions. The maximum marks are 300.
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry
having 30 questions in each part of equal weightage. Each part (subject) has two sections.
(i) Section-A: This section contains 20 multiple choice questions which have only one correct
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer.
(ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be
rounded off to the nearest integer.
MATHEMATICS
SECTION - A
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4 choices
(1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer:
1. Let
2 1 0
1 2 1 .
0 1 2
A
??
??
=-
??
?? -
??
If ( ) ( ) ( ) 2 16 ,
n
adj adj adj A =
then n is equal to
(1) 8 (2) 10
(3) 9 (4) 12
Answer (2)
Sol. |A| = 2(3) – 1(2) = 4
? Now |adj(adj(adj(2A)))|
?
3
( 1)
2
n
A
-
8
24 8
2 2 4 A = = ?
= 2
4
= 16
10
? 10 n =
2. If the points with position vectors
ˆ ˆ ˆ
10 13 , i j k ? + +
9
ˆ ˆ ˆ ˆ ˆ ˆ
6 11 11 , 8
2
i j k i j k + + + ? - are collinear, then
(19? – 6?)
2
is equal to
(1) 36 (2) 25
(3) 49 (4) 16
Answer (1)
Sol. • A(?, 10, 13)
• B(6, 11, 11)
•
9
, , 8
2
C
??
?-
??
??
Since, A, B, C are collinear
8 13
11
1
k
k
-+
=
+
? 11k + 11 = –8k + 13
? 19k = 2
?
2
19
k =
? Ratio = 2 : 19
19 9
6
21
?+
=
? 19? = 117
?
117
19
?=
2 190
11
21
?+
=
?
41
2
?=
? (19? – 6?)
2
= (117 – 123)
2
= 36
3. Let R be the focus of the parabola y
2
= 20x and the
line y = mx + c intersect the parabola at two points
P and Q. Let the points G(10, 10) be the centroid of
the triangle PQR. If c – m = 6, then (PQ)
2
is
(1) 296 (2) 325
(3) 317 (4) 346
Answer (2)
Sol.
( )
22
12
51
10
3
tt ++
=
?
22
12
5 tt += ...(i)
( )
12
10
10
3
tt +
=
?
12
3 tt += ...(ii)
? t
1
= 1, t
2
= 2
? P ? (5, 10) Q ? (20, 20)
? Equation of
10
10 ( 5)
15
PQ y x = - = -
3y – 30 = 2x – 10
2 20
33
yx =+
? PQ
2
= 225 + 100 = 325
4. The number of arrangements of the letters of the
word “INDEPENDENCE” in which all the vowels
always occur together is
(1) 16800 (2) 33600
(3) 18000 (4) 14800
Answer (1)
Sol. Vowels : I, E, E, E, E
Consonants : N N N D D P C
3 , 2 , , I E E E E N D P C
Number of required words
8! 5!
3! 2! 4!
=?
= 16800
5. The shortest distance between the lines
4 2 3
4 5 3
x y z - + +
== and
1 3 4
3 4 2
x y z - - -
== is
(1) 63 (2) 26
(3) 62 (4) 36
Answer (4)
Sol.
12
ˆ ˆ ˆ
4 5 3
3 4 2
i j k
ll ?=
ˆ ˆ ˆ
2i j k = - + +
( ) ( )
12
12
a b l l
d
ll
- ? ?
=
?
( ) ( )
ˆ ˆ ˆ ˆ ˆ ˆ
3 5 7 2
6
i j k i j k - - ? - + +
=
657
6
- - -
=
36 =
6. Let ? ?
sin cos 2
( ) , 0, .
sin cos 4
x
f x x
xx
+ - ? ??
= ? ? -
??
-
??
Then
77
12 12
ff
?? ? ? ? ?
??
? ? ? ?
? ? ? ?
is equal to
(1)
2
9
(2)
2
3
-
(3)
1
33
-
(4)
2
33
Answer (1)
Sol.
sin cos 2
()
sin cos
xx
fx
xx
+-
=
-
2 sin 2
4
()
2 sin
4
x
fx
x
? ??
+-
??
??
=
?? ? ??
-
?? ??
?? ??
sin 1
4
()
sin
4
x
fx
x
? ??
+-
??
??
=
? ??
-
??
??
cos 1
4 sin
x
fx
x
?- ??
+=
??
??
= tan
2
x
-
? ( ) tan
28
x
fx
? ??
= - -
??
??
( )
2
1
sec
2 2 8
x
fx
-? ??
? = -
??
??
2
1
( ) sec tan
2 2 8 2 8
xx
fx
?? ?? ? ? ? ?
?? = - -
?? ? ? ? ?
? ? ? ? ??
77
tan
12 24 8
f
? ? ? ? ? ? ?
= - -
? ? ? ?
? ? ? ?
4
tan
24
? ??
=-
??
??
1
tan
6
3
? ??
= - = -
??
??
2
7 1 2 1
12 2
33
f
?? ? ??
?? = - ? ?
?? ??
?? ??
2
33
-
=
7 7 2
12 12 9
ff
?? ? ? ? ?
?? =
? ? ? ?
? ? ? ?
7. In a bolt factory, machines A, B and C manufacture
respectively 20%, 30% and 50% of the total bolts.
Of their output 3, 4 and 2 percent are respectively
defective bolts. A bolts is drawn at random from the
product. If the bolt drawn is found the defective,
then the probability that it is manufactured by the
machine C is
(1)
5
14
(2)
9
28
(3)
3
7
(4)
2
7
Answer (1)
Sol. Using Bayes’ Theorem
Required probability
50 2
20 3 30 4 50 2
?
=
? + ? + ?
10
6 12 10
=
++
10
28
=
5
14
=
8. Let P =
31
2 2 1 1
,
0 1 3 1
22
A
??
??
??
??
=
??
??
??
-??
??
and Q = PAP
T
. If P
T
Q
2007
,
ab
P
cd
??
=
??
??
then 2a + b – 3c – 4d equal to
(1) 2004
(2) 2005
(3) 2007
(4) 2006
Answer (2)
Sol.
3 1 3 1
0 2 2 2 2 1
0 1 3 1 3 1
2 2 2 2
T
PP
? ? ? ?
-
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
- ? ?
? ? ? ?
? ? ? ?
Similarly P
T
P = I
Now, Q
2007
= (PAP
T
) (PAP
T
) … 2007 times
= PA
2007
P
T
11
0 1
A
??
=
??
??
2 12
0 1
A
??
=
??
??
3
2007
3 1
0 1
.
.
.
2007 1
0 1
A
A
??
=
??
??
??
=
??
??
2007 2007 2007 1
0 1
T
P Q P A
??
==
??
??
? a = 1, b = 2007, c = 0, d = 1
2a + b – 3c – 4d = 2005
9. The number of ways, in which 5 girls and 7 boys
can be seated at a round table so that no two girls
sit together, is
(1) 720
(2) 126(5!)
2
(3) 7(360)
2
(4) 7(720)
2
Answer (2)
Sol.
7
5
6! · 5! C ?
=
76
720 5!
2
?
??
? ( )
2 766
5!
2
??
?
? 126 × (5!)
2
Page 5
Answers & Solutions
for
JEE (Main)-2023 (Online) Phase-2
(Mathematics, Physics and Chemistry)
08/04/2023
Morning
Time : 3 hrs. M.M. : 300
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) The Test Booklet consists of 90 questions. The maximum marks are 300.
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry
having 30 questions in each part of equal weightage. Each part (subject) has two sections.
(i) Section-A: This section contains 20 multiple choice questions which have only one correct
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer.
(ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be
rounded off to the nearest integer.
MATHEMATICS
SECTION - A
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4 choices
(1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer:
1. Let
2 1 0
1 2 1 .
0 1 2
A
??
??
=-
??
?? -
??
If ( ) ( ) ( ) 2 16 ,
n
adj adj adj A =
then n is equal to
(1) 8 (2) 10
(3) 9 (4) 12
Answer (2)
Sol. |A| = 2(3) – 1(2) = 4
? Now |adj(adj(adj(2A)))|
?
3
( 1)
2
n
A
-
8
24 8
2 2 4 A = = ?
= 2
4
= 16
10
? 10 n =
2. If the points with position vectors
ˆ ˆ ˆ
10 13 , i j k ? + +
9
ˆ ˆ ˆ ˆ ˆ ˆ
6 11 11 , 8
2
i j k i j k + + + ? - are collinear, then
(19? – 6?)
2
is equal to
(1) 36 (2) 25
(3) 49 (4) 16
Answer (1)
Sol. • A(?, 10, 13)
• B(6, 11, 11)
•
9
, , 8
2
C
??
?-
??
??
Since, A, B, C are collinear
8 13
11
1
k
k
-+
=
+
? 11k + 11 = –8k + 13
? 19k = 2
?
2
19
k =
? Ratio = 2 : 19
19 9
6
21
?+
=
? 19? = 117
?
117
19
?=
2 190
11
21
?+
=
?
41
2
?=
? (19? – 6?)
2
= (117 – 123)
2
= 36
3. Let R be the focus of the parabola y
2
= 20x and the
line y = mx + c intersect the parabola at two points
P and Q. Let the points G(10, 10) be the centroid of
the triangle PQR. If c – m = 6, then (PQ)
2
is
(1) 296 (2) 325
(3) 317 (4) 346
Answer (2)
Sol.
( )
22
12
51
10
3
tt ++
=
?
22
12
5 tt += ...(i)
( )
12
10
10
3
tt +
=
?
12
3 tt += ...(ii)
? t
1
= 1, t
2
= 2
? P ? (5, 10) Q ? (20, 20)
? Equation of
10
10 ( 5)
15
PQ y x = - = -
3y – 30 = 2x – 10
2 20
33
yx =+
? PQ
2
= 225 + 100 = 325
4. The number of arrangements of the letters of the
word “INDEPENDENCE” in which all the vowels
always occur together is
(1) 16800 (2) 33600
(3) 18000 (4) 14800
Answer (1)
Sol. Vowels : I, E, E, E, E
Consonants : N N N D D P C
3 , 2 , , I E E E E N D P C
Number of required words
8! 5!
3! 2! 4!
=?
= 16800
5. The shortest distance between the lines
4 2 3
4 5 3
x y z - + +
== and
1 3 4
3 4 2
x y z - - -
== is
(1) 63 (2) 26
(3) 62 (4) 36
Answer (4)
Sol.
12
ˆ ˆ ˆ
4 5 3
3 4 2
i j k
ll ?=
ˆ ˆ ˆ
2i j k = - + +
( ) ( )
12
12
a b l l
d
ll
- ? ?
=
?
( ) ( )
ˆ ˆ ˆ ˆ ˆ ˆ
3 5 7 2
6
i j k i j k - - ? - + +
=
657
6
- - -
=
36 =
6. Let ? ?
sin cos 2
( ) , 0, .
sin cos 4
x
f x x
xx
+ - ? ??
= ? ? -
??
-
??
Then
77
12 12
ff
?? ? ? ? ?
??
? ? ? ?
? ? ? ?
is equal to
(1)
2
9
(2)
2
3
-
(3)
1
33
-
(4)
2
33
Answer (1)
Sol.
sin cos 2
()
sin cos
xx
fx
xx
+-
=
-
2 sin 2
4
()
2 sin
4
x
fx
x
? ??
+-
??
??
=
?? ? ??
-
?? ??
?? ??
sin 1
4
()
sin
4
x
fx
x
? ??
+-
??
??
=
? ??
-
??
??
cos 1
4 sin
x
fx
x
?- ??
+=
??
??
= tan
2
x
-
? ( ) tan
28
x
fx
? ??
= - -
??
??
( )
2
1
sec
2 2 8
x
fx
-? ??
? = -
??
??
2
1
( ) sec tan
2 2 8 2 8
xx
fx
?? ?? ? ? ? ?
?? = - -
?? ? ? ? ?
? ? ? ? ??
77
tan
12 24 8
f
? ? ? ? ? ? ?
= - -
? ? ? ?
? ? ? ?
4
tan
24
? ??
=-
??
??
1
tan
6
3
? ??
= - = -
??
??
2
7 1 2 1
12 2
33
f
?? ? ??
?? = - ? ?
?? ??
?? ??
2
33
-
=
7 7 2
12 12 9
ff
?? ? ? ? ?
?? =
? ? ? ?
? ? ? ?
7. In a bolt factory, machines A, B and C manufacture
respectively 20%, 30% and 50% of the total bolts.
Of their output 3, 4 and 2 percent are respectively
defective bolts. A bolts is drawn at random from the
product. If the bolt drawn is found the defective,
then the probability that it is manufactured by the
machine C is
(1)
5
14
(2)
9
28
(3)
3
7
(4)
2
7
Answer (1)
Sol. Using Bayes’ Theorem
Required probability
50 2
20 3 30 4 50 2
?
=
? + ? + ?
10
6 12 10
=
++
10
28
=
5
14
=
8. Let P =
31
2 2 1 1
,
0 1 3 1
22
A
??
??
??
??
=
??
??
??
-??
??
and Q = PAP
T
. If P
T
Q
2007
,
ab
P
cd
??
=
??
??
then 2a + b – 3c – 4d equal to
(1) 2004
(2) 2005
(3) 2007
(4) 2006
Answer (2)
Sol.
3 1 3 1
0 2 2 2 2 1
0 1 3 1 3 1
2 2 2 2
T
PP
? ? ? ?
-
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
- ? ?
? ? ? ?
? ? ? ?
Similarly P
T
P = I
Now, Q
2007
= (PAP
T
) (PAP
T
) … 2007 times
= PA
2007
P
T
11
0 1
A
??
=
??
??
2 12
0 1
A
??
=
??
??
3
2007
3 1
0 1
.
.
.
2007 1
0 1
A
A
??
=
??
??
??
=
??
??
2007 2007 2007 1
0 1
T
P Q P A
??
==
??
??
? a = 1, b = 2007, c = 0, d = 1
2a + b – 3c – 4d = 2005
9. The number of ways, in which 5 girls and 7 boys
can be seated at a round table so that no two girls
sit together, is
(1) 720
(2) 126(5!)
2
(3) 7(360)
2
(4) 7(720)
2
Answer (2)
Sol.
7
5
6! · 5! C ?
=
76
720 5!
2
?
??
? ( )
2 766
5!
2
??
?
? 126 × (5!)
2
10. The area of the region
? ?
22
( , ) : 8 , 7 x y x y x y ? ? - ?
is
(1) 27
(2) 18
(3) 20
(4) 21
Answer (3)
Sol.
Required area =
47
04
28 ydy ydy
??
?? +-
??
??
??
( )
7 4
2
3
2
2
0
4
8
2
33
22
y
y
??
??
-
??
=-
??
??
??
( ) ( )
4
8 1 8
3
= - -
= 20 sq. units
11. Let
1 2 ...
K
K
S
K
+ + +
= and
22
1
()
n
j
j
n
S Bn Cn D
A
=
= + +
?
, where A, B, C, D ?
and A has least value. Then
(1) A + C + D is not divisible by B
(2) A + B = 5(D – C)
(3) A + B + C + D is divisible by 5
(4) A + B is divisible by D
Answer (4)
Sol.
·( 1) 1
22
K
K K K
S
K
++
==
( ) ( )
( )
2
2
2 2 2
11
1
2 3 4 ... 1
4
nn
j
jj
Sn
==
= + + +
??
( )( )( ) 1 2 2 3 1
1
46
n n n ?? + + +
?-
??
??
??
( )
( )
2
3 2 2 3 – 6
1
46
n n n
??
+ + +
??
=
??
??
??
3 2 2
1 2 6 4 3 9 6 6
46
n n n n n
??
+ + + + + -
?? =
??
??
32
1 2 9 13
46
n n n
??
++
?? =
??
??
( )
2
2 9 13
24
n
nn = + +
A = 24, B = 2, C = 9, D = 13
26
2
13
AB
D
+
==
12. Negation of ( ) ( ) p q q p ??? is
(1) (~ ) pp ?
(2) (~ ) qp ?
(3) (~ ) qp ?
(4) (~ ) pq ?
Answer (2)
Sol. ( ) ( ) p q q p ???
( ) ( ) p q q p ??? ? ? ? ?
( ) ( ) p q q p ?? ? ? ? ?
pq? =?
Now () pq?? ?
pq ? =?
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