ASHWIN PATEL’S PRIVATE TUTORIALS
CENTERS : MALAD (EAST) , KANDIVALI (WEST)
CONTACT: 98199 80268 / 99307 39289
Std. XI Marks : 90
Test – 49 (PCM) - JEE Time – 90 Min.
Date: 13/07/2025
ANSWERS
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 D 27 C 28 C 29 D 30 C
31 B 32 C 33 A 34 B 35 A 36 A 37 B 38 C 39 D 40 B
41 B 42 C 43 C 44 D 45 B 46 B 47 D 48 C 49 A 50 A
51 C 52 B 53 A 54 A 55 D 56 C 57 C 58 B 59 B 60 B
61 C 62 C 63 B 64 D 65 B 66 B 67 A 68 D 69 D 70 B
71 C 72 B 73 C 74 D 75 B
HINTS & SOLUTIONS
51. Case I When only one flag is used
Number of signals made = 3P1 = 3
Case II When only two flags are used
Number of signals made = 3P2 = 3 2 = 6
Case III When three flags are used Number of signals made= 3P3 = 3! = 3 2 1 = 6
Hence, required number of signals = 3 + 6 + 6 = 15
52. Number of numbers from 1000 to 9999 (both inclusive) using 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 when
Case I Repeatition is allowed
We cannot place zero at thousand place. Thus, thousand place I can be filled in 9 ways.
And unit's, ten's and hundreds place can be filled in 10 ways each.
Number of numbers = 9 10 10 10 = 9000
Case II Repeatition is not allowed We cannot place zero at thousand place. Thus, thousand place can
be filled in 9 ways.
Since, one digit one of ten digits assign at thousand place, i therefore hundreds place can be filled in 9
ways. Similarly, unit's and ten's place can be filled in 7 and 8 ways.
Number of numbers = 9 9 8 7 = 4536
Required number of numbers that do not have all 4 different digit = 9000 4536 = 4464
1
ASPT (PCM) - JEE Test - 49
53. If all the girl sit together, then consider all the three girls as one group.
Now, we have to arrange 5 + 1 = 6 persons
Number of ways arranging 6 persons is 6!.
And the three girls arranged themselves in 3! way.
Number of ways of arranging all the girls together is 6! 3!.
Total number of arrangement of 5 boys and 3 girls are 81
Total number of ways in which ail the girls are never together
= Total number of arrangement
= Total number of arrangement in which all the girls are always together = 8! 613! = 40320 4320 =
36000
54. Excluding two specified guest, 12 persons can be divided into two groups are containing 6 and other
(12 2)! 10!
containing (6 2) = 4 in ways i.e. ways.
6!4! 6!4!
And can sit either side of master and mistress in 2! ways and can arrange themselves in 6! 4'ways
Now, the two specified guest where 4 guest are seated have 5 gaps and can arrange themselves in 2!
ways. The number of ways, when G1, G2 will always be together is
10!
× 2!6!4!×5× 2!
6!4!
= 10! 4 5 = 20 10!
55. Let the city be represented by a rectangle whose sides are of length a and b North-South and West-
East, respectively. Man has to go from P to Q. For this, he will have to travel a distance ‘a’ vertically
downward and a distance ‘b’ horizontally from left to right.
Let a1, a2 ... am1 denote the distances between consecutive street drawn horizontally beginning with the
street passing through P and P b2, b2 ,.., bn1 denote the distance between consecutive streets drawn
vertically beginning with the street passing through P.
Each arrangement of (m + n 2) things a1 a2...am1, b1, b2 ...bn1 in a row, so that order of ai’ s does not
change and order of bi’ s does not change corresponds to one path go from P to Q.
Therefore, the required number is equal to the number of arrangements of (m + n 2) things such that
order of ai’s order of ai’ s dose not change and order of bi’ s does not change and order of bi' s dose not
(m n 2)!
change which is equal to
(m 1)!(n 1)!
2
ASPT (PCM) - JEE Test - 49
Alternate Method
(Let the city be represented by a rectangle whose sides are of length a and b.
For each path total distance covered in horizontal direction is a and that in vertical direction is b. a is
the sum of length of (m 1) horizontal line segments and b is the sum of length of (n 1) vertical line
segments.
Each path to go from P to Q will be a arrangement of (m + n 2 lines segment of which (m 1) are
horizontal and (n 1) are vertical.
Therefore, the required number is equal to number of arrangement of (m + n 2) different things of
(m n 2)!
which (m 1) are one kind and (n 1) are of another kind which is given by
(m 1)!(n 1)!
1
56. Clearly, probability of chosen any month out of 12 month =
12
There are 7 possible ways in which the month can start and it will be a Friday on 13th day, if the first
day of the month is Sunday.
1
So, its probability =
7
1 1 1
Thus, required probability =
12 7 84
1
Hence, the probability that the 13th day of a randomly selected month is Friday =
84
57. Total number of ways of putting 4 letters in 4 envelopes = 4!
Number of ways of putting in right envelopes = 1
Now, required probability = 1 P (all of them are in correct envelopes)
1 1 23
= 1 1
4! 24 24
58. There are 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The last two digits can be dialed in 10P2
= 90ways cut of which only one way is favourable.
1
Required probability =
90
4 1
59. Let p = Probability of throwing 9 with two dice =
36 9
8
and q = Probability of not throwing 9 with two dice =
9
Let A be the event that A win the game and B be the event that 6 win He game.
New as A start the game, therefore 6 win, if he throw 9 in 2nd or 4th or 6th and so on attempt.
Probability that B win the game
3
ASPT (PCM) - JEE Test - 49
= P(ABorAB ABor ABABA or...)
= P(AB) P(AB AB) P( ABABAB) ...
= P(A)P(B) P(A)P(B)P( A)P(B) ...
= qp + q3 p + …
1 8
qp 9 9 8
=
1 q 2 1 64 17
81
60. Clearly, matches played by India are four and maximum points in any match are 2.
Maximum point in four matches can be 8 only.
Required probability = Probability of India getting 7 points
+ Probability of India getting 8 points …(i)
Note that, 7 points can be get in 4 matches, if India get 1point in one of the 4 matches and 2 points in
each of rest 3 matches and 8 points can be get only if India get 2 point in each of the 4 matches.
From Eq. (i), we have
Required probability = 4C1 (0.05) (0.5)3 + 4C4 (0.5)4
= 4(0.05) (0.5)3 + (0 5)4
= 0.0250 + 0.0625
= 0.0875
61. So B(2 3, 0)
and C(2 3, 0).
Clearly, AB = BC = CA
th the circumcentre. Coordinates of the circumcentre
are O' (0, 2) and radius = O'A = 4.
Hence, the equation of the circumcircle is
(x 0)2 + (y 2)2 = 42
x2 + y2 4y = 12
62. Intersecting points of the line y = 2x and x2 + y2 = 10xare
x2 + 4x2 = 10x
5x2 10x = 0
5x(x 2) = 0
x = 0, 2
Put x = 0 in y = 2x,
y=0
and x = 2 in y = 2x,
y=4
Hence, intersecting points are (0, 0), (2, 4).
Now, equation of the circle with (0, 0) and
(2, 4) as ends of the diameter is
(x 0)(x 2) + (y 0)(y 4) = 0
x2 + y2 2x 4y = 0
63. The given equation of lines are
x sin α y cos α = a ... (i)
x cos α + y sin α = a ... (ii)
4
ASPT (PCM) - JEE Test - 49
On squaring and adding both the equations,
we get
x2 + y2 = 2a2
64. If the straight line y = mx is outside
the circle
x2 + y2 20y + 90 = 0, then there will be no
point of intersection.
Putting y = mx in equation of circle,
x2 + m2x2 20(mx) + 90 = 0
x2(1 + m2) 20 mx + 90 = 0
If there is no point of intersection, then D < 0
b2 4ac < 0
400 m2 4(1 + m2) (90) < 0
400m2 < 4(1 + m2) (90)
10m2 < 9 + 9 m2
m<9
m<±3
|m| < 3
65. The given equations of the circle are
S1 x2 + y2 + 5x 8y + 1 = 0
S2 x2 + y2 3x + 7y 25 = 0
The equation of the common chord is
S1 S2 = 0
(5x 8y + 1) (3x + 7y 25) = 0
8x 15y + 26 = 0
The another equation of circle
x2 + y2 = 2x
x2 + y2 2x = 0
having centre (1, 0) and radius 1. The distance of 8x 15y + 26 = 0 from the (1, 0) is
8 26 34
2 units
64 225 17
g(x)
Let I
a
66. dx …(i)
0 f (x) f (a x)
g(a x)
I
a
dx …(ii)
0 f (a x) f (x)
On adding Eqs. (i) and (ii), we get
a g(x) g(a x)
2I
f (a x) f (x)
dx
0 f (x) f (a x)
I = 0, if g(x) = g(a x)
2
{a (4 4a)x 4x3}dx 12
2
66. We have,
1
[a 2 x (2 2a)x 2 x 4 ]12 12
(2a2 + 8 8a + 16) (a2 + 2 2a + 1) ≤ 12
a2 6a + 21 ≤ 12
a2 6a + 9 ≤ 0
(a 3)2 ≤ 0
5
ASPT (PCM) - JEE Test - 49
a=3
a
68. We have, a
{f (x) f (x)}{g(x) g(x)}dx
Integral is an odd function.
a
a
{f (x) f (x)}{g(x) g(x)}dx 0
sin t 1
69. We have,
1 t
dt
0
4 sin/ 2
Let I dt
4 2 4 2 t
Put t/2 = x dt = 2dx
Now,
2 2sin x 2 sin x
I dx dx
2 4 2 2x 2 1 2 1 x
2 sin(2 x)
I dx
2 1 2 1 x
Let 2 x = z dx = dz
0 sin z
I dz
1 z 1
1 sin t
I dt
0 t 1
I=
1 cos2 t
70. Let I1 xf{x(2 x)}dx
sin 2 t
1
I1 (1 cos2 t sin 2 t x)f (2 x)(x)dx
sin 2 t
1 cos2 t
I1 (2 x)f{x(2 x)}dx
sin 2 t
1 cos2 t
I1 2f{x(2 x)}dx
sin 2 t
1 cos2 t
xf{x(2 x)}dx
sin 2 t
I1 = 2I2 I1
I1
2I1 = 2I2 1
I2
71. We have,
x2
0 1,if 8 x 8
64
x2
2 2 3,if | x | 8
64
x2
y 2 2,if | x | 8
64
6
ASPT (PCM) - JEE Test - 49
The graph of given curves is as shown in figure.
Required area = Area of the shaded region
2
= 0
x dy
2
= (y 1)dy
0
1
= [(y 1) 2 ]02
2
9 1
= 4sq units
2 2
Alternate Method
Required area = Area of trapezium OABC
1
= (OA BC) OC
2
1
= (1 3) 2
2
= 4 sq units
72. Given curves are y = log x …(i)
y = log | x |, … (ii)
y = | log x | … (iii)
and y = | log |x|| …(iv)
The graph of the given curves could be plotted as
Required area = Area of the shaded region
= 2 (Area in first quadrant)
1
= 2 | log x | dx
0
1
= 2 log x dx
0
7
ASPT (PCM) - JEE Test - 49
= 2[x(log x 1)]10 2 sq units
73. We, have, | x 1| ≤ 2 1 ≤ x ≤ 3
The graph of | x 1 | ≤ 2 and x 2 y2 = 1 could be plotted as
Required area = Area of the shaded region
3
= 2 x2 1dx
1
3
= 2 x x 2 1 log(x x 2 1
1
= {3 8 log(3 8)}
= 6 2 log | 3 2 2 |
74. Clearly, f : [0, 2] → [0, 2] given by
f(x) = x + sin x is a bijection. So, its inverse exists, the graph of f 1 (x) is the mirror image of the
graph of f(x) in the line y = x.
Required area = 4 (Area of one loop)
= 4 (x sin x)dx x dx
0 0
= 4 sin x dx
0
= 4[cos x]0
= 4 [cos cos 0]
= 8 sq units
1
75. We have, f (x) max sin x, cos x,
2
8
ASPT (PCM) - JEE Test - 49
cos x, 0 x
4
5
= sin x, x
4 6
1 5 5
2, x
6 3
5 / 3
Required area = 0
f (x)dx
/ 4 5 / 6 5 / 3 1
=
0
cos x dx
/4
sin x dx
5 / 6 2
dx
1 5 / 3
= [sin x]0 / 4 [cos x]5/ 4/ 6 [x]5 / 6
2
1 3 1 1 5 5
=
2 2 2 2 3 6
1 3 1 5
=
2 2 2 12
3 5
= 2
2 12
