r

i to
TARGET : JEE (MAIN + ADVANCED) 2025
TEST DATE: 30TH DECEMBER, 2024

Ed
ACADEMIC SESSION: 2024-25

COURSE COURSE BATCH

JP 01JP CLASS XII
DETAILS NAME CODE(S)
TEST TEST TEST REVISION CUMULATIVE TEST CODE &
JEE (MAIN) TEST (RMCT) SEQUENCE RMCT 03
DETAILS PATTERN TYPE

F
PD
ANSWER KEY
Q.No. 1 2 3 4 5 6 7 8 9 10

Ans. 1 2 2 1 3 1 4 1 4 2

PART-A: Q.No. 11 12 13 14 15 16 17 18 19 20
MATHS
te r
Ans. 3 3 4 3 3 2 3 2 1 2

Q.No. 21 22 23 24 25

Ans. 0003 0002 0014 0019 0001

Q.No. 26 27 28 29 30 31 32 33 34 35
as

Ans. 1 2 4 1 3 3 1 1 1 2

PART-B: Q.No. 36 37 38 39 40 41 42 43 44 45
nM

PHYSICS
Ans. 4 1 2 3 1 2 1 3 2 3

Q.No. 46 47 48 49 50

Ans. 0012 0020 0177 0005 0020

Q.No. 51 52 53 54 55 56 57 58 59 60

Ans. 2 3 3 4 3 4 3 1 2 4
di

PART-C: Q.No. 61 62 63 64 65 66 67 68 69 70
CHEMISTRY Ans. 3 1 2 3 3 1 1 4 3 3
a te

Q.No. 71 72 73 74 75

Ans. 0004 0003 0006 0025 0005

re

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 r
TEXT SOLUTIONS (TS)
yiˆ + ĵx

i to
PAPER k̂ × r = –
Now, x(  zjˆ + k̂y ) + y ( îz – xkˆ ) + z
PART-A: MATHEMATICS
(  yiˆ  ĵx )=  xzjˆ + xykˆ + yziˆ –
1. (300)
xykˆ  zyiˆ + xzjˆ = 0

Ed
2.  xy – 2x – y + 2 = 0 (3) ( a  b )2 = (– c )2
 x(y – 2) – (y – 2) = 0  | a |2 + | b |2 + 2 a.b = | c |2
 centre is (1, 2)  9 + 25 + 2 a.b = 49
 Required circle is orthogonal to  2 a.b = 15
x2 + y2 + 2x + 4y – 4 = 0
 2| a | | b | cos  = 15
 radius = length of tangent drawn
from (1, 2) to the given circle  2(C)(5) cos  = 15

F
= 1 4  2  8 – 4  11 1  
cos  = = cos   =
 required circle is 2 3 3
(x – 1)2 + (y – 2)2 = ( 11 )2 (4) We have

PD
 x2 + y2 – 2x – 4y – 6 = 0

221
0 + C1 + C2 + …… + 21C10 =
3. 21C 21 21 = 220
2

4. For product of the digits to be divisible by 3 a + b + c = o

atleast one digit should be either 0,3 6 or 9  a × ( a +b +c ) = a × o
Therefore, possible four digit numbers  a × a + a × b + a × c = o
te r
= total – (when none of the digit is 0, 3, 6 ( a ×a =o )
or 9)  a × b = c × a
Total = 9000
4 digit numbers when none of digit is 0,3,6 Similarly a × b = c × a = b × c Hence (B)
or 9 = 64 = 1296
 Number of possible four digit 6. Let c  xiˆ  yjˆ  zkˆ
as

numbers = 9000 – 1296 = 7704 2 1 1

1 2 –1 = 0  x=y+z
5. (1) r a  ba x y z
 r  a – b  a  0 or
c   y  z  ˆi  yjˆ  zkˆ
(r – b)  a  0
nM

 c.a0
r  b  a ... (i)
y = –z
Similarly other line is
r  a  b  c  –zjˆ  zkˆ
Clearly common point is c  1  1
z=±
a  b  3iˆ  ˆj – kˆ 2
(2) Let r = xiˆ  yjˆ  zkˆ , r  î  x ,  c
1
2
 – ˆj  kˆ 
di

r  ĵ  y , r  k̂ = z
i j k 7. LCM of p and q is r2 s2 t3 . Therefore at
î × r = 1 0 0 = – ĵ (z) + k̂ (y) least one of p and q is divisible by r 2 . If
p is divisible by r2 and q is not divisible by
x y z
r3, therefore possible ways are (2, 0), (2, 1),
a te

= – zjˆ + k̂y (2, 2), (1, 2), (0, 2)

which are 5 in number. Similarly 5 ways for
ˆi ˆj kˆ
s and 7 ways for t
ĵ × r = 0 1 0 = î(z) + k̂(  x) total number of ways = 5 × 5 × 7 = 175.
x y z
8.  + m + n = 0 , 2mn + 2m – n = 0
= – x k̂ + îz  2m(– – m) + 2m – (– – m) = 0
re

 –2m2 + 2 + m = 0
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 

r
2 + m – 2m2 = 0 70
when 43 – r =  r = 43 – 35 = 8
 2 + 2m – m – 2m2 = 0 2

i to
 ( + 2m) – m( + 2m) = 0
 ( + 2m) ( – m) = 0 14. abc = 0
 = m,  = –2m  b  c  –a
when  = m, m + m + n = 0  n = –2m
 48 + c2 + 48 = 144
direction ratio

Ed
m, m, –2m  1, 1, –2  c2 = 48  c2  4 3
1 1 2 2
Direction cosine , , c
6 6 6  – a = 24 – 12 = 12
 –2m, m, m–2, 1, 1 2
Direction cosines R
2 1 1 
, , a
6 6 6 

F
b
9. Number are of form
 x + y + z = 7 ; 0  x, y z  7 P  Q

PD
Number of ways = 9C2 (9 identical balls in c
3 different boxes, empty allowed) Further
a  b  –c
10. 8x – ky + (k2 – 8h) = 0  144 + 48 + 2a.b = 48
2x + y – p = 0
 a.b = –72
Comparing coefficients of x, y and constant
term, we get  abc = 0
k  8h
2  ab  ac = 0
4 = –k = k = –4
p  a  b  c  a = 2 a  b
te r
16 – 8h = –4p
= 2. 144.48 –  72  = 48 3 Ans.(C)
2
4 – 2h = –p  p = 2h – 4

11. To the obtain answer take n=5 in the 15. There are three choices for each element
following solution (a) Present in Set B but not in Set C
as

Total n-digit numbers using 1, 2 or 3 = 3n (b) Present in Set C but not in Set B
total n-digit numbers using any two digits (c) Not present in any Set
out of 1, 2 or 3 = 3C2 × 2n – 6 = 3 × 2n – 6 So total ways = 3n
total n-digit numbers using only one digit of Every pair is being repeated twice except 
1, 2 or 3 = 3
3n – 1
 the numbers containing all three of in last part. Hence 1
nM

the digits 2
1, 2 and 3 at least once = 3n–(3×2n–6)–3
= 3n – 3 . 2n + 3 16. Let the point of intersection with the line be
(–3– 1, 2+ 3, –+ 2)
1 0 1 Hence direction ratio of the line is
12. [ab c] = x 1 1 x (–3+ 3, 2– + 1). Since this line is
parallel to
y x 1 x  y
x + 2y – z = 5
1 0 0 so v1 . v2 = 0
di

 c3 c3 + c1 = x 1 1  –3+ 3 +  +  – 1 = 0
y x 1 x  2+ 2 = 0  = –2
Hence direction ratio 6, –2, 2 so line
1 0 0
x  4 y  3 z 1
c3  c3 – c2 = x 1 0  [ab c] = 1  
3 1 1
a te

y x 1

So [ab c] does not depends on x and y . 17. (c.a) b – (c.b) a  (a.c)b – (a.b)c
13. 40C 30
r . C3 +
40C
r+1 . C4 + ….. +
30 40C
r+7
30C
10
 (a·b)c  (c.b)a
mean coefficient of x43–r in (1 + x)70
that is 70C43–r  will be maximum
re

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 r
18. x number of 1 unit steps a2 – b2
y number of 2 unit steps = 2b

i to
x + 2y = 7
x = 7 y = 0 Number of ways = 1 23. Straight line and circle intersect in
x = 5 y = 1 Number of ways = 6 maximum (8 × 2 × 6) points = 96 points
x = 3 y = 2 Number of ways = 10 Straight lines and straight line intersect in
x = 1 y = 3 Number of ways = 4 maximum 8C2 points = 28 points
Total number of ways = 21 ways Circle and circle intersect maximum 2×6C2

Ed
Now he cannot go on 8th step. Now he points = 30 points
reach 9th step in 1 ways and then again Total maximum intersection points =
move from 9th to 15th steps x + 2y = 7 96 + 28 + 30 = 154 point
Number of ways = 21   . k = 154  k = 14
Total number of ways = 21.21 = 441
1 2 3
19. AB = 4  1 4 = 3 24. 3 1 3 =

F
AC = 144  16  9 = 13 2 3 1
We know from geometry that internal  = (1 – 9) – 2(3 – 6) + 3(9 – 2)
bisector of BAC divides BC in the ratio  = – 8 + 6 + 21

PD
AB : AC i.e., 3 : 13. Hence  = 19 mat
 27  65 18  39 –9  26  25.
D  , ,
 16 16 16 
= (19/8, 57/16, 17/16).

20. Let x i  2n i  1 where n1 , n 2 , n 3 , n 4  1

Hence we have
2n1  1  2n 2  1  2n 3  1  2n 4  1  98
te r
n1  n 2  n 3  n 4  51
Given n1 , n 2 , n 3 , n 4 each equal to 1 Equation PQ
x 1 y  2 z – 3
We have n1  n 2  n 3  n 4  47   
2 3 6
 C3 
50.49.48
 19600  n Let Q  (2 + 1, 3 – 2, –6 + 3)
as

50
6 Q lies on x –y + z = 5
n   (2 + 1) – (3 – 2) + (– 6 + 3) = 5
  196
100 1  9 11 15 
 =  Q  ,  , 
7 7 7 7 
2
nM

3x – 4y  7 2 2 2
21. (x – 3) + (y – 2) = PQ= 1 – 9     2  11   3  15  = 1
2 2
5  7  7  7 
focus is (3, 2) & directrix is 3x – 4y+7=0
latus rectum = 2 × r distance from focus to --------------------------------------------------------------
16
directrix =
5
PART-B: PHYSICS
22.
26. Minimum work done to accelerate the truck
di

from speed 0 to v and from v to 2v are

1 1 1
W 1 = mv2 – m(0)2 = mv2
2 2 2
1 1 3
and W 2 = m(2v)2 – m(v)2 = mv2
2 2 2
a te

 W 1 < W 2
1
area of PF1F2 = A = (2 2ae) 2 bsin 
2 1
27. Vd
 A = 2abesin r2
Maximum area A = 2abe r
2ab a2 – b2 r1 =
= 2
re

a Ans. (B)
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 0 qv sin  v sin  2m1

r
28. B= and 
4 r 2 r T1 1 qB  m  q B 
 = 1 1 =  1  2 2

i to
36.
T2 2 2 m 2  q1B1   m2 
29. Let the charge on the other plate be q q2 B 2
Electric field between the plates
Time period is independent of kinetic
q  10 energy.
= where A = area of plate
2o A

Ed
Potential difference between the plates 37.
q  10 r 
= .d where d = distance between the
2o A – +
plates A B
q  10 2t  10 2r
=   0.2 t  1
2 C 2 (5) + – i
2

F
0 I1  I q0
30. B1  , B2  0 2 , B1 B2 
2a 2a 38. i0 = A

PD
K 0
B B12  B22 2 2
q0 q0
u= =
2C K A
31. N = m2r 2 0
f = mg d
s.m2r = mg d = 5 mm.
g 10
s = 2 = = 0.2 mv sin  2m v cos 
r 25  2 39. & P
qB qB
te r
2r r tan  1
32. I= = 2rn.   
1/ n P 2 2
 r2 P
B = 0 2 (2rn) r
2
2y
as

32
33. No charge will flow in the circuit after the 40. In series  =
switch is closed. Hence no heat is produced 3r  2
in the resistor. 2
In parallel  =
r /3  2
34. The expression of magnetic energy density 6 6
nM

but = r = 2
B2 3r  2 r6
is .
20 32
then  = = 0.75 Amp.
The magnitude of magnetic field at end 32 2
region is half that of at middle region.
Hence the ratio of magnetic field energy 2 2
densities at middle region to the end region 41. i=  VAB = × 10
10  R 10  R
is 4:1
2 10
x =   xmax = 0.2 V/m.
di

10  R 10
  /2
35. R=  R1 =
A A 42. Potential difference across ‘C1’
R1 is stretched to double length but volume C2 V V
is constant. V1 = =
C1  C 2 1  C1 C2
A
a te

A. = A’ ×   A’ = When dielectric is inserted C2 will increase.

2 2
 2
R’ = = = 2R q02
A' A 43. U0 =
2C
q02 e 2t1 /  U0 q2
U= = = 0
2C 2 4C
re

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 r
1 90 V
 e2t1 /   = V1 + 2
2 4 4

i to
 V1 = 20 m/s.
t1 = n2 ....(1)
2
1
and o q = q0 e t2 /  48. Emax = 2
CVmax
2
q0
= q0 e t2 /  , K. 0 A

Ed
4 Vmax = Emax.d and C =
d
1
e t 2 /   1 K. 0 A 2
4 Emax = . Emax. . d2
2 d
t2 = 2 ln 2 ....(2) 2
t1 1 = K.Î0. Emax. (Ad) = 177 × 10–7 J

t2 4
49. Pres = 2 Prod
(12 – V) i = 2 V i

F
44. W=Zit V = 4 volt
E 32 / 4 I2 R = 2 vi
Z 
F 96500 (0.2)2 V3 R = 2 V (0.2) V3/2

PD
w = 1 gm R=5
l=2A
Solving t  6000sec dx
50. = v0cos
t = 100min. dt

C3
45. q3 = .Q
C 2  C3
3 3  
te r
q3 = × 80 = × 80 = 48 C
32 5 dv t
= 10–1x
dx
0 q1v 0 q 2 v
46. B  .
4 r 2 4 r 2 2  1 20
xmax = = 20 cm.
101  1
as

+2e
v

--------------------------------------------------------------

PART-C: CHEMISTRY
nM

v –1e
0 v
  q1  q 2  51. According to CFT theroy.
4 r 2
 v 52. Primary valency = oxidation no. of central
 0 . 2  2   1  atom = 3
4 I This compound contains Cl¯ and C2O42– as
 3 ligand. Cl¯ is a unidented ligand and C2O42–
 0 .v  0  v  12m / s
4  is a bidented ligand. so secondary valency
=6
di

53. Oxidation state of cobalt in Co2(CO)8 = 2x

47. + 8 (–0) = or x = 0
So EAN = 27 + 1 + 8 = 36.
As it will be the atomic number of Kr and
a te

according to EAN rule, the complex having

V2 = V1 cos60º EAN equal to noble gas is more stable.
V
V2 = 1 54. (4)
2
Chelating ligand form chelate rings which
1 1 1 1 are comparatively much stable.
× 90 × = 1V1 + V2
2 2 2 2
re

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 r
55. (3) 70. Number of sp3 hybridised carbon atom is 6.
Ligands are named in alphabetical order.

i to
56. [Ni(H2O)4]2+ will be paramagnetic with 2
unpaired electrons.

57. (3) Potassium hexacyanidoferrate(III) 71.

............... K4[Fe(CN)6]

Ed
58. Hyb  dsp2, sq. Planar. 72. [Mabcd] 3 (ab) (cd) aTb
(ac) (bd) aTc
59. N is central atom. (ad) (bc) aTd

60. Fact 73. 4FeCr2O4 + 16NaOH + 7O2 

8Na2CrO4 + 2Fe2O3 + 8H2O.

F
61. Chromyl chloride confirmatory test for ionic Let the oxidation state of Cr in CrO42– is x.
chlorides which forms CrO2Cl2 (deep red) So x + 4 (–2) = –2 or x=6

PD
62. V2O5 and Cr2O3 are amphoteric in nature. 74. =q×d
Mn2O7 and CrO3 are acidic in nature. 1.2 × 10–18 = q × 10–8 cm
V2O3, CrO and FeO are basic in nature. 1.2  1018
q = 1.2 × 10–10
108
63. 2Na2S2O3 + AgBr  q
% charge =  100
Un exp osed e
Na3 [Ag(S2O3)2] + NaBr 1.2  10 10

This property is used for fixing in =  100 = 25%

4.8  10 10
te r
photography.

64. Np and Pu shows maximum oxidation H

number + 7. 75. CCCCC CCCCC
65. NaNO3 does not react with AgCl. OH OH
(d  )
as

AgCl + 2Na2S2O3  Na3[Ag(S2O3)2] + NaCl

AgCl + 2NH4OH  [Ag(NH3)2]Cl + 2H2O C
3AgCl + Na3AsO3  Ag3AsO3 + 3NaCl CCCC
OH
66.  = n(n  2) ; 3d5 has maximum,
nM

(d  )
5 unpaired electrons so it will have highest
magnetic moment.
---- TEXT SOLUTIONS (TS) END ----
67. (1) VO 2 
2 < Cr2O7 < MnO4

68. (4) Oxidation state of iron in mohr’s salt

FeSO 4 ( NH 4 )2 SO 4 . 6 H 2 O is + 2.
di

69. (2)
a te

(4) Species Number of * electrons

O2 2
O2– 3
O22– 4
Total = 9 electrons
re

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