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1
Answer Keys
General Aptitude
1 C 2 B 3 B 4 24 5 15876 6 D 7 D
8 4 9 54 10 C
Electronics and Communication Engineering
1 B 2 -1 3 0.5 4 B 5 B 6 -0.1538 7 0.333
8 D 9 20 10 0.4 11 D 12 C 13 A 14 C
15 2 16 D 17 C 18 B 19 B 20 2000 21 A
22 B 23 D 24 0 25 A 26 6 27 B 28 D
29 D 30 C 31 C 32 15.62 33 1.231 34 D 35 A36 D 37 1.53 38 C 39 C 40 B 41 A 42 46
43 C 44 B 45 22.89 46 7 47 1.95 48 1.15 49 1393
50 -165.3 51 50 52 B 53 0.001 54 A 55 53.31
Explanations:-
General Aptitude
4.
4
5
6
5
(?) 12
48(?)
4 6
5 5
6
25 5
(?) (?) 48 12
(?) 48 12 (?) 48 12
? 48 12 12 2 24
5. Let the side of the square be x cms
4x 504
504
x 126cm4
2 2Area of the Square x 126 126 15876cm
8. While we could use formulas, a lot of combination questions are easier to answer through a bit of common sense and brute force. On this question, it's much quicker to just list the possible committees than it is to use a formula approach.
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2
No Paul means no Jane, which leaves us with only Joan/Stuart/Jessica. Sadly, Stuart cannotappear in any other committees.
Paul then needs 2 co-members and has 3 possible buddies for those 2 spots. We could use3C2=3 to calculate or we could just brute force:
Paul/Joan/Jessica
Paul/Joan/JanePaul/Jane/Jessica
So, as others have noted, there are 4 possible committees
9. Let's start by breaking 210 down into primes:
210 = 10 * 21 = 2*5*3*7
So, to start, we know that all permutation of 2,3,5,7 will fulfill the requirements: 4! = 4*3*2 =
24
however, we have to recognize that we may be able to use 1 and another number in place of a
pair of our numbers.
Since each digit has to be less than or equal to 9, the only substitution we can do is to use 1*6
instead of 2*3 (2*5, the next smallest pair of primes, would give us 10).
So, we can also use the digits 1,5,6,7. Same as last time, this gives us 4! = 24 possible
Arrangements
So, 24+24=48 nowhere does it say we need 4 digits. So, we can also make some 3 digit
numbers using 5, 6 and 7.
That's 3! more possibilities.. Add another 6 to our 48=54
10. You have 20 shoes, and are looking for a matching pair. The first shoe you take could be
anyone, so probability of getting one shoe is 1. Now, you took 1 shoe out of the 20 original,
and you have 19 left, but there is only 1 shoe in the 19 that matches the first one. So the
probability to get the matching shoe is 1/19.
So multiply your probabilities: 1 * 1/19 = 1/19
Electronics and Communication Engineering
1. We have
2x y
2 2
x y
1 r
tan r
For r 1
tan 0
0
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3
2. Let y xf x, y x y C Implict function
Then y 1 x
y x 1
f yx y log ydy x
f dx x log x x.yy
Atdy
x 1, y 1 ; 1dx
3. For final value, finding,s 0limsF(s)
s 0
1 1lims. 0.5
s(s 2) 2
4. Taking Laplace transform of given equation
2 10s y(s) sy(0) y (0) 4 sy(s) y(0) 5y(s)
s 1
Substituting initial conditions
2 2
s 6 10y(s)
s 4s 5 (s 1)(s 4s 5)
Using partial fractions
2
1 1 2t t
2
1 1y(s)
(s 2) 1 s 1
1 1L [y(s)] L e sin t e
(s 2) 1 s 1
5. y 3 yD.E e dx y xe dy … (1)
(1) Can be written as
y 3dxx e .y linear equation in x
dy
Also, (1) can be written as
y y 3e dx xe y dy 0 isexact
M N
M NSince
y x
Equation (1) is not a homogeneous D.E.
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4
6. Using source transformations:
x x x
x
x
10 5i 50i 12.5i 5
5 32.5i
i 0.1538A
8. The circuit is ambiguous and hence cannot exist. Because two voltage sources of different
values are connected in paralleled across the same points
9. 10 5 2 K 550 2K 10
K 20
10.
11.
C G G
R 1 ( 1) G G 2 1 G
E
F 20E
10 5
F 50xE 20E
x 20 50 0.4
x5 i
x5 i
5 5 2A
10
10V
5 x50 i
5V
10
10V
2.5
xi
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5
12.
t2t 2t
2 0
0
e 1 eu t+ e u d
2 2
18.
For SR flip-flop S=R=1, will never occur, hence it can be used as slave
19.
0111 0011
RAR 0011 1001
0100 1010
20. From Einstein’s Relation
n
n
2
n
D KT
q
502000cm V s
0.025
21.D A
N N
So, the sample is n-type
23. Synchronous Detector uses LPF.
24. for r 2,
2
120 0
r r
25.
O OO
broad
114.6 114.6
FNBW 6.37L 18
O O O
end
2 2FNBW 114 114 38.2
L 18
J nQ J nQ
nQnQK K
Clock
J nQ S nQ
nQnQK R
Clock
JK JK
(1)(2)
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6
26.
6 2 2 2
AX x 2 a 1 1
2 1 b 1
6 2 2 2 2
2 a 1 1 12 1 b 1 1
16 2
4 a 1 1 2 16 8
5 b 1
5 a 5 a 8
a 3 a 3
5 b b 3
a b 6
27: Given
2
0 0
dy0.025y 1; x 0; y 1;h 1
dx
Here f(x, y) =0.025y2+1.
By implicit Euler’s method; we have
n 1 n n 1 n 1
2
1 0 1 1 1 1
2
1 12
1 1
y y hf x , y
y y hf (x ,y ) y 1 1[0.025 y 1]
y 0.025y 2
0.025y y 2 0
1y 20 8 5 37.88,2.12
28. Given that 0 0
dyx y & x 0; y 1
dx
1 0 1 2 3
11 0 0 2 0 0
Let h 0.2; f(x, y) x y
By3 rd order runge kutta method; we have
1y y k 4k k ;
6k h
where k hf (x , y ), k hf x , y2 2
12 0 0
3 0 0 2 1
k hk hf x ,y
2 2
k hf x h, y 2k k
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7
1 0 0
2
3
k 0.2[x y ] 0.2[0 1] 0.2
0.2 0.2k 0.2 0 1 0.2[1.2] 0.24
2 2
k 0.2 0 0.2 1 2(0.24) 0.2 0.296
11y 1 0.2 4(0.24) 0.296 1.2436
29. z = 0,1 (singularities) of f(z), lies inside
C : z 1 3
2z 2z2z
z z 1 e ef z e
z z 1 z 1 z
Also, e2z is analytic in and on ‘C’
2z 2z
C C C
2z 2z
z 1 z 0
2
e e
f z dz dz dzz 1 z
2 j e 2 j e using cachy's integral formula
2 j e 1 , where j 1
30.c
(3x 4 y)dx (2x 3y)dy
2 2
(2 4)dxdy (by Green 's theorem)
2 dxdy 2( r ) 2( 2 ) 8
31.O
O
L C
4 0I I I 4 90
1 j10
2
4 1 5 1 16 j j4 j 1 4 j 4 I j A
5 5 5 5
32. Applying KCL at Node A
OA A
V V5 30 4 j3 4 j3
O
AV 15.625 30
33.
t 1
L 1 1 L
t t
L 1 L L L
8I t Ke 10sin 100t tan , I 3A
6
I t 0 I I t I e
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8
If transients go out
L LI t I
So, 1 o
11000t tan 8 6 17.45
1000t 70.587
t 1.2319 m sec
34. O
pc : GH 180
1 O
pc90 tan 1803
2
2 2
3M 0
9
pc1GMM
35.2
3 1M 1 1
4
4 2 24 9 0 1.60, 5.60
gc 1.26
O 1
pc
180180 3 90 tan 2 0.449 rads sec
36.
1
1
dMslope
d log
6 3220 2
log log 0.1
2
2
0 620 5
log 10 log
37. 3 1 1 1Y s = . =s s+3 s s+3
3tY t = 1 e u t
Steady state value y 1
3t1 e 0.99 1
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9
t 1.53sec
38. 2
1 1H s = =
s +5s+6 s+3 s+2
1/2 1 3 / 2Y s = +s s+3 s 2
3t 2t1 3Y t = e e u t
2 2
39.
41.
' inin
'
out out
R R
1 A
R R (1 A )
42. In –ve cycle capacitor will charged to 26V.
In +ve cycle, after capacitor got charged diode will never turn ON.
and
0 iV V 26V
0 peak inV V 26 20 26 46V .
43. Let P be even parity bit and d be difference output.
a b c d
1 -1
a b c d -a -b -c -d
a b-a c-b d-c -d
1 0 -1 0 0
0
1
1
d
a
c b
b
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10
2 1 0
Input p d
d d d
0 0 0 0 0
0 0 1 1 1
0 1 0 1 1
0 1 1 0 0
1 0 0 1 1
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1
So p = d
' ' ' ' '
2 1 0 1 0 2 1 0 1 0
'
2 1 0 2 1 0
2 2 2
p d d d d d d d d d d d
d d d d d d
d d d
F
44.
3 2 1 0
Input
x x x x
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
3 2 1 0
Output
y y y y
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
So y3 y2 y1 y0 is excess-3 equivalent of x3 x2 x1 x0
45. a R nn 16
N 65535V V 3 V
2 2
Step size R 16
1V
2
00 01 11 10
0
1
2d1 0d d
1
1 1
1
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11
3
Step size V 45.77 V65535
step sizeQuantisation error 22.89 V
2
46.
A B C D
B C
Serial inQ Q Q Q
Q Q
X 1 0 0 0 1
1 1 1 0 0 0
2 0 1 1 0 1
3 1 0 1 1 0
4 0 1 0 1 0
5 0 0 1 0 0
6 0 0 0 1 1
7 1 0 0 0
47. At punch through,
1
2 bi PE c
0
B c B
26 V V N 1x
e N N N
PE bi
14 164
19 16B B
16 3
B
V 25V& neglect V
2 11.7 8.85 10 25 10 1
0.75 10 1.602 10 N N 10
N 1.95 10 cm
48. Since,ec e b c d
b e
4
dd 7
s
100ps 25ps
x 1.2 1012ps
V 10
c c c
ec
T 12ec
T
r C 10 0.1pF 1ps
so, 100 25 12 1
138ps
1 1f
2nT 2n 138 10
f 1.15GHz
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12
49. PT
P T
23
P
DV
p
D p V
cmD 500 25.9 10 V
V sec
2
P
cmD 12.95
sec.
So, diffusion length of holes
P P
9
P
3
P
P
L D
L 12.95 150 10 cm
L 1.393 10
L 1393 microns
50. Thermal Noise Power = KTB
23
23
1.38 10 306 7000
2955960 10
Converting into dB scale
2310log 2955960 10
165.3dBw
51. si sf f 2IF
SS
f f 2
2
si sf 2f
si s s s
s si s s
f f 2f f
f f f 2f
2 0.5
1.5
2 2
RR I 1 Q
2 2 21 Q
2 2275 1 Q 1.5
5625 1Q 50
2.25
52. The AGC allows listening to station without constantly monitoring the volume control.
53. OO
I dl cos t Br A
4 r
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13
6
8
2 100.02 rad m
c 3 10
67 10 0.1 cos 2 10 t 2A 10A
4 100
6cos 2 10 t 2 wb m 0.001 wb m
55. 2avg 0
1P H
2
f x,z x z 1 0
x zn n
f a aa ; ds dsa
f 2
x z2avg xt 0
1 a aP P .ds H a .
2 2
53.31 mW.
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