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GATE SOLVED PAPER
Electrical Engineering
20142
Copyright By NODIA & COMPANY
Information contained in this book has been obtained by authors, from sources believes to be reliable. However,neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor itsauthors shall be responsible for any error, omissions, or damages arising out of use of this information. This bookis published with the understanding that Nodia and its authors are supplying information but are not attemptingto render engineering or other professional services.
NODIA AND COMPANYB8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039
Ph : +91  141  2101150
www.nodia.co.in
email : [email protected]

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ia.co.in
General Aptitude
Q. 1  Q. 5 Carry one mark each.
Q. 1 Choose the most appropriate phrase from the options given below to complete
the following sentence.India is a postcolonial country because
(A) it was a former British colony
(B) Indian Information Technology professionals have colonized the world
(C) India does not follow any colonial practices
(D) India has helped other countries gain freedom
Sol. 1 Correct option is (A).A pronoun (it) is used after a conjunction (because), so the complete sentence is:India is a postcolonial country because it was a former British colony.
Q. 2 Who ______ was coming to see us this evening ?(A) you said (B) did you say
(C) did you say that (D) had you said
Sol. 2 Correct option is (B).Who did you say was coming to see us this evening ?This is a question and who in that sentence is an interrogative pronoun. For thatreason we need to invert subject and verb and write did you say ?
Q. 3 Match the columns
Column 1 Column 2
1. eradicate P. misrepresent
2. distort Q. soak completely
3. saturate R. use
4. utilize S. destroy utterly
(A) 1 : S, 2 : P, 3 : Q, 4 : R (B) 1 : P, 2 : Q, 3 : R, 4 : S
(C) 1 : Q, 2 : R, 3 : S, 4 : P (D) 1 : S, 2 : P, 3 : R, 4 : Q
Sol. 3 Correct option is (A).
Eradicate to remove or destroy utterlyDistort to give a false, perverted, or disproportionate meaning to; misrepresent

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(C) 110 (D) 120
Sol. 4 Correct option is (B).
Average value 19200 9 100#
= +
100=
Q. 5 The value of ...12 12 12+ + + is(A) 3.464 (B) 3.932
(C) 4.000 (D) 4.444
Sol. 5 Correct option is (C).Given expression can be written as
x ...12 12 12= + + +
x x12= + ...x 12 12a = + + x2 x12= +
x x 122   0=
x x x4 3 122  +  0=
x x x4 3 4 + ^ ^h h 0= x x3 4+ ^ ^h h 0= x 3= , 4
x 4= (xcan not be negative)
Q. 6  Q. 10 Carry two marks each.
Q. 6 The old city of Koenigsberg, which had a German majority population beforeWorld War 2, is now called Kaliningrad. After the events of the war, Kaliningradis now a Russian territory and has a predominantly Russian population. It isbordered by the Baltic Sea on the north and the countries of Poland to the southand west and Lithuania to the east respectively. Which of the statement belowcan be inferred from this passage ?(A) Kaliningrad was historically Russian in its ethnic make up
(B) Kaliningrad is a part of Russia despite it not being contiguous with the rest
of Russia
C Koeni sber was renamed Kalinin rad as that was its ori inal Russain

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Sol. 6 Correct option is (A).
Ethnic cleansing or makeup is the process of using violent methods to force
certain groups of people out of a particular area or country. We can infer that
Kaliningrad was historically Russian in its ethnic make up.
Q. 7 The number of people diagnosed with dengue fever (contracted from the bite ofa mosquito) in north India is twice the number diagnosed last year. Municipalauthorities have concluded that measures to control the mosquito populationhave failed in this region.Which one of the following statements, if true, does notcontradict this
conclusion ?
(A) A high proportion of the affected population has returned fromneighbouring countries where dengue is prevalent
(B) More cases of dengue are now reported because of an increase in theMunicipal Offices administrative efficiency
(C) Many more cases of dengue are being diagnosed this year since theintroduction of a new and effective diagnostic test
(D) The number of people with malarial fever (also contracted from mosquitobites) has increased this year
Sol. 7 Correct option is (D).
The number of people with malarial fever (also contracted from mosquito bites)
has increased this year.
Q. 8 If xis real and x x2 3 112  + = , then possible values of x x x3 2 +  include(A) 2, 4 (B) 2, 14
(C) 4, 52 (D) 14, 52
Sol. 8 Correct option is (D). x x2 32  + 11=So we can have two possible equalities
x x2 32  + 11= or x x2 8 02   = ...(i)and x x2 32  + 11= or x x2 14 02  + = ...(ii)For eq (ii),
Discriminant T b ac42
=  ( )( )4 4 1 14 01= So, roots of equation (ii) will be imaginary, but given xis real. Therefore wedont consider this equation.
For eq(i)Discriminant T b ac42=  ( )( )4 4 1 8 01=  
So, roots of equation (i) will be real.
x x2 82   0=
x x x4 2 82  +  0=
x x4 2 +^ ^h h 0=

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x x x3 2 +  64 16 4 52 +  =^ ^h hQ. 9 The ratio of male to female students in a college for five years is plotted in the
following line graph. If the number of female students doubled in 2009, by whatpercent did the number of male students increase in 2009 ?
Sol. 9 Correct answer is 140.Let male students in 2006 be x1and female students be y1. From the given graphwe can see that in 2008 ratio of male to female students is 2.5. So
yx
1
1 .2 5=
x1 . y2 5 1= Let number of male student in 2009 be x2and number of female students be y2
. From the graph we can see that in 2009, ratio of male to female students is 3.So
yx
2
2 3=
Given that number of female students is double in 2009, so we have y y22 1= . The
above ratio now becomes as
y
x2 1
2 3=
x2 y6 1=
%of No. of male students increased in 2009
x3
x
x x 1001
2 1#=

x3 .
.y
y y2 5
6 2 5100 140
1
1 1#=
=
Q. 10 At what time between 6 a.m. and 7 a.m. will the minute hand and hour hand ofa clock make an angle closest to 60c?(A) 6 : 22 a.m. (B) 6 : 27 a.m.
(C) 6 : 38 a.m. (D) 6 : 45 a.m.
Sol. 10 Correct option is (A).

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As shown in Figure above, at 6:00 a.m. initial angle between minute and hour
hand is 180c. As we know that hour hand completes 30c in every hour (60
minutes), so angle moved by hour hand in 1 minute is 6030 . Similarly, minute hand
complete 360cin every hour, so angle moved by minute hand in 1 minute is60
360 .
Let us assume that after xminutes the angle is 60c. Let us assume that after x
minutes the angle is 60c.
60c x x18060360
6030
angle moved
by hour hand m
cc
c
c
c= + 
angle moved
by in hand
b bl l
60c .x x180 6 0 5c= + 
. x5 5 120=
x.
. min6 5
120 21 8c= =
Therefore approximately at 6.22 am the angle will close to 60cbetween the
hour hand and minute hand.
END OF THE QUESTION PAPER

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Electrical Engineering
Q. 1  Q. 25 Carry one mark each.
Q. 1 Which one of the following statements is true for all real symmetric matrices ?(A) All the eigenvalues are real
(B) All the eigenvalues are positive
(C) All the eigenvalues are distinct
(D) Sum of all the eigenvalues is zero
Sol. 1 Correct option is (A).Corollary: Every real symmetric matrix has eigenvalues which are all real numbers.
Q. 2 Consider a dice with the property that the probability of a face with n dotsshowing up proportional to n. The probability of the face with three dots showingup is____.
Sol. 2 Correct answer is 0.14 .On the first dot probability 1\On the second dot probability 2\
.
.
.
On the sixth dot probability 6\ Total probability 1 2 3 4 5 6= + + + + +
21=
Probability of 3rddot .213
71 0 14= = =
Q. 3 Minimum of the real valued function f x x 1 /2 3= ^ ^h h occurs at xequal to(A) 3 (B) 0
(C) 1 (D) 3
Sol. 3 Correct option is (C).
f x^ h x10 1/2 3
= ^ hSince f x^ his a real valued function x 1^ hcan not be negative, because in thatcase fractional power of negative number is imaginary. So at x 1= , the value of
f x^ hwill be minimum.Q. 4 All the values of the multivalued complex function 1i, where i 1=  , are
(A) purely imaginary (B) real and nonnegative
(C) on the unit circle (D) equal in real and imaginary parts
Sol. 4 Correct option is (B).
We can write
1 ( ) ( )cos sinn i n2 2p p= + , where nis an

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1i e n2= p
Therefore all values of 1iare nonnegative and real number.
Q. 5 Consider the differential equation xdxd y x
dxdy y 02 2
2
+  = . Which of the following is
a solution to this differential equation for x 0> ?(A) ex (B) x2
(C) /x1 (D) lnx
Sol. 5 Correct option is (C).
This is a CauchyEuler equation, so we substitute x et= . Then assuming x 0>
, we have
t ln x=
dxdy
dtdy
dxdt
=
ordxdy
dtdy
x1= b l lnt x dxdt x1&= =: D
or xdxdy
dtdy
= ...(i)
Now,dx
d y2
2
x dt
d ydxdt
x dtdy1 1
2
2
2= c m
dx
d y2
2
x dt
d ydtdy1
2 2
2
= c mor x
dx
d y22
2
dt
d ydtdy
2
2
=  ...(ii)
Substituting eq (i) and (ii) in given differential equation, we get
dt
d ydtdy
dtdy
y2
2
 +  0=
dt
d yy2
2
 0=
Characteristic equation
m 12  0=
m 1!=Roots are distinct real, so two possible solution
y et= and y e t= 
Since x et=
So y x= orx1 are two possible solutions.
Q. 6 Two identical coupled inductors are connected in series. The measured inductancesfor the two possible series connections are H380 m and H240 m . Their mutualinductance in Hm is _____.
Sol. 6 Correct answer is 35.
There are two possible connectionsi Series adding connection

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Equivalent inductance
Leq L L M21 2= + +
380 L L M2= + +or L M+ 190= ...(i)(ii) Series opposing connection
Equivalent inductance
Leq L L M21 2= + 
240 L L M2= + or L M 120= ...(ii)
Substituting eq (ii) from eq (i), we get
L2 70=
or L mH35=
Q. 7 The switch SW shown in the circuit is kept at position 1 for a long duration.At t 0= +, the switch is moved to position 2. Assuming V V>o o2 1 , the voltagev tc^ hacross the capacitor is
(A) v t V e V 1 /c ot RC
o22
1=  ^ ^h h
(B) v t V e V1 /c ot RC
o22
1=  +^ ^h h
(C) v t V V e V 1 /c o ot RC
o2 12
1= +  ^ ^ ^h h h
(D) v t V V e V 1 /c o ot RC
o2 12
1=   +
^ ^ ^h h hSol. 7 Correct option is (D).For t 0< , the switch was at position 1 and circuit is in steady state. In steady
state capacitor behaves as open circuit. So
Therefore v 0C ^ h Vo1=Step 2 : For t 0> , the switch moves to position 2. Again in steady state capacitor

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vC 3^ h Vo2=Time constant t R CTh=
RThis the Thevenin resistance across the capacitor terminal after t 0= . So
t RC2=
Now, at any time t 0> , capacitor voltage is given by
v tC h v v v e 0/
C C Ct
3 3= + t
^ ^ ^h h h6 @ V V V e /o o o t RC2 1 2 2= +   6 @or, V V V e /o o o
t RC2 1 2
2= +   6 @or, v tC h V e V e 1o RCt o RCt2 2 1 2=   ^ h v tC h
V e V e V e V e 1 1 1 /o RCt
o RCt
o RCt
ot RC
2 2 1 2 1 2 12=     +    ^ ^ ^h h h
or, v tC h V V e V 1o o RCt o2 1 2 1=   +^ ^h hQ. 8 A parallel plate capacitor consisting two dielectric materials is shown in the figure.
The middle dielectric slab is placed symmetrically with respect to the plates.
If the potential difference between one of the plates and the nearest surface ofdielectric interface is 2 Volts, then the ratio :1 2e e is(A) 1 : 4 (B) 2 : 3
(C) 3 : 2 (D) 4 : 1
Sol. 8 Correct option is (C).

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As shown in Figure above this arrangement is equivalent to series combination
of three capacitors C1, C2and C3. We have
C1 /dA
dA
441 1e e
= =
C2 /dA
dA
222 2e e= =
C3 /dA
dA
441 1e e= =
Now given that voltage across first capacitor is C1is 2 V, so we write
VC1/
j C j C j C
j CV
1 1 11
2s
1 2 3
1#
w w w
w=
+ +=
C C C C C C C C
102 3 1 3 1 22 3 #+ + 2= V 10 Vs =^ h
Substituting values of C1, C2and C3, we get
8 16 8
8 101 2 1
21 2
1 2#
e e e e ee e
+ + 2=
16 16
8
1 2 12
1 2
e e ee e
+
51=
2 22 1
2
e ee+
51
=
2 22 1e e+ 5 2e=
2 1e 3 2e=
2
1
ee
23
=
:1 2e e :3 2=
Q. 9 Consider an LTI system with transfer function
H s^ hs s 4
1=
+^ hIf the input to the system is cos t3^ hand the steady state output is sinA t3 a+^ h, then the value of Ais(A) 1/30 (B) 1/15
(C) 3/4 (D) 4/3
Sol. 9 Correct option is (B).
Let input x t^ h cosk t3= , k 1= y t^ h sinA t3 a= +^ hWe can see that output differs only in phase. In Laplace domain
Y s^ h X s H s = ^ ^h h Y s^ h X s H s = ^ ^h h Y jw^ h X j H j w w= ^ ^h h A 1
1612# w w
=+

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the system is y t e u t e u t t t3 5=  ^ ^ ^h h hthen the input, x t^ h, is given by(A) e u tt3 ^ h (B) e u t2 t3 ^ h(C) e u tt5
^ h (D) e u t2 t5
^ hSol. 10 Correct option is (B).Impulse response h t^ h e u tt5=  ^ hLaplace transform, H s^ h
s 51=+
Output, y t^ h e u t e u t t t3 5=  ^ ^h hLaplace transform, ( )Y s
s s31
51
=+
++
s s3 5
2=
+ +^ ^h hWe know that, ( )Y s X s H s = ^ ^h hSo, X s^ h H sY s= ^ hhSubstituting Y s^ hand H s^ h, we get X s^ h
s 32=+
Taking inverse Laplace, transform
x t^ h e u t2 t3=  ^ hQ. 11 Assuming an ideal transformer, the Thevenins equivalent voltage and impedance
as seen from the terminals xand yfor the circuit in figure are
(A) ,sin t2 4w W^ h (B) ,sin t1 1w W^ h(C) ,sin t1 2w W^ h (D) , .sin t2 0 5w W^ h
Sol. 11
Correct option is (A).
Referring to secondary side

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VTh NN
Vs1
2= b l VTh sin sint t
1
2 2w w= =
^ ^h h RTh R1
2P
2
= b l R
12 4 1 4P
2
# W= = =b lQ. 12 A single phase, 50 kVA, 1000 V/100 V two winding transformer is connected as
an autotransformer as shown in the figure.
The kVA rating of the autotransformer is ________.
Sol. 12 Correct answer is 550.
From the given information for the two winding transformer, we have
V1 V1000= , VV 1002 = , kVAS 500 =
So, I1 A50= , AI 5002 =
For the autotransformer as shown in figure
Vi VV 10001= =
Vo VV V 1000 100 11001 2= + = + =
Rating of auto transformer
S a0 V I a0 2#= I Ia2 2=^ h kVA1100 500 550#= =
Q. 13 A threephase, 4pole, self excited induction generator is feeding power to a loadat a frequency f1. If the load is partially removed, the frequency becomes f2. Ifthe speed of the generator is maintained at 1500 rpm in both the cases, then
(A) f1, Hzf 50>2 and f f>1 2 (B) Hzf 502
(C) f1, Hzf 502 1 (D) Hzf 50>1 and Hzf 50

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As the load decreases, the speed increases. Relative speed between stator rmf and
rotor speed reduced which decrease the slip consequently emf induced in the rotor
and current decreases. Hence frequency decreases.
So ,f f1 2 Hz50 1
Q. 14 A single phase induction motor draws 12 MW power at 0.6 lagging power. Acapacitor is connected in parallel to the motor to improve the power factor of thecombination of motor and capacitor to 0.8 lagging. Assuming that the real andreactive power drawn by the motor remains same as before, the reactive powerdelivered by the capacitor in MVAR is ______.
Sol. 14 Correct answer is 7.
Let complex power (kVA) is given as
S P jQ= +
P"Real power W
^ h Q"Reactive Power VR^ hReal power P cosS q=
Reactive power Q sinS q=
or Qcos
sin tanP Pq
q q= =
Initially when the real power is MW12 and power factor is 0.6, the reactive power
is
Q1 tanP1 1q=
.tan12 53 13c=
^ h . .cos 0 6 53 131 cq = =
^ h MVR12 34 16#= =Now, when capacitor is added, improved power factor angle
Q2 . .cos 0 80 36 861
c= = ^ hTotal reactive power after adding capacitor
Q2 tanP1 2q= (P1is same)
Q2 .tan12 36 86# c= ^ h MAR12
43 9#= =
From the power triangle (before and after capacitor is added)

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Reactive power delivered by capacitor
QC Q Q1 2= 
MVR16 9 7=  =
Q. 15 A three phase starconnected load is drawing power at a voltage of 0.9 pu and 0.8power factor lagging. The three phase base power and base current are 100 MVAand 437.38 A respectively. The linetoline load voltage in kV is _____.
Sol. 15 Correct answer is 118.80 .
p.u. valueBase value in same unitsAcual value in same units
=
Base voltage MVABase CurrentBase
=
Base voltage. A437 38
100 106#= . kV228 63=
0.9 p.u.. kV
Actual value228 63
=
Vactual . . kV228 63 0 9#= . kV205 77=
As load is star connected
VL L . kVV
3 3205 77acutal= =
VL L . kV118 80=
Q. 16 Shunt reactors are sometimes used in high voltage transmission systems to(A) limit the short circuit current through the line
(B) compensate for the series reactance of the line under heavily loadedcondition
(C) limit overvoltages at the load side under lightly loaded condition.
(D) compensate for the voltage drop in the line under heavily loaded condition.
Sol. 16 Correct option is (C).
Shunt reactors are used in high voltage transmission system to limit over voltagesat the load side under lightly loaded condition.

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Closed loop T.F.
T s^ h.s s0 4 44
2= + +
G s
G s
1 + ^^ hh .s s0 4 44
2= + +
G s
G s1 +
^^hh
.s s4
0 4 42=
+ +
G s
1 1+ ^ h .s s4
0 4 42=
+ +
G s
1^ h
.s s4
0 4 4 42= + + 
Open loop. T.F. G s^ h.s s 0 4
4=
+
^ hError constant, Kp limG ss 0= " ^ h
.lim
s s 0 44
s 03=
+ =
" ^ hSteady state error ess K1
1 0p
=+
=
Q. 18 The state transition matrix for the system
x
x
1
2
o
o> H x
x u
1
1
0
1
1
11
2= +> > >H H H is
(A)
e
e e
0t
t t> H (B)e
t e e
0t
t t2> H(C)
e
te e
0t
t t> H (D) e tee0t t
t> HSol. 18 Correct option is (C).
x
x
1
2
o
o> H x
x U
1
1
0
1
1
11
2= +> > >H H H
Y AX BU = +
State transmission matrix
tf
^ h sI AL 1 1=  
^ h SI A6 @ s s11 0 1=  > H
So, tf^ hs
s
s11 1
1
0
1L
12= 


^ h> H
0L
s
s s
1 11
11
11
2
=  
 ^ h> H
tf^ h ete e
0t
t t= > HQ. 19 The sawtooth voltage waveform shown in the figure is fed to a moving iron
voltmeter. Its reading would be close to __________

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Sol. 19 Correct answer is 57.74 .
From the graph, we write mathematical expression of voltage v t^ h v t^ h t
20 10100
3#
=  ,
So, v t^ h Voltt5 103#= ,A moving iron voltmeter reads rms value of voltage, so
Vrms T v t dt 1
T 2
0= ^ h#
t dt20 10
1 5 1033 2
0
20 10 3
# # #=
#


^ h# t dt
20 1025 10
3
62
0
20 10 3
#
#=#


#
t20 1025 10
336 3
0
20 10 3
#
#=#


: D
20 10
25 103
20 103
6 3 3
#
##
#= 
^ h
3
25 10 400 106 6# # #=
. A3
100 57 74= =
Q. 20 While measuring power of a threephase balanced load by the twowattmetermethod, the readings are 100 W and 250 W. The power factor of the load is______.
Sol. 20 Correct answer is 0.8029.
In two wattmeter method, power factor angle is given by
f tanW W
W W311 2
1 2=
+ ^ h= G
250 1003 250 100
=+
^ h
f .tan 0 74231=  ^ h f .36 586c=
Power factor cosf=
. .cos 36 586 0 8029c= =^ h laggingQ. 21 Which of the following is an invalid state in an 8421 Binary Coded Decimal
counter(A) 1 0 0 0 (B) 1 0 0 1

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equal to 9.
(A) 1000 8 9#=
(B) 1001 9 9#
=(C) 0011 3 9#=
(D) 1100 12 9>= Invalid
Q. 22 The transistor in the given circuit should always be in active region. Take. VV 0 2CE sat =^ h , . VV 0 7BE= . The maximum value of Rcin W which can be used,
is _____.
Sol. 22 Correct answer is 22.38 .
Given: V satCE h .0 2= VBE .0 7=
b 100=
Writing KVL around the input loop
.I R5 0 7B S  0= IB
. . .k
mA2
5 0 72 10
4 3 2 153#W
= 
= =
IC . .mA AI 100 2 15 0 215B #b= = =
Writing KVL around the output loop
I R V5 C C CE   0=
VCE . R5 0 215 C= 
For active region VCE V> satCE hor, VCE . V0 2>
So, . R5 0 215 C .0 2> . R0 215 C .5 0 2< 

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Sol. 23 Correct answer is 300.
The given circuit is a bridge full wave rectifier with zener regulator. The voltage
across the regulator is a full wave rectified output as shown below.
IS I IZ L= +
IZ I IS L= 
I maxZ I Imax minS L=  ...(i)
Because when ILwill be minimum IZwill be maximum.
We have
I maxS RV V
R20 5
min mins
o Z
s=
=

I minL 0= (for RL"3)
I maxZ / AVP 51 4 201ZZ= = =^ h
Substituting these values in eq (i), we get
201
R20 5
mins= 
R mins 15 20 300# W= =
Q. 24 A stepup chopper is used to feed a load at 400 V dc from a 250 V dc source.The inductor current is continuous. If the off time of the switch is s20 m , theswitching frequency of the chopper in kHz is _____.
Sol. 24 Correct answer is 31.2 .
As step up chopper

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In stepup chopper output voltage is given by
VoV
1s
a=
 where a is duty cycle
TTONa =
1 a VV
400250
o
s= =
1 a /8
5 3 8& a= =
So,T
TON 83
=
T
T TOFF 83
= (T T TON OFF = + )
T
T1 OFF 83
=
T
TOFF 85
= where, sT 20OFF m= and HzT
f1
=
So, f 20 10 6# # 
85
=
f .8 20 10
5 31 2 kHz6# #
= =
Q. 25 In a constant /V f control of induction motor, the ratio /V f is maintainedconstant from 0 to base frequency, where Vis the voltage applied to the motorat fundamental frequency f. Which of the following statements relating to lowfrequency operation of the motor is TRUE ?(A) At low frequency, the stator flux increases from its rated value.
(B) At low frequency, the stator flux decreases from its rated value.
(C) At low frequency, the motor saturates.
(D) At low frequency, the stator flux remains unchanged at its rated value.
Sol. 25 Correct option is (B).
fV =Constant
At low frequency the stator plan decrease from its rated value as at low frequency
will increase the flux density which results in saturation of stator are therefore
stator flux decreases.
Q. 26  Q. 55 Carry two marks each.
Q. 26 To evaluate the double integralx y
dx dy2
2/y
2
2 1
0
8 +^ h## , we make the

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(C) udu dv0
1
0
4c m## (D) udu dv0
2
0
4c m##Sol. 26 Correct option is (B).
x y
dx dy2
2
/y
y
2
21
0
8 + bf l p## ?=
x yu
22 
= &x x y
u2
=  =
dx du=
At x y
2= u
/y y2
2 20#=
=
at x y
21= + u
yy
2
22
11=
+ =
a kThus integral becomes udu dy
0
1
0
8; E## v
y2
=
dvdy2
= dy dv 2& =
at y 0= & v 0=
at y 8= & v 4=
udu dv20
1
0
4
=
; E##
Reduced form udu dv20
1
0
4
= ; E##Q. 27 Let Xbe a random variable with probability density function
f x^ h. ,
. ,
,
for
for
otherwise
x
x
0 2 1
0 1 1 4
0
 and Vv 2  and Vv
22

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A1 . AV10120 1 99C1= = =
Now, consider current source sini t t10 5=^ h . Short circuit voltage source andredrawing circuit in frequency domain with /secrad52w = .
IC2
j1
51
1 10 0c=+
(Using current division)
j
j1 5
5 10 0# c= +
VC2 I j51
C2 #=
VC2 j1 510 0c
=+
VC2 tan26
10 5=  
A2 . AV26
10 1 96C2= = =
Q. 31 The total power dissipated in the circuit, shown in the figure, is 1 kW.
The voltmeter, across the load, reads 200 V. The value of XLis _____.
Sol. 31 Correct answer is 17.34 .
Total power consumed, P KW W1 1000= =
We can see that power will be consumed by only 1 W resistor and load resistor
R . From the figure we can notice that current in 1 W resistor is 2 A and in load
resistorRis 10 A. So we can write total power equal to sum of power dissipated
in these two resistors.
P ( ) ( ) ( ) R2 1 102 2= +

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VL I R XL2 2= +
200 . X10 9 96 L2 2= +^ h
202
^ h . X9 96 L2 2
#= ^ h XL .400 99 20=  XL .17 34 W=
Q. 32 The magnitude of magnetic flux density B^ hat a point having normal distancedmeters from an infinitely extended wire carrying current of IA is d
I
20
p
m (in SIunits). An infinitely extended wire is laid along the xaxis and is carrying currentof 4 A in the ve+ xdirection. Another infinitely extended wire is laid along theyaxis and is carrying 2 A current in the ve+ ydirection. 0m is permeability offree space. Assume i, j, kto be unit vectors along x, yand zaxes respectively.
Assuming right handed coordinate system, magnetic field intensity, H atcoordinate , ,2 1 0^ hwill be(A) /weber mk2
3 2
p (B) /A mi3
4
p
(C) /A mk23p
(D) /A m0
Sol. 32 Correct option is (C).
Hby wire on xaxisd
Ik
2p=
k14
21
# p=
Hby wire on yaxisd
Ik
2p= ^ h
k22
21
# p= ^ h
Hon (2, 1, 0) k k24 21p p= 
3

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It has initial conditions ;X X0 1 0 01 2= =^ ^h h . The pole locations of the systemfor a 1= , are(A) j1 0! (B) j1 0!
(C) j1 0! + (D) j0 1!
Sol. 33 Correct option is (A).
Characteristic equation
zI A 0=
where Aa
a
a
a1
1=
+
> HSo, zI A
z z a
a
a
a0 0 1
1= 
+
> >H H z aa az a1 1=    +> HFor a 1= , zI A
z
z
1
2
0
1=

 > H
Characteristic equation
zI A 0=
z 1 2^ h 0= z j1 0!=
The roots of characteristic equation gives the system poles.
Q. 34 An input signal sinx t t2 5 100p= +
^ ^h his sampled with a sampling frequency of
Hz400 and applied to the system whose transfer function is represented by
X z
Y z
^^
hh
N zz1
11 N
1= 


c mwhere, Nrepresents the number of samples per cycle. The output y n^ hof thesystem under steady state is(A) 0 (B) 1
(C) 2 (D) 5
Sol. 34 Correct option is (C).
Input signal, x t
^ h sin t2 5 100p= +
^ hSample signal, x nTs^ h sin nT2 5 100 s#p= + ^ h Ts"Sampling period, T 400
1s =
x nTs^ h sin n2 5 100 4001# #p= + b l x nTs^ h sin n2 5 4p= + a kTransfer function
H zX z
Y z=^ ^
^h hh
N zz1
11 N
1= 
c m
N z1 nn
N
0
1
= =

/1

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So h n^ hotherwise
n81 0 7
0
# #
= * y n^ h x n h n = *^ ^h h x n^ h , , , , .......2 2 2
5 2 5 22
5 2= + + +^ ch m) 3By solving the convolution
y n^ h x k h n k n
N
0
1
= =
 ^ ^h h/The integration sum of sine term will be zero and for constant term
y n^ h 2 81 2n 0
7
#= ==
/Q. 35 A 10 kHz evensymmetric square ware is passed through a bandpass filter with
centre frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is(A) a highly attenuated square wave at 10 kHz
(B) nearly zero
(C) a nearly perfect cosine wave at 30 kHz
(D) a nearly perfect sine wave at 30 kHz
Sol. 35 Correct option is (C).
Q. 36 A 250 V dc shunt machine has armature circuit resistance of .0 6 W and fieldcircuit resistance of 125 W. The machine is connected to 250 V supply mains. The
motor is operated as a generator and then as a motor separately. The line currentof the machine in both the cases is 50 A. The ratio of the speed as a generator tothe speed as a motor is _____.
Sol. 36 Correct answer is 1.27 .
When the machine is running as a generator
Field current Isn A120250 2= =
Load current IL A50=
Armature current Ia AI I 50 2 52L sn= + = + =
Induced EmfE
g . . V250 52 0 6 281 2= + =
^ hWhen the machine is running as a motorField current snIl A125
250 2= =
Load current LIl A50=
Armature current aIl L sn AI I 50 2 48=  =  =l l
Induced Emf Em . . V250 48 0 6 221 2=  =^ hFor DC machine, f is constant, so
EN
g
g EN
m
m=
Speed RatioNN
m
g .. .
EE
221 2281 2 1 27
m
g= = =

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If the rotor winding is open circuited and the system is made to run at rotationalspeed fr with the help of primemover in anticlockwise direction, then thefrequency of voltage across slip rings is f1 and frequency of voltage acrosscommutator brushes is f2. The values of f1and f2respectively are(A) f fr+ and f (B) f fr and f
(C) f fr and f fr+ (D) f fr+ and f fr
Sol. 37 Correct option is (A).
Given that Rotor winding is open circuit and rotor is rotating anticlockwise,
therefore
Slip sN
N Ns
s r= +
or f1 f f f fs r r= + = +
So frequency of voltage across slip rings is Hzf fr+^ h .At the same time, frequency of voltage across commutator brushes is same as
rotor frequency f2 f=
Q. 38 A 20pole alternators is having 180 identical stator slots with 6 conductors ineach slot. All the coils of a phase are in series. If the coils are connected to realizesinglephase winding, the generated voltage is V1. If the coils are reconnected torealize three phase star connected winding, the generated phase voltage is V2.Assuming full pitch, singlelayer winding, the ratio /V V1 2is
(A)3
1 (B)21
(C) 3 (D) 2
Sol. 38 Correct option is (D).
For single phase winding
no. of coils in the phase
n20 1
180 9#
= =
The slop span in electric degree
r. /no of slots per pole
180180 20
180 20c= = =
Distribution factor kd1
/
/
sin
sin
n r
nr
2
2=
^^
hh
/
.sin 9 20 2
0 6398#
= =^ h

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Now for three phase winding
Slot span r 20c=
No. of coils in a phase group
n20 3
180 3#
= =
Distribution factor kd2/
/
sin
sin
n r
nr
2
2= ^
^hh
/
/.
sin
sin
3 20 2
3 20 20 9597
#
#= =^
^hh
no. of turns, N2
. ..
no of slots no of conductorno of phase2
1# #=
180 26 31# #=
180=
The rms value of generator voltage
For single, phase, V1 . k fN4 44 d1 1f=
For three phase, V2 . k fN4 44 d2 2f=
VV
2
1 k Nk N
d
d
2 2
1 1
#
#=
..
0 9597 1800 6398 540
#
#=
2
Q. 39 For a single phase, two winding transformer, the supply frequency and voltage areboth increased by 10%. The percentage changes in the hysteresis loss and eddycurrent loss, respectively, are(A) 10 and 21 (B) 10 and 21
(C) 21 and 10 (D) 21 and 10
Sol. 39 Correct option is (A).
Hysteris loss in transformer
P
hk fB
maxn
n
=Where, the kn is the proportionality constant dependent on the characteristics
and volume of material, nis known as steinmetz exponents.
Bmax fV
\
Change in voltage and frequency is same, so Bmaxis constant.
So, Ph f\
Change in voltage and frequency is same, so Bmaxis constant.
So, Ph f\
Therefore percentage change in hysteris loss will be same as that of change in
frequency
PhT %f 10T= =

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Bmax=maximum flux density
f=frequency
ke=Proportionality constantAgain, Bmax f
V\ (Constant)
So, Pe f2
\
PP
e
e
2
1 f
f
2212
=
ff
2
1 . .f f
f0 1 1 1
1=
+ =
SoPP
e
e
2
1 .1 11 2
= b lor Pe2 . P1 1 e2 1= ^ h Pe2 . P1 21 e1=
Now, percentage change in eddy current loss
PeT .
PP P 100
11 21 1 100
e
e e
1
2 1# #=
=

%21=
Q. 40 A synchronous generator is connected to an infinite bus with excitation voltage. puE 1 3f= . The generator has a synchronous reactance of 1.1 pu and is delivering
real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0 pu.
Neglect stator resistance. The reactive power (Q) in pu supplied by the generatorto the bus under this condition is _____.
Sol. 40 Correct answer is 0.1092 .
As active power in synchronous generator is given by
P sinX
E V
s
fd=
d =Load angle
0.6.
. sin1 1
1 3 1# d= { . puE 1 3f= , puV 1= ,
. puX 1 1s = , . puP 0 6= }
sind.
. .1 3 1
0 6 1 1#
#=
sind .0 5077=
d .30 504c=
Reactive power is given by
Q cosX
E VXV
s
f
s
2
d= 
.
. ..
cos1 1
1 3 1 30 5041 112# c= ^ h
Q . . .1 1818 0 8616 0 9090#=  Q . pu0 1092=

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system in Hz in the steady state is ______.
Sol. 41 Correct answer is 50.
Since the generators are in parallel, they will operate at the same frequencyf
^ hat steady loss. Let load on generator G1is P1and on G2is P2. Then we can writefor generator G1
.P
f51 51
=droop constant
.P
f51 51
 1=
. f51 5  P1= ...(i)
Now in steady state total load is . MW2 5 , so load by second generator is . P2 5 1^ h. For G2we write
. Pf2 551 1 1=
f51  . P2 5 1=  ...(ii)
Adding eq (i) and (ii), we get
. f f51 5 51 +  .P P2 51 1= + 
. f102 5 2 .2 5=
f2 100=
f Hz50=
Q. 42 The horizontally placed conductors of a single phase line operating at 50 Hz
are having outside diameter of 1.6 cm, and the spacing between centers of theconductors is 6 m. The permittivity of free space is . /F m8 854 10 12#
 . Thecapacitance to ground per kilometer of each line is(A) . F4 2 10 9#
 (B) . F8 4 10 9# 
(C) . F4 2 10 12#  (D) . F8 4 10 12#

Sol. 42 Correct option is (B).
Cln
dd
2
1
2
0pe= b ld2 =distance b/w two sphere, d1 =radius of first sphere
.
.
ln0 8600
2 8 85 10 12# # #p=
b l C . /F m8 39 10 12#=

C . F8 39 10 1012 3# #=  
Ceq . F8 4 109
#. 
Q. 43 A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. Thepositive, negative, and the zero sequence reactances of the generator are 0.2 pu,0.2 pu and 0.05 pu, respectively, at the machine base quantities. If a bolted single
phase to ground fault occurs at the terminal of the unloaded generator, the faultcurrent in amperes immediately after the fault is_____.

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The magnitude of fault current
I I3f a1= . . pu3 2 22 6 67#= =
Base value of current
Ibase . A3 25 10100 10 2309 40
3
6
# #
#= =
So actual value of fault current
If . .6 67 2309 40#=
. A15403 698=
Q. 44 A system with the open loop transfer function
G s^ hs s s s
K2 2 22
=+ + +^ ^h h
is connected in a negative feedback configuration with a feedback gain of unity.
For the closed loop system to be marginally stable, the value of Kis ______
Sol. 44 Correct answer is 5.
G s^ hs s s s
K2 2 22
=+ + +^ ^h h
Closed loop T.F.G s
G s
1=
+ ^^
hh
s s s s K
s s s s K
12 2 2
2 2 2
2
2
=+
+ + +
+ + +
^ ^
^ ^
h h
h h
Closed loop T.F.s s s s K
K2 2 22
=+ + + +^ ^h h
Characteristics equation
s s s s K 2 2 22+ + + +^ ^h h 0= s s s s K 2 2 22 2+ + + +^ ^h h 0= s s s s K 4 4 44 3 2+ + + + 0=
Routh array
s4 1 6 k
s3
4 4 0s2 5 k 0
s1 k5
20 4 0
s0 k
For marginally stable
k20 4 0=
or k 5=
Q. 45 For the transfer function
G s^ h.s s s s
s
0 25 4 25
5 42= + + +
+^ h

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Sol. 45 Correct option is (A).
G s^ h.s s s
s
0 25 4 25
5 42= + + +
+
^ ^
^
h h
h
.
.
/
s s s
s0 25 25
5 4
10 25
1254
25
1 42
#
#=+ + +
+
a b^k l
h
.
.
/
s s s
s3 2
10 25
1254
25
1 42#=
+ + +
+
a b^k l
h
So constant gain term, K .3 2=
Highest corner frequency Hw 5=
Q. 46
The second order dynamic system
dtdX PX Qu = +
y RX=has the matrices P, Qand Ras follows :
P1
0
1
3=

> H Q 0
1= > H R 0 1= 6 @
The system has the following controllability and observability properties:(A) Controllable and observable
(B) Not controllable but observable
(C) Controllable but not observable(D) Not controllable and not observable
Sol. 46 Correct option is (C).
For controllability, we check controllability matrix
CM B AB= 6 @or, CM Q PQ= 6 @ PQ
1
0
1
3
0
1
1
3=

 =
> > >HH H
So, CM
0
1
1
3= > H det CM^ h 1 0!=So, the system is controllable.
For observalibility, we check observalibility matrix
OMC
CA= > H
or OMR
RP= > H
RP 0 1
1
0
1
3 0 3=

 = 8 > 8B H B0 1

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Q. 47 Suppose that resistors R1and R2are connected in parallel to give an equivalentresistor R. If resistors R1 and R2 have tolerance of 1% each, the equivalentresistor Rfor resistors R 3001 W= and R 2002 W= will have tolerance of
(A) 0.5% (B) 1%
(C) 1.2% (D) 2%
Sol. 47 Correct option is (B).
Equivalent resistance when connected in parallel is
Req R RR R1 2
1 2=+
Let R R1 2+ Rsum=
% Tolerance in Req % % %R R Rsum1 2T T T= + ^ ^ ^h h h ...(i) RsumT % %300 1 200 1! != +
^ ^h h 300 300 1001 200 200 1001! # ! #= +b bl l; ;E E 300 3 200 2! != +^ ^h h 500 5!=
%RsumT ^ h 5005 100#= b l %1=
Now, from eq (i)
%ReqT ^ h % % %1 1 1= +  %ReqT ^ h %1=
Q. 48 Two ammeters X and Yhave resistances of .1 2 W and .1 5 W respectively andthey give fullscale deflection with 150 mA and 250 mA respectively. The rangeshave been extended by connecting shunts so as to give full scale deflection with15 A. The ammeters along with shunts are connected in parallel and then placedin a circuit in which the total current flowing is 15 A. The current in amperesindicated in ammeter Xis_____.
Sol. 48 Correct answer is 10.16 .
For ammeter X, Let shunt resistor be RmX
I RmX mX I I RmX snX = ^ h . .0 15 1 2# . R15 0 15 snX= ^ h RsnX .
. .14 850 18 0 0121 W= =
For ammeter Y, Let shunt resistor be RMY

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I RMY MY I I RMY snY = ^ h . .0 25 1 5# . R15 0 25 snY= ^ h RsnY .
. .14 750 375 0 0254 W= =
Equivalent resistance of ammeter X
RX R RmX snX =
.  .1 2 0 0121=
.0 0121 W
Similarly RY .0 0254 W=
When connected in parallel, current in the circuit is A15 , so current through
ammeter X
IX R RR 15
X Y
Y#= +
(Using current division)
. .
.0 0121 0 0254
0 0254 15#= +
. A10 16=
Q. 49 An oscillator circuit using ideal opamp and diodes is shown in the figure.
The time duration for ve+ part of the cycle is t1T and for ve part is t2T . The
value of e RCt t1 2T T
will be ______.
Sol. 49 Correct answer is 1.2.
Q. 50 The SOP (sum of products) form of a Boolean function is , , , ,0 1 3 7 11S^ h, whereinputs are A, B, C, D(Ais MSB, and Dis LSB). The equivalent minimizedexpression of the function is
(A) B C A C A B C D + + + +^ ^ ^ ^h h h h(B) B C A C A C C D + + + +^ ^ ^ ^h h h h

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SOP Form, F , , , ,0 1 3 7 11S= ^ hPOS Form, F , , , , , , , , , ,2 4 5 6 8 9 10 12 13 14 15P= ^ hNow we represent the POS form on the kmap
Here we enter 0s in those cells correspond to the max terms present in the given
POS function.
So, Fmin B C A C A B C D = + + + +^ ^ ^ ^h h h hQ. 51 A JK flip flop can be implemented by T flipflops. Identify the correct
implementation.

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Sol. 51 Correct option is (B)
Here, Tflipflop is available and we want to convert it into JKflipflop. Hence
Jand Kare the external input and Tis the actual input to the existing flipflop.
So, we have to find expression of Tin terms of J, Kand Qn.
Step 1:Construct the presentstateandnextstate table of a JK flipflop as
shown in table below.
Step 2:We write the values of input Tthat is required to change the state of
the flipflop from Qnto Qn 1+ . The complete conversion table is as shown in table
below.
External Inputs Present State Next State Flipflop Input
J K Qn Qn 1+ T
0 0 0 0 0
0 0 1 1 0
0 1 0 0 0
0 1 1 0 1
1 0 0 1 1
1 0 1 1 0
1 1 0 1 1
1 1 1 0 1
Step 3:Draw Kmap for T
Writing expression of Tin POS form
T J Kn nq q= + +^ ^h hThus the logic diagram showing the conversion of Tflipflop to JKflipflop is
show below.
Q. 52 In an 8085 microprocessor, the following program is executed

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2005H RAR2006H DCR B2007H JNZ 2005
200AH HLTAt the end of program, register A contains(A) 60H (B) 30H
(C) 06H (D) 03H
Sol. 52 Correct option is (A).
XRA A"A H00! ExOR (operation with
accumulate)2001H MVI B, 04H B H04! (04H started in B)2003H MVI A, 03H A H03! (03H started in A)
2005H RAR rotate accumulater right withcarry2006H DCR B B H03! (B decremented by one)2007H JNZ 2005 Jump to 2005H till B not comes equal to zeroSo
B C A
03H 1 01H
02H 1 80H 01H 0 COH
00H 0 60H
Q. 53 A fully controlled converter bridge feeds a highly inductive load with ripple freeload current. The input supply (vs) to the bridge is a sinusoidal source. Triggeringangle of the bridge converter is 30ca = . The input power factor of the bridge is
Sol. 53 Correct answer is 0.779 .
For a fully controlled bridge converter input power factor is given by
Input P.F. cos2 2p
a=
cos2 2 30cp
=
2 223
#p=

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all throughout and the rms value of current through SCR, under this operatingcondition, are(A) 30cand 46 A (B) 30cand 23 A
(C) 45cand 23 A (D) 45cand 32 A
Sol. 54 Correct option is (C).
The maximum firing angle at which the voltage across device becomes zero is the
minimum firing angle to converter. It is given by
f tanR
L1 w=  b l f tan
52 50 16 10 451
3# # # c
p= 
b lVoltage across the device is zero is v vo i=
If input and output voltage are identical then VV 230rmso =
I rmso ZV
5 2 50 16 10
230rmso2 3 2
# # #p= =
+ ^ ^h h
5 2230
=
rms value of current through SCR
rmsIl AI
2 5 2 2230 23rmso
#= = =
Q. 55 The SCR in the circuit shown has a latching current of 40 mA. A gate pulse ofs50 m is applied to the SCR. The maximum value of Rin W to ensure successful
firing of the SCR is ______.
Sol. 55 Correct answer is 5882.35 .
Let us assume SCR is conducting mAI 40L =^ h i ti h RV e1s L
Rt=  8 B
. e0 2 1 200 10500 50 10
36
=  # # # 
8 B

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Rs .17 10
100 5882 353#
W= =
END OF THE QUESTION PAPER
**********

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ANSWER KEY
General Aptitude
1 2 3 4 5 6 7 8 9 10
(A) (B) (A) (B) (C) (B) (D) (D) (140) (A)
Electrical Engineering
1 2 3 4 5 6 7 8 9 10
(A) (0.130.15)
(C) (B) (C) (35) (D) (C) (B) (B)
11 12 13 14 15 16 17 18 19 20
(A) (545555)
(C) (6.977.03)
(117120)
(C) (0) (C) (5659) (0.780.82)
21 22 23 24 25 26 27 28 29 30
(D) (2223) (299301)
(31.031.5)
(B) (B) (0.350.45)
(B) (B) (A)
31 32 33 34 35 36 37 38 39 40
(17.317.4)
(C) (A) (C) (C) (1.221.32)
(A) (D) (A) (0.100.12)
41 42 43 44 45 46 47 48 49 50
(49.950.1)
(B) (1530015500)
(5) (A) (C) (B) (9.910.3)
(1.21.3) (A)
51 52 53 54 55
(B) (A) (0.740.82)
(C) (60556065)