nodia and company - wordpress.com · 2015-2.in general aptitude q. 1 - q. 5 carry one mark each. q....

No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, photocopying, or otherwise without the prior permission of the author.

GATE SOLVED PAPERElectronics & Communication Engineering2015-2

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 its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineering or other professional services.

NODIA AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039Ph : +91 - 141 - 2101150www.nodia.co.inemail : [email protected]

Buy Online all GATE Books: shop.nodia.co.in *Shipping Free* *Maximum Discount*


GATE SOLVED PAPER - EC2015-2

© ww

w.no

dia.co

.in

General Aptitude

Q. 1 - Q. 5 Carry one mark each.

Q. 1 What is the adverb for the given word below?Misogynous(A) Misogynousness (B) Misogynity

(C) Misogynously (D) Misogynous

Sol. 1 Correct option is (C).

Q. 2 Choose the appropriate word-phrase out of the four options given below, to complete the following sentenceDhoni, as well as the other team members of Indian team ______ present on the occasion(A) Were (B) Was

(C) Has (D) Have

Sol. 2 Correct option is (B).Here, the focus of sentence is on Dhoni, not team members. Because the team phrase can be omitted from the sentence, it will be was and not were.

Q. 3 Ram and Ramesh appeared in an interview for two vacancies in the same department. The probability of Ram’s selection is 6

1 and that of Ramesh is .81

What is the probability that only one of them will be selected?

(A) 4847 (B) 4

1

(C) 4813 (D) 48

35

Sol. 3 Correct option is (B).

PRam (Select) 61=

PRamesh (Select) 81=

PRam (not select) 65=

PRamesh (not select) 87=

So, P (only one) P= (Ramesh select # Ram not) P+ (Ram select # Ramesh not)

81

6561

87

# #= +

485

487

4812

41= + = =

GATE SOLVED PAPER - EC 2015-2

© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 4 Choose the word most similar in meaning to the given word:Awkward(A) Inept

(B) Graceful

(C) Suitable

(D) Dreadful

Sol. 4 Correct option is (A).

Q. 5 An electric bus has on board instruments that report the total electricity consumed since the start of the trip as well as the total distance covered. During a single day of operation, the bus travels on stretches M, N, O and P, in that order. The cumulative distances traveled and the corresponding electricity consumption are shown in the Table below

Stretch Cumulative distance (km) Electricity used (kWh)

M 20 12

N 45 25

O 75 45

P 100 57The stretch where the electricity consumption per km is minimum is(A) M

(B) N

(C) O

(D) P

Sol. 5 Correct option is (D).We obtain the electric consumption per km for different stretched as

M .2012 0 6= =

N .45 2025 12

2513 0 52= −

− = =

O .75 4545 25

3020 0 66= −

− = =

P .100 7557 45

2512 0 48= −

− = =

Hence, for stretch P the electricity consumption per km is minimum.

Q. 6 - Q. 10 Carry two marks each.

Q. 6 Given below are two statements followed by two conclusions. Assuming these statements to be true, decide which one logically follows.Statements:I. All film stars are playback singers.

II. All film directors are film stars

Conclusions:I. All film directors are playback singers.

II. Some film stars are film directors.


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

(A) Only conclusion I follows

(B) Only conclusion II follows

(C) Neither conclusion I nor II follows

(D) Both conclusions I and II follow

Sol. 6 Correct option is (D)

Hence, from Venn diagram above, both conclusions I and II follow.

Q. 7 Lamenting the gradual sidelining of the arts in school curricula, a group of prominent artists wrote to the Chief Minister last year, asking him to allocate more funds to support arts education in schools. However, no such increase has been announced in this year’s Budget. The artists expressed their deep anguish at their request not being approved, but many of them remain optimistic about funding in the futureWhich of the statement(s) below is/are logically valid and can be inferred from the above statements?(i) The artists expected funding for the arts to increase this year

(ii) The Chief Minister was receptive to the idea of increasing funding for the arts

(iii) The Chief Minister is a prominent artist

(iv) Schools are giving less importance to arts education nowadays

(A) (iii) and (iv) (B) (i) and (iv)

(C) (ii) (iii) and (iv) (D) (i) and (iii)


Q. 8 A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deer’s 4. If the tiger and the deer cover 8 metre and 5 metre per leap respectively. What distance in metres will the tiger have to run before it catches the deer?

Sol. 8 Correct answer is 800.

One tiger leap 8= mSo, Tiger Speed 5= leap/min 40= m/min One deer leap 5= mSo, Dear Speed 4= leap/min 20= m/minAfter time t tiger catches the deer. Equating the distances, we obtain

Initial gap 50= leap of time m m50 8 400#= =or m t400 20+ t40#=


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

t min200400 20= =

Hence, total distance 400 20 20#= + 800= m

Q. 9 If a b c 12 2 2+ + = , then ab bc ac+ + lies in the interval

(A) ,1 32: D (B) ,2

1 1-: D

(C) ,1 21-: D (D) [ , ]2 4-


We know a b c 2+ +^ h o b c ab bc ca22 2 2= + + + + +^ hGiven a b c2 2 2+ + 1=So, a b c 2+ +^ h ab bc ca1 2= + + +^ hSince, square is always positive quantity, so

ab bc ca1 2+ + +^ h 0$

ab bc ca+ + 21$-

Q. 10 In the following sentence certain parts are underlined and marked P, Q and R. One of the parts may contain certain error or may not be acceptable in standard written communication. Select the part containing an error. Choose D as your answer if there is no error.The student corrected all the errors that the instructor marked on the P Qanswer book. R

(A) P

(B) Q

(C) R

(D) No error

Sol. 10 Correct option is (B).In the part Q, the is not required.

END OF THE QUESTION PAPER


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Electronics & Communication Engineering

Q. 1 - Q. 25 Carry one mark each.

Q. 1 Let the signal ( )f t 0= outside the interval T1, T2, where T1 and T2 are finite. Furthermore, ( )f t


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Sol. 3 Correct answer is 7.For finding the modular of counter, if the input of NAND gate are 1,1 then the counter will be clear

Q. 4 By performing cascading and/or summing/differencing operations using transfer function blocks ( )G s1 and ( )G s2 , one CANNOT realize a transfer function of the form

(A) ( ) ( )G s G s1 2 (B) ( )( )

G sG s2

1

(C) ( )( )

( )G sG s

G s111

2+c m (D) ( ) ( ) ( )G s G s G s1

11

2-c m

Sol. 4 Correct option is (B)Division of two transfer function can’t be performed by performing cascading and/or summing/differencing operations.

Q. 5 The electric field of a uniform plane electromagnetic wave is

Ev ( ) [ ( . )]expa j a j t z4 2 10 0 2x y 7#π = + −v vThe polarization of the wave is(A) Right handed circular

(B) Right handed elliptical

(C) Left handed circular

(D) Left handed elliptical


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in


Ev ( ) [ ( . )]expa j a j t z4 2 10 0 2x y 7#π = + −v vFor finding polarization, put Z 0= so that polarization can be seen in my plane. Let

Ex cos tω = ω t2 107#π = Ey cos sin t4 42ω ω= + =−

π ` j

As y direction component is multiplied by j , so it is 2π shifted. Hence, it is left

hand elliptical polarization.

Q. 6 A piece of silicon is doped uniformly with phosphorous with a doping concentration of /cm1016 3. The expected value of mobility versus doping concentration for silicon assuming full dopant ionization is shown below. The charge of an electron is .1 6 10 19# - C. The conductivity (in S-cm 1- ) of the silicon sample at 300 K is _______.

Sol. 6 Correct answer is 1.92Semiconductor is doped with phosphorous, so it is N type semiconductor.Given, donor concentration /cm1016 3=At this donor concentration, mobility of electron, obtained from the graph, is

cm v s1200 2 1 1= − −

So, Conductivity N e N6 n Dµ = = .1 6 10 1200 1019 16# # #= −

. cm1 92 1= −

Q. 7 In the figure shown, the output Y is required to be Y AB CD= + . The gates G1 and G2 must be, respectively,


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

(A) NOR, OR (B) OR, NAND

(C) NAND, OR (D) AND, NAND

Sol. 7 Correct option is (A)

We need Y AB C D= +As Y is sum of two literals, so G2 should be OR gate.Again, to get AB literal G2 should be NOR-gate, i.e.

A B= + AB=

Q. 8 In the bistable circuit shown, the ideal opamp has saturation level of 5! V. The value of ( )inR k1 Ω that gives a hysteresis width of 500 mV is _______,

Sol. 8 Correct answer is 1.We have the graph for threshold values as

Again, from the op-amp circuit,

RV 0in

1

- RV0 out2

= −

or RVin1 R

Vout2

= −

or Vin RR Vout2

1= −

Given, hysteresis width 500= mVSince, width V VTH TL= −

or 500 mV kR

kR5 20 5 20

1 1=− − +^ b bh l l

or 500 mV kR21=

So, R1 500 10 2 103 3# # #= −

1000Ω = k1=


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 9 Two causal discrete-time signals [ ]x n and [ ]y n are related as [ ] [ ]y n x mm

n

0=

=/

If the z -transform of [ ]y n is ( )z z 12

2- , the value of [ ]x 2 is _______.


Given y n6 @ X mn

n

0

==6 @/

This is accumulation property given as

y n6 @ u n x n= 6 6@ @By z -transformation,

Y z^ h z

Y z1 1

=− −^^

h

h

Since, Y z^ h z z 12

2= −^ h

So, z z 12

2-^ h zX z z

1=

−^^

h

h

or X z^ h z

z1

2 2=−

−

^ h

z

z12

1

3

=− −

−

^ h

Again, taking inverse z -transform

x n6 @ u n2 3= −6 @Hence, x 26 @ 0=

Q. 10 The bilateral Laplace transform of a function if

otherwisef t

a t b1

0

# #=^ h )

(A) sa b- (B) ( )s

e a b2 -

(C) se eas bs-- - (D) s

e ( )s a b-

Sol. 10 Correct option is (C)

Given f t^ h ,, otherwise

t b100# #

= *

So, F s^ h e dt1 st0

=3 −#

se st

a

b

= −−; E

se eas bs= +

−− −

Q. 11 The 2-port admittance matrix of the circuit shown is given by

(A) ....

0 30 20 20 3> H (B) 155 515> H

(C) .

.3 335

53 33> H (D) .. ..0 30 4 0 40 3> H


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Sol. 11 Correct option is (*).The admittance matrix is defined as

II1

2> H yy yy VV l1121 1222 12= > >H H

I1 y V y V11 1 12 2= + I2 y V y V21 2 22 2= +Now, from the given circuit we have

I1 V V V10 51 1 2= + −

I1 . .V V0 3 0 21 2= −

Also, I2 V V V10 52 2 1= + −

I2 V V V10 5 52 2 1= + −

I2 . .V V0 2 0 31 2=− +Hence, the admittance matrix is

Y ..

..

0 30 2

0 20 3= −

−> HThus, we can observe that any of the options is not correct.

Q. 12 The value of x for which all the eigen-values of the matrix given below are real is

xj10

4

5202

4210

+

−

R

T

SSSS

V

X

WWWW

(A) j5 + (B) j5 -(C) j1 5- (D) j1 5+

Sol. 12 Correct option is (B)

Given A xj10

4

5202

4210

=+

−

R

T

SSSS

V

X

WWWW

and eigen values are real.For real eigen value, A is Hermitian matrix, i.e.

A A T= ^ h (1)

Here, A xj10

4

5202

4210

=−

−

R

T

SSSS

V

X

WWWW

and A T] g jx10

54202

4210

= −−

R

T

SSSS

V

X

WWWW

Using equation (1), we get x j5= −


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 13 The signal cos t10 4π +π

^ h is ideally sampled at a sampling frequency of 15 Hz.

The sampled signal is passed through a filter with impulse response ( )sint

tpp

b l cos

t40 2pp-a k. The filter output is

(A) cos t215 40 4p

p-a k (B) ( )sin

costt

t215 10 4p

p p p+c dm n

(C) cos t215 10 4p

p-a k (D) ( )sin

costt

t215 40 2p

p p p-b al k

Sol. 13 Correct option is (A).Given continuous time signal,

x t^ h cos t10 4pp= +d n

Here, we neglect the phase-shift /4π , as it can be inserted in the final result. So, we have the Fourier transform pair,

x t1^ h cos t X f f f10 21 5 5L 1π δ δ= = − + +^ ^ ^h h h7 A

Given filter impulse response,

h t^ h sin costt t40 2p

p p p= −b bl l

sin sinct t40π = ^ ^h hTaking its Fourier transform,

H f^ h rect f j f f21 20 20d d= − − +* ^ ^h h7 A

rect rectj f f21 20 20= − − +^ ^h h7 A

Now, X f1^ h repeats with a value f 150 = Hz and each impulse value is /15 2. Thus, the sampled signal spectrum and the spectrum of the filter is

Hence, we get

X f H fS ^ ^h h j f f415 20 20d d= − − +^ ^h h7 A

x tr ^ h sin t215 40π = ^ h

cos t215 40 2p

p= −c m

This is recovered signal. Hence, applying phase shift /4π , we get

x tr ^ h cos cost t215 40 2 4 2

15 40 4pp p p p= − + = −d dn n


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 14 A sinusoidal signal of amplitude A is quantized by a uniform quantizer Assume that the signal utilizes all the representation levels of the quantizer. If the signal to quantization noise ratio is 31.8 dB, the number of levels in the quantizer is _______


SNRq . dBn1 76 6= +^ hwhere n = number of bitsGiven SNR .31 8= dBSo, . n1 76 6+ .31 8= dBor n6 30=or n 5=Hence, Levels 2 2 32n 5= = = 32. levels

Q. 15 The magnitude and phase of the complex Fourier series coefficients ak of a periodic signal ( )x t are shown in the figure. Choose the correct statement from the four choices given.Notation: C is the set of complex numbers, R is the set of purely real numbers, and P is the set purely imaginary numbers.

(A) ( )x t Rd

(B) ( )x t Pd

(C) ( ) ( )x t C Rd -(D) The information given is not sufficient to draw any conclusion about ( )x t

Sol. 15 Correct option is (A)Since the magnitude spectrum is even then the corresponding time-domain signal is real signal.

Q. 16 The general solution of the differential equation coscos

dxdy

xy

1 21 2= −

+ is

(A) tan coty x c− = (c is a constant)(B) tan cotx y c− = (c is a constant)(C) tan coty x c+ = (c is a constant)(D) tan cotx y c+ = (c is a constant)


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Sol. 16 Correct option is (C)

Given, dxdy cos

cosxy

1 21 2= −

+

cos ydy

1 2+ cos xdx

1 2= −

cos ydy2 2

sin xdx2 2

=

sec ydy2# cosec ydy2= # tany k+ cotx=− tan coty x+ c=

Q. 17 An n-type silicon sample is uniformly illuminated with light which generates 1020 electron hole pairs per cm3 per second. The minority carrier lifetime in the sample is 1 sµ . In the steady state, the hole concentration in the sample is approximately 10x , where x is an integer. The value of x is _______.

Sol. 17 Correct answer is 14.Rate of generation is

1020= electron hole pairs per cm3 per second.At steady state (at the end of lifetime) sect 1µ = , concentration of hole-electron pair in sec1 µ is 10 10 1020 6 14#= =−

So, x 14=

Q. 18 If the circuit shown has to function as a clamping circuit, which one of the following conditions should be satisfied for sinusoidal signal of period T ?

(A) RC T>


Time constant RCτ = =If RC T>> = period of sinusoidThen the capacitor will not play its role and clamping will take place.

Q. 19 In a source free region in vacuum, if the electrostatic potential ,x y cz2 2 2 2ϕ = + + the value of constant must be _______

Sol. 19 Correct answer is 3- .Given electrostatic potential

φ x y cz2 2 2 2= + +So, the electric field is obtained as

Ev dφ =−


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

xa ya Cza4 2 2x y z=− + +v v v^ h (1)In source free region,

E$d v 0=Substituting equation (1), we get

xa ya Cza4 2 2x y z$d - - -v v v^ h 0=or C4 2 2- - - 0=or C 3=−

Q. 20 In an 8085 microprocessor, which one of the following instructions changes the content of the accumulator?(A) MOV B, M

(B) PCHL

(C) RNZ

(D) SBI BEH

Sol. 20 Correct option is (D)SBI BE (H)This instruction subtract the immediate data from accumulator and store the result in accumulator. So, accumulator is affected.

Q. 21 The voltage (Vc) across the capacitor (in Volts) in the network shown is______.

Sol. 21 Correct answer is 100.For the given circuit, we have

So, V V V VR C L2 2= + −^ h 100 2^ h V80 40C

2 2= + −^ ^h h 100 802 2-^ ^h h V 40C 2= −^ h 180 20^ ^h h V 40C

2= −^ h V 40C - 180 20! #= ^ h 10 136!= V 40C - 60!= VC 40 60!= VC 100= V


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 22 Let ( )f z cz daz b= +

+ . If ( ) ( )f z f z1 2= for all z z1 2! , a 2= , b 4= and c 5= , then d should be equal to _______.


Given f z1^ h f z2= ^ h

f z^ h cz daz b= +

+

z dz52 41

1

++ z d

z52 42

2= ++

z z d10 41 2+ z z d z d z z d10 4 20 22 20 21 2 1 2 2 1= + + + + + z z20 2 1-^ h d z z2 2 1= −^ h d 10=

Q. 23 In the circuit shown the average value of the voltage Vab (in Volts) in steady state condition is _______.

Sol. 23 Correct answer is 5.We have the circuit

Now, we have to determine the average value of voltage Vab . Here, the circut consists two voltage sources (one A.C and other D.C.). So, we may use superposition theorem to obtain the desired average value of the voltage.Firstly we consider ac voltage ( )sin t5 500π , the circuit becomes

In steady state, the voltage across capacitor is given by

Vab1 ( ) ( )k j k j1 10 25 0

101°

( )j 101 3 6

6

#+

ωπ

ω=

+ + +ω − −

−

Here, we have ω = 5000 rad/sec. Solving the above equation, we will get the voltage across capacitor in the form

Vab1 V0+φ =or ( )V tab ( )sinV t0 ω φ= +


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Since, the time average of sinusoidal signal is zero. So, we get the average value of the voltage across capacitor (due to ac voltage only) as

Vav1 ( )T V t1

abbT

= # ( )sinT V t1 T 0 ω φ= +# 0=Again, we consider the dc voltage 5V. In this case, circuit becomes

In steady state, capacitor will be fully charged and behave as open circuit. Also, the inductor will be short circuited in steady state. Hence, the dc voltage across capacitor in steady state is

Vab2 5= VSo, the average value of the voltage across capacitor (due to dc voltage only) is

Vav2 Vab2= 5= VHence, applying the super position theorem, we get the net average value of voltage Vab as

Vav V Vav av1 2= + 5 0= + 5= VNote: Average value of dc voltage is same as its instantaneous value.

Q. 24 For the signal flow graph shown in the figure, the value of ( )( )R sC s

is

(A) G G H G G H G G H G G G G H HG G G G

1 1 2 1 3 4 2 2 3 3 1 2 3 4 1 21 2 3 4

− − − +

(B) G G H G G H G G H G G G G H HG G G G

1 1 2 1 3 4 2 2 3 3 1 2 3 4 1 21 2 3 4

+ + + +

(C) G G H G G H G G H G G G G H H11

1 2 1 3 4 2 2 3 3 1 2 3 4 1 2+ + + +

(D) G G H G G H G G H G G G G H H11

1 2 1 3 4 2 2 3 3 1 2 3 4 1 2− − − +Sol. 24 Correct option is (B)

We have the signal flow graph as


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

By Mason’s gain formula,

T s^ h px xTTΣ =

Single loops,

Q1 G G H1 2 1=− Q2 G G H3 4 2=− Q3 G G H2 3 3=−Non touchy loops,

P11 G G G G H H1 2 3 4 1 2= T G G H G G H G G H G G G G H H1 1 2 1 3 4 2 2 3 3 1 2 3 4 1 2= + + + + Forward path P G G G G11 1 2 3 4= = 1T 1=

Hence, T s^ h G G H G G H G G H G G G G H HG G G G

1 1 2 1 3 4 2 2 3 3 1 2 3 4 1 21 2 3 4= + + + +

Q. 25 In the circuit shown, V Vo oA= for switch SW in position A and V Vo oB= for SW

in position B . Assume that the opamp is ideal. The value of VVoA

oB is_______.

Sol. 25 Correct answer is 1.5When SW is at position A

V+ k kk V1 11 1= +c m .0 5= V


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

V+ .V 0 5= =− V

.k15 0 5- . k

V1

0 5 A0= −

.4 5 . V0 5 A0= − V A0 V4=−When SW is at position B

VoB kk

kk5 1

1 1 11=− −b bl l

VoB V6=−

Hence, VVoA

oB 46= −

− .1 5=

Q. 26 - Q. 55 Carry two marks each.

Q. 26 Let { , }X 0 1d and { , }Y 0 1d be two independent binary random variables. If ( )P X p0= = and ( )P Y q0= = , then ( )P X Y 1$+ is equal to

(A) ( )( )pq p q1 1+ − −(B) pq

(C) ( )p q1-(D) pq1 -


Given P x 0=^ h p=So, P x 1=^ h p1= − P y 0=^ h q=and P y 1=^ h q1= −Let Z X Y= +

X Y Z

0 0 0

0 1 1

1 0 1

1 1 2

So, P z 1$^ h and andP x y P x y1 0 1 1 0= − = = + = =" ", , andP x y1 1+ = =" , andP x y1 0 0= − = =" , pq1= −


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 27 An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R . The resonant frequency of the tank circuit is

(A) LC21

π (B)

LCR LC

21 1 2

π -

(C) LC R C

L21 1 2π

- (D) LC

R LC

21 1 2

π -

Sol. 27 Correct option is (B)

Total admittance, Y Y Yc LR= +

Y j Cj L R

1ww

= ++^ h

Y j CR j L R j L

R j L1w

w ww

= ++ −

−^ ^

^

h h

h

Y j CR LR j L

2 2 2w ww= +

+−

Y R LR j C

R LL

2 2 2 2 2 2ww

ww=

++ −

+c m

For Resonance, Im Ŷ h 0=

Cω R L

L2 2 2ww=+

R C L C2 2 2ω + L= L C2 2ω L R C2= −

2ω L CL R

2

2

= −

2ω LL C1 L

R C

2

2

=−_ i

ω LC L

R C1 1 2= −

Hence, f LC

R LC

21 1 2

π = −

Q. 28 Xn nn

33

=−=

" , is an independent and identically distributed (i, i, d) random process with Xn equally likely to be 1+ or 1- . Yn n

n33

=−=

" , is another random process obtained as .Y X X0 5n n n 1= + − . The autocorrelation function of Yn n

n33

=−=

" , denoted by [ ]R kY is


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Sol. 28 Correct option is (B).The autocorrelation function is defined as

R kY ^ h ,R n n ky= +^ h E Y n Y n k$= +^ ^h h7 ANow, we have

Y n^ h .x n x n0 5 1= + −^ ^h hSo, R ky ^ h . .E x n x n x n k x n k0 5 1 0 5 1= + − + + + −^ ^^ ^ ^_h hh h hi7 A .E x n x n k x n x n k0 5 1$= + + + −^ ^ ^ ^h h h h6 . .x n x n k x n x n k0 5 1 0 25 1 1$+ − + + − + −^ ^ ] ^h h g hA .E x n x n k E x n x n k0 5 1$= + + + −^ ^ ^ ^h h h h6 7@ A . .E x n x n k E x n x n k0 5 1 0 25 1 1+ − + + − + −^ ^_ ^ ^h hi h h7 8A B . . .R k R k R k R k0 5 1 0 5 1 0 25X x x x= + − + + +^ ^ ^ ^h h h h R ky ^ h . . .R k R k R k1 25 0 5 1 0 5 1x x x= + − + +^ ^ ^h h h R kx ^ h E x n x n k$= +^ ^h h7 AFor k 0= , we obtain

R 0x ^ h E x n2= ^ h8 B 1 2

1 1 212 2$ #= + −^ h

1=Again, for k 0! , we have

R kx ^ h E x n E x n k= +^ ^h h7 7A A 0= ,SinceE x n E x n k0 0= + =^ ^h h7 7A A# -Hence, we get

R 0y ^ h . . .R R R1 25 0 0 5 1 0 5 1x x x= + − +^ ^ ^h h h .1 25= R 1y ^ h . . .R R R1 25 1 0 5 0 0 5 2x x x= + +^ ^ ^h h h .0 5= R 1y -^ h . . .R R R1 25 1 0 5 2 0 5 0x x x= − + − +^ ^ ^h h h .0 5=R ky ^ h for k other than 0, 1 and 1 0− = . Thus, the autocorrelation function R ky ^ h is plotted as


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 29 In a MOS capacitor with an oxide layer thickness of 10 mm, the maximum depletion layer thickness is 100 mm. The permittivities of the semiconductor and the oxide layer are sε and oxε respectively. Assuming / 3s oxe e = , the ratio of the maximum capacitance to the minimum capacitance of this MOS capacitor is _______.

Sol. 29 Correct answer is 4.33The maximum capacitance per unit area is given by

Cmax toxoxε =

Cmin occurs at maximum value of xd (width). When both capacitance are in parallel. So, we have

Cmin

t x

t x

max

max

ox

ox

d

s

ox

ox

d

s

e e

e e=

+a a

a a

k k

k k

Hence, we obtain the ratio as

CCmin

max tt x

t x

max

max

ox

ox

ox

ox

d

s

ox

ox

d

s

ee e

e e=

+a

a a

a ak

k k

k kR

T

SSSS

V

X

WWWW

tx tmax

ox

ox

ox s

ox d s oxee e

e e= +

tx1 maxox

d

s

ox

ee= +< F

.1 10100

31 4 33#= + =; E

Q. 30 Let the random variable X represent the number of times a fair coin needs to be tossed till two consecutive heads appear for the first time. The expectation of X is_______.

Sol. 30 Correct answer is 1.5

Let X be random variable which denote number of tosses to get two heads.

P X 2=^ h HH 21

21

#= =

P X 3=^ h THH 21

21

21

# #= =

P X 4=^ h TTHH 21 4= = b l

So, E X^ h XP XΣ = ^ h

.....2 21 3 2

1 4 212 3 4= + +^ b b bh l l l

Again, let s .....2 21 3 2

1 4 212 3 4= + +^ b b bh l l l (1)

s2 .....2 21 3 2

1 4 213 4 5= + +^ b b bh l l l (2)

Subtracting equation (2) from (1),

s2 .....2 21

21 3 2 2

1 4 32 3 4

= + − + −b b b ^l l l h6 @ s2 .....2 2

121

21

212 3 4 5= + + +b b b bl l l l


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

s2 21- .....2

121

213 4 5= + +b b bl l l

s2 21-

1 21

21

21 3 2

=−

=_

bi

l

s2 2141= +

s 1 21= +

s 23=

Hence, E X^ h .23 1 5= =

Q. 31 In the circuit shown, the Norton equivalent resistance (in Ω ) across terminals a b- is_______.


To find Norton equivalent, an external source V0 is applied and current through it is I0. So, we have

Req IV

0

0=

I0 V V V V4 2 2 2 40 0 0 0= + + − b l

I0 V 2141

0= +b l

I0 V 43

0= b l

IV

0

0 .34 1 333= =

Req .R 1 333norton= = .

Q. 32 The figure shows a binary counter with synchronous clear input. With the decoding logic shown, the counter works as a


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

(A) mod-2 counter

(B) mod-4 counter

(C) mod-5 counter

(D) mod-6 counter

Sol. 32 Correct option is (C).From figure, it can be observed that once the Ex-NOR gate output is ‘0’ login counter will be reset to initial stage.

Hence, it is modulo 5 counter.

Q. 33 In the ac equivalent circuit shown, the two BJTs are biased in active region and have identical parameters with 1>>β . The open circuit small signal voltage gain is approximately________.

Sol. 33 Correct answer is 1- .


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

VBE .0 7= V Vin .V 0 7E=− = V Vin .0 7= V

So, VVin

out ..0 70 7 1= − =−

Q. 34 The state variable representation of a system is given as

xo x0011= −> H ; ( )x 0 10= > H

y [ ]x0 1=The response ( )y t is(A) ( )sin t

(B) e1 t-(C) ( )cos t1-(D) 0

Sol. 34 Correct option is (D).

xo x0011= −> H ;

x 0^ h 10= > H

y x0 1= 6 @Since, X s^ h s X 0φ = ^ ^h hwhere sφ ̂ h is state transition matrix given by sφ ̂ h sI A 1= − −^ hSo, X s^ h sI A 01#= − −^ ^h h

s

s01110

1

=−+

−> >H H

s ss

s11 1

01 10= +

+^ h

> >H H

s ss

s11 1

01 10= +

+^ h

> >H H

s ss

11 1

0= ++

^ h> H

X s^ h 0s1

= > H x t^ h

10= > H


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Hence, y t^ h 0 110 0= =6 >@ H

Q. 35 Consider the differential equation . x10 0 2dtdx = − with initial condition ( )x 0 1= .

The response ( )x t for t 0> is(A) e2 . t0 2- -

(B) e2 . t0 2-(C) e50 49 . t0 2- -

(D) e50 49 . t0 2-


dtdx . x10 0 2= −

.dtdx x0 2+ 10=

For the differential equation, we have

I.F. e . dt0 2= #

I.F. e . t0 2=

xe . t0 2 e dt C10 . t0 2= +# xe . t0 2 .

e C10 0 2. t0 2

= +; E xe . t0 2 e C2

10 10 . t0 2#= +

x t^ h Ce50 . t0 2= + −

At t 0= , x 0^ h Ce50 10= + = C50+ 1= C 49=−Hence, x t^ h e50 49 . t0 2= − −

Q. 36 For the voltage regulator circuit shown, the input voltage ( )Vin is %V20 20! and the regulated output voltage ( )Vout is 10V. Assume the opamp to be ideal. For a load RL drawing 200 mA, the maximum power dissipation in Q1 (in Watts) is _______.



© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Power dissipation in Q1 V I maxCE C#= ^ h R2 10= k

IE I RV

RVout

CL

A

2= = +

By virtual ground property,

VA 4= V

IE mA kI 200 104

C= = +

IC .mA mA200 0 4= + IC . mA200 4=Current through RL is

IL RV 200outL

= = mA

So, V maxCE^ h V Vin 0= − 24 10= − 14= VHence, Power .V mA14 200 4#= ^ ^h h .2 8056= watts

Q. 37 Input ( )x t and output ( )y t of an LTI system are related by the differential equation ( ) ( ) ( ) ( )y t y t y t x t6− − =ll l . If the system is neither causal nor stable, the impulse response ( )h t of the system is

(A) ( ) ( )e u t e u t51

51t t3 2− + + −− (B) ( ) ( )e u t e u t5

151t t3 2− − + −−

(C) ( ) ( )e u t e u t51

51t t3 2- - - (D) ( ) ( )e u t e u t5

151t t3 2- - - -


y t y t y t6- -m l^ ^ ^h h h x t= ^ hGiven system is neither causal nor stable. Taking the Laplace transform,

s y s sy s y s62 - -^ ^ ^h h h X s= ^ h (Given initial condition 0= ) Y s s s 62- -^ ^h h X s= ^ h

X sY s^

^

h

h H s

s s s s61

3 21

2= = − −=

− +^ ^ ^h

h h

s s5

131

21=

−−

+^ ^h h= G

Given that h t^ h is non causal so ROC should be left side of plane. Also, h t^ h is instable, so ROC should not contain jω axis


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Hence H s^ h s s5

131

21=

−−

+^ ^h h= G

and h t^ h e u t e u t51 t t3 2= − − + −− −^ ^h h8 B

e u t e u t51

51t t3 2= − − + −− −^ ^h h

Q. 38 The diode in the circuit given below has .V 0 7ON = V but is ideal otherwise. The current (in mA) in the 4kΩ resistor is ________.

Sol. 38 Correct answer is 0.6Given circuit is

Here, we have

kk42 k

k63=

So, bridge is balanced, and hence, no current will flow through diode. The equivalent circuit is shown below.


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Current through k4 Ω resistor is

k kk mA6 99 1= +c m

kk mA mA159 1 5

3#= =

.0 6= mA

Q. 39 A zero mean white Gaussian noise having power spectral density N20 is passed

through an LTI filter whose impulse response ( )h t is shown in the figure. The variance of the filtered noise at t 4= is

(A) A N23 2

0 (B) A N43 2

0

(C) A N2 0 (D) A N21 2

0

Sol. 39 Correct option is (A).Let N t^ h be the noise at the output of filter.

Variation of N t^ h ( )E N t E N t2 2= −^ h8 6B @$ .Since the input noise is zero mean.Output noise mean is also zero.

E N t^ h6 @ E W t h t dt$=3

3

−^^ ^bhh h l#

E W t^ h6 @ 0=W t^ h is white noise, so

var N t^^ hh E N t2= ^^ hh] R 0N= ^ hSince RN τ ^ h h h Rτ τ τ= − ω **^ ^ ^h h h

and RN τ ^ h N20 $ δ τ= ^ h

RN τ ^ h *h h N20$t t= −^ ^h h7 A

RN τ ^ h N h k h k dk20 $ τ = +

3

3

−^ ^h h#


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

R 0N ^ h N h k dk N A2 2 30 2 0 2= =

3

3

−^ ^h h#

A N23 2

0$ $=

Q. 40 Assuming that the op-amp in the circuit shown below is ideal, the output voltage V0 (in volts) is_______.

Sol. 40 Correct answer is 12.For the given op-amp,

V+ V> −So, Vout V 12saturation= = Volts

Q. 41 A 1-to-8 demultiplexer with data input Din, address inputs S0, S1, S2 (with S0 as the LSB) and Y0 to Y7 as the eight demultiplexed output, is to be designed using two 2-to-4 decoders (with enable input E and address input A0 and A1) as shown in the figure Din, S0, S1 and S2 are to be connected to P , Q , R , and S, but not necessarily in this order. The respective input connections to ,P Q , R and S terminals should be

(A) S2, Din, S0, S1(B) S1, Din, S0, S2(C) Din, S0, S1, S2(D) Din, S2, S0, S1

Sol. 41 Correct option is (D).We need to implement 1:8 DEMUX


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

As input to both the decoder should be same. So from figure only line P is acting same to both 2 4# decoder. Hence, P is mapped with Din . Again, we have

Here, we observe that S2 is ‘0’ in 4 cases then ‘1’ logic.From figure, it can be seen only line Q is connected to NOT gate to OR gate.So Q is mapped to S2 and remaining two line should be mapped in same order because select lines of 1:8 DEMUX should be mapped with address line of decoder. Hence, the mapping is

P Din"

R S0"

Q S2"

S S1"

Q. 42 The value of the integral ( )( )

cossint t

tdt12 2 4

4p pp

3

3

-# is_______.


We solve the given integral as

I cos sint tt dt12 2 4

4p pp=

3

3

−#

cos sintt t dt4

12 2 2 4p

p p=3

3

−#

sin sinttdt

ttdt3 6 2

00pp p= +

33; E##Since, sin cos sin sinA B A B A B− = + + −^ ^h h, so we can rewrite the integral as

I sin sine tt dt e t

t dt3 6 6 2t t00π

π π= +33 θ θ< F##

The integral can be considered as the Laplace transform with s 0= , i.e.

I sin sinL tt L t

t3 6 2p

p p= +; E& &0 0or I

sds

sds3

366

42

ss2 2 2 2p p

pp

p=+

++

33< F## with s 0=


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

tan tans s3 6 61

6 2 21

2s

1 1$ $p p p p p p p= +3

− −d cn m> H with s 0=

tan tan tan tans s3 6 21 1 1 13 3p p p= − + −

− − − −d ^ cn h m< F

tan tan3 2 0 2 01 1

pp p= − + −− −< F

3 2 0 2 03 3#p

p pp p= − + − = =< F

Q. 43 A function of Boolean variables X , Y and Z is expressed in terms of the minterms as

( , , )F X Y Z ( , , , , )1 2 5 6 7Σ =Which one of the product of sums given below is equal to the function ( , , )F X Y Z ?(A) ( ) ( ) ( )X Y Z X Y Z X Y Z$ $+ + + + + +(B) ( ) ( ) ( )X Y Z X Y Z X Y Z$ $+ + + + + +(C) ( ) ( ) ( ) ( ) ( )X Y Z X Y Z X Y Z X Y Z X Y Z$ $ $ $+ + + + + + + + + +(D) ( ) ( ) ( ) ( ) ( )X Y Z X Y Z X Y Z X Y Z X Y Z$ $ $ $+ + + + + + + + + +


Given minterm is , ,F X Y Z^ h , , , ,1 2 5 6 7Σ = ^ hMaxterm or POS is complement of SOP. So, in POS form, we obtain

, ,F X Y Z^ h , ,0 3 4π = ^ h X Y Z X Y Z X Y Z= + + + + + +^ ^ ^h h h

Q. 44 The transfer function of a mass-spring damper system is given by

( )G s Ms Bs K

12= + +

The frequency response data for the system are given in the following table.

ω in rad/s ( )G jω in dB ( ( ))arg G jω in deg

0.01 .18 5- .0 2-

0.1 .18 5- .1 3-

0.2 .18 4- .2 6-

1 16- .16 9-

2 .11 4- .89 4-

3 .21 5- 151-

5 .32 8- 167-

10 .45 3- .174 5-

The unit step response of the system approaches a steady state value of________.


G s^ h Ms Bs K

12= + +

Now, we have to obtain unit step response. So,


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

input s1=

Therefore, Y s^ h G s X s= ^ ^h h

Ms Bs K s

1 12= + +

d n

At steady state value,

Y ω ^ h lim limy t sY st s 0

= =" "3^ ^h h

limMs Bs K

1s 0 2

=+ +"

Y ω ^ h K1=

Now, from given table

At .0 01ω = G jdB

ω ^ h .18 5=−

So, log G j20 ω ^ h .18 5=−

log K201 .18 5=−

log K1 .20

18 5= −

K1 10

.2018 5

=−

Hence, y ω ^ h .K1 10 0 1188

.2018 5

= = =−

Q. 45 Two half-wave dipole antennas placed as shown in the figure are excited with sinusoidally varying currents of frequency 3 MHz and phase shift of 2

π between them (the element at the origin leads in phase). If the maximum radiated E -field at the point P in the x -y plane occurs at an azimuthal angle of 60° the distance d (in meters) between the antennas is _______.

Sol. 45 Correct answer is 50.For maximum electric field, we have

ψ cosd 0β θ α= + = (1)where

β ms

f23 102

8#λ

π π= =

3 10

2 3 108

6

#

# #π = ^ h 1002π =

θ = Azimuthal angle 60c=

α = Phase shift 2π =−

Substituting these values in equation (1), we get


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

cosd1002

60 2cp p+ −^ ^ ch

h m 0=

d 2100#p

p=

50= m

Q. 46 An air-filled rectangular waveguide of internal dimensions a cm # b cm ( )a b> has a cut-off frequency of 6 GHz for the dominant TE10 mode. For the same waveguide, if the cut-off frequency of the TM11 mode is 15 GHz, the cut-off frequency of the TE01 mode in GHz is _______.

Sol. 46 Correct answer is 13.74We have rectangular waveguide with cm cma b a b># ^ hFor TE10, f 6C = GHzFor TM11, fC 15= GHzSince, we have

fC C

am

bn

2

2 2

= +c dm n

So, 15 109# a b2

3 10 1 182 2

#= +

a b1 12 2+ 3 10

15 10 28

9

## #=

a b1 12 2+ 100

2= ^ h

and 6 GHz a23 10 18 2#= b l

6 109# a23 10 18# #=

100 a b1 12 2= +b bl l

a 2 6 103 10

9

8

# ##=

Therefore, a 401= m

Again, 100 2^ h b4012 2= +^ bh l

100 402 2+^ ^h h b1 2= b l

140 60^ ^h h b1 2= b l

b .91 65=

For TE01, fC b23 10 18 2#= b l

b23 10 18# #=

.23 10 91 65

8# #=

.13 7442= GHz


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 47 Consider two real sequences with time-origin marked by the bold value

[ ]x n1 { , , , }1 2 3 0= , [ ] { , , , }x n 1 3 2 12 =Let ( )X k1 and ( )X k2 be 4-point DFTs of [ ]x n1 and [ ]x n2 , respectivelyAnother sequence [ ]x n3 is derived by taking 4-point inverse DFT of ( ) ( ) ( ) .x k x k x k3 1 2=

The value of [ ]x 23 is_______.


Given x n16 @ , , ,1 2 3 0= " ,, , , ,x n 1 3 2 12 =6 @ " ,and X k3^ h X k X k1 2= ^ ^h hBy convolution circular property of DFT

Circular convolution in time domain = Multiplication in fourier domainSo x n x n1 276 6@ @ X k X k x n1 2 3= =^ ^h h 6 @

x 036 @ 2 6 1 9= + + =Now rotate for second term

So, x 136 @ 6=

Again, x 236 @ 2 6 3 11= + + =Q. 48 Let ( ) ( ) ( )x t s t s tα = + − with ( ) ( )s t e u tt4β = − , where ( )u t is unit step function.

If the bilateral Laplace transform of ( )x t is

( )X s s 16162= − { }Re s4 4<


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

s t^ h e u tt4β = − ^ hSo, x t^ h e u t e u tt t4 4αβ β= + −− +^ ^h hNow, we have ROC Re s4 4<


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

(A) Right Hand Circular

(B) Left Hand Elliptical

(C) Right Hand Elliptical

(D) Linear


,E z tv^ h cos cost z t za a5 2 10 3 2 10 2x y9 9

# #π β π β π= + + + −^ ch m

So, Ex cos z5 2 109#π β= +^ h

Ey cos z3 2 10 29

#π β π= + −c m

Since phase difference between Ex and Ey is 2π , and magnitudes are not equal. So,

this is elliptical polarization. Now, we have to determine direction.

At z 0= Ex cos t5 2 109#π = ^ h

Ey cos t3 2 10 29

#p p= −c m

For t 1= , t 2= ,...... we have the circulation as shown below.

Hence, the polarization is left hand elliptical.

Q. 51 The energy band diagram and electron density profile ( )n x in a semiconductor

are shown in the figure. Assume that ( )n x e10 kTq x

cm15 3=α −b l , with .0 1α = V/cm

and x expressed in cm. Given .0 026qkT = V, D 36n = cm s2 1- , and D qkT=µ . The electron current density (in A/cm2) at x 0= is

(A) .4 4 10 2#- -

(B) .2 2 10 2#- -

(C) 0

(D) .2 2 10 2# -



© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

Q. 52 A dc voltage of 10V is applied across an n -type silicon bar having a rectangular cross-section and a length of 1 cm as shown in figure. The donor doping concentration ND and the mobility of electrons nµ are 1016 cm 3- and 1000 cm2 V s1 1- - , respectively. The average time (in sµ ) taken by the electrons to move from one end of the bar to other end is _______.


Ev cmV

dV

110 10= = =

v V/cm

Given Vapplied 10= V d 1= cm (length)

Vd = drift velocity Eµ = v

cm V S1000 102 1 1#= − − V/cm

/cm s104=

time /

seclengthcm scm

V 101 10

d4

4= = = −

sec100 µ =

Q. 53 In the circuit shown, the initial voltages across the capacitors C1 and C2 are 1V and 3V, respectively. The switch is closed at time t 0= . The total energy dissipated (in Joules) in the resistor R until steady state is reached is_______.


The capacitor can be represented in Laplace domain as

where V0 is initial voltage. So, the given circuit result in


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

From the circuit,

I s^ h / // /

s ss s

10 1 3 13 1

=+ +

−^^

hh

//

ss

10 4 32=

+ ^ h

/

/s s

s30 4 32=+^ h

ss

s30 43 2= + b l

s s30 46

15 23= + = +

s15 23= +

s15

3 1152= +d n

s5

1 1152= +d n

Therefore, i t^ h e u t51 t152= − ^ h

Hence, energy dissipated t t Rdt20

=3−^ h#

e dt251 10/ t4 15

0=

3 − ^^ hh# e dt25

10 / t4 150

=3 −^ h#

.2510

415 1 5#= = Joules

Q. 54 The output of a standard second-order system for a unit step input is given as ( ) cosy t e t1 3t

32

6= − −π − _ i. The transfer function of the system is

(A) ( )( )s s2 3

2+ +

(B) s s2 11

2+ +

(C) s s2 3

32+ +

(D) s s2 4

42+ +

Sol. 54 Correct option is (D).

Given y t^ h cose t132 3 6

t π = − −− c m

In standard form, we define

y t^ h cosAe dt1 /t ω φ= − −τ − ^ hFor standard equation,

τ nξω=

dω 1 3n 2ω ξ= − =


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

or 1ξ 132ξ

=−

( nξω 1= , or nω /1 ξ = )

or 12ξ 132ξ

=−

or 1 2ξ - 3 2ξ =or 1 4 2ξ =

or 2ξ 41=

So, ξ .0 5=

Again, nω 1ξ = 121=

2=So, the characteristic equation is

s s2 n n2 2ξω ω= + +

s s2 21 2 42# # #= +

s s2 42= + +This denominator term is present only in option (D).

Q. 55 If C denotes the counter clockwise unit circle, the value of the contour integral

{ }Rej z dz21

Cπ # is_______.


Rej z dz21

Cπ ^ h#

Let z ej= θ

dz je dj θ = θ

Taking limit from 0 to 2π

Rej e je d21 j j

0

2

π θ=θ θ

π^ h#o

cos sin cosj j d21 2

0

2

π θ θ θ θ= +π ^ h#

cos sin cosd j d21 2

0

2

0

2

π θ θ θ θ= +ππ

< F## n2

1 0π = −6 @

.21 0 5= =

END OF THE QUESTION PAPER


© ww

w.no

dia.co

.in



© ww

w.no

dia.co

.in

ANSWER KEY

General Aptitude

1 2 3 4 5 6 7 8 9 10

(C) (B) (B) (A) (D) (D) (B) (800) (B) (B)

Electronics & Communication

1 2 3 4 5 6 7 8 9 10

(C) (400) (7) (B) (D) (1.92) (A) (1) (0) (C)

11 12 13 14 15 16 17 18 19 20

(*) (B) (A) (32) (A) (C) (14) (D) (–3) (D)

21 22 23 24 25 26 27 28 29 30

(100) (10) (5) (B) (1.5) (D) (B) (B) (4.33) (1.5)

31 32 33 34 35 36 37 38 39 40

(1.33) (C) (–1) (D) (C) (2.806) (B) (0.6) (A) (12)

41 42 43 44 45 46 47 48 49 50

(D) (3) (B) (0.12) (50) (13.74) (11) (–2) (A) (B)

51 52 53 54 55

(C) (100) (1.5) (D) (0.5)

nodia and company - wordpress.com · 2015-2.in general aptitude q. 1 - q. 5 carry one mark each. q....

Documents