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2/13

Classroom Study Course (CSC) Postal Study Course (PSC) Interview Guidance Course (IGC)

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1.1.1.1.1. (d)

2.2.2.2.2. (a)

3.3.3.3.3. (a)

4.4.4.4.4. (a)

5.5.5.5.5. (d)

6.6.6.6.6. (b)

7.7.7.7.7. (b)

8.8.8.8.8. (a)

9.9.9.9.9. (b)

10.10.10.10.10. (c)

11.11.11.11.11. (d)

12.12.12.12.12. (d)

13.13.13.13.13. (a)

27.27.27.27.27. (b)

28.28.28.28.28. (b)

29.29.29.29.29. (c)

30.30.30.30.30. (c)

31.31.31.31.31. (a)

32.32.32.32.32. (a)

33.33.33.33.33. (d)

34.34.34.34.34. (c)

35.35.35.35.35. (a)

36.36.36.36.36. (c)

37.37.37.37.37. (c)

38.38.38.38.38. (b)

39.39.39.39.39. (a)

40.40.40.40.40. (c)

41.41.41.41.41. (a)

42.42.42.42.42. (d)

43.43.43.43.43. (c)

44.44.44.44.44. (d)

45.45.45.45.45. (c)

46.46.46.46.46. (d)

47.47.47.47.47. (a)

48.48.48.48.48. (c)

49.49.49.49.49. (c)

50.50.50.50.50. (c)

51.51.51.51.51. (c)

52.52.52.52.52. (a)

53.53.53.53.53. (a)

54.54.54.54.54. (a)

55.55.55.55.55. (c)

56.56.56.56.56. (b)

57.57.57.57.57. (a)

58.58.58.58.58. (((((c)))))

59 .59.59.59.59. (c)

60.60.60.60.60. (b)

61.61.61.61.61. (b)

62.62.62.62.62. (d)

63.63.63.63.63. (a)

64.64.64.64.64. (c)

65.65.65.65.65. (c)

14.14.14.14.14. (a)

15.15.15.15.15. (d)

16.16.16.16.16. (a)

17.17.17.17.17. (b)

18.18.18.18.18. (d)

19.19.19.19.19. (a)

20.20.20.20.20. (d)

21.21.21.21.21. (a)

22.22.22.22.22. (d)

23.23.23.23.23. (c)

24.24.24.24.24. (c)

25.25.25.25.25. (b)

26.26.26.26.26. (a)

GATE-2012

(TEST SERIES-6)

EC: Electronics Engineering

(Full Syllabus Test - I)

SlSlSlSlSl..... NNNNNooooo .: 150112.: 150112.: 150112.: 150112.: 150112

A N S W E R SA N S W E R SA N S W E R SA N S W E R SA N S W E R S

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EXPLANATIONS

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1.1.1.1.1. (d )(d )(d )(d )(d )

If all the elements of a determinant located

on one side of principle diagonal are 0, then

= Product of the diagonal elements = 1

2.2.2.2.2. (a )(a )(a )(a )(a )

Since the eigen values of a real skew-

symmetric matrix are purily imaginary or

zero.Let = bi (where i2 = 1)

2

2

(1 )

(1 )

+

=2

2

(1 bi)

(1 bi)

+

=2

2

(1 b ) 2bi

(1 b ) 2bi

+

2

2

(1 )

(1 )

+

=2 2

2 2

(1 b) (2b)

(1 b) ( 2b)

+ +

= 1

3.3.3.3.3. (a )(a )(a )(a )(a )

f(x) = 3x cos x 1

f(x) = 3 + sin x

xn + 1

= xn

n

n

f(x )

f (x )

= n n n

n

x sinx cosx 1

3 sinx

+ ++

Let x0

= 0.6

First approximation

x1

= 0 0 0

0

x sinx cosx 1

3 sinx

+ ++

=0.6sin(0.6) cos(0.6) 1

3 sin(0.6)

+ ++

x1 = 0.6664Second approximation :

x2

=

0.6664sin(0.6664)

cos(0.6664) 1

3 sin(0.6664)

++

+

= 0.6666

4.4.4.4.4. (a)(a )(a)(a )(a )

dB0

S

N

= 1.76 + 6.02 n

= 1.76 + 6.02 16 = 98.08 dB

5.5.5.5.5. (d)(d )(d)(d )(d )

T0

= 273 + 20 = 293 K

The effective noise temperature of the

receiverT

e= (F 1) T

0

= (1005 1) 293 633K

The overall effective noise temperature

TIN

= Ta

+ Te

= 1050 + 633 = 1683 K

Overall noise figure

F = IN

0

T1

T+

=

1683

1 293+ = 674F(dB) = 10 log 674 = 829 dB.

6.6.6.6.6. (b)(b )(b)(b )(b )

It is an FM signal, because the message

signal is integrated to produce a phase

variation of the carrier.

7.7.7.7.7. (b)(b )(b)(b )(b )

x0

(t) =jt jt1(e e )

2

= j sin t

8.8.8.8.8. (a)(a )(a)(a )(a )

x(t) = et u(t)

X(s) =1

s 1+

y(t) = e2t cos t u(t)

Y(s) =2

s 2

(s 2) 1

++ +


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ELECTRONICS ENGINEERING FULL SYLLABUS TEST - I [23]

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Therefore,

H(s) =Y(s)

X(s)=

2

(s 2)

(s 2) 11

(s 1)

++ +

+

H(s) =2

(s 1) (s 2)

(s 2) 1

+ ++ +

H(s) =2 2

s 2 11

(s 2) 1 (s 2) 1

+

+ + + +Taking inverse Laplace transform

h(t) = (t) (e2t cos t + e2t sin t) u(t)

9.9.9.9.9. (b )(b )(b )(b )(b )

= 120 = hfeV

A= 150 V

ICQ

= 0.25 mA

At room temperature

VT

= 0.0259 V

gm

=CQ

T

I

V=

30.25 10

0.0259

= 9.65

mA/V

10.10.10.10.10. (c )(c )(c )(c )(c )

x(t) = u(t)

X(s) =1

s

y(t) = (t) Y(s) = 1.

H(s) = Y(s) 1 s1X(s)

s

= =

11.11.11.11.11. (d )(d )(d )(d )(d )

O

R=

A(B C) AB AC

1 A(B C) 1 AB AC

+ +=

+ + + +

12.12.12.12.12. (d )(d )(d )(d )(d )

A zero exists at f = f1.

14.14.14.14.14. (a)(a )(a)(a )(a )Inside a perfect conductor electric field

intensity and charge density both are zero.

17.17.17.17.17. (b)(b )(b)(b )(b )

With IB

= 0, only emitter-to-collector leakage

current flows, and

ICEO

= ( + 1) ICBO= (100 + 1) (5 106)

= 505 106 A = 505 A.

18.18.18.18.18. (d)(d )(d)(d )(d )

p

Na

q pp = 10

16 1.6 1019 400

= 0.64(-cm)1

Therefore,

resistivity =1

0.64= 1.56 -cm.

19.19.19.19.19. (a)(a )(a)(a )(a )

h12

=

1

1

2 I 0

V

V =

When I1

= 0

V1

= 2I2

and

V2

= (1 + 2 + 2)I2

= 5I2

h12 =1 2

2 2

V 2I0.4

V 5I= =

21.21.21.21.21. (a)(a )(a)(a )(a )

Skin depth

= 1f

= 6.44 105 m.

Because = = 1, = 1.55 104 + j 1.55 104

= 2.20 10445(m1)

22.22.22.22.22. (d)(d )(d)(d )(d )

Number of modes supported by

Ms

=2 2V (24)

2882 2

= =


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23.23.23.23.23. (c )(c )(c )(c )(c )

x

1

1

Y

A

B

Y = A B+Y = 0 B B+ =

B = Y 1 Y =So Y = B Y Y= =So it remains at its initial state.

24.24.24.24.24. (c )(c )(c )(c )(c )

R =3

12 1 5

25 10

= 420 .

25.25.25.25.25. (b )(b )(b )(b )(b )

x[n]h[n] =

1 2 3 3 2

1 1 2 3 3 22 2 4 6 6 4

1 1 2 3 3 2

x[n] h[n] = { }1, 4, 8, 11, 11, 7, 2

26.26.26.26.26. (a )(a )(a )(a )(a )

f(x, y) = 2f(x,y) f(x, y) is homogeneous function of order2.

So by Eulers theorem,

2 2 22 2

2 2

f f fx 2xy y

x yx y

+ +

= 2 (2 1) f(x, y)= 2f(x, y)

27.27.27.27.27. (b )(b )(b )(b )(b )

Now

cos [(2p + 1) ( x)]= cos [(2p + 1) (2 + 1)x]

= cos (2p + 1)xalso, cos2( x)

= cos2x

f(x) =2cos xe cos2 (2p + 1)x

f( x)=2cos x 2e cos (2p 1)x +

= f(x)

Therefore,

I =2cos x 2

0

e cos (2p 1)x dx

+ = 0

28.28.28.28.28. (b)(b )(b)(b )(b )

Assuming Nyquist ideal bandwidth, the

maximum bit rate can be supported in a

transmission bandwidth of 3 2 kHz

(bandpass), isb

T bR

B 2 R2

= = is 32

kbps.

BPSK can give a maximum spectral

efficiency of 1 bps/Hz. BPSK can not support 48 kbps.QPSK is twice spectrally efficient with a

spectral efficiency of 2 bps/Hz.

It can support 2 32 = 64 kbps.8-PSK has a spectral efficiency of 3 bps/

Hz and hence can support 3 32 = 96kbps.

So the best choice would be to use QPSK

and to make the system more efficient

some good pulse shaping can be used.

29.29.29.29.29. (c )(c )(c )(c )(c )

H(X) =16

2i 1

1 1log

16 16= = 4b/element

r = 2 106 (32) = 64 106

elements/s

Hence average rate of information conveyed

R = rH(X)

R = 64 106 4= 256 106 b/s

= 256 Mb/s


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30.30.30.30.30. (c )(c )(c )(c )(c )

=8

8

3 103m

10

=

Rrad

= rad2

2P

I

Rrad

= 7902

dl

= 878 103

I =3

200 151A8 78 10

This extremely high current illustrates that

an antenna with a length much less than a

wavelength is not an efficient radiator.

31.31.31.31.31. (a )(a )(a )(a )(a )

For the given input x[n] = cos 2n

Response

y[n] = | X || H (ej) | cos(1n + + )

H (ej)| = 2 = j0 2

j0 2

0 1e

e 0 5

=0 1 11 3

0 098 j0 0196 0 5

+

= 025 141Therefore,

y[n] = 025 cos (02n + 10 + 141)

y[n] = 025 cos (02n + 241)32.32.32.32.32. (a )(a )(a )(a )(a )

(4 + j) X(j) = A(3 j )+

X(j) = A1

(4 j ) (3 j )

+ +

x(t) = A[e3t e4t] u(t)

Also 2x(t) dt

= 1,

2A

168= 1

A= 168

33.33.33.33.33. (d)(d )(d)(d )(d )

Deactivating all the energy sources.

10

10

10 10

20

20

ba

10

10

10

10

10

a

b

Converting into Y circuit.

3.33

3.33

10

10

3.33

a

bRTh

RTh = 10


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34.34.34.34.34. (c )(c )(c )(c )(c )

Sweep error =

s

v

cc

V

A

V=

101100

30 300=

35.35.35.35.35. (a )(a )(a )(a )(a )

BC BC BC BC

A

A 1

1

1 1

1 1

Y = BC AC AB+ +

B

C

A

C

A

B

Y

orB

C

A

C

A

B

Y

or

B

C

A

C

A

B

Y

So total 7-NAND gates required.

36.36.36.36.36. (c )(c )(c )(c )(c )Charging time constant

Tc

= (R1

+ R2

+ R3)C

Tc

= 52 103 0.01 106

Tc

= 52 105 sec

Discharging time constant varies.

It's maximum value = (50 + 1)C

Td

= 51 C

So Tmax

= 0.69 Tc

+ 0.69 Td

Tmax

= 0.69[Tc

+ Td]

= 0.69 108 [103] 103

fmin

=max

11.407 kHz

T=

Minimum value of discharging time constant

Td

= R2C

= 105 sec

Tmin

= 0.69[Tc

+ Td

]

= 0.69[53 103] 108

fmax

=5

min

1 1

T 0.69 53 10=

fmax

= 2.734 kHz

37.37.37.37.37. (c )(c )(c )(c )(c )

Y = ( )PQ R ST U +

Y = PQ R ST U+ + +

Y = ( )PQ R S T U+ + +

Y = PQ RS RT U+ + +

38.38.38.38.38. (b)(b )(b)(b )(b )

First of all, we will check whether Zener

diode is in breakdown region or not.

For this we will find VOC

across Zener diode

and equivalent circuit for this is

RL

1 k

+

6 V

2 k


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ion.VOC =

26 4 V

2 1 =

+ VOC < VZSo it is not in breakdown region so it is open

circuited.

Therefore

IL

=36 10 2 mA

1 2

=+

39.39.39.39.39. (a )(a )(a )(a )(a )

The characteristics equation of the system

can be approximated as

1 +2

K(1 s)

s(s 8s 9)

+ +

= 0

s3 + 8s2 + (9 K)s + K = 0The Rouths array is g iven below

3

2

1

0

s 1 (9 K)

s 8 K

72 9Ks 8

s K

For the system to be stable K > 0 and (72

9K) > 0

Therefore, range of gain K for the system

to be stable is 0 < K < 8.

40.40.40.40.40. (c )(c )(c )(c )(c )

Putting s = 1 + j 3

GH = Kj 3(1 j 3) (3 j 3)+ +

GH = 90 60 30= 180The angle criterion is satisfied, hence point

P1

= 1 + j 3 is located on root-locus.

41.41.41.41.41. (a )(a )(a )(a )(a )

The order of A is 2, which means the power

series of eAt contains only two terms.

Therefore (t) = eAt = 0(t)I + 1(t) A, where0(t) and 1(t) are scalar coeffic ients.The eigen values of the system can be

obtained from the characteristics equation

|I A| = 0i.e., ( + 2) = 0i.e., 1 = 0, 2 = 2Therefore,

e0t = 0(t) + 01(t)and e2t = 0(t) 21(t)

i.e. 0(t) = 1 and 1(t) =2t1 e

2

So the state transition matrix is given by

(t) = eAt = 0(t)I + 1(t) A

=2t1 0 0 11 e

20 1 0 2

+

i.e. (t) =

2t

2t

1 e1

2

0 e

42.42.42.42.42. (d)(d )(d)(d )(d )

Nothing is said about the permeabilities of

the two regions; however, since 1H

is

entirely tangential, a change in permeability

would have no effect. Since1 2n n

B 0,B 0= =

and therefore2n

H 0=

121 2 n(H H ) a

= K

2y y y x (10a H a ) a = z6.5a

2y z(10 H ) ( a ) = z6.5a

2yH = 16.5(A|m)

Thus,2H

= y16.5a A|m


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43.43.43.43.43. (c )(c )(c )(c )(c )Using JFET equation,

15 = 40

2GS

p

V1

V

Vp

= 5V

1 + GsV5

= 0.612

VGs

= 8.06 V, 1.94 Volts.

Since |VGs| should not be greater than |VP|and hence, V

Gs= 1.94 Volt.

44.44.44.44.44. (d )(d )(d )(d )(d )

q = q0(1 et/RC) = 5(1 et/8)

i = t / 8dq 5

edt 8

=

Power in 4 resistor

P = i2R = t / 425

4 e64

= t / 425

e16

= t / 4

0 0

25Pdt e dt

16

=

=t / 4

0

25e

16

=100

Joules16

45.45.45.45.45. (c )(c )(c )(c )(c )

The network reduction at first step is shown

below:

2F

4F

2F

3F

C

x

y

1F 1F

= 2F

| |

1F 3F

= 4F

||

a

b

m

Converting star into equivalent delta with

terminals a-b-m within the main circuit

3F

C

x

y

C1 C22F

2F

a

b

m

C3

Here

C1

=4 4

1.6F;

4 4 2

=

+ +C

2=

4 20.8F;

4 4 2

=

+ +

C3

=4 2

0.8F4 4 2

=

+ +Subsequent reductions are shown below:

3F

C

x

y

2F C

= 3.6 F

|| 1

2F C= 2.8 F

||3

C2b

x

y

0.8F

3.6 2.8

3.6 2.8

+

= 1.575F

3C

3 C+

As given equivalent capacitance across x-y

is to be 5F ; so

5 =3C

1.575 0.83 C

+ ++

3C

3 C+= 2.625

3C = 3 2.625 + 2.625 C

C =3 2.625

0.375

C 21F=


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47.47.47.47.47. (a )(a )(a )(a )(a )It is 1 byte instruction and having only 1

machine cycle as opcode fetch.

48.48.48.48.48. (c )(c )(c )(c )(c )

I

I

I

I

0

1

2

3

C

C

C

C

Y

BA

Let A = D

This circuit implements

Y = D B C

Y = ( )A B C A B C = Y = A B C

A B = AB AB A B+ =

49.49.49.49.49. (c )(c )(c )(c )(c )In case of (c)

Y = ( )A B C , take ( )C D=

Y = ( )A B C A B C = Y = A B C A B C =

50.50.50.50.50. (c )(c )(c )(c )(c )

For DC analysis, all the capacitors will be

open circuited and equivalent circuit will be

V = 10 VDD

8 k

100

250

50 k

VGS = 0 350 IDS = 350 IDS

IDS

=

2

GSDSS

P

VI 1

V

IDS

=( )

2

DS3 350 I2 10 12

500 I

DS= (1 175 I

DS)2

IDS = 0.0265 A, 1.23 mA

IDS

can't be greater than IDSS

so IDS

= 1.23 mA

Now, gm

= DSS DSp

2 I I 1.568 mA/VV

=

For small signal AC analysis we can assume

gm

to be constant and equal to 1.568 mA/V

AC equivalent circuit is,

All capacitors will be shorted and DC

sources will be replaced by their internal

resistances

Vi

iVG

RG

g Vm GS RD RL

io

V = VD o

100

S

Vo

= gm

VGS

(RD|| RL) ...(i)

Vi= V

G

VGS

= VG

VS

= VG

gm

VGS

100

VG

= (1 + 100 gm

)VGS

= (1 + 100 1.568 103)VGS

VG

= 1.1568 VGS

Vi= 1.1568 V

GS...(ii)

ov

i

VA

V= =

( )m D Lg R R1.1568

||

=3 31.568 10 2.67 10

1.1568

Av

= 3.62


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51.51.51.51.51. (c )(c )(c )(c )(c )

AI= o o G

3i

i V R

i V4 10=

AI=

o G

3i

V R

V 4 10

AI=

3

3

50 103.62

4 10

A

I= 45.25

52.52.52.52.52. (a )(a )(a )(a )(a )

Normal to x2 + y2 + z = 16 at ( 1, 1, 2) is :1 = (x

2 + y2 + z)

= 2xi 2yj k+ +

1 ( 1,1,2) = 2i 2j k+ +

Normal to x2 + 2y + z2 = 9 at ( 1, 1, 2) is :

2 = 2xi 2j 2zk+ +

2 ( 1,12) = 2i 2j 4k+ +So

1 + 2 = 2i 2j k 2i 2j 4k+ + + += 4i 4j 5k+ +

53.53.53.53.53. (a )(a )(a )(a )(a )

The required angle is the angle between

the normals to the surfaces at point (1, 1,

2).

Let be the angle.(1) (1) = |1| |2| cos

12 = 9 24 cos

cos =12 2

3 24 6= = 3526

54.54.54.54.54. (a )(a )(a )(a )(a )

The modulation index can be computed by

= max minmax min

A A

A A

+

=150 30 2

150 30 3

=

+

so AM(t) = Ac (1 + cos mt) cos ctmax [AM(t)] = Ac(1 + ) = 150

Ac =150

21

3+

= 90 (volt)

So

AM(t) = m c2

90 1 cos t cos t3

+

AM(t) = 90 cos ct +60 cos mt cos ct

55.55.55.55.55. (c )(c )(c )(c )(c )

AM(t) = 90 m2

1 cos t3

+ cos ct + A

cos ct AM(t) = 90 cos ct + 60 cos mt cosct + A cos ct

AM(t) = (90 + A) m60

1 cos t(90 A)

+ +

cos ct

Given = 08 = 6090 A+

A = 15 = 15 180.

56.56.56.56.56. (b)(b )(b)(b )(b )

(a2 + b2 + c2) p2 (ab + bc + cd) p + (b2 +

c2 + d 2) 0 (a2 p2 2abp + b2) + (b2 p2 2bcp + c2)+ (c2 p2 2cd + d2) 0 (ap b)2 + (bp c)2 + (cp d)2 0Now, sum of three perfect squares can

never be negative. Only possibility remains

their sum will be equal to zero, which will

be when each of them is independently

equal to zero.

i.e. (ap b)2 = (bp c)2 = (cp d)2 = 0

ap = b, bp = c and cp = d

b c dp

a b c= = =

hence a, b, c, d are in G.P.


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57.57.57.57.57. (a )(a )(a )(a )(a )

Parsimony means thrift or stinginess,

therefore generositygenerositygenerositygenerositygenerosity is the opposite.

58.58.58.58.58. (((((c)))))

Pensive means moodily or dreamily

thoughtful.thoughtful.thoughtful.thoughtful.thoughtful.

59.59.59.59.59. (c )(c )(c )(c )(c )

To gesticulate(v.) means to use gestures

or make motions; express through motion,

especially while speaking.

60.60.60.60.60. (b )(b )(b )(b )(b )

The correct answer choice is (b). Diaphanous

is the opposite of opaque. Distraught is the

opposite of calm.

61.61.61.61.61. (b )(b )(b )(b )(b )

3f(x + 2) +1

4f2

+x

= 4x ...(1)

put x = 2 in (1),

3 f(4) + 4f1

4

= 4 2 ...(2)

also let1

42

=+x

4x + 8 = 1

x = 7/4put x = 7/4 in (1)

13f

4

+ 4 f(4) = 4

7

4

...(3)

let f(4) = a and f(1/4) = b in equation (2)

and (3)

(2) becomes 3a + 4b = 8 ...(4) 3

(3) becomes 3b + 4a = 7 ... (5) 4

multiply (4) by 3 and (5) by 4 and subtract

9a 16a = 52, 7a = 52, a = 52/7

Hence f(4) =52

7

62.62.62.62.62. (d)(d )(d)(d )(d )2!2! 4!4! 6!6! 8!8! 10!10!

2!2! does not contain any five so it does not

give any zero.

4!4! will also not give any zero.

6! contains only one five. So 6!6! ends with

6! zeroes.

8! also contains single five. So 8!8! ends

with 8! zeroes.

10! contains double five so 10!10! ends with

2 10! zeroes.

As all numbers are multiplied so given

expression ends with

6! + 8! + 2 10! zeroes.

63.63.63.63.63. (a)(a )(a)(a )(a )

Number of ascending order arrangements

(a < b < c) will be equal to 6c3

because

every selection of three number from six

numbers will give one and only one

ascending order arrangement.

while total sample space will have 6 6 6

= 216 possibilities.

Required probability =favourable chances

total chances

=6

3

6 5 4c 6

216 216

=

=20 5

216 54=

64.64.64.64.64. (c )(c )(c )(c )(c )

Refer to the table given in question.

As we need to maximize number of people

more than thirty five years of age.

In salary range (0 - 5 Lakhs) 1 male and 4

female can be above 35 years of age. Only

one male having age 38 its mentioned in

this salary bracket. Also in (0 - 5 L) salary 1

female have to be of 34 years, as minimum

age mentioned is 34 and in order to

maximize number of people we have to

consider remaining 4 females out of five to


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be more than 35 years of age.Following table can be reconstructed to

maximize people above 35 yeas of age.

Annual salary Male Female Total

Less than 5 Lakhs 1 4 5

5 - 10 Lakhs 0 7 7

10 - 15 Lakhs 7 3 10

More than 15 Lakhs 0 1 1

Total 8 15 23

Maximum number of people above 35 years

of age = 23.

Required % =23

10030

= 76.66%

= 76.67%.

!!!!

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