GATE-2016-PAPER-01 General Aptitude

29 Pages·2016·1.08 MB·English

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General Aptitude
Q. No. 1 – 5 Carry One Mark Each

1. The man who is now Municipal Commissioner worked as _____________.
(A) the security guard at a university
(B) a security guard at the university
(C) a security guard at university
(D) the security guard at the university
Key: (B)

2. Nobody knows how the Indian cricket team is going to cope with the difficult and seamer-friendly
wickets in Australia.
Choose the option which is closest in meaning to the underlined phase in the above sentence.
(A) put up with (B) put in with (C) put down to (D) put up against
Key: (D)

3. Find the odd one in the following group of words.
Mock, deride, praise, jeer
(A) mock (B) deride (C) praise (D) jeer
Key: (C)

4. Pick the odd one from the following options.
(A) CADBE (B) JHKIL (C) XVYWZ (D) ONPMQ
Key: (D)
5. In a quadratic function, the value of the product of the roots , is 4. Find the value of
n n

n n
(A) n4 (B) 4n (C) 22n1 (D) 4n1
Key: (B)
Exp: Given 4
n n n n

n n 1 1

n n
n nnn

n n
()n 4n

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Q. No. 6 – 10 Carry Two Mark Each

6. Among 150 faculty members in an institute, 55 are connected with each other through Facebook
and 85 are connected through WhatsApp. 30 faculty members do not have Facebook or WhatsApp
accounts. The number of faculty members connected only through Facebook accounts is ______.
(A) 35 (B) 45 (C) 65 (D) 90
Key: (A)
Exp: F Facebook, WWhatsApp, ETotal faculties
given
E,n(E)150
 
n(E)150,n FW 30
nFWn(E)NFW15030 55 85
nFW120
nf wnfn(w)n(Fw F Fw W
120n(F)85 n(F)35 nFw n(w)60
20
n(F)1208535

55n(F)nFW
30
nFw55n(F)553520
 
FW
n(w)852065

7. Computers were invented for performing only high-end useful computations. However, it is no
understatement that they have taken over our world today. The internet, for example, is
ubiquitous. Many believe that the internet itself is an unintended consequence of the original
invention with the advent of mobile computing on our phones, a whole new dimension is now
enabled. One is left wondering if all these developments are good or more importantly, required.
Which of the statement(s) below is/are logically valid and can be inferred from the above paragraph?
(i) The author believes that computers are not good for us
(ii) Mobile computers and the internet are both intended inventions
(A) (i) (B) (ii) only
(C) both (i) and (ii) (D) neither (i) nor (ii)
Key: (D)

8. All hill-stations have a lake. Ooty has two lakes.
Which of the statement(s) below is/are logically valid and can be inferred from the above
sentences?
(i) Ooty is not a hill-station
(ii) No hill-station can have more than one lake.
(A) (i) Only (B) (ii) Only (C) both (i) and (ii) (D) neither (i) nor (ii)
Key: (D)
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9. In a 24 rectangle grid shown below, each cell is a rectangle. How many rectangles can be
observed in the grid?
(A) 21 (B) 27
(C) 30 (D) 36
Key: (C)
Exp: 1: (AEOK)
2: (AEJF), (FJOK)
A B C D E
4: (ABLK), (BCML), (CDNM), (DEON)
2: ACMK, ADNK 2:ECMD,EBLO2:ACHF,ADIF
F G H I J
2: ECHJ, EBGJ 2:FHMK,FINK2:JHMD,JGLO
1:BDNL 2:BDIG,GINL
8: ABGF, BCHJ, CDIH, EDI, FGLK, GHML, HINM K L M N O
Total = 1+2+4+2+2+2+2+2+2+1+2+8=30

10.
25
fx
2

1.5

1
0.5
0
4 3 2 1 0.50 1 2 3 X4

1

1.5
2
25

Chose the correct expression for f(x) given in the graph.
(A) fx1 x1 (B) fx1 x1
(C) fx 2  x1 (D) fx 2  x1
Key: (C)
Exp: Substituting the coordinates of the straight lines and checking all the four options given, we get
the correct option as C which is f(x)= 2 x1

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Electrical Engineering
Q. No. 1 –25 Carry One Mark Each

1. The maximum value attained by the function fx xx1 x2 in the interval 1, 2 is ____.
Key: 0
Exp: f(x) x(x1)(x2) f(x) x33x2 2x f1(x)03x2 6x20
1
 x1
3
1
But x1 only lies on the interval [1,2]
3

1  1 
At x1 ;f11(x)6x6 6 1 60
 
3  3

1
 x1 is a point of minimum
3

 f(x) = x(x-1)(x-2) = 0 at either ends x =1 & x =2

 Max value = 0

2. Consider a 33 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
Key: 3
1 1 1
 
Exp: Consider A  1 1 1
33  
1 1 1
 
It’s only non-zero Eigen value λ = 1order of the matrix = 133

3. The Laplace Transform of ft e2tsin5tut is
5 5 s2 5
(A) (B) (C) (D)
s24s29 s25 s24s29 s5
Key: (A)
Exp: Consider
 5
x(t) = sin5t u(t) X(S)
s2+25

By frequency shifting property, 1X(S-S )x(t)eS0t
0
thus at S 2
0
 5
F(s)
f(t) = e2tsin5tu(t) (s-2)2+25
5
F(s)
s2 4s29
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4. A function yt,such that y0 1 and y13e1, is a solution of the differential equation
d2y dy
 2  y 0. Then y2 is
dt2 dt
(A) 5e1 (B) 5e2 (C) 7e1 (D) 7e2
Key: (B)
Exp: The operator function of given D.E is (D2+2D+1)y = 0
A.E is D2+2D+1 = 0D = -1,-1
y = e-xC + C  (1)
1 2
Given y(0)=1 & y(1) =3e-1

From ; y(0) = 1 i.e , y = 1 at x = 0 1 = C
1
y(1) = 3e-1i.e, y = 3/e at x=1

3
e-11C (1)31C  C 2
e 2 2 2
y = e-x[1+2x] y(2) = e-2[1+4] = 5e-2

2z5
5. The value of the integral  dz over the contour z 1. taken in the anti-
C z1z24z5
 2
clockwise direction, would be
24i 48i 24 12
(A) (B) (C) (D)
13 13 13 13
Key: (B)
Exp: Singular points are obtained by
z1z2-4z+50 z =1&z =2i
 2 2
1
Out of these z = only lies inside C. z =1
2

By Cauchy’s integrated formula,

 1 
 2 5 
 2z+5 dz =  2z+5 dz = 2πi 2   48πi
C z -12z2-4z+50 C z2z- 4-z1+5 122 4125 13
2

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Ys s
6. The transfer function of a system is  , The steady state output x(t) is A cos2tfor
Rs s2
the input cos2t. The value of A and , respectively are
1 1
(A) , 45O (B) , 45O (C) 2, 45O (D) 2, 45O
2 2
Key: (B)
S
Exp: H(s) =
S+2
 
H() = 900 tan1 
2+4  2 

If input is COS 2t i.e.,  = 2
1
H() = 450
2
1
y(t) = cos(2t450)
2
1
by comparision A= & 45
0
2
100
7. The phase cross-over frequency of the transfer function G(s) = in rad/s is
s13
1
(A) 3 (B) (C) 3 (D) 3 3
3
Key: (A)
Exp: Phase Crossover frequency  : GH 1800
PC =
PC
GH 3tan-11800 3tan-1w W tan600  3
PC PC

8. Consider a continuous-time system with input x(t) and output y(t) given by
yt xtcost.This system is
(A) linear and time-invariant (B) Non-linear and time-invariant
(C) linear and time-varying (D) Non-linear and time-varying
Key: (C)
Exp: Linearity
y (t)= x (t)cost; y (t)= x (t)cost
1 1 2 2
y (t) + y (t) = x (t)cost + x (t)cost
1 2 1 2
y (t) + y (t) = [x (t) + x (t)]cost
1 2 1 2
If x (t) + x (t) = x(t) then y (t) + y (t) = y(t)
1 2 1 2
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Thus the system is linear
Time-invariance
considery (t) x (t) cost ; If x (t)x(t - τ)
1 1 1
y (t) x(t - τ)cost
1
Define y(t - τ)x(t - τ)cos(t - τ)
y (t) y(t - τ) system is time-varient
1

9. The value of et 2t2 dt. where t is the Dirac delta function, is

1 2 1 1
A B C D
2e e e2 2e2
Key: (A)

10. A temperature in the range of -40OC to 55OC is to be measured with a resolution of 0.1OC. The
minimum number of ADC bits required to get a matching dynamic range of the temperature
sensor is
(A) 8 (B) 10 (C) 12 (D) 14
Key: (B)
V -V
Exp: Usually when voltage information is given we use the formulaR = max min
2n
Here based on the given information we can think the systems as following block diagram

Temperature
nbit
Temp volttoage ADC .
analog
converter voltage .

.
 Here the V andV will be the equivalent of T andT re spectively
max min max min
So we can still use the above relation
T -T 5-(-40)
Resolution = max min 2n= 2n= 950
2n 0.1
n log (950)9.8910
2
So minimum requirement is 10 bits

11. Consider the following circuit which uses a 2-to-1 multiplexer
0
as shown in the figure below. The Boolean expression for Y F
output F in terms of A and B is S
1
(A) A B (B) A  B
A B
(C) A  B (D) A  B
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Key: (D)
Exp: We can redraw the max circuit as follows

A 0

F

A 1

B

So the Boolean expression of F(A, B)=BABA=AB=AB

12. A transistor circuit is given below. The Zener diode breakdown voltage is 53 V as shown The
base to emitter voltage droop to be 0.6V. The value of the current gain  is ________.
10V

4.7k 220

0.5mA

5.3V 470

Key: 19
Exp: 5.3 = 0.6 + 470 I
E
I 0.01A
E
10 - 5.3
I = 1mA
X 4.7103

I =10.50.5mA
B
I (1β)I
E B
0.01
0.01(1β)0.5103 (1β) 20 β19
0.5103

13. In cylindrical coordinate system, the potential produced by a uniform ring charge is given

fr,z, where f is a continuous function of r and z. Let E be the resulting electric field. Then

the magnitude of E
(A) increases with r (B) is 0
(C) is 3 (D) decreases with z
Key: (B)
 
Exp: Since the charge is not varying with time the field E is static So E0
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14. A soft-iron toroid is concentric with a long straight conductor carrying a direct current I. If the
relative permeability  of soft-iron is 100, the ratio of the magnetic flux densities at two adjacent
r
points located just inside and just outside the toroid, is ________.
Key: 100
Exp: The field inside and outside the toroid is due to long straight conductor only
Let the two points almost at same distance
μ μ I
The flux density inside toroid = 0 r
2πr
μ I
The flux density outside toroid = 0 T
2πr
μ μ I
0 r
2πr
Ratio = μ  100
μ I r
0
2πr

15. R and R are the input resistances of circuits as shown below. The circuits extend infinitely in
A B
the direction shown. Which one of the following statements is TRUE?

2 2 2

R
A 1 1 1

2 2 2

R
B 1 1 1 1

(A) R R (B) R R 0 (C) R R (D) R R 1R 
A B A B A B B A A
Key: (D)
Exp: By comparing 2 network on the input side
R
we can say that R =1//R R = A
B A B 1R
A

16. In a constant V/f induction motor drive, the slip at the maximum torque
(A) is directly proportional to the synchronous speed
(B) remains constant with respect to the synchronous speed
(C) has an inverse relation with the synchronous speed
(D) has no relation with the synchronous speed
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Key: (C)
r r r
Exp: SmT  2  2  2
x jL j2L .f
2 2 2
1
SmT 
f
120f
N 
S P
1
 SMT 
NS

17. In the portion of a circuit shown, if the heat generated in 5 resistance is 10 calories per second
then heat generated by the 4 resistance, the calories per second, is ______.
4 6

5
Key: 2
Exp: Here the power information regarding the resistor is given because
E Calorie
P =  =watt
t sec
P = 10
5
V
 5Ω = 10 V = 50
5 5Ω

P is asked
4Ω
V 2 1 4 2
P = 4Ω  50
4Ω 4 446 

1 16 calorie
  502
4 100 sec

18. In the given circuit, the current supplied by the battery, in ampere, is ________.

` l 1 1 l
1 2

1


1V l
2

Key: 0.5
Exp: If we write KCL at node × then
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