Question 1
In the following truth table, V = 1 if and only if the input is valid. gatecs20133 What function does the truth table represent?
A
Priority encoder
B
Decoder
C
Multiplexer
D
Demultiplexer
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Question 1 Explanation: 
Since there are more than one outputs and number of outputs is less than inputs, it is a Priority encoder V=1 when input is valid and for priority encoder it checks first high bit encountered. Except all are having at least one bit high and ‘x’ represents the “don’t care” as we have found a high bit already. So answer is (A).
Question 2
Which one of the following expressions does NOT represent exclusive NOR of x and y?
A
xy + x' y'
B
x ^ y' where ^ is XOR
C
x' ^ y where ^ is XOR
D
x' ^ y' where ^ is XOR
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Question 2 Explanation: 
It is a simple De Morgan's laws question.  
Question 3
The truth table truthtable represents the Boolean function
A
X
B
X+Y
C
X xor Y
D
Y
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Question 3 Explanation: 
The value of f(X, Y) is same as X for all input pairs. We see from truth table – Column x= f(x,y) So , f(x,y)=x Ans is (A) part.
Question 4
W hat is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term.
gatecs2012Kmap
A
b'd'
B
b'd' + b'c'
C
b'd' + a'b'c'd'
D
b'd' + b'c' + c'd'
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Question 4 Explanation: 
gatecs2012KmapSolution
Question 5
Which one of the following circuits is NOT equivalent to a 2-input XNOR (exclusive NOR) gate? GATECS201113
A
A
B
B
C
C
D
D
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Question 5 Explanation: 
All options except D produce XOR. See following image (Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html) gate201113
Question 6
The simplified SOP (Sum Of Product) form of the boolean expression (P + Q' + R') . (P + Q' + R) . (P + Q + R') is
A
(P'.Q + R')
B
(P + Q'.R')
C
(P'.Q + R)
D
(P.Q + R)
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Question 6 Explanation: 
Question 7
Consider the following circuit involving three D-type flip-flops used in a certain type of counter configuration. GATECS2011Q50 If at some instance prior to the occurrence of the clock edge, P, Q and R have a value 0, 1 and 0 respectively, what shall be the value of PQR after the clock edge?
A
000
B
001
C
010
D
011
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Question 7 Explanation: 
P' = R Q' = (P + R)' R' = QR' Given that (P, Q, R) = (0, 1, 0), next state P', Q', R' = 0, 1, 1 ----------------------------------------------------------------------------------------------- D flip flop truth table
D Q(t+1)
0 0
1 1
Initially (p,q,r) =(0,1,0) D for p=R D for q=NOT(p xor r) D for r= (not)r.q So Q(t+1) for(p,q,r) p=>r=0 so p=0 q=> NOT(p xor r) => 1      so q=1 r=>(not)r.q => 1         so r=1 (p,q,r)=(0,1,1)
Question 8
Consider the data given in previous question. If all the flip-flops were reset to 0 at power on, what is the total number of distinct outputs (states) represented by PQR generated by the counter?
A
3
B
4
C
5
D
6
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Question 8 Explanation: 
There are four distinct states, 000 → 010 → 011 → 100 (→ 000) so the answer is B
Question 9
The minterm expansion of f(P, Q, R) = PQ + QR' + PR' is
A
m2 + m4 + m6 + m7
B
m0 + m1 + m3 + m5
C
m0 + m1 + m6 + m7
D
m2 + m3 + m4 + m5
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Question 9 Explanation: 
CSE_2010_06_ans
Question 10
The Boolean expression for the output 'f' of the multiplexer shown below is CSE_201009
A
(P(XOR)Q(XOR)R)'
B
P(XOR)Q(XOR)R
C
(P+Q+R)'
D
P+Q+R
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Question 10 Explanation: 
For 4 to 1 mux truth table SEL                  INPUT                 O/P
Q P R R’ R’ R F
0 0 X X X 1 1
0 1 X X 1 X 1
1 0 X 1 X X 1
1 1 1 X X X 1
p’q’r+p’qr’+pq’r’+pqr pXORqXORr
Question 11
What is the Boolean expression for the output f of the combinational logic circuit of NOR gates given below? CSE_201031
A
(Q+R)'
B
(P+Q)'
C
(P+R)
D
(P+Q+R)'
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Question 11 Explanation: 
Answer is Option A. The above question contains the NOR gates. Let's see what NOR gate does. If A and B are the two inputs to the NOR gate, the NOR gate gives (A+B)' as the output. Let's assign numbers to the Gates for the easy understanding.
In the 1st column there are 4 NOR Gates, 
number them as 1 to 4 ( top to down).

In the 2nd column there are 2 NOR Gates, 
number them as 5 and 6 ( top to down).

In the 3rd column there is only 1 NOR Gate, 
number it as 7.

1st numbered Gate gives output as : ( P + Q )'
2nd numbered Gate gives output as : ( Q + R )'
3rd numbered Gate gives output as : ( P + R )'
4th numbered Gate gives output as : ( R + Q )'
5th numbered Gate gives output as :
(( P + Q )' + ( Q + R )')'
= ((P + Q)'' . ( Q + R )'') ( De Morgan's law)
= (P + Q ) . ( Q + R ) ( Idempotent Law, A'' = A)
= (PQ + PR + Q + QR )
= (Q(1 + P + R) + PR) = Q + PR ( as, 1 + " any 
                                boolean expression" = 1 )

Similarly 6th numbered Gate gives output as : R + PQ 
                             (as this time R is common here)

Now 7th numbered Gate gives output as :
((Q + PR) + (R + PQ))'
= (Q( 1+P) + R(1+P))'
= (Q+R)' 
Question 12
In the sequential circuit shown below,if the initial value of the output Q1Q0 is 00,what are the next four values of Q1Q0? CSE_201032
A
11, 10, 01, 00
B
10, 11, 01, 00
C
10, 00, 01, 11
D
11, 10, 00, 01
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Question 12 Explanation: 
We have t flip flop Truth table of t flip flop  
t q
0 q
1 q’
So q0 always inverted as t=1 always So 1)q10=1 2)q0 =0 3)q0 =1 4)q0 =1 For q1 also t=1 always but clock is so we have to observe positive edge of clock I.e. when is q0 going from 0 -> 1 1)q1 =1 2)q1 =1 3)q1 =0 4)q1 =0 So final combination q0q1->(11,10,01,00) Ans (A)
Question 13
What is the minimum number of gates required to implement the Boolean function (AB+C)if we have to use only 2-input NOR gates?
A
2
B
3
C
4
D
5
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Question 13 Explanation: 
AB+C = (A+C)(B+C) = ((A+C)' + (B+C)')' So, '3' 2-input NOR gates are required.
Question 14
In the Karnaugh map shown below, X denotes a don't care term. What is the minimal form of the function represented by the Karnaugh map?
7

A) 8  


B) 9

C) 10

D) 11 
A
A
B
B
C
C
D
D
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Question 14 Explanation: 
107
Question 15
Let r denote number system radix. The only value(s) of r that satisfy the equation 12
A
decimal 10
B
decimal 11
C
decimal 10 and 11
D
any value > 2
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Question 15 Explanation: 
As we can see 121 contains digit ‘2’ which can’t be represented directly in base ‘2’ (as digits should be less than base) so number system must have   “r>2”. Ans (D) part.
Question 16
Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit 14 15
A) sigma_Em(4, 6)

B) sigma_Em(4, 8)

C) sigma_Em(6, 8)

D) sigma_Em(4, 6, 8)
A
A
B
B
C
C
D
D
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Question 16 Explanation: 
From logic diagram  we have f=f1.f2+f3 f=m(4,5,6,7,8).f2+m(1,6,15)----(1) from eq(1) we need to find such f2 so that we can get  f=m(1,6,8,15) eq(1)  says we can get m(1,6,15) from f3 ,so only  8 left now from option (a,b,d) we get (4,6), (4,8) ,(4,6,8) respectively for m(4,5,6,7,8)f2 which is not required as m4 is undesired. But option (D) m(4,5,6,7,8)(6,8)+(1,6,15)
  • m(6,8)+m(1,6,15)
  • m(1,6,8,15)an
s is  ( C) part.
Question 17
If P, Q, R are Boolean variables, then (P + Q')(PQ' + PR)(P'R' + Q') simplifies
A
PQ'
B
PR'
C
PQ' + R
D
PR'' + Q
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Question 17 Explanation: 
112
Question 18
How many 3-to-8 line decoders with an enable input are needed to construct a 6-to-64 line decoder without using any other logic gates?
A
7
B
8
C
9
D
10
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Question 18 Explanation: 
18 So total signals in=a, b, c, x, y, z  i.e. 6 And total output =8*8=64 hence required decoders (from fig.)=9  so ans is ( C) part.
Question 19
Consider the following Boolean function of four variables: f(w,x,y,z) = ∑(1,3,4,6,9,11,12,14) The function is:
A
independent of one variables.
B
independent of two variables.
C
independent of three variables.
D
dependent on all the variables.
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Question 19 Explanation: 
19 On solving K-MAP we get ZX’+XZ’ so  it is independent of w,y Ans (B) part.
Question 20
Let f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15). Which of the following expressions are NOT equivalent to f?
A
x'y'z' + w'xy' + wy'z + xz
B
w'y'z' + wx'y' + xz
C
w'y'z' + wx'y' + xyz + xy'z
D
x'y'z' + wx'y' + w'y
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Question 20 Explanation: 
20_0 Solving this k-map we get x'y'z' + w'xy' + wy'z + xz which is (A) part 20_1 Solving this k-map we get w'y'z' + wx'y' + xz which is (B) part. 20_2 Solving this k-map we get w'y'z' + wx'y' + xyz + xy'z which is ( C) part . But we can’t get (D) part from any combination so Ans is (D) part.
Question 21
Define the connective * for the Boolean variables X and Y as: X * Y = XY + X' Y'. Let Z = X * Y.
Consider the following expressions P, Q and R.
P: X = Y⋆Z 
Q: Y = X⋆Z 
R: X⋆Y⋆Z=1
Which of the following is TRUE?
A
Only P and Q are valid
B
Only Q and R are valid.
C
Only P and R are valid.
D
All P, Q, R are valid.
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Question 21 Explanation: 
* is nothing but working as EX NOR here.Explanation: P:

X= Y * Z

=(Y XOR Z)’

=YZ + Y’Z’

=Y(XY + X’Y’)+Y’(XY+X’Y’)’

=XY+Y’((Y XOR X)’)’

=XY+Y’(Y XOR X)

=XY+Y’(Y’X+X’Y)

=XY+Y’X

=X(Y+Y’)

=X 
Q:

Y=X*Z

=(X XOR Z)’

=X(XY + X’Y’) + X’(XY + X’Y’)’

=XY+X’(X’Y+XY’)

=XY+X’Y

=Y 
R:

X * Y *Z

WE HAVE SEEN FROM P Y*Z =X

SO X * X
  • NOT(X XOR X)=X’X’+XX
  • 1
SO ALL P,Q,R ARE CORRECT ANS IS (D)
Question 22
Suppose only one multiplexer and one inverter are allowed to be used to implement any Boolean function of n variables. What is the minimum size of the multiplexer needed?
A
2n line to 1 line
B
2n+1 line to 1 line
C
2n-1 line to 1 line
D
2n-2 line to 1 line
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Question 22 Explanation: 
We can use n-1 selection lines , and using 0,1 and nth variable and its compliment to realize the function So ans is 2^(n-1):1  Part-(C )
Question 23
In a look-ahead carry generator, the carry generate function Gi and the carry propagate function Pi for inputs Ai and Bi are given by:
Pi = Ai ⨁ Bi and Gi = AiBi 
The expressions for the sum bit Si and the carry bit Ci+1 of the look-ahead carry adder are given by:
Si = Pi ⨁ Ci and Ci+1 = Gi + PiCi , where C0 is the input carry. 
Consider a two-level logic implementation of the look-ahead carry generator. Assume that all Pi and Gi are available for the carry generator circuit and that the AND and OR gates can have any number of inputs. The number of AND gates and OR gates needed to implement the look-ahead carry generator for a 4-bit adder with S3, S2, S1, S0 and C4 as its outputs are respectively:
A
6, 3
B
10, 4
C
6, 4
D
10, 5
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Question 23 Explanation: 
let the carry input be c0 Now,
c1 = g0 + p0c0 = 1 AND, 1 OR
c2 = g1 + p1g0 + p1p0c0 
   = 2 AND, 1 OR

c3 = g2 + p2g1 + p2p1go + p2p1p0c0 
   = 3 AND, 1 OR
c4 = g3 + p3g2 + p3p2g1 + p3p2p1g0 + p3p2p1p0c0 
   = 4 AND, 1 OR
So, total AND gates = 1+2+3+4 = 10 , OR gates = 1+1+1+1 = 4 So as a general formula we can observe that we need a total of " n(n+1)/2 " AND gates and "n" OR gates for a n-bit carry look ahead circuit used for addition of two binary numbers.
Question 24
The control signal functions of a 4-bit binary counter are given below (where X is “don’t care”) The counter is connected as follows: GATECS2007Q36 Assume that the counter and gate delays are negligible. If the counter starts at 0, then it cycles through the following sequence:
A
0, 3, 4
B
0, 3, 4, 5
C
0, 1, 2, 3, 4
D
0, 1, 2, 3, 4, 5
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Question 24 Explanation: 
Initially A1 A2 A3 A4 =0000 Clr=A1 and A3 So when A1 and A3 both are 1 it again goes to 0000 Hence 0000(init.) -> 0001(A1 and A3=0)->0010 (A1 and A3=0) -> 0011(A1 and A3=0) -> 0100 (A1 and A3=1)[ clear condition satisfied] ->0000(init.) so it goes through 0->1->2->3->4 Ans is ( C) part.
Question 25
You are given a free running clock with a duty cycle of 50% and a digital waveform f which changes only at the negative edge of the clock. Which one of the following circuits (using clocked D flip-flops) will delay the phase of f by 180°? cs20068a cs20068b
A
A
B
B
C
C
D
D
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Question 25 Explanation: 
We assume the D flip-flop to be negative edge triggered.
 
In option (A), during the negative edge of the clock, first flip-flop inverts complement of ‘f’. But, the output of first flip-flop has the same phase as ‘f’. Now, we give this output as input to the second flip-flop, which is enabled by ‘clk’. Thus, we get a double inverted output having same phase as the input. So, A is not the correct option.
 
In option (B) and (D), the output is inverted ‘f’. But, we want ‘f’ as the output. So, (B) and (D) can’t be the answer.
 
In option (C), the first flip-flop is activated by ‘clk’. So, the output of first flip-flop has the same phase as ‘f’. But, the second flip-flop is enabled by complement of ‘clk’. Since the clock ‘clk’ has a duty cycle of 50% , we get the output having phase delay of 180 degrees.
 
Therefore, (C) is the correct answer.
 
Please comment below if you find anything wrong in the above post.
Question 26
GATECS2014Q17
A
A
B
B
C
C
D
D
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Question 26 Explanation: 
F= PQ+P’QR+P’QR’S =Q(P+P’+P’R’S) USE PROP a+a’b=a+b =Q(P+R+P’R’S) =Q(P+P’R’S+R) USE PROP a+a’b=a+b =Q(P+R’S+R) =Q(P+R+R’S) USE PROP a+a’b=a+b =Q(P+R+S) =PQ+QR+QS Ans (a)
Question 27
Consider a 4-to-1 multiplexer with two select lines S1 and S0, given below GATECS2014Q55 The minimal sum-of-products form of the Boolean expression for the output F of the multiplexer is
A
P'Q + QR' + PQ'R
B
P'Q + P'QR' + PQR' + PQ'R
C
P'QR + P'QR' + QR' + PQ'R
D
PQR'
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Question 27 Explanation: 
For 4 to 1 mux =p’q’(0)+p’q(1)+pq’r+pqr’ =p’q+pq’r+pqr’ =q(p’+pr’)+pq’r =q(p’+r’)+pq’r =p’q+qr’+pq’r Ans (a)
Question 28
Let k = 2n. A circuit is built by giving the output of an n-bit binary counter as input to an n-to-2n bit decoder. This circuit is equivalent to a
A
k-bit binary up counter.
B
k-bit binary down counter.
C
k-bit ring counter.
D
k-bit Johnson counter.
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Question 28 Explanation: 
For output of a decoder , only single output will be ‘1’ and remaining will be ‘0’ at the same time. So high output  will give the count of the ring counter. Hence Ans is ( C) part.
Question 29
Consider the equation (123)5 = (x8)y with x and y as unknown. The number of possible solutions is _____ .
A
1
B
2
C
3
D
4
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Question 29 Explanation: 
Changing (123) base 5 into base 10= 1*25+2*5+3*1=38 Changing x8 base y in decimal= x*y+8 Equating both we get xy+8=38
  • xy=30
  • possible combinations =(1,30),(2,15),(3,10)
but we have ‘8’ present in x8 so base y>8 as all three are satisfying the conditions so total solutions =3 hence ans is ( C) part
Question 30
GATECS2014Q17
A
A
B
B
C
C
D
D
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Question 30 Explanation: 
30 On solving k-map we get qs+q’s’ So Ans is (B) part.
Question 31
Consider the following combinational function block involving four Boolean variables x, y, a, b where x, a, b are inputs and y is the output.
f (x, y, a, b)
{
   if (x is 1) y = a;
   else y = b;
}
Which one of the following digital logic blocks is the most suitable for implementing this function?
A
Full adder
B
Priority encoder
C
Multiplexer
D
Flip-flop
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Question 31 Explanation: 
This function can be interpreted as having two inputs a, b and select signal x. Output y will depend on the select signal x. Function will be like (ax+bx’) Its implementation will be like 31 So ans is ( C) part.  
Question 32
GATECS2014Q61 The above sequential circuit is built using JK flip-flops is initialized with Q2Q1Q0 = 000. The state sequence for this circuit for the next 3 clock cycle is
A
001, 010, 011
B
111, 110, 101
C
100, 110, 111
D
100, 011, 001
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Question 32 Explanation: 
JK ff truth table---
j k Q
0 0 Q0
1 0 1
0 1 0
1 1 Q0’
Initially Q2Q1Q0=000 Present state FF input                   Next state
Q2 Q1 Q0 J2 K2 J1 K1 J0 K0 Q2 Q1 Q0
0 0 0 1 0 0 1 0 1 1 0 0
1 0 0 1 0 1 0 0 1 1 1 0
1 1 0 0 0 1 0 1 1 1 1 1
                       
So ans is ( C) part.
Question 33
Let X denote the Exclusive OR (XOR) operation. Let ‘1’ and ‘0’ denote the binary constants. Consider the following Boolean expression for F over two variables P and Q:
F(P, Q) = ( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) ) 
The equivalent expression for F is
A
P + Q
B
(P + Q)'
C
P X Q
D
(P X Q)'
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Question 33 Explanation: 
We need to simplify the above expression. As the given operation is XOR, we shall see property of XOR. Let A and B be boolean variable. In A XOR B, the result is 1 if both the bits/inputs are different, else 0. Now,

( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) )

( P' X P X Q ) X ( P X Q X Q ) ( as 1 X P = P' and Q X 0 = Q )

(1 X Q) X ( P X 0) ( as P' X P = 1 , and Q X Q = 0 )

Q' X P ( as 1 X Q = Q' and P X 0 = P )

PQ + P'Q' ( XOR Expansion, A X B = AB' + A'B )

This is the final simplified expression.

Now we need to check for the options.

If we simplify option D expression.

( P X Q )' = ( PQ' + P'Q )' ( XOR Expansion, A X B = AB' + A'B )

((PQ')'.(P'Q)') ( De Morgan's law )

( P'+ Q).(P + Q') ( De Morgan's law )

P'P + PQ + P'Q' + QQ'

PQ + P'Q' ( as PP' = 0 and QQ' = 0 ) 

Hence both the equations are same. Therefore Option D. 
Question 34
GATE2006Q35 Consider the circuit above. Which one of the following options correctly represents f (x, y, z)?
A
xz' + xy + y'z
B
xz' + xy + (yz)'
C
xz + xy + (yz)'
D
xz + xy' + y'z
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Question 34 Explanation: 
Output from MUX 1=> Z’X+ZY’ Output from MUX2=> Y’(Z’X+ZY’)+YX =>Y’Z+Y’Z’X+YX =>Y’Z+X(Y’Z’+Y) =>Y’Z+X(Y+Z’)        USING A+A’B=(A+B) =>Y’Z+XY+XZ’ So Ans is (A).
Question 35
Given two three bit numbers a2a1a0 and b2b1b0 and c, the carry in, the function that represents the carry generate function when these two numbers are added is: GATE2006Q36
A
A
B
B
C
C
D
D
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Question 35 Explanation: 
For carry look ahead adder we know carry generate function--- Where   As we are having two 3 bits number to add so final carry out will be C3- Putting value of Pi,Gi in 3 C3=(A2.B2)+(A1.B1)(A2+B2)+(A0.B0)(A1+B1)(A2+B2)                       (TAKING C0=0) C3=A2.B2 +A1A2B1+A1B2B1+(A0B0)(A1A2+A1B2+B1A2+B1B2) C3=A2B2+A1A2B1+A1B2B1+A0A1A2B0+A0A1B0B2+A0A2B1B0+A0B0B1B2 SO ANS IS (A) PART.
Question 36
Consider the circuit in the diagram. The ⊕ operator represents Ex-OR. The D flipflops are initialized to zeroes (cleared). GATE2006Q37 The following data: 100110000 is supplied to the “data” terminal in nine clock cycles. After that the values of q2q1q0 are:
A
000
B
001
C
010
D
101
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Question 36 Explanation: 

The D flipflops are initialized to zeroes. This implies q0 = 0, q1 = 0 and q2 = 0 initially.
Clock cycle 1 : q0 = data = 1 , q1 = q0before XOR q2before = 0 XOR 0 = 0 , q2 = q1before = 0
Clock cycle 2 : q0 = data = 0 , q1 = q0before XOR q2before = 1 XOR 0 = 1 , q2 = q1before = 0
Clock cycle 3 : q0 = data = 0 , q1 = q0before XOR q2before = 0 XOR 0 = 0 , q2 = q1before = 1
Clock cycle 4 : q0 = data = 1 , q1 = q0before XOR q2before = 0 XOR 1 = 1 , q2 = q1before = 0
Clock cycle 5 : q0 = data = 1 , q1 = q0before XOR q2before = 1 XOR 0 = 1 , q2 = q1before = 1
Clock cycle 6 : q0 = data = 0 , q1 = q0before XOR q2before = 1 XOR 1 = 0 , q2 = q1before = 1
Clock cycle 7 : q0 = data = 0 , q1 = q0before XOR q2before = 0 XOR 1 = 1 , q2 = q1before = 0
Clock cycle 8 : q0 = data = 0 , q1 = q0before XOR q2before = 0 XOR 0 = 0 , q2 = q1before = 1
Clock cycle 9 : q0 = data = 0 , q1 = q0before XOR q2before = 0 XOR 1 = 1 , q2 = q1before = 0
 
Thus, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 37
Consider a Boolean function f (w, x, y, z). suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors i1 = (w1, x1, y1, z1) and i2 = (w2, x2, y2, z2) we would like the function to remain true as the input changes from i1 to i2 (i1 and i2 differ in exactly one bit position), without becoming false momentarily. Let f (w, x, y, z) = ∑(5,7,11,12,13,15). Which of the following cube covers of f will ensure that the required property is satisfied?
A
w'xz, wxy', xy'z, xyz,wyz
B
wxy,w'xz,wyz
C
wx(yz)', xz, wx'yz
D
wzy, wyz, wxz, w'xz, xy'z, xyz
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Question 37 Explanation: 

We construct the K - Map as :
2006-38
After simplification of this K - Map we get answer as : w'xz, wxy', xy'z, xyz, wyz
 
Thus, option (A) is the answer.
 
Please comment below if you find anything wrong in the above post.
Question 38
We consider the addition of two 2’s complement numbers bn-1bn-2...b0 and an-1an-2...a0. A binary adder for adding unsigned binary numbers is used to add the two numbers. The sum is denoted by cn-1cn-2...c0 and the carry-out by cout. Which one of the following options correctly identifies the overflow condition? GATECS2006Q39
A
A
B
B
C
C
D
D
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Question 38 Explanation: 
When two signed 2's complement numbers are added, overflow is detected if:
  1. both operands are positive and the result is negative, or
  2. both operands are negative and the result is positive.
  • INPUTS OUTPUTS
    Asign Bsign CARRY IN CARRY OUT Cn OVERFLOW
    0 0 0 0 0 0
    0 0 1 0 1 1
    0 1 0 0 1 0
    0 1 1 1 0 0
    1 0 0 0 1 0
    1 0 1 1 0 0
    1 1 0 1 0 1
    1 1 1 1 1 0
We may note here that overflow occurs only when CARRYin ≠ CARRYout Overflow = Cin XOR Cout or Cn-1 XOR Cout
Question 39
Consider numbers represented in 4-bit gray code. Let h3h2h1h0 be the gray code representation of a number n and let g3g2g1g0 be the gray code of (n + 1) (modulo 16) value of the number. Which one of the following functions is correct? GATECS2006Q40
A
A
B
B
C
C
D
D
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Question 39 Explanation: 
Write gray code (n) numbers from 0 to 15 and make another column for (n+1) numbers by shifting the next number on top. As shown in the table: gray-code Now to determine the min terms for g3, g2, g1, g0, we see the '1s' in the corresponding columns. The digit they correspond to give the min terms of the function. Thus the answers we will be: G3(h3h2h1h0)= ∑(4,12,13,15,14,10,11,9) G2(h3h2h1h0)= ∑(2,6,7,5,4,12,13,15) G1(h3h2h1h0)= ∑(1,3,2,6,13,15,14,10) G0(h3h2h1h0)= ∑(0,1,6,7,12,13,10,11) Thus g2 is given correctly in the options. Read about K-Maps to know about mapping, SOP and POS forms: K-Map (Karnaugh Map) Watch NPTEL videos to learn more about: Code converters Logic Minimization Using Karnaugh Maps Karnaugh Map Minimization Using Maxterms This explanation has been contributed by Kriti Kushwaha.
Question 40
Consider the following circuit. GATECS2005Q14 Which one of the following is TRUE?
A
f is independent of X
B
f is independent of Y
C
f is independent of Z
D
None of X, Y, Z is redundant
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Question 40 Explanation: 
[(XY’)’ NAND (YZ)’]= [(XY’)’ .(YZ)’]’ =XY’+YZ SO NONE OF X,Y,Z IS REDUNDANT Ans (D)
Question 41
The hexadecimal representation of 6578 is
A
1AF
B
D78
C
D71
D
32F
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Question 41 Explanation: 
We can first convert to Binary, we get 110 101 111. Then convert binary to base 16, we get 1AF (0001 1010 1111). (657)base 8= Writing binary of each digit=>  110=6 => 101=5 => 111=7 Adding extra 0’s I beginning to make groups of 4 binary digits each 000110101111= 0001 1010 1111 In octal
  • 0001 =1
  • 1010 =A
  • 1111 =F
So Ans is (A) part.
Question 42
The switching expression corresponding to f(A, B, C, D) = Σ (1, 4, 5, 9, 11, 12) is
A
BC'D' + A'C'D + AB'D
B
ABC' + ACD + B'C'D
C
ACD' + A'BC' + AC'D'
D
A'BD + ACD' + BCD'
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Question 42 Explanation: 
42   On solving above k-map we get AB’D+C’D+BC’D’ So Ans is (A) part.
Question 43
Consider the following circuit involving a positive edge triggered D FF. gatecs20051 Consider the following timing diagram. Let Ai represent the logic level on the line A in the i-th clock period. gatecs2005 Let A' represent the complement of A. The correct output sequence on Y over the clock periods 1 through 5 is
A
A0 Al A1' A3 A4
B
A0 Al A2' A3 A4
C
Al A2 A2' A3 A4
D
Al A2' A3 A4 A5'
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Question 43 Explanation: 
The Flip Flop used here is a Positive edge triggered D Flip Flop, which means that only at the "rising edge of the clock" flip flop will capture the input provided at D and accordingly give the output at Q. And at other times of the clock the output doesn't change. The output of D flip flop is same as input, i.e. Y=Q=D ( at the rising edge ). Now, in the question above, 5 clock periods are given, and we have to find the output Q or Y in those clock periods. First, let's derive the boolean expression for the Logic gate. which is : D = AX + X' Q' Now, In the 1st clock period, (i.e. when t = 0 to 1 ) here the clock has rising edge at t= 0, hence at this moment only, D flip flop will change its state. In the 1st clock,  X = 1, So,  D = A. Now A logic line may have different levels at different clock periods, i.e. may be high or low, therefore we have to answer with respect to the ith clock period where Ai is the logic level ( high or low ) of logic line A in the ith clock. So in the 1st clock period, A logic value should be A1 ( i.e. value of A in 1st clock period), but due to the delay provided by the Logic Gates ( Propagation Delay) the value of A used by Flip Flop is previous value of A only, i.e.it will capture the value of D resulted by using the logic line A in the 0th clock period, which is A0. Same happens with the value of X, i.e. instead of Xi, previous value of X  is used in the in the ith clock period, which is Xi-1. Now, In the 1st clock period value of X is same as in the 0th clock, i.e. logic 1. So, X = 1 ,and A = A0, therefore, D = A0, and hence Q = Y = A0 Similarly we have to do for other clock periods, i.e. instead of taking Ai and Xi,  Ai-1 and Xi-1 need to be taken for getting the output in the ith clock period. In the 2nd clock period, (i.e. when t = 1 to 2 ) X = 1 ( value in the previous clock), So, D = A1 ( value of A in the previous clock)  , therefore Q = Y = A1 In the 3rd clock period, (i.e. when t = 2 to 3 ) X = 0 ( value in the previous clock,see the timing diagram), So, D = Q' = A1' , therefore Q = Y = A1'   ( because of the feedback line ) In the 4th clock period, (i.e. when t = 3 to 4 ) X = 1 ( value in the previous clock,  ), So, D = A3 , therefore Q =  Y = A3 In the 5th clock period, (i.e. when t = 4 to 5 ) X = 1 ( value in the previous clock ), so, D = A4 , therefore Q = Y = A4 Hence the output sequence is : A0 A1 A1' A3 A4
Question 44
Consider the following circuit gatecs2005 The flip-flops are positive edge triggered D FFs. Each state is designated as a two bit string Q0Q1. Let the initial state be 00. The state transition sequence is:
 A) digi64A 
B) digi64B
C)digi64
D)digi64D
A
A
B
B
C
C
D
D
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Question 44 Explanation: 
Q0 will toggle in every cycle because Q0' (Q0 complement) is fed as input to the D0 flip flop. For the D1 flip flop, D1 = Q0 ⊕ Q1' , i.e., Q0 XOR Q1'. So, the bit pattern Q0 Q1 will be :
Q0     Q1

0      0

1      1

0      1

1      0

0      0

.      .

.      .

.      .
Thus, the transition sequence will be GATECS2005Q64D So, D would be the correct choice.   Please comment below if you find anything wrong in the above post.
Question 45
Consider the following floating point format GATECS2005Q84A Mantissa is a pure fraction in sign-magnitude form. The decimal number 0.239 × 213 has the following hexadecimal representation (without normalization and rounding off :
A
0D 24
B
0D 4D
C
4D 0D
D
4D 3D
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Question 45 Explanation: 
ans_45
Question 46
Consider the following floating point format GATECS2005Q84A Mantissa is a pure fraction in sign-magnitude form. The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0's are padded in while shifting a field. The normalized representation of the above number (0.239 × 213) is:
A
0A 20
B
11 34
C
4D D0
D
4A E8
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Question 46 Explanation: 
ans_46
Question 47
The Boolean function x'y' + xy + x'y is equivalent to
A
x' + y'
B
x + y
C
x + y'
D
x' + y
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Question 47 Explanation: 
x’y’+xy+xy’ =(x’y’+y(x+x’)) =x'y’+y =x'+y Ans (d)
Question 48
In an SR latch made by cross-coupling two NAND gates, if both S and R inputs are set to 0, then it will result in
A
Q = 0, Q' = 1
B
Q = 1, Q' = 0
C
Q = 1, Q' = 1
D
Indeterminate states
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Question 48 Explanation: 
If both R reset and S set inputs are inactivated. It means made ‘0’, both Q and Q’ tend to be ‘1’. [NAND gate characteristic is that only when both inputs are 1,the output is 0.]  The logic of the circuit (Q’ is complement of Q) not satisfied, and the final state at an instant after inputs R and S changed to ‘0’, is only a matter of chance. Logic state is said to be indeterminate state or racing state. Each state, Q =‘1’and Q =‘0’, and Q =‘0’, Q=‘1’ trying to race through so “RACE CONDITION”  occurs and output become unstable. So ans is (D) part.
Question 49
A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000, 1 by 0001, ..., 9 by 1001. A combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit ≥ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?
A
2
B
3
C
4
D
5
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Question 49 Explanation: 
We should get output 1 for values>=5 Making truth table for problem
A B C D Op
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 1
1 0 0 1 1
1 0 1 0 X
1 0 1 1 X
1 1 0 0 X
1 1 0 1 X
1 1 1 0 X
1 1 1 1 X
Putting this in kmap and solving   49 Here crucial point is that we need to make pair of 8 elements using don’t cares also…so final expression is A+BD+BC
  • A+B(C+D)
Hence we’ll use two OR gate and one AND gate so total 3 gates. Ans (B) part.
Question 50
Which are the essential prime implicants of the following Boolean function? f(a, b, c) = a'c + ac' + b'c
A
a'c and ac'
B
a'c and b'c
C
a'c only
D
ac' and bc'
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Question 50 Explanation: 
Essential prime implicants are prime implicants that cover an output of the function that no combination of other prime implicants is able to cover i.e., if we delete such elements then the property of group, quad, octet is destroyed. By drawing k map and by putting c on left side and a, b on right side we can find that a'c and ac' are essential prime implicants.
Question 51
Consider a multiplexer with X and Y as data inputs and Z as control input. Z = 0 selects input X, and Z = 1 selects input Y. What are the connections required to realize the 2-variable Boolean function f = T + R, without using any additional hardware ?
A
R to X, 1 to Y, T to Z
B
T to X, R to Y, T to Z
C
T to X, R to Y, 0 to Z
D
R to X, 0 to Y, T to Z
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Question 51 Explanation: 
For mux we will be having equation like z’x+zy-----(1) Now putting X=R , Y=1 and Z=T  in eq(1) we find----
  • T’R+T(1)
  • T+T’R
  • T+R (SINCE a+a’b=a+b)
So ans is (A) part.
Question 52
Consider the partial implementation of a 2-bitt counter using T flip-flops following the sequence 0-2-3-1-0, as shown below GATECS20014Q61 To complete the circuit, the input X should be 1) 2) 3) 4)
A
Q2'
B
Q2 + Q1
C
(Q1 ⊕ Q2)'
D
Q1 ⊕ Q2
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Question 52 Explanation: 
From circuit we see T1=XQ1’+X’Q1----(1) AND T2=(Q2+Q1)’----(2) AND DESIRED OUTPUT IS 00->10->11->01->00 SO X SHOULD BE Q1Q2’+Q1’Q2 SATISFYING 1 AND 2. SO ANS IS (D) PART.
Question 53
A 4-bit carry lookahead adder, which adds two 4-bit numbers, is designed using AND, OR, NOT, NAND, NOR gates only. Assuming that all the inputs are available in both complemented and uncomplemented forms and the delay of each gate is one time unit, what is the overall propagation delay of the adder? Assume that the carry network has been implemented using two-level AND-OR logic.
A
4 time units
B
6 time units
C
10 time units
D
12 time units
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Question 53 Explanation: 
Let the input carry to the first adder be denoted by C1. Now, to calculate C2 we need = P1C1 + G1 = 4 gate levels (P1 takes 2 gate levels) to calculate S1 we need = P1 XOR C1 = 2 + 2 = 4 gate levels. Since it is a Carry look ahead adder, computing C3 , S2 doesn't have to wait for carry output C2 from the previous adder as C2, C3 etc will get computed at the same time. Now, S2 is computed as = P2 XOR C2 = P2.C2' + P2'.C2 = P2 (P1.C1 + G1 )' + P2' (P1.C1 + G1) [ notice that we are not using the output carry from first adder C2 anywhere here ] which can be implemented using 4 gate levels. also C3 can be computed by using 4 gate levels and so on... so the overall propagation delay is 4 gate level as the outputs at Si , Ci are available at the respective full adders after 4 gate levels = 4 time units. To understand it with more clarity draw the carry look ahead adder circuit and then check it.
Question 54
Consider an array multiplier for multiplying two n bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is
A
Θ(1)
B
Θ(log n)
C
Θ(n)
D
Θ(n2)
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Question 54 Explanation: 
Number of gates used in ‘n’ bit array multiplier (n * n) = (2n – 1)
Each gate in the circuit has a unit delay.
Total delay = 1 * (2n – 1 ) = O(2n – 1) = O(n)
 
Thus, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 55
A 1-input, 2-output synchronous sequential circuit behaves as follows : Let zk, nk denote the number of 0's and 1's respectively in initial k bits of the input (zk + nk = k). The circuit outputs 00 until one of the following conditions holds.
    zk - nk = 2. In this case, the output at the k-th and 
                 all subsequent clock ticks is 10.
    nk - zk = 2. In this case, the output at the k-th and
                 all subsequent clock ticks is 01.
What is the minimum number of states required in the state transition graph of the above circuit?
A
5
B
6
C
7
D
8
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Question 55 Explanation: 
The answer must be 5 to this question as we just need to count the difference of the number of 0's and 1's in the first k bit of a number. And we just need to count till this count reaches 2 or -2 (negative when number of 0's is less than number of 1's) . So, the possibilities are -2, -1, 0, 1 and 2 which represents the five states of the state transition diagram. For state -2, the output of the circuit will be 01, for state 2, output will be 10 (both these states not having any outgoing transitions) and for other 3 states, output will be 00 as per the given description of the circuit.
Question 56
The literal count of a boolean expression is the sum of the number of times each literal appears in the expression. For example, the literal count of (xy + xz') is 4. What are the minimum possible literal counts of the product-of-sum and sum-of-product representations respectively of the function given by the following Karnaugh map ? Here, X denotes "don't care" GATECS2009Q45
A
(11, 9)
B
(9, 13)
C
(9, 10)
D
(11, 11)
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Question 56 Explanation: 
Analyzing both POS as well as SOP 56 Sop- wy+w’y’+z’wx’+xyz’ Literal count=10 Pos--- (w’+Z’)(z’+y)(z’+y’) Literal count=9 Ans is ( c) part.
Question 57
Consider the ALU shown below. GATECS2009Q46 If the operands are in 2's complement representation, which of the following operations can be performed by suitably setting the control lines K and C0 only (+ and - denote addition and subtraction respectively) ?
A
A + B, and A - B, but not A + 1
B
A + B, and A + 1, but not A - B
C
A + B, but not A - B, or A + 1
D
A + B, and A - B, and A + 1
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Question 57 Explanation: 
We can set value of k and c as 0 or 1 Two things we need to know---
  1. If we take xor of any number with 1 we get it in its complement form.
  2. If we take xor of any number with 0 we get that number itself.
So on setting k=1 we can get –B and c will work like select signal Like c=0 means add C=1 means subtract Hence with k=1 c=1 we get A-B With K=0 c=0 we get A+B We need b=1 to get A+1 which we can’t set with the help of k and c only So Ans is (A) part.
Question 58
Consider the following circuit composed of XOR gates and non-inverting buffers. GATECS2009Q47 The non-inverting buffers have delays d1 = 2 ns and d2 = 4 ns as shown in the figure. Both XOR gates and all wires have zero delay. Assume that all gate inputs, outputs and wires are stable at logic level 0 at time 0. If the following waveform is applied at input A, how many transition(s) (change of logic levels) occur(s) at B during the interval from 0 to 10 ns ? GATECS2009Q47B
A
1
B
2
C
3
D
4
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Question 59
Minimum sum of product expression for f(w, x, y, z) shown in Karnaugh-map below is  
yz\wx 00 01 11 10
00 0 1 1 0
01 x 0 0 1
11 x 0 0 1
10 0 1 1 x
A
xz + y'z
B
xz' + zx'
C
x'y + zx'
D
None of these
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Question 59 Explanation: 
59 On solving this we get xz’+zx’ so ans is (B) part
Question 60
Consider the following logic circuit whose inputs and function and output is f.
f(dx, y, z)
f1(dx, y, z)
f2(dx, y, z)
f3(dx, y, z) = ?
GATECS2002Q26 Given that
f1(dx, y, z) = ∑(d0, 1, 3, 5),
f1(dx, y, z) = ∑(d6, 7) and
f1(dx, y, z) = ∑(d1, 4, 5),
f3 is :
A
∑(1, 4, 5)
B
∑(6, 7)
C
∑(0, 1, 3, 5)
D
None of these
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Question 60 Explanation: 
ans_60
Question 61
Consider the following multiplexor where 10, 11, 12, 13 are four data input lines selected by two address line combinations A1A0 = 00, 01, 10, 11 respectively and f is "the output of the multiplexor. EN is the enable input. GATECS200227 The function f(x, y, z) implemented by the above circuit is :
A
xyz'
B
xy + z
C
x + z
D
None of these
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Question 61 Explanation: 
F = (A1’A0’I0 + A1’A0I1 + A1A0’I2 + A1A0I3) * EN F = (XYZ' +XYZ + Y'ZY + ZY')Z' F = (XYZ' +XYZ + ZY'(Y + 1))Z' F = (XYZ' +XYZ + ZY' * 1)Z' F = (XY(Z’ + Z) + ZY’)Z’ F = (XY + ZY’)Z’ F = XYZ’ + 0 F = XYZ' Thus, option (A) is correct. Please comment below if you find anything wrong in the above post.
Question 62
Let f(A, B) = A' + B. Simplified expression for function f(f(x + y, y)z) is :
A
x' + z
B
xyz
C
xy' + z
D
None of these
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Question 62 Explanation: 

Simplified expression for given function 'g' is :
g = f( f(x + y, y), z) g = f( ((x + y)’ + y), z) g = f( (x’y’ + y), z) g = f( ((y + y’) * (x’ + y)), z) g = f( (1 * (x’ + y)), z) g = f( (x’ + y), z) g = (x’ + y)’ + z) g = xy’ + z
 
Thus, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 63
Given the following Karnaugh map, which one of the following represents the minimal Sum-Of-Products of the map? GATECS2001Q11
A
xy + y'z
B
wx'y' + xy + xz
C
w'x + y'z + xy
D
xz + y
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Question 63 Explanation: 
ans_63 On solving we get xy+y’z so ans is (A) part.
Question 64
Consider the following circuit with initial state Q0 = Q1 = 0. The D Flip-flops are positive edged triggered and have set up times 20 nanosecond and hold times 0. GATECS2001Q33 Consider the following timing diagrams of X and C; the clock period of C <= 40 nanosecond. Which one is the correct plot of Y? GATECS2001Q33B
A
a
B
b
C
c
D
d
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Question 64 Explanation: 

Q0 = 0 (given) Therefore, Q0' = 1
During the first clock cycle nothing will change since input doesn't come before rising edge of the clock cycle Q0' becomes ’0’ on rising edge of next clock cycle. Therefore, for only one clock cycle D1 will be ‘1’.
 
2001-33
 
Thus, option (C) is the answer.
 
Please comment below if you find anything wrong in the above post.
Question 65
Consider the circuit given below with initial state Q0 =1, Q1 = Q2 = 0. The state of the circuit is given by the value 4Q2 + 2Q1 + Q0
GATECS2001Q39
Which one of the following is the correct state sequence of the circuit?
A
1,3,4,6,7,5,2
B
1,2,5,3,7,6,4
C
1,2,7,3,5,6,4
D
1,6,5,7,2,3,4
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Question 65 Explanation: 
Initial value Q0 Q1 Q2 2Q1+4Q2+Q0
clk 1 0 0 1
1 0 1 0 2
2 1 0 1 5
3 1 1 0 3
4 1 1 1 7
5 0 1 1 6
6 0 0 1 4
So ans is (B) part.
Question 66
The simultaneous equations on the Boolean variables x, y, z and w,
GATECS200033
have the following solution for x, y, z and w, respectively.
A
0 1 0 0
B
1 1 0 1
C
1 0 1 1
D
1 0 0 0
Digital Logic & Number representation    GATE-CS-2000    
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Question 66 Explanation: 
We solve this question by putting in the options in the statements.   Statement 1 :  x + y + z = 1
OPTION    x  y  z  w    LHS     LHS=1
-------------------------------------- 
  A       0  1  0  0     1       Yes
  B       1  1  0  1     1       Yes
  C       1  0  1  1     1       Yes
  D       1  0  0  0     1       Yes
Till now, all the options are possible.   Statement 2 : x y = 0
OPTION    x  y  z  w    LHS     LHS=0
-------------------------------------- 
  A       0  1  0  0     0       Yes
  B       1  1  0  1     1       No
  C       1  0  1  1     0       Yes
  D       1  0  0  0     0       Yes
Since LHS ≠ 0, B is not possible.   Statement 3 : x z + w = 1
OPTION    x  y  z  w    LHS     LHS=1
-------------------------------------- 
  A       0  1  0  0     0       No
  C       1  0  1  1     1       Yes
  D       1  0  0  0     0       No
Since LHS ≠ 1, A and D are not possible.   Statement 4 : x y + z' w' = 0
OPTION    x  y  z  w    LHS    LHS=0
-------------------------------------- 
  C       1  0  1  1     0      Yes

Thus, C is the correct option. Please comment below if you find anything wrong in the above post.
Question 67
Which function does NOT implement the Karnaugh map given below? GATECS200034
A
a
B
b
C
c
D
d
Digital Logic & Number representation    GATE-CS-2000    
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Question 67 Explanation: 
As we know option (A) and (B) are same so they can be represented by 1st K-MAP But (C ) is in POS form so we need to make group of 0’s as in 2nd K-MAP- ans_67 So all (A),(B) ,(C) can be realized hence ans (D) part.
Question 68
The following arrangement of master-slave flip flops has the initial state of P, Q as 0, 1 (respectively). After three clock cycles the output state P, Q is (re­spectively),
gatecs2000Q35
A
1, 0
B
1, 1
C
0, 0
D
0,1
Digital Logic & Number representation    GATE-CS-2000    
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Question 68 Explanation: 
Given P = 0 ,  J = 1 and k = 1
  • JK flipflop toggles the input in ‘11’ state. Therefore, output of first flipflop at P is ‘1’.
  • Initial value of P is input for D flipflop. So, D = 0 .Therefore, output of second flipflop at Q is ‘0’. Thus, option (A) is the answer. Please comment below if you find anything wrong in the above post.
Question 69
Consider a 4 bit Johnson counter with an initial value of 0000. The counting sequence of this counter is:
A
0, 1, 3, 7, 15, 14, 12, 8, 0
B
0, 1, 3, 5, 7, 9, 11, 13, 15, 0
C
0, 2, 4, 6, 8, 10, 12, 14, 0
D
0, 8, 12, 14, 15, 7, 3, 1, 0
Digital Logic & Number representation    GATE-CS-2015 (Set 1)    
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Question 69 Explanation: 
Refer http://en.wikipedia.org/wiki/Ring_counter#Johnson_Counter_.284-bits.29 The four bit Johnson's counter connects the complement of the output of the last shift register to the input of the first register with shift distance=1 i.e 1 bit will shift/cycle It will work as follows: 0000 //Last 0 complemented and fed as input to first register 1000 1100 1110 1111 //Last 1 complemented and fed as input to first register 0111 0011 0001 0000
Question 70
A positive edge-triggered D flip-flop is connected to a positive edge-triggered JK flipflop as follows. The Q output of the D flip-flop is connected to both the J and K inputs of the JK flip-flop, while the Q output of the JK flip-flop is connected to the input of the D flip-flop. Initially, the output of the D flip-flop is set to logic one and the output of the JK flip-flop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flip-flop when the flip-flops are connected to a free-running common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the state-holding mode of the JK flip-flop. Both the flip-flops have non-zero propagation delays.
A
0110110...
B
0100100...
C
011101110...
D
011001100...
Digital Logic & Number representation    GATE-CS-2015 (Set 1)    
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Question 70 Explanation: 
jk Initially Q output of D - FF  = 1 Initially Q output of JK - FF  = 0 Now with the help of present state and next state table we can see what is happening in circuit.
  • Toggle: J = K = 1
  • Hold : J = K = 0
jk1 We see from table Q output of D-FF is going to next state input of JK-FF and the bits sequence produced is like 110110….. Including initial condition (0) we get output as 0110110110…. Hence answer is (A) part. Another Explanation: Here, it is given that JK flip flop will toggle when J = K = 1 and it will retain the output if J = K = 0. Also, the output of the D flip flop would remain the same as the input. So, we have Initial output : D = 1

JK = 0

After clock 1 : D = 0 (D gets 0 as input 
from initial output of JK, so output = 0)

JK = 1 (J = K = 1 from initial output of D, so output would be toggled from 0 to 1)

After clock 2 : D = 1 (D gets 1 as input 
from previous output of JK, so output = 1)

JK = 1 (J = K = 0 from previous output of D, so output would be retained to 1)

After clock 3 : D = 1 (D gets 1 as input
from previous output of JK, so output = 1)

JK = 0 (J = K = 1 from previous output of D, so output would be toggled from 1 to 0)

After clock 4 : D = 0 (D gets 0 as input
from previous output of JK, so output = 0)

JK = 1 (J = K = 1 from previous output of D, so output would be toggled from 0 to 1)

After clock 5 : D = 1 (D gets 1 as input
from previous output of JK, so output = 1)

JK = 1 (J = K = 0 from previous output of D, so output would be retained to 1)


Thus, the bit sequence generated at the Q output of the JK flip flop will be 0110110...

 
Question 71
Consider the operations f(X, Y, Z) = X'YZ + XY' + Y'Z'  and  g(X′, Y, Z) = X′YZ + X′YZ′ + XY Which one of the following is correct?
A
Both {f} and {g} are functionally complete
B
Only {f} is functionally complete
C
Only {g} is functionally complete
D
Neither {f} nor {g} is functionally complete
Digital Logic & Number representation    GATE-CS-2015 (Set 1)    
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Question 71 Explanation: 
A function is considered as functionally complete if it does not belong to T0,T1,L,M,S which are Property 1: We say that boolean function f preserves zero, if on the 0-input it produces 0. By the 0-input we mean such an input, where every input variable is 0 (this input usually corresponds to the first row of the truth table). We denote the class of zero-preserving boolean functions as T0 and write f ∈ T0. Property 2: Similarly to T0, we say that boolean function f preserves one, if on 1-input, it produces 1. The 1-input is the input where all the input variables are 1 (this input usually corresponds to the last row of the truth table). We denote the class of one-preserving boolean functions as T1 and write f ∈ T1. Property 3: We say that boolean function f is linear if one of the following two statements holds for f:
  • For every 1-value of f, the number of 1’s in the corresponding input is odd, and for every 0-value of f, the number of 1’s in the corresponding input is even.
or
  • For every 1-value of f, the number of 1’s in the corresponding input is even, and for every 0-value of f, the number of 1’s in the corresponding input is odd.
If one of these statements holds for f, we say that f is linear1. We denote the class of linear boolean functions with L and write f ∈ L. Property 4: We say that boolean function f is monotone if for every input, switching any input variable from 0 to 1 can only result in the function’s switching its value from 0 to 1, and never from 1 to 0. We denote the class of monotone boolean functions with M and write f ∈ M. Property 5: We say that boolean function f(x1,...,xn) is self-dual if f(x1,...,xn) = ¬f(¬x1,...,¬xn). The function on the right in the equality above (the one with negations) is called the dual of f. We will call the class of self-dual boolean functions S and write f ∈ S. As in our case we can see  on giving all i/p to 0 (g )produce 0 so it preserving 0 and can’t be functionally complete. But f is neither preserving 0 nor 1.
  • F is not linear(see defn. of linear above)
  • F is not monotone(see defn. of monotone above)
  • F is not self dual as f(x,y,z) is not equal to –f(-x,-y,-z)
So f is functionally complete. Hence ans is (B) part
Question 72
The minimum number of JK flip-flops required to construct a synchronous counter with the count sequence (0, 0, 1, 1, 2, 2, 3, 3, 0, 0,...) is ________
A
0
B
1
C
2
D
3
Digital Logic & Number representation    GATE-CS-2015 (Set 2)    
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Question 72 Explanation: 
Ans = 3, mod 8 up counter using 3 JK flip flops. Ignore the output of LSB
Question 73
The number of min-terms after minimizing the following Boolean expression is _________.
      [D′ + AB′ + A′C + AC′D + A′C′D]′
A
1
B
2
C
3
D
4
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Question 73 Explanation: 
Given Boolean expression is: 
              [D′ + AB′ + A′C + AC′D + A′C′D]′

Step 1 : [D′ + AB′ + A′C + C′D ( A + A')]′ 
( taking C'D as common )

Step 2 : [D′ + AB′ + A′C + C′D]′ 
( as, A + A' = 1 )

: [D' + DC' + AB' + A'C]' (Rearrange)

Step 3 : [D' + C' + AB' + A'C]' 
( Rule of Duality, A + A'B = A + B )

: [D' + C' + CA' + AB']' (Rearrange)

Step 4 : [D' + C' + A' + AB']' 
(Rule of Duality)

: [D' + C' + A' + AB']' (Rearrange)

Step 5 : [D' + C' + A' + B']' 
(Rule of Duality)

:[( D' + C' )'.( A' + B')'] 
(Demorgan's law, (A + B)'=(A'. B'))

:[(D''.C'').( A''.B'')] (Demorgan's law)

:[(D.C).(A.B)] (Idempotent law, A'' = A)

: ABCD

Hence only 1 minterm after minimization. 
Question 74
A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate. The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit ripple-carry binary adder is implemented by using full adders. The total propagation time of this 4-bit binary adder in microseconds is
A
19.2 microseconds
Digital Logic & Number representation    GATE-CS-2015 (Set 2)    
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Question 74 Explanation: 
A Ripple Carry Adder allows to add two n-bit numbers. It uses half and full adders. Following diagram shows a ripple adder using full adders.
rca-4bit_png
Let us first calculate propagation delay of a single
1 bit full adder.

Propagation Delay by n bit full adder is (2n + 2) 
gate delays.
[See this for formula].

Here n = 1, so total delay of a 1 bit full adder 
is (2 + 2)*1.2 = 4.8 ms

Delay of 4 full adders is = 4 * 4.8 = 19.2 ms
Question 75
The total number of prime implicants of the function f(w, x, y, z) = Σ(0, 2, 4, 5, 6, 10) is ________.
A
2
B
3
C
4
D
5
Digital Logic & Number representation    GATE-CS-2015 (Set 3)    
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Question 75 Explanation: 
ans_751 Red,Blue and Green together make total three prime implicant.
As we know "A prime implicant of a function is an implicant that cannot be covered by a more general (more reduced - meaning with fewer literals) implicant. (Wikipedia)" we've to see only prime implicants . We can make group of four 1's as in the figure(green group) , a group of two 1's in blue and red .As we can't reduce it further so this is minimal representation of problem and hence total number of prime implicant =3. So answer (B) part.
Question 76
Given the function F = P′ + QR, where F is a function in three Boolean variables P, Q and R and P′ = !P, consider the following statements.
  S1: F = Σ (4, 5, 6)
  S2: F = Σ (0, 1, 2, 3, 7)
  S3: F = Π (4, 5, 6)
  S4: F = Π (0, 1, 2, 3, 7) 
Which of the following is true?
A
S1-False, S2-True, S3-True, S4-False
B
S1-True, S2-False, S3-False, S4-True
C
S1-False, S2-False, S3-True, S4-True
D
S1-True, S2-True, S3-False, S4-False
Digital Logic & Number representation    GATE-CS-2015 (Set 3)    
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Question 76 Explanation: 
After drawing K map of F = P` + QR ,we can find out S2 and S3 are TRUE. Alternate Explanation :
P Q R = (!P + QR)

0 => 0 0 0 => 1+0.0 =1 =>Σ

1 => 0 0 1 => 1+0.1 =1 =>Σ

2 => 0 1 0 => 1+1.0 =1 =>Σ

3 => 0 1 1 => 1+1.1 =1 =>Σ

4 => 1 0 0 => 0+0.0=0 =>Π

5 => 1 0 1 => 0+0.1=0 =>Π

6 => 1 1 0 => 0+1.0=0 =>Π

7 => 1 1 1 => 0+1.1=1 =>Σ

as sigma means 1 and pi means 0
therefore Σ = (0,1,2,3,7)
Π = (4,5,6)
Question 77
What is the minimum number of NAND gates required to implement a 2-input EXCLUSIVE-OR function without using any other logic gate?
A
3
B
4
C
5
D
6
Digital Logic & Number representation    GATE-IT-2004    
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Question 77 Explanation: 
itq Pic Courtesy: http://www.electronics-tutorials.ws/logic/logic_7.html Other way around: x XOR y = x’y+xy’ = x’y+xy’+xx’+yy’ = (x+y) (x’+y’) Using NAND gates F= (x+y)(xy)’ = x. (xy)’ + y. (xy)’ Taking compliment F’= ( x. (xy)’ + y. (xy)’ )’ = (x. (xy)’)’. (y. (xy)’) Compliment again F=( (x. (xy)’)’. (y. (xy)’) )’ So Answer is B
Question 78
Using a 4-bit 2’s complement arithmetic, which of the following additions will result in an overflow? (i) 1100 + 1100 (ii) 0011 + 0111 (iii) 1111 + 0111
A
(i) only
B
(ii) only
C
(iii) only
D
(i) and (iii) only
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Question 78 Explanation: 
When two signed 2's complement numbers are added, overflow is detected if:
  • both operands are positive and the result is negative, or
  • both operands are negative and the result is positive
When two unsigned numbers are added, overflow occurs if
  • there is a carry out of the leftmost bit
With these conditions only (ii) will get an overflow Therefore, B is the answer  
Question 79
The number (123456)8 is equivalent to        
A
(A72E)16 and (22130232)4
B
(A72E)16 and (22131122)4
C
(A73E)16 and (22130232)4
D
(A62E)16 and (22120232)4
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Question 79 Explanation: 
(123456)8 =(001 010 011 100 101 110)2 To convert it into hexadecimal,make groups of 4 (1010 0111 0010 1110) =(A72E)16 To convert it into base 4,make groups of 2 (10 10 01 11 00 10 11 10) =(22130232)4  
Question 80
The function AB’C + A’BC + ABC’ + A’B’C + AB’C’ is equivalent to
A
AC’+AB+A’C
B
AB’+AC’+A’C
C
A’B+AC’+AB’
D
A’B+AC+AB’
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Question 80 Explanation: 

(A'BC + A'B'C) + (ABC' + AB'C') + AB'C A'C(B + B') + AC'(B + B') + AB'C A'C * 1 + AC' * 1 + AB'C A'C + AC' + AB'C A'C + A(C' + B'C) A'C + A(C' + B') A'C + AC' + AB'
 
Thus, option (B) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 81
Which of the following expressions is equivalent to (A⊕B)⊕C
A
(A+B+C)(A¯+B¯+C¯)
B
(A+B+C)(A¯+B¯+C)
C
ABC+A¯(B⊕C)+B¯(A⊕C)
D
None
Digital Logic & Number representation    Gate IT 2005    
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Question 81 Explanation: 
(A ⊕ B) ⊕ C

By Solving, We get
= (A ⊕ B)′ C + (A ⊕ B) C′

The above expression can be written as:
= (A ⊙ B) C + (A ⊕ B) C′

Now,

= (AB + A′B′) C + (A ⊕ B) C′

= ABC + A′B′C + AB′C′ + A′BC′	     [As: X + X = X]				     
			         [So, A′B′C + A′B′C = A′B′C]
= ABC + (A′B′C + A′B′C) + AB′C′ + A′BC′

This can be written as:

= ABC + A′ (B ⊕ C) + B′ (A ⊕ C)

This explanation has been provided by Saksham Seth.
Question 82
Using Booth's Algorithm for multiplication, the multiplier -57 will be recoded as
A
0 -1 0 0 1 0 0 -1
B
1 1 0 0 0 1 1 1
C
0 -1 0 0 1 0 0 0 (A+B+C)(A¯+B¯+C) ABC+A¯(B⊕C)+B¯(A⊕C)
D
0 1 0 0 -1 0 0 1
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Question 83
How many pulses are needed to change the contents of a 8-bit up counter from 10101100 to 00100111 (rightmost bit is the LSB)?
A
134
B
133
C
124
D
123
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Question 83 Explanation: 
8 bit Counter range 0-255 To go from 10101100 (172) to 00100111 (39)
  • first counter will move from 172 to 255(255-172=83)
  • 255 to 0=1 pulse
  • and then  0 to 39(39-0=39).
Total=83+1+39=123 So answer is D  
Question 84
Which of the following input sequences will always generate a 1 at the output z at the end of the third cycle?d1
A
0 0 0 1 0 1 1 1 1
B
1 0 1 1 1 0 1 1 1
C
0 1 1 1 0 1 1 1 1
D
0 0 1 1 1 0 1 1 1
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Question 84 Explanation: 
Let's take cycles as 1,2,3.

There are two flip-flops, let's call them D1 and D2.
And Q1, Q2 to be the output for flip-flop D1, D2 respectively.
here Q11 means output of D1 flip-flop for 1st cycle, similarly Q12 
is the output of D1 flip-flop for 2nd Cycle.
A1, A2, A3 means input for A at cycles 1, 2, 3 respectively. Similarly for B and C.

Let's check for the option D where:

A1=0 A2=1 A3=1
B1=0 B2=1 B3=1
C1=1 C2=0 C3=1

Q11=0 
Q12 = A1.B1 = 0.0 = 0 
Q13 = A2.B2 = 1.1 = 1 
~Q11=1
~Q12 = ~(A1.B1) = 1 
~Q13 = ~(A2.B2) = 0 
Q21 = 0 
Q22 = C1.(~Q11) = 1.1 = 1 
Q23 = C2.(~Q12) = 0.1 = 0

Z1 = 0 
Z2 = Q12.Q21 = 0.0 = 0 
Z3 = Q13.Q22 = 1.1 = 1 

Hence, sequence given in option D is generating a 1 at the output z at the 
end of the third cycle (Z3).
This explanation has been contributed by Harshit Sidhwa.
Question 85
The circuit shown below implements a 2-input NOR gate using two 2-4 MUX (control signal 1 selects the upper input). What are the values of signals x, y and z?di
A
1, 0, B
B
1, 0, A
C
0, 1, B
D
0, 1, A
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Question 85 Explanation: 

In MUX1, equation is : g = Az + Bz’ In MUX2, equation is : f = xg + yg’ = x(Az + Bz’) + y (Az + Bz’)’
Function ‘f’ should be equal to (A + B)’ .
Using hit and try method, put x = 0, y = 1 and z = A. f = 0(AA + BA’) + 1(AA + BA’)’ f = (A + A’B)’ f = (A + B)’
 
Thus, option (D) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 86
We want to design a synchronous counter that counts the sequence 0-1-0-2-0-3 and then repeats. The minimum number of J-K flip-flops required to implement this counter is   Note : This question was asked as Numerical Answer Type.
A
1
B
2
C
4
D
5
Digital Logic & Number representation    GATE-CS-2016 (Set 1)    
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Question 86 Explanation: 
Total 4. 2 J-K flip- flops for synchronous counter + 2 J-K flip-flop to make 2 bit counter. Actually, when we are repeating again then after 3 we don't know that our Synchronous counter will go to which zero. To make it work right, we need to move it to 1st zero after 3 2nd zero after 1 3rd zero after 2 i.e. 0 -> 1 -> 0 -> 2 -> 0 -> 3 (from here it again go to 1st zero). In order to decide which zero to move on we use counter from 1 to 3, I have attached truth table of normal 2 bit synchronous counter using JK flip flop. bit-counter // This Explanation has been contributed by Mohit Gupta
Question 87
Consider the two cascaded 2-to-1 multiplexers as shown in the figure. GT20161 The minimal sum of products form of the output X is gt162
A
A
B
B
C
C
D
D
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Question 87 Explanation: 
Assume P=0 and Q=0 The output should be R' Only option D satisfies this case.   Hence, D is the correct choice.
Question 88
Consider a carry lookahead adder for adding two n-bit integers, built using gates of fan-in at most two. The time to perform addition using this adder is
A
Θ(1)
B
Θ(Log (n))
C
Θ(√ n)
D
Θ(n)
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Question 88 Explanation: 
Look ahead carry generator gives output in constant time if fan-in = number of inputs. For Example:
It will take O(1) to calculate 
c4 = g3 + p3g2 + p3p2g1 + p3p2p1g0 + p3p2p1p0c0c4 
   = g3 + p3g2 + p3p2g1 + p3p2p1g0 + p3p2p1p0c0, 
              if OR gate with 5 inputs is present.
And, if fan-in != number of inputs then we will have delay in each level, as given below. If we have 8 inputs, and OR gate with 2 inputs, to build an OR gate with 8 inputs, we will need 4 gates in level-1, 2 in level-2 and 1 in level-3. Hence 3 gate delays, for each level. Similarly an n-input gate constructed with 2-input gates, total delay will be O(log n). // This Explanation has been provided by Saksham Raj Seth.
Question 89
Consider an eight-bit ripple-carry adder for computing the sum of A and B, where A and B are integers represented in 2’s complement form. If the decimal value of A is one, the decimal value of B that leads to the longest latency for the sum to stabilize is _____________ [This Question was originally a Fill-in-the-blanks Question]
A
-1
B
2
C
1
D
-2
Digital Logic & Number representation    GATE-CS-2016 (Set 2)    
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Question 89 Explanation: 
Here "longest latency for the sum to stabilize" means maximum delay that 
ripple carry adder would take to add A and B, we are given value of A and 
need to find the value of B.
The Delay in Ripple Carry Adder is as follows 
- For sum there are 2 XOR gates.
- For carry there is 1 XOR,1 AND and 1 OR gate.
i.e total 3 gate delays in case of carry and 2 gate delays in sum.
If we do 2's complement of 1 in 8 bit we get "00000001".
same we do for each option 
-1 : "11111111"
2 : "00000010"
1 : "00000001"
-2 : "11111110"
So in case of -1 the carry bit will change and thus it will take 1 extra gate delay, hence we could see that the maximum delay we could get when input at B will be -1, i.e. add "00000001" with "11111111" and would get Maximum delay. This explanation has been provided by Harshit Sidhwa.
Question 91
Let X be the number of distinct 16-bit integers in 2’s complement representation. Let Y be the number of distinct 16-bit integers in sign magnitude representation. Then X −Y is _________ [This Question was originally a Fill-in-the-blanks Question]
A
1
B
2
C
3
D
0
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Question 91 Explanation: 
For n bits, Distinct values represented in 2's complement is -2^n-1 to 2^n-1 -1 Distinct values represented in Signed Magnitude is -(2^(n-1) -1) to 2^(n-1) -1 For example if n = 8, we can represent numbers from -128 to 127 in 2's complement representation and numbers from -127 to 127 in signed magnitude representation. Difference is 1. The difference of 1 is there because there are two different representations of +0 and -0 in signed magnitude representation. But in 2's complement representation, there is one representation of 0.
Question 92
The addition of 4-bit, two's complement, binary numbers 1101 and 0100 results in  
A
0001 and an overflow
B
1001 and no overflow
C
0001 and no overflow
D
1001 and an overflow
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Question 92 Explanation: 
Its  -3+4=1, so no overflow So Answer is C.
Question 93
Which of the following input sequences for a cross-coupled R-S flip-flop realized with two NAND gates may lead to an oscillation ?
A
11, 00
B
01, 10
C
10, 01
D
00, 11
Digital Logic & Number representation    Gate IT 2007    
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Question 93 Explanation: 
RS flip flop using NAND gates. So, 00 input causes indeterminate state which MAY lead to oscillation.
Question 94
The following circuit implements a two-input AND gate using two 2-1 multiplexers.

What are the values of X1, X2, X3?
A
X1=b, X2=0, X3=a
B
X1=b, X2=1, X3=b
C
X1=a, X2=b, X3=1
D
X1=a, X2=0, X3=b
Digital Logic & Number representation    Gate IT 2007    
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Question 94 Explanation: 
2007_8_2
F = (bX1' + aX1)X3 + X2X3'
By putting following values: X1 = b, X2 = 0, X3 = a
We will get F = ab.
Question 95
The following bit pattern represents a floating point number in IEEE 754 single precision format 1 10000011 101000000000000000000000 The value of the number in decimal form is
A
-10
B
-13
C
-26
D
None of these
Digital Logic & Number representation    Number Representation    Gate IT 2008    
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Question 95 Explanation: 
  To convert the floating point into decimal, we have 3 elements in a 32-bit floating point representation: i. Sign ii. Exponent iii. Mantissa Sign bit is the first bit of the binary representation. '1' implies negative number and '0' implies positive number Exponent is decided by the next 8 bits of binary representation. 131-127=4 Hence the exponent of 2 will be 4. i.e 24=16. 127 is the unique number for 32 bit floating point representation. It is known as bias. It is determined by 2k-1-1 where 'k' is the number of bits in exponent field. Thus bias = 3 for 8 bit conversion and 127 for 32 bit. (28-1-1 = 128-1=127) Mantissa is calculated from the remaining 24 bits of the binary representation. It consists of '1' and a fractional part which is determined by: The fractional part of mantissa is given by: 1*(1/2) + 0*(1/4) + 1*(1/8) + 0*(1/16) +.........=0.625 Thus the mantissa will be 1+0.625=1.625 The decimal number hence given as Sign*Exponent*Mantissa = (-1)*(16)*(1.625) = -26. Related : https://www.youtube.com/watch?v=03fhijH6e2w http://quiz.geeksforgeeks.org/number-representation/ This solution is contributed by Kriti Kushwaha.,
Question 96
What Boolean function does the circuit below realize?
2008_9
A
xz+x'z'
B
xz'+x'z
C
x'y'+yz
D
xy+y'z'
Digital Logic & Number representation    Gate IT 2008    
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Question 97
A processor that has carry, overflow and sign flag bits as part of its program status word (PSW) performs addition of the following two 2's complement numbers 01001101 and 11101001. After the execution of this addition operation, the status of the carry, overflow and sign flags, respectively will be:
A
1, 1, 0
B
1, 0, 0
C
0, 1, 0
D
1, 0, 1
Digital Logic & Number representation    Computer Organization and Architecture    Gate IT 2008    
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Question 97 Explanation: 
01001101
+11101001
------------
100110110
carry flag =1
overflow happens only when two same sign numbers are added and carry generated is different from both added numbers.
so, overflow flag = 0,
sign bit = 0
Question 98
Consider the following state diagram and its realization by a JK flip flop
2008_37
The combinational circuit generates J and K in terms of x, y and Q.
The Boolean expressions for J and K are :
A
(x⊕y)'and x'⊕y'
B
(x⊕y)'and x⊕y
C
x⊕y and (x⊕y)'
D
x⊕yand x⊕y
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Question 99
The Boolean function f implemented in the figure shown below, using two input multiplexers is: mock_1
A
AB’C + ABC’
B
A’B’C + A’BC’
C
A’BC + A’B’C’
D
ABC + AB’C’
Digital Logic & Number representation    GATE 2017 Mock    
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Question 99 Explanation: 
The output of a multiplexer (2X1) can be expressed in the form of: mock_16
f= S’. L1 + S. L2
For the above diagram the output from the first multiplexer would be B’C + BC’ which acts a select line for the second multiplexer. Hence,
f= (B’C + BC’)’A + (B’C + BC’).0
= ((B’C)’ . (BC’)’)A + 0
= (B+C’)(B’ + C)A
= (BC + B’C’) A
= ABC + AB’C’
Question 100
The complement of the function F = (A + B’)(C’ + D)(B’ + C) is:
A
A’B + CD’ + BC’
B
AB’ + C’D + B’C
C
AB’ + CD’ + BC
D
AB + BC + CD
Digital Logic & Number representation    GATE 2017 Mock    
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Question 100 Explanation: 
Using Demorgan’s Law (Break the line, change the sign)
Complement of
F = F’ = ((A + B’)(C’ + D)(B’ + C))’
F’ = ((A + B’))’ + ((C’ + D))’ + ((B’ + C))’
F’ = A’B + CD’ + BC’
Question 101
Which of the following is false:
A
Digital signature is used to verify that a message is authentic.
B
Digital certificate is issued by a third party.
C
Digital certificate ensures integrity of the message.
D
Digital signature ensures non-repudiation.
Digital Logic & Number representation    GATE 2017 Mock    
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Question 101 Explanation: 
Digital certificate only verifies the identity of the user for whom the digital certificate is issued.
There are 101 questions to complete.

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