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Test: Boolean Operations - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test - Test: Boolean Operations

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Test: Boolean Operations - Question 1

Consider a carry look-ahead 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:

Detailed Solution for Test: Boolean Operations - Question 1

Look-ahead carry generator gives output in constant time if the fan in = number of inputs.  

For example, it will take  O(1) to calculate   , if the OR gate with 5 inputs is present.
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, the total delay will be O(log n).
Hence answer is option B.

Test: Boolean Operations - Question 2

Consider an eight-bit ripple-carry adder for computing the sum of A and B, where A and B are integers represented in '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 ___________

Detailed Solution for Test: Boolean Operations - Question 2

in case of -1  we get bit sequence 11111111 adding this we get a carry upto carry flag,so largest time to ripple !

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Test: Boolean Operations - Question 3

An N-bit carry lookahead adder, where N is a multiple of 4, employs ICs 74181 (4 bit ALU) and 74182 ( 4 bit carry lookahead generator).
The minimum addition time using the best architecture for this adder is

Detailed Solution for Test: Boolean Operations - Question 3

For N = 64 bits
Suppose you want to build a 64 bit adder then you need 16 4-bit ALU and 16 4-bit carry generator, at this point there will be 16 carries that will ripple through these 16 ALU modules, to speed up the adder we need to get rid of these 16 rippling carries, now we can again use 4 4-bit carry generator to generate these 16 carries, now we have only 4 carries to ripple through, again we can use the same trick to minimize the rippling of these 4 carries, we can use an additional 4-bit carry generator which will generate these carry and we are done :) there will be no more propagation of carry among the ALU modules.
So the we have used 3 level of 4-bit carry generator, and the time taken to add 64 bits will be proportional to 3 which is log464.
So in general to add N-bits it takes Log4N time.

Test: Boolean Operations - Question 4

The number of full and half-adders required to add 16-bit numbers is:

Detailed Solution for Test: Boolean Operations - Question 4

For LSB addition we do not need a full adder for addition of subsequent bits we need full adders since carry from previous addition has to be fed into the addition operation

Test: Boolean Operations - Question 5

A 4-bit carry look ahead 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.

Detailed Solution for Test: Boolean Operations - Question 5

It would take 6 time units.


XOR can be implemented in 2 levels; level-1 ANDs and Level-2 OR. Hence it would take 2 time units to calculate  Pi and Si 
The 4-bit addition will be calculated in 3 stages


Level-2 we get the carries by computing the disjunction (OR).
3. (2 time units) Finally we compute the Sum in 2 time units, as its an XOR operation.
Hence the total is 2 + 2 + 2 = 6 time units.

Test: Boolean Operations - Question 6

The maximum gate delay for any output to appear in an array multiplier for multiplying two-bit numbers is:

Detailed Solution for Test: Boolean Operations - Question 6

In a N X M array multiplier we have N * M AND gates and (M-1) N bit adders used.
Total delay in N X M (N>=M) array multiplier due to AND gate in partial products at all level is just 1 unit AND gate delay as the operation is done parallel wise at each step. Now delays at level 1 to (M-1) is = (M-1)*delay due to 1 unit of N bit adder.  Therefore the maximum gate delay is O(M) but here M=N therefore O(N).

Test: Boolean Operations - Question 7

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:

Detailed Solution for Test: Boolean Operations - Question 7

Take A = A1 A2 A3 A4
B =   B1 B2 B3 B4 

Now to multiply these two numbers.  
1 and gate require b1 multiply with a1 a2 a3 a4. 
1 and gate require b2 multiply with a1 a2 a3 a4. 
1 and gate require b3 multiply with a1 a2 a3 a4.
1 and gate require b4 multiply with a1 a2 a3 a4.  
 Now 3 or gate require.
Total 7 gates are required for 4 bit take n bit u find 2n-1.
So time complexity will be = ϴ(n)

Test: Boolean Operations - Question 8

are Boolean variables,  operator.

Which one of the following must always be TRUE?

Detailed Solution for Test: Boolean Operations - Question 8

Test: Boolean Operations - Question 9

The total number of Boolean functions which can be realised with four variables is:

Detailed Solution for Test: Boolean Operations - Question 9

A Boolean function of 4 variables is a function from a set 24 = 16 of elements (all combinations of 4 variables) to a set of 2  elements. So, number of such functions will be 216 = 65,536

Test: Boolean Operations - Question 10

The simultaneous equations on the Boolean variables x, y, z and w,

Lightbox

have the following solution for x, y, z and w, respectively:

Detailed Solution for Test: Boolean Operations - Question 10

Statement 1 :  x + y + z = 1
Till now, all the options are possible.

Statement 2 : x y = 0
Since LHS ≠ 0, B is not possible.

Statement 3 : x z + w = 1
Since LHS ≠ 1, A and D are not possible.

Statement 4 : x y + z’ w’ = 0
Thus, C is the correct option.

Test: Boolean Operations - Question 11

Consider the Boolean operator # with the following properties :

Detailed Solution for Test: Boolean Operations - Question 11

These are properties of XOR function. 

Test: Boolean Operations - Question 12

Simplify Y = AB’ + (A’ + B)C

Detailed Solution for Test: Boolean Operations - Question 12

Y = AB’ + (A’ + B)C = AB’ + (A’ + B)C = AB’ + (AB’)’C = AB’ + C.

Test: Boolean Operations - Question 13

Find values of Boolean variables A,B,C which satisfy the following equations:

  • A+ B = 1
  • AC = BC
  • A + C = 1
  • AB = 0
Detailed Solution for Test: Boolean Operations - Question 13

From A + B = 1 and AB = 0
We get either of A,b is 1 and another is 0.
Now AC = BC ,here C has to be 0 (because A,b  has different values)
C = 0
Now A + C = 1
So, A = 1 and Ab = 0, So,  B = 0
So we get
A = 1, B = 0 , C = 0
These are the values.

Test: Boolean Operations - Question 14

Complement of the expression A’B + CD’ is:

Detailed Solution for Test: Boolean Operations - Question 14

(A’B + CD’)’ = (A’B)'(CD’)’ = (A” + B’)(C’ + D”) = (A + B’)(C’ + D).

Test: Boolean Operations - Question 15

In boolean algebra, the OR operation is performed by which properties?

Detailed Solution for Test: Boolean Operations - Question 15

The expression for Associative property is given by A+(B+C) = (A+B)+C & A*(B*C) = (A*B)*C.
The expression for Commutative property is given by A+B = B+A & A*B = B*A.
The expression for Distributive property is given by A+BC=(A+B)(A+C) & A(B+C) = AB+AC.

Test: Boolean Operations - Question 16

Simplified expression for function

Detailed Solution for Test: Boolean Operations - Question 16

Test: Boolean Operations - Question 17

A Boolean function 

Detailed Solution for Test: Boolean Operations - Question 17

x'y' + x'y = x'(y+y') = x'
x' + xy = x' + y

Test: Boolean Operations - Question 18

The function AB’C + A’BC + ABC’ + A’B’C +  AB’C’ is equivalent to

Detailed Solution for Test: Boolean Operations - Question 18


So, the equivalent expression will be A'C + AC' + AB' 

Test: Boolean Operations - Question 19

Which of the following expressions is equivalent to 

Detailed Solution for Test: Boolean Operations - Question 19




Test: Boolean Operations - Question 20

Define the connective

 Consider the following expressions

Which of the following is TRUE?

Detailed Solution for Test: Boolean Operations - Question 20

P:
Y * Z = Y * (X*Y)
= Y * (XY + X'Y')
= Y (XY + X'Y') + Y' (XY + X'Y')'
= XY + Y' ((X' + Y') (X+ Y))
= XY + Y' (X'Y + XY')
= XY + XY'
= X(Y + Y')
= X
So, P is valid.
Q: 
X * Z = X * (X*Y)
= X * (XY + X'Y')
= X (XY + X'Y') + X' (XY + X'Y')'
= XY + X' ((X' + Y') (X+ Y))
= XY + X' (X'Y + XY')
= XY + X'Y
= Y(X + X')
= Y
So, Q is also valid.
R:
X * Y * Z = (X * Y) * (X * Y)
= (XY + X'Y') * (XY + X'Y')
= (XY + X'Y') (XY + X'Y') + (XY + X'Y')' (XY + X'Y')'
= (XY + X'Y') + (XY + X'Y')' (Since, AA = A)  = 1 (Since A + A' = 1)
So, R is also valid. 
Hence, D choice.

Test: Boolean Operations - Question 21

If P,Q,R are Boolean variables, then

Detailed Solution for Test: Boolean Operations - Question 21

Test: Boolean Operations - Question 22

In boolean algebra, the OR operation is performed by which properties?

Detailed Solution for Test: Boolean Operations - Question 22

The expression for Associative property is given by A + (B + C) = (A + B) + C & A * (B * C) = (A * B) * C. 
The expression for Commutative property is given by A + B = B + A & A * B = B * A. 
The expression for Distributive property is given by A + BC = (A + B)(A + C) & A(B + C) = AB + AC.

Test: Boolean Operations - Question 23

Which one of the following expressions does NOT represent exclusive NOR of x and y?

Detailed Solution for Test: Boolean Operations - Question 23

A: means both are either true OR both are false. then it will be true = ExNOR
B & C: Whenever any one of the literal is complemented then ExOR can be turned to ExNOR and complement sign on the literal can be removed. So these two also represents ExNOR operation of x and y.
Hence, the correct answer is option D it is the ExOR operation b/w the two.

Test: Boolean Operations - Question 24

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:
The equivalent expression for  is

The equivalent expression for F is

Detailed Solution for Test: Boolean Operations - Question 24

XOR is associative and commutative. Also,

Test: Boolean Operations - Question 25

If are Boolean variables, then which one of the following is INCORRECT?

Detailed Solution for Test: Boolean Operations - Question 25

Option C (wx'(y+xz')+w'.x')y = (wx'y+wx'xz'+w'x')y  =
xx' = 0.
So
Wx'y+w'x'y = x'y(w+w') = x'y

Test: Boolean Operations - Question 26

Choose the correct alternatives (more than one may be correct) and write the corresponding letters only: The operation which is commutative but not associative is:

Detailed Solution for Test: Boolean Operations - Question 26
  1. Every logic gate follows Commutative law.
  2. AND, OR, Ex-OR, EX-NOR follows Associative law also. NAND, NOR doesn’t follow Associative law.
Test: Boolean Operations - Question 27

Which of the following expressions is not equivalent

Detailed Solution for Test: Boolean Operations - Question 27

x OR 1 = 1 and so x NOR 1 = 0.

Test: Boolean Operations - Question 28

When multiplicand Y is multiplied by multiplierusing bit-pair recoding in Booth's algorithm, partial products are generated according to the following table.

The partial products for rows 5 and 8 are

Detailed Solution for Test: Boolean Operations - Question 28

Partial product is calculated by using bit pair recording in booths algorithm, which is improvement technique used in booths algorithm. Here we consider 3 bits at a time for getting the partial product. This eliminates the worst case behaviour of normal Booth's algorithm.
"100" corresponds to -2M and "111" corresponds to 0 -2 X(i+1) + x (i) + X(i-1)

Test: Boolean Operations - Question 29

Booth’s algorithm for integer multiplication gives worst performance when the multiplier pattern is

Detailed Solution for Test: Boolean Operations - Question 29

The worst case of an implementation using Booth’s algorithm is when pairs of 01s or 10s occur very frequently in the multiplier.

Test: Boolean Operations - Question 30

Booth's coding in 8 bits for the decimal number -57 is

Detailed Solution for Test: Boolean Operations - Question 30

There are 2 ways for answering This questions.
Way 1 -> Convert 57 to Binary & Get 2's complement. It is "11000111" & Attach one extra 0 to right of it
110001110

To calculate booth code subtract right digit from left digit in every consecutive 2 digits.
So 11-> 0 , 10 -> +1, .. finally 10 -> +1

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