- The Boolean function [~ (~p˄q)˄~(~p˄~q)]˅(p˄r) is equal to the Boolean function:
(A) q (B) p˄r
(C) p˅q (D) p
Answer: D
- Let us assume that you construct ordered tree to represent the compound proposition (~(p˄q))↔(~p˅~q).
Then, the prefix expression and post-fix expression determined using this ordered tree are given as ……….. and …………. respectively.
(A) ↔~˄pq˅ ~~pq, pq˄~p~q~˅↔
(B) ↔~˄pq˅ ~p~q, pq˄~p~q~˅↔
(C) ↔~˄pq˅ ~~pq, pq˄~p~ ~q˅↔
(D) ↔~˄pq˅ ~p~q, pq˄~p~~q˅↔
Answer: B
3 Let A and B be sets in a finite universal set U. Given the following:
|A – B|, |AÅB|, |A|+|B| and |AÈB|
Which of the following is in order of increasing size ?
(A) |A – B| ≤ |AÅB| ≤ |A| + |B| ≤ |AÈB|
(B) |AÅB| ≤ |A – B| ≤ |AÈB| ≤ |A| + |B|
(C) |AÅB| ≤ |A| + |B| ≤ |A – B| ≤ |AÈB|
(D) |A – B| ≤ |AÅB| ≤ |AÈB| ≤ |A| + |B|
Answer: D
- What is the probability that a randomly selected bit string of length 10 is a palindrome?
(A) 1/64 (B) 1/32
(C) 1/8 (D) ¼
Answer: B
Explanation:
select 10 bits and with every bit we have two choice either 0 or 1 so the total no of 10 length bit strings are 2^10
now in palindrome if we chose first 5 bits then our job is done as next 5 are fixed ( first 5 in reverse order).
So, for five bits can be chosen in 25 (for every bit either 0 or 1).
Probability = 2^5/2^10= 1/ 32
- Given the following graphs :
Which of the following is correct?
(A) G1 contains Euler circuit and G2 does not contain Euler circuit.
(B) G1 does not contain Euler circuit and G2 contains Euler circuit.
(C) Both G1 and G2 do not contain Euler circuit.
(D) Both G1 and G2 contain Euler circuit.
Answer: C
Explanation:
- The octal number 326.4 is equivalent to
(A) (214.2)10 and (D6.8)16 (B) (212.5)10 and (D6.8)16
(C) (214.5)10 and (D6.8)16 (D) (214.5)10 and (D6.4)16
Answer: C
Explanation:
(326.4)8 = 82 x 3 + 81 x 2 + 80 x 6 . 8-1 x 4 = ( 214.5)10
another way convert into decimal
To convert in hexa decimal i assume you already familiar with the conventional way there is also a shortcut which is only applicable if one base can be written in the power of another base eg.( r1)m = r2
(326.4)8 = (011010110.100)2 here we expanded every digit in three bits
| 3 | 2 | 6 | . | 4 |
| 011 | 010 | 110 | . | 100 |
(011010110.100)2 = (D6.8)16 here we grouped every 4 bits
| 0000 | 1101 | 0110 | . | 1000 |
| 0 | D | 6 | . | 8 |
- Which of the following is the most efficient to perform arithmetic operations on the numbers?
(A) Sign-magnitude (B) 1’s complement
(C) 2’s complement (D) 9’s complement
Answer: C
check video lecture on how to find sign magnitude,1’s compliment and 2’s compliment
- The Karnaugh map for a Boolean function is given as:
The simplified Boolean equation for the above Karnaugh Map is
(A) AB + CD + AB’ + AD (B) AB + AC + AD + BCD
(C) AB + AD + BC + ACD (D) AB + AC + BC + BCD
Answer: B
- Which of the following logic operations is performed by the following given combinational circuit ?
(A) EXCLUSIVE-OR (B) EXCLUSIVE-NOR
(C) NAND (D) NOR
Answer: A
Explanation:
(X+(X+Y)’)’ + (Y+(X+Y)’)’
Apply DE MORGAN theorem
X'(X+Y)+Y'(X+Y)
XX’+X’Y+Y’X+Y’Y
0+X’Y+Y’X+0
X XOR Y
- Match the following:
| List – I | List – II |
| 1. Controlled Inverter | i. a circuit that can add 3 bits |
| 2. Full adder | ii. a circuit that can add two binary numbers |
| 3. Half adder | iii. a circuit that transmits a binary word or its 1’s complement |
| 4. Binary adder | iv. a logic circuit that adds 2 bits |
Codes :
a b c d
(A) iii ii iv i
(B) ii iv i iii
(C) iii iv i ii
(D) iii i iv ii
Answer: D
Explanation:
Half adder-> Its is used to add 2 bits.
Full adder-> It is used to add 3 bits.
controlled inverter-> a circuit that transmits a binary word or its 1’s complement
Binary adder -> It is used to add 2 binary numbers