Skip to content
Related Articles

Related Articles

Improve Article
GATE | GATE-CS-2004 | Question 90
  • Last Updated : 19 Nov, 2018

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


Answer: (B)

Explanation: We should get output 1 for values>=5

Making truth table for problem

ABCDOp
00000
00010
00100
00110
01000
01011
01101
01111
10001
10011
1010X
1011X
1100X
1101X
1110X
1111X

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.

Quiz of this Question

Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :