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GATE | GATE-CS-2004 | Question 90

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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

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                        

 

 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.


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Last Updated : 19 Nov, 2018
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