GATE CSE 2004 | Question: 58

Question

A circuit outputs a digit in the form of $4$ bits. $0$ is represented by $0000, 1$ by $0001, \ldots, 9$ by $1001$. A combinational circuit is to be designed which takes these $4$ bits as input and outputs $1$ if the digit $\geq$ $5$, and $0$ otherwise. If only $\textsf{AND}, \textsf{OR}$ and $\textsf{NOT}$ gates may be used, what is the minimum number of gates required?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Comments

Prateek's answer is right, Since min no. of gates have been asked so we can assume 3-i/p gate is available and for the don't care part , he has taken 1s in place of don't care.

or A + BC + BD => A + B(C + D) only 2 i/p gates are required.

---

$f(A,B,C,D) = A+BD+BC = A+B(C+D)$

---

when the output of circuit is till 9 (1001). then why the combinational circuit is taking 16 combination of inputs?

means i know from 4 bits 2^4 =16 combination possible

Answer 1

answer - B

using K map simplified SOP AC' + B

values 10-15 are treated as dont care

Comments

Please show the k-map for this solution

---

How are you getting AC' + B ? A+ BD +BC is correct !