edited by
1,130 views
17 votes
17 votes

How many minimum number of two input AND gates and two input OR gates are required to realize $Y = BD+CE+AB$ ?

  1. $2, 2$
  2. $4, 2$
  3. $3, 2$
  4. $2, 3$
edited by

2 Answers

Best answer
5 votes
5 votes

Given : Y = BD+CE+AB here 3 and gate and 2 or gate [ we can minimize 1 AND gate]

Y = B(D + A) + CE

Now we see that 2 AND Gate i.e.1st B & (D+A), 2nd C & E..

selected by
0 votes
0 votes
BD+AB+CE

= B(D+A)+CE

 1 or gate for X= D+A

1 and gate for  Y=B.(D+A)

1 and gate for  Z=CE

1 or gate for Y and Z

hence 2   AND gate and 2 OR gate
Answer:

Related questions

2 votes
2 votes
1 answer
1
Bikram asked Nov 26, 2016
929 views
The excess $3$ code of decimal number $26$ is:$0100 1001$$01011001$$1000 1001$$01001101$
3 votes
3 votes
1 answer
2
Bikram asked Nov 26, 2016
272 views
$1$’s complement representation of decimal number of -$17$ by using $8$ bit representation is:$1110 \ 1110$$1101 \ 1101$$1100 \ 1100$$0001 \ 0001$
3 votes
3 votes
1 answer
3
Bikram asked Nov 26, 2016
2,858 views
If signed numbers are used in binary arithmetic, then which one of the following notations would have unique representation for zero?Sign-magnitude$1$’s complement$2$�...