Digital-GrandTest1-Q15

Given two binary numbers M of 5 bits and N of 4 bits, an array mul-

tiplier takes these two numbers and multiplies M to N where M is the

multiplicand and N is the multiplier. If number of AND gates required is

P and number of one-bit adders required is R then twice of the value of

expression  is:

(A) 138

(B) 276

(C) 350

(D) 175 asked Jun 19 in Digital
reshown Jun 20

+1 vote

Ans: C

Number of AND gate required   5*4=20

And number of 1 bit adder required  5*(4-1) =15

2*(P^2-R^2)=350 answered Jun 20 by (1,800 points)
selected Jun 21 by gbmentor
please explain how you got 15 1-bit adders
+1 vote
Number of AND gate required   P= 5*4=20

And number of 1 bit adder required  R= 5*(4-1) =15

So 2(P^2-R^2)=350

Hence C will be the correct answer. answered Jun 20 by (340 points)
edited Jun 20
"twice"

So, u have to double the value.
thanks...............
Oh  u re correct... Ans is option C.  My mistake
can anyone explain how is the number of adders is 15? answered Jun 21 by (4,190 points)
+1 vote

Answer : Option C)->350

Solution:

Consider a 2 bit multiplier : Circuit: Similarly Here....

Number of AND gate required   5*4=20

And number of 1 bit adder required  5*(4-1) =15
Hence P=20, R=15.
2*(P^2-R^2)=350 answered Jun 21 by (73,870 points)
sir we don't have to consider the FA In the first Row?
Because some figures depict FA in first row and some depict HA in the first row
So we are solving this question based on considering HA in the first row of array multiplier.
Which one to consider in the examination?