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Directions to Solve
In the following questions choose the word which is the exact OPPOSITE of the given words.
ANS:C - Discomfort
No answer description is available.
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B Error: Rvalue required for t
C Error: invalid function show() call
D Error: invalid function display() call
ANS:A - Error: invalid function display() call
The call to show() is improper. This is not the way to pass variable argument list to a function.
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ANS:D - 16
Should it be done by 2 to the power 3?
Because input goes like 2 to the power 0 to 3 which gives us 4 columns of input.
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A Whenever a 0 is present at an input
B Only when all inputs = 0
C Whenever a 1 is present at an input
D Only when all inputs = 1
ANS:C - Whenever a 1 is present at an input
No answer description is available
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A 
B 
C 
D 
ANS:B -
When you find the maxterms corresponding to the question you will obtain ( 0,1,2,3,6) and when you do same for the possible answers.
A. (7,5,4)
B. (0,2,3)
C. (0,2,2)
D. (0,5,2)
The answer is B because it's the only one that satisfies the SOP values of the question.
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A A = 1, B = 0, C = 0, D = 0
B A = 1, B = 0, C = 1, D = 0
C A = 0, B = 1, C = 0, D = 0
D A = 1, B = 0, C = 1, D = 1
ANS:B - A = 1, B = 0, C = 1, D = 0
When you add a variable with 1 result will be equivalent to 1 as the Boolean logic A+1 = 1. Again, the result will be equivalent to variable when you'd add any variable with 0 according to Boolean 2nd logic A+0 = A (if A = 1, A + 0 = 1).
So it is proved that if we make any sum term equal to 0, value of every variable will must be 0 in that sum term. If we determine A = 0, it will be inverted into 1. But value 1 is not allowed for this term. So, the values will be A = 1, B = 0, C = 1 & D = 0.
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ANS:D - AB + CD
SOP is sum of minterms.here no of variable=4(A,B,C,D)
so AB+CD is not a SOP form.for example ABCD+ABCD' is a SOP form.which can be written as sum(14,15).because 14=1110; and 15=1111 in binary which makes the corresponding minterm logic high.
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A 
B 
C 
D 
ANS:A -
OR GATE OP=(A+B)
1ST AND OP=(A+B)C
2ND AND OP=[(A+B)C]D
3RD AND OP=[(A+B)C]D)E
ANS: [(A+B)C]D)E ....{AFTER SIMPLIFICATION}
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A 
B 
C 
D 
ANS:C -
3 main steps to solve such questions are:
1. Consider those outputs that have 1 in it.
2. In sop zero(0) is compliment of a variable eg:- A(bar) = 0
Take AND of a variable.
3. Add OR to the variables.
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A a NOR and an AND gate with inverted inputs
B a NAND and an OR gate with inverted inputs
C an AND and a NOR gate with inverted inputs
D a NOR and a NAND gate with inverted inputs
ANS:A - a NOR and an AND gate with inverted inputs
Since the LHS of the expression is the logical expression of two inputs NOR Gate and the RHS of the expression consists of two inverted inpuuts of A and B, that is two inverted input are ANDed to get the RHS of the expression and since according to Demorgan Laws, the complement of the sum of two or more variables are equivalent to the product of the complement of the individual variables, we can say that a NOR gate is equivalent to an AND gate with inverted inputs.
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