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Directions to Solve

In the following questions choose the word which is the exact OPPOSITE of the given words.

Q1: DEAR

A Cheap

B Free

C Worthless

D Priceless

Q2:

Point out the error in the following program.

#include<stdio.h>
#include<stdarg.h>
void varfun(int n, ...);

int main()
{
    varfun(3, 7, -11.2, 0.66);
    return 0;
}
void varfun(int n, ...)
{
    float *ptr;
    int num;
    va_start(ptr, n);
    num = va_arg(ptr, int);
    printf('%d', num);
}

A Error: invalid access to list member

B Error: ptr must be type of va_list

C Error: too many parameters

D No error

Q3: The systematic reduction of logic circuits is accomplished by:

A using Boolean algebra

B symbolic reduction

C TTL logic

D using a truth table

Q4: Which output expression might indicate a product-of-sums circuit construction?

Q5: An AND gate with schematic "bubbles" on its inputs performs the same function as a(n)________ gate.

A NOT

B OR

C NOR

D NAND

Q6: For the SOP expression , how many 1s are in the truth table's output column?

A 1

B 2

C 3

D 5

Q7: A truth table for the SOP expression has how many input combinations?

A 1

B 2

C 4

D 8

Q8: How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z)

A 1

B 2

C 4

D 5

Q9: Determine the values of A, B, C, and D that make the product term equal to 1.

A A = 0, B = 1, C = 0, D = 1

B A = 0, B = 0, C = 0, D = 1

C A = 1, B = 1, C = 1, D = 1

D A = 0, B = 0, C = 1, D = 0

Q10: What is the primary motivation for using Boolean algebra to simplify logic expressions?

A It may make it easier to understand the overall function of the circuit.

B It may reduce the number of gates.

C It may reduce the number of inputs required.

D all of the above

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Q1: DEAR

Q2:
Point out the error in the following program.
`#include<stdio.h> #include<stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11.2, 0.66); return 0; } void varfun(int n, ...) { float *ptr; int num; va_start(ptr, n); num = va_arg(ptr, int); printf('%d', num); }`

Q3: The systematic reduction of logic circuits is accomplished by:

Q4: Which output expression might indicate a product-of-sums circuit construction?

Q5: An AND gate with schematic "bubbles" on its inputs performs the same function as a(n)________ gate.

Q6: For the SOP expression , how many 1s are in the truth table's output column?

Q7: A truth table for the SOP expression has how many input combinations?

Q8: How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z)

Q9: Determine the values of A, B, C, and D that make the product term equal to 1.

Q10: What is the primary motivation for using Boolean algebra to simplify logic expressions?

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Q1: DEAR

Solution

Q2: Point out the error in the following program. #include<stdio.h> #include<stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11.2, 0.66); return 0; } void varfun(int n, ...) { float *ptr; int num; va_start(ptr, n); num = va_arg(ptr, int); printf('%d', num); }

Solution

Q3: The systematic reduction of logic circuits is accomplished by:

Solution

Q4: Which output expression might indicate a product-of-sums circuit construction?

Solution

Q5: An AND gate with schematic "bubbles" on its inputs performs the same function as a(n)________ gate.

Solution

Q6: For the SOP expression , how many 1s are in the truth table's output column?

Solution

Q7: A truth table for the SOP expression has how many input combinations?

Solution

Q8: How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z)

Solution

Q9: Determine the values of A, B, C, and D that make the product term equal to 1.

Solution

Q10: What is the primary motivation for using Boolean algebra to simplify logic expressions?

Solution

Q2:
Point out the error in the following program.
`#include<stdio.h> #include<stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11.2, 0.66); return 0; } void varfun(int n, ...) { float *ptr; int num; va_start(ptr, n); num = va_arg(ptr, int); printf('%d', num); }`