SOLVED CS101 Assignment 2 Solution and Discussion Fall2019

Assignment No. 02
Semester: Fall 2019
Introduction to Computing – CS101
Total Marks: 20
Due Date: 05122019
Please carefully read the following instructions before attempting assignment
Objective of Assignment
Objective of this assignment is to increase the learning capabilities of the students about:
• Decimal to Binary Conversion
• Boolean Logical Operations
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• The assignment is submitted before or on the due date.
• The submitted assignment file is not corrupted or damaged.
• The assignment is not copied (from another student or internet).
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Question No.1 8 Marks
Convert a decimal number 724 into its equivalent binary numbers. You are required to show the complete steps of conversion and also write the final answer in its equivalent binary number.
Question No.2 12 Marks
Complete the following truth table by using the stated Boolean logical operations.
A B C A + B A . C (A + B) ⊕ C
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1For any query, feel free to email at:
[email protected]Best of Luck

@zareen
Solution:
Question # 01 8 Marks
Convert a decimal number 724 into its equivalent binary numbers. You are required to show the complete steps of conversion and also write the final answer in its equivalent binary numbers.2 724 2 362 0 2 181 0 2 90 1 2 45 0 2 22 1 2 11 0 2 5 1 2 2 1 1 0 Answer: (724)2 =1011010100
Question # 02 12 Marks
Complete the following truth table by using the stated Boolean logical operations.A B C A + B A . C (A + B) ⊕ C 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 1 0 1 0 1 1 1 1 1 1 0 
@zareen
Solution:
Question # 01 8 Marks
Convert a decimal number 724 into its equivalent binary numbers. You are required to show the complete steps of conversion and also write the final answer in its equivalent binary numbers.2 724 2 362 0 2 181 0 2 90 1 2 45 0 2 22 1 2 11 0 2 5 1 2 2 1 1 0 Answer: (724)2 =1011010100
Question # 02 12 Marks
Complete the following truth table by using the stated Boolean logical operations.A B C A + B A . C (A + B) ⊕ C 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 1 0 1 0 1 1 1 1 1 1 0 
