in this post we are going to discuss cs101 assignment solution
Assignment No. 01 (Graded) Due Date : 21/11/20018
Semester Fall 2018 Total Marks : 20
CS101 Introduction to Computing
Objective
The objective of this assignment is:
This assignment has been designed to elaborate the following concepts:
 To understand Number System
 Number System conversion
 To develop a truth table for logic operation.
cs101 assignment solution
Question#1: [Marks 10]
Convert the following decimal numbers into equivalent binary numbers and then convert the binary answer back into equivalent decimal. Show all the steps of conversions.
 3242
Decimal to binary Conversion
Divider  Quotient  Reminder 
2  3242  0 
2  1621  1 
2  810  0 
2  405  1 
2  202  0 
2  101  1 
2  50  0 
2  25  1 
2  12  0 
2  6  0 
2  3  1 
2  1  1 
Binary to Decimal Conversion
Before Solving Question 2 you must have basic knowledge about
AND
OR
NOT
XOR OPERATOR
For your understanding I’m briefly explaining them (remember they are not part of assignment just for understanding )
AND operator
In AND operator if both values are 1 then answer will be 1 otherwise if any value or both values are 0 then ans will be 0.
AND operator symbol → A.B its means A AND B
OR Operator
In OR operator if one value is 1 then answer will be 1 and if all values are 0 then ans will be 0.
OR Operator Symbol → A+B it means A OR B
NOT Operator
Not operator is reverse operator
If value is 1 then ans will be 0 and if value is 0 then asn will be 1.
NOT Operator symbol → A^{/ }
XOR Operator
In XOR operator if both values are same like (1 ,1) or (0,0) then ans will be 0 and if values are not same then ans will be 1.
XOR operator Symbol → AÅB .
Question#2:
[Marks 10]
Solve the given Boolean Expression by using truth table.
(AÅB) . ((A^{/} + C^{/}) Å (A^{/}+B))
(Here . is for And Sign)
Sample Solution:
A  B  C  AÅB  A^{/}  C^{/}  (A^{/} + C^{/})

(A^{/}+B)  (A^{/} + C^{/}) Å (A^{/}+B)

(AÅB).((A^{/} + C^{/}) Å (A^{/}+B))

0  0  0  0  1  1  1  1  0  0 
0  0  1  0  1  0  1  1  0  0 
0  1  0  1  1  1  1  1  0  0 
0  1  1  1  1  0  1  1  0  0 
1  0  0  1  0  1  1  0  1  1 
1  0  1  1  0  0  0  0  0  0 
1  1  0  0  0  1  1  1  0  0 
1  1  1  0  0  0  0  1  1  0 
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