CS302 Assignment 1 Solution Fall 2018
CS302 Assignment 1 Solution Fall 2018

CS302 Assignment

Question no. 1:

Perform the Following Operations with
complete steps.

1. (725157)0+ (F12ADE) H
— (101101010101011111001110101)2 — (9925)0

 Answer.

(725157)0

Covert this octal numb into decimal number

 (725157)0 = 7 x 85+2 x 84+5
x 83+1 x 82+5 x 81+7 x 80

(725157)0
=229376+8192+2560+64+40+7

(725157)0
= (240238) D

(F12ADE)H

Convert this hex
number into decimal number

(F12ADE)H=F
x 165+1 x 164+2 x 163+A x 162+D x
161+E x 16°

(F12ADE)H=
(15) x 1048576+1 x 65536+2 x 4096+ (10) x 256+ (13) x 164(14) x 1

(F12ADE)H
=15728640+65536+8192+2560+208+14

 (F12ADE)H= (15805150) D

(101101010101011111001110101)2

Convert binary into decimal number


















First we converting it into Octal Number
System.

= (552537165)0

Now converting it into Decimal Numbers
System.

(552537165)0
= 5 x 88+5 x 87+2 x 86+5 x 85+3 x 84+7
x 83+1 x 82+6 x 81+5 x 8°

(552537165)0
= 83886080+10485760+524288+163840+12288+3584+64+48+5 (552537165)0= (95075957)
D

Now add all these four values

= (240239) D-(15805150)
D
-(95075957) D-(9925) D

=-(79040493) D Answer

Question no. 2:

Step l: Convert the Hexadecimal
number to binary by replacing the each hexadecimal symbol with appropriate four
bits binary number. So we get.

Hexadecimal

E

A

C

1

2

9

number Binary

1110

1010

1100

0001

0010

1001

 Step 2: Now convert the each binary number to gray code by
the grayest digit is sane as the left most binary digit and then add the
adjacent pair of binary digits to find the next gray code.

number Binary

1110

1010

1100

0001

0010

1001

Gray Code

1001

1111

1010

0001

0011

1101

 FOR FULL QUESTION DOWNLOAD THIS FILE




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