Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

Answer

List the characters (digits) for the following bases.

1) Decimal: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

2) Octal: 0, 1, 2, 3, 4, 5, 6, 7

3) Binary: 0, 1

4) Hexadecimal: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

What base(s) could these numbers represent? (circle those that apply)

5) 4 3 6 7 5 Decimal Octal Binary Hexadecimal

6) 1 7 C 2 D Decimal Octal Binary Hexadecimal

7) 1 0 1 1 1 0 0 1 0 Decimal Octal Binary Hexadecimal

8) 6 1 7 2 8 5 Decimal Octal Binary Hexadecimal

Add these decimal numbers. (show work)

13) 1 7 9 4 8 2 7 3-------------

1 8 2 2 1

14) 1 5 6 8 3424 6 7 5----------6 5 8 5

What base(s) do these numbers represent? (circle those that apply)

9) 4 3 6 9 510 Decimal Octal Binary Hexadecimal

10) 1 7 C 2 D16 Decimal Octal Binary Hexadecimal

11) 1 0 1 1 1 0 0 1 02 Decimal Octal Binary Hexadecimal

12) 6 1 7 2 4 58 Decimal Octal Binary Hexadecimal

40

40

40

40

40

40

40

40

10

10

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerAdd these binary numbers. (show work)

15) 1 0 1 1 0 1 1 0 1 1 -------------

1 1 0 1 1 0 1 1 1 0 1 1 1--------------- ---- 0 10 0 Carry 1

1 1 1 1 0 1 1 0 1 0 1 0 1 1 1--------------- ---- 0 0 10 0 Carry 1

1 1 1 1 1 0 1 1 0 1 1 1 0 1 1 0--------------- ---- 0 0 0 10 0 Carry 1

1 1 1 1 1 1 0 1 1 0 1 1 1 0 1 1 1--------------- ---- 1 0 0 0 11 1 Carry 1

11---10 10

1---11 1 Carry 1

1 1 1 1 1 1 0 1 1 0 1 0 1 0 1 1 --------------- ---- 1 1 0 0 0 1 No Carry

1 1 1 1 1 0 1 1 0 1 1 1 0 1 1 --------------- ---- 1 1 1 0 0 0 1 No Carry

1 2

3

4

5 6

10

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerAdd these octal numbers.

16) 2 6 6 75 3 4 1----------

12 6 6 75 3 4 1--------- 0

1 12 6 6 7 6 5 3 4 1 4 --------- --- 0 13

1 1 12 6 6 7 6 5 3 4 1 3 --------- --- 3 0 12

1 1 1 12 6 6 7 2 5 3 4 1 5 --------- --- 2 3 0 10

1 1 1 1 2 6 6 7 5 3 4 1 ------------1 0 2 3 0

1 2 3

4 5

0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 etc.

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerAdd these hexadecimal numbers.

17) 2 A 3 45 3 F 6----------

2 A 3 4 45 3 F 6 6---------- --

A

0 1 2 3 4 5 6 7 8 9 A B C D E F

0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 etc.

2 A 3 4 35 3 F 6 F---------- -- A 12

1 2 A 3 4 35 3 F 6 F---------- -- 2 A 12

1 12 A 3 4 A5 3 F 6 3---------- -- 2 A E

1 2 A 3 4 25 3 F 6 5---------- -- E 2 A 7

1 2 A 3 45 3 F 6 ---------- 7 E 2 A

1 2 3

4 65

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerConvert this decimal number to binary.

18)

4 3 2 7 7

0 1 0 1 0 1 0 0 1 0 0 0 0 1 1 0 11481632

64

2128

256

512

1024

2048

4096

8192

16384

32768

65536

4 3 2 7 7 (remaining number)- 3 2 7 6 8 (largest number not exceeding remainder - 43277) ------------ (mark this bit as ‘on’ = ‘1’) 1 0 5 0 9 (new remaining number)

1 0 5 0 9 (remaining number) - 8 1 9 2 (largest number not exceeding remainder - 10509) ------------ (mark this bit as ‘on’ = ‘1’) 2 3 1 7 (new remaining number)

Skip 16384 because it is too large to go into 1 0 5 0 9 (remaining number)Mark position as ‘off’ = ‘0’

Skip 4096 because it is too large to go into 2317 (remaining number)Mark position as ‘off’ = ‘0’

2 3 1 7 (remaining number) - 2 0 4 8 (largest number not exceeding remainder - 2317) ------------ (mark this bit as ‘on’ = ‘1’) 2 6 9 (new remaining number)

Skip 1024 because it is too large to go into 269 (remaining number)Mark position as ‘off’ = ‘0’

Skip 512 because it is too large to go into 269 (remaining number)Mark position as ‘off’ = ‘0’

Positional weight values

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

Answer 2 6 9 (remaining number) - 2 5 6 (largest number not exceeding remainder - 269) ------------ (mark this bit as ‘on’ = ‘1’) 1 3 (new remaining number)

Skip 128 because it is too large to go into 13 (remaining number)Mark position as ‘off’ = ‘0’

Skip 64 because it is too large to go into 13 (remaining number)Mark position as ‘off’ = ‘0’

Skip 32 because it is too large to go into 13 (remaining number)Mark position as ‘off’ = ‘0’

Skip 16 because it is too large to go into 13 (remaining number)Mark position as ‘off’ = ‘0’

1 3 (remaining number) - 8 (largest number not exceeding remainder - 13) ------------ (mark this bit as ‘on’ = ‘1’) 5 (new remaining number)

5 (remaining number) - 4 (largest number not exceeding remainder - 5) ------------ (mark this bit as ‘on’ = ‘1’) 1 (new remaining number)

Skip 2 because it is too large to go into 1 (remaining number)Mark position as ‘off’ = ‘0’

1 (remaining number) - 1 (largest number not exceeding remainder - 1) ------------ (mark this bit as ‘on’ = ‘1’) 0 (new remaining number)

Process Completed

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerConvert this binary number to decimal.

19)

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

1 0 1 0 1 0 0 1 0 0 0 0 1 1 0 11481632

64

2128

256

512

1024

2048

4096

8192

16384

32768

Positional weight values

Add the values of the bits that are ‘on’ = ‘1’

3 2 7 6 8 8 1 9 22 0 4 8

2 5 6841

----------4 3 2 7 7

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerConvert this binary number to octal.

20)

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

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

1) Separate bits into groups of three (3)

2) Determine octal number

0 0001 0012 0103 0114 1005 1016 1107 111

1 2 4 4 1 5

Cis303a_chapt03_exam1_answer.ppt

CIS303A: System ArchitectureExam 1: Chapter 3

AnswerConvert this binary number to hexadecimal.

21)

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

1) Separate bits into groups of four (4)

2) Determine hex number

0 00001 00012 00103 00114 01005 01016 01107 01118 10009 1001A 1010B 1011C 1100D 1101E 1110F 1111

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

A 9 0 D