1. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 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 8 342 4 6 7 5 ---------- 6 5 8 5 What base(s) do these numbers represent? (circle those that apply) 9) 4 3 6 9 5 10 Decimal Octal Binary Hexadecimal 10) 1 7 C 2 D 16 Decimal Octal Binary Hexadecimal 11) 1 0 1 1 1 0 0 1 0 2 Decimal Octal Binary Hexadecimal 12) 6 1 7 2 4 5 8 Decimal Octal Binary Hexadecimal 40 10

2. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Add 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 1 1 --- 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 12 3 4 56 10

3. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Add these octal numbers. 16)2 6 6 7 5 3 4 1 ---------- 1 2 6 6 7 5 3 4 1 --------- 0 1 1 2 6 6 7 6 5 3 4 1 4 --------- --- 013 1 1 1 2 6 6 7 6 5 3 4 1 3 --------- --- 3 0 12 1 1 1 1 2 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 0 123 45 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.

4. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Add these hexadecimal numbers. 17)2 A 3 4 5 3 F 6 ---------- 2 A 3 4 4 5 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 3 5 3 F 6 F ---------- -- A 12 1 2 A 3 4 3 5 3 F 6 F ---------- -- 2 A 12 1 1 2 A 3 4 A 5 3 F 6 3 ---------- -- 2 A E 1 2 A 3 4 2 5 3 F 6 5 ---------- -- E 2 A 7 1 2 A 3 4 5 3 F 6 ---------- 7 E 2 A 123 46 5

5. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Convert 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 1 1481632 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

6. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 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

7. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Convert this binary number to decimal. 19) 1 0 1 0 1 0 0 1 0 0 0 0 1 1 0 1 1481632 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 2 2 0 4 8 2 5 6 8 4 1 ---------- 4 3 2 7 7

8. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Convert this binary number to octal. 20) 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 0000 1001 2010 3011 4100 5101 6110 7111 1 2 4 4 1 5

9. Cis303a_chapt03_exam1_answer.ppt CIS303A: System Architecture Exam 1: Chapter 3 Answer Convert 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 00000 10001 20010 30011 40100 50101 60110 70111 81000 91001 A1010 B1011 C1100 D1101 E1110 F1111 1 0 1 0 1 0 0 1 0 0 0 0 1 1 0 1 A 9 0 D