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Number Systems Practice Paper: Answer Sheet

Model answers, working and marking points for the Number Systems Practice Paper.

How to Use This Answer Sheet

Try the practice paper first, then compare. For calculations, check each step, not just the final answer. For written answers, the model answers show the points a marker looks for — your wording can be different, and other sensible choices with good reasons also earn marks.

Worked example

Working for 1101011011

To octal — groups of 3 from the right, adding leading zeros to the last group: 0011010110111533 So 110101101 1_{2} = 153 3_{8} . To hexadecimal — groups of 4 from the right: 001101011011 35 B (11) So 110101101 1_{2} = 35 B_{16} .

Worked example

Working for 2741

To binary — replace each octal digit with 3 bits : 2741010111100001 274 1_{8} = 01011110000 1_{2} = 10111100001 (drop the leading zero). To hexadecimal — regroup the same bits in 4s from the right: 010111100001 5 E (14) 1 So 274 1_{8} = 5 E 1_{16} .

Worked example

Working for D2F4

To binary — replace each hex digit with 4 bits : D (13) 2 F (15) 4 1101001011110100 D 2 F 4_{16} = 110100101111010 0_{2} . To octal — regroup in 3s from the right (add two leading zeros): 001101001011110100151364 So D 2 F 4_{16} = 15136 4_{8} .

Worked example

Working for 186 to binary (repeated division by 2)

Division Quotient Remainder 186 \div 2 930 93 \div 2 461 46 \div 2 230 23 \div 2 111 11 \div 2 51 5 \div 2 21 2 \div 2 10 1 \div 2 01 Read the remainders from the bottom up : 18 6_{10} = 1011101 0_{2} .

Worked example

Working for 2749 to octal (repeated division by 8), then binary

Division Quotient Remainder 2749 \div 8 3435 343 \div 8 427 42 \div 8 52 5 \div 8 05 Bottom up: 274 9_{10} = 527 5_{8} . Shortcut to binary: each octal digit becomes 3 bits — 5 \to 101, 2 \to 010, 7 \to 111, 5 \to 101 — so 274 9_{10} = 10101011110 1_{2} . (Dividing by 2 twelve times gives the same answer, but takes longer.)

Part A — Binary to Octal and Hexadecimal

Worked example

Working for 1101011011

To octal — groups of 3 from the right, adding leading zeros to the last group: 0011010110111533 So 110101101 1_{2} = 153 3_{8} . To hexadecimal — groups of 4 from the right: 001101011011 35 B (11) So 110101101 1_{2} = 35 B_{16} .

Part B — Octal to Binary and Hexadecimal

Worked example

Working for 2741

To binary — replace each octal digit with 3 bits : 2741010111100001 274 1_{8} = 01011110000 1_{2} = 10111100001 (drop the leading zero). To hexadecimal — regroup the same bits in 4s from the right: 010111100001 5 E (14) 1 So 274 1_{8} = 5 E 1_{16} .

Part C — Hexadecimal to Binary and Octal

Worked example

Working for D2F4

To binary — replace each hex digit with 4 bits : D (13) 2 F (15) 4 1101001011110100 D 2 F 4_{16} = 110100101111010 0_{2} . To octal — regroup in 3s from the right (add two leading zeros): 001101001011110100151364 So D 2 F 4_{16} = 15136 4_{8} .

Part D — Decimal to Binary and Octal

Worked example

Working for 186 to binary (repeated division by 2)

Division Quotient Remainder 186 \div 2 930 93 \div 2 461 46 \div 2 230 23 \div 2 111 11 \div 2 51 5 \div 2 21 2 \div 2 10 1 \div 2 01 Read the remainders from the bottom up : 18 6_{10} = 1011101 0_{2} .

Worked example

Working for 2749 to octal (repeated division by 8), then binary

Division Quotient Remainder 2749 \div 8 3435 343 \div 8 427 42 \div 8 52 5 \div 8 05 Bottom up: 274 9_{10} = 527 5_{8} . Shortcut to binary: each octal digit becomes 3 bits — 5 \to 101, 2 \to 010, 7 \to 111, 5 \to 101 — so 274 9_{10} = 10101011110 1_{2} . (Dividing by 2 twelve times gives the same answer, but takes longer.)

Part E — Written Questions