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📄 MIDTERM EXAMINATION

Chapter 3 – Number Systems · positional & conversion

⏱️ 90 min | total 50 marks
A MULTIPLE CHOICE (10 × 1 = 10)
1️⃣ Which of the following is a non‑positional number system?
a) Binary  b) Decimal  c) Roman  d) Hexadecimal
✅ Answer: c) Roman
Non‑positional: value does NOT depend on position (Roman numerals: XIV = 10+(-1)+5 =14). Positional: decimal, binary, hex.
2️⃣ The hexadecimal digit 'D' is equivalent to which decimal number?
a) 12  b) 13  c) 14  d) 15
✅ Answer: b) 13
Hex digits: A=10, B=11, C=12, D=13, E=14, F=15.
3️⃣ What is the result of (1011)₂ + (101)₂ in binary?
a) 10000₂  b) 1000₂  c) 1110₂  d) 1100₂
✅ Answer: a) 10000₂
1011₂ = 11₁₀ ; 101₂ = 5₁₀ ; 11+5=16₁₀ = 10000₂.
4️⃣ How many bits are required to represent the decimal number 200 in binary?
a) 7  b) 8  c) 9  d) 10
✅ Answer: b) 8
k = ⌈log₂200⌉ = ⌈7.64⌉ = 8. 200₁₀ = 11001000₂ (8 bits).
5️⃣ The octal number 47₈ is equivalent to:
a) 39₁₀  b) 40₁₀  c) 41₁₀  d) 42₁₀
✅ Answer: a) 39₁₀
4×8 + 7 = 32+7 = 39.
6️⃣ Binary equivalent of hexadecimal 2F₁₆ is:
a) 101111₂  b) 101111₂  c) 101111₂  d) All the same (trick)
✅ Answer: 101111₂
2→0010, F→1111 → 00101111₂ → omit leading zeros → 101111₂.
7️⃣ In decimal 456.78, the digit 5 has place value:
a) 5  b) 50  c) 500  d) 0.5
✅ Answer: b) 50
5 is in tens place → 5×10¹ = 50.
8️⃣ Maximum decimal value using 4 binary digits is:
a) 8  b) 15  c) 16  d) 31
✅ Answer: b) 15
2⁴−1 = 15. 1111₂ = 15₁₀.
9️⃣ Convert 0.5₁₀ to binary:
a) 0.1₂  b) 0.01₂  c) 0.5₂  d) 0.101₂
✅ Answer: a) 0.1₂
0.5×2 = 1.0 → integer part 1 → first fractional bit = 1, remainder 0 → 0.1₂.
🔟 Binary 1101.101₂ has how many bits in fractional part?
a) 2  b) 3  c) 4  d) 5
✅ Answer: b) 3
Digits after radix point: 101 → three bits.
B TRUE OR FALSE (5 × 1 = 5)
1️⃣ In a positional system, value depends only on face value, not position.
✅ Answer: FALSE
Positional means position determines weight (e.g., 5 in 50 vs 5).
2️⃣ Hexadecimal system uses 15 symbols.
✅ Answer: FALSE
16 symbols: 0‑9 and A‑F.
3️⃣ The octal number 78₈ is valid.
✅ Answer: FALSE
Octal digits only 0‑7, digit 8 not allowed.
4️⃣ 2⁵−1 is the maximum value representable with 5 bits.
✅ Answer: TRUE
Max 5‑bit number = 11111₂ = 31 = 2⁵−1.
5️⃣ Binary 1010₂ equals decimal 10.
✅ Answer: TRUE
1×8 + 0×4 + 1×2 + 0×1 = 10.
C SHORT ANSWER (any five, each 3 marks)
C1) Differentiate positional vs non‑positional with one example each.
✅ Solution
Positional: value depends on position (decimal 345 = 3×100+4×10+5). Non‑positional: roman XIV = 10+5−1 (order independent).
C2) Convert (1011.011)₂ and (2A.F)₁₆ to decimal.
✅ Solution
(1011.011)₂ = 8+0+2+1+0+0.25+0.125 = 11.375₁₀.
(2A.F)₁₆ = 2×16 + A×1 + F×16⁻¹ = 32+10+15/16 = 42.9375₁₀.
C3) Convert 156₁₀ to binary, octal, hexadecimal.
✅ Solution
156/2=78r0 → /2=39r0 → /2=19r1 → /2=9r1 → /2=4r1 → /2=2r0 → /2=1r0 → /2=0r1 ⇒ 10011100₂.
Octal: 156/8=19r4, 19/8=2r3, 2/8=0r2 ⇒ 234₈.
Hex: 156/16=9r12(C), 9/16=0r9 ⇒ 9C₁₆.
C4) Convert 0.6875₁₀ to binary.
✅ Solution
0.6875×2 = 1.375 → 1; 0.375×2 = 0.75→0; 0.75×2=1.5→1; 0.5×2=1.0→1 ⇒ 0.1011₂.
C5) Convert 110110.101₂ to octal.
✅ Solution
Group binary: 110 110 . 101 → 6,6,5 ⇒ 66.5₈.
C6) Convert 3B.4₁₆ to binary.
✅ Solution
3→0011, B→1011, 4→0100 ⇒ 00111011.0100₂ = 111011.01₂.
D LONG ANSWER (any four, 5 marks each)
D1) a) (567.24)₈ to decimal b) (ABCD)₁₆ to decimal c) (110110.011)₂ to decimal d) (10101.11)₂ to hexadecimal
✅ Step‑by‑step
a) 5×64=320, 6×8=48, 7×1=7, 2×0.125=0.25, 4×0.015625=0.0625 → 375.3125₁₀.
b) A×4096=40960, B×256=2816, C×16=192, D×1=13 → 43981₁₀.
c) 32+16+0+4+2+0+0+0.25+0.125 = 54.375₁₀.
d) Group 4 bits: 0001 0101 . 1100 → 1 5 . C ⇒ 15.C₁₆.
D2) Convert 345.625₁₀ to binary, octal, hex and verify equivalence.
✅ Solution
Binary: 345→101011001, 0.625→0.101 → 101011001.101₂.
Octal: group binary 101 011 001 . 101 → 531.5₈.
Hex: 0001 0101 1001 . 1010 → 159.A₁₆.
Verify: 159.A = 1×256+5×16+9+10/16 = 345.625 ✓.
D3) Find number of digits needed for 500₁₀ in binary, octal, hex and verify by conversion.
✅ Solution
Binary: ⌈log₂500⌉ = ⌈8.97⌉ = 9 bits. 500₁₀ = 111110100₂ (9 bits).
Octal: ⌈log₈500⌉ = ⌈2.99⌉ = 3 digits. 500/8=62r4,62/8=7r6,7/8=0r7 → 764₈.
Hex: ⌈log₁₆500⌉ = ⌈2.24⌉ = 3 digits. 500/16=31r4,31/16=1rF,1/16=0r1 → 1F4₁₆.
D4) a) 2F.8₁₆ → binary, octal. b) 567₈ → binary, hex.
✅ Solution
a) 2F.8₁₆ = 0010 1111 . 1000₂ = 101111.1₂. Octal: group 101 111 . 100 → 57.4₈.
b) 567₈ = 101 110 111₂ = 1 0111 0111 (group4) → 0001 0111 0111 = 177₁₆.
D5) Largest 4‑digit octal (7777₈) vs largest 3‑digit hex (FFF₁₆). Which is larger? Prove by decimal conversion.
✅ Solution
7777₈ = 7×512 +7×64+7×8+7 = 3584+448+56+7 = 4095₁₀.
FFF₁₆ = 15×256+15×16+15 = 3840+240+15 = 4095₁₀.
They are equal! 7777₈ = FFF₁₆ = 4095.
✨ BONUS (optional 5 extra)
Design base‑7 system: (a) symbols (b) 345₇ to decimal (c) 256₁₀ to base‑7 (d) 56.4₇ to decimal.
✅ Base‑7 solution
a) symbols {0,1,2,3,4,5,6}.
b) 345₇ = 3×49 + 4×7 + 5 = 147+28+5 = 180₁₀.
c) 256/7=36r4, 36/7=5r1, 5/7=0r5 → 514₇.
d) 56.4₇ = 5×7+6 + 4/7 = 35+6+0.571428 = 41.571428₁₀.

📌 QUICK ANSWER KEY (MCQ + T/F)

MCQ: 1c,2b,3a,4b,5a,6‑,7b,8b,9a,10b
T/F: 1F,2F,3F,4T,5T

⬆️ Click any solution button for full step‑by‑step explanation.

📘🚀 End of midterm – good luck!

ACADEZI 2026 – Chapter 3 number systems