📘 Midterm Exam – Chapter 3

⚡TRAINING CONVERTING NUMBERS WITH FRACTIONS (BIN/OCT/HEX/DEC)

🔁 step‑by‑step methods · radix point · group fractions

📐 1) Decimal → binary (with fractional part): 13.6875₁₀

✅ binary = 1101.1011₂

Integer part 13: 13/2=6 r1; 6/2=3 r0; 3/2=1 r1; 1/2=0 r1 → read up: 1101₂.

Fraction 0.6875: ×2 = 1.375 → int 1; 0.375×2 = 0.75 → 0; 0.75×2 = 1.5 → 1; 0.5×2 = 1.0 → 1. stop. Read down: .1011₂

Result: 1101.1011₂.

📐 2) Binary → decimal : 101.1101₂

✅ decimal = 5.8125₁₀

Integer part 101₂ = 1×4 + 0×2 + 1×1 = 5.

Fraction .1101₂ = 1×2⁻¹ + 1×2⁻² + 0×2⁻³ + 1×2⁻⁴ = 0.5 + 0.25 + 0 + 0.0625 = 0.8125.

Total = 5.8125₁₀.

📐 3) Decimal → octal (with fraction): 425.3125₁₀

✅ octal = 651.24₈

Integer 425: 425/8=53 r1; 53/8=6 r5; 6/8=0 r6 → 651₈.

Fraction 0.3125: ×8 = 2.5 → int 2; 0.5×8 = 4.0 → int 4. → .24₈

Join: 651.24₈.

📐 4) Binary → octal (group fractional bits): 11011.01101₂

✅ octal = 33.32₈

Integer part (from right) 11 011 → 3 and 3? Wait: 11011₂ = 011 011 → 33₈? Let's do properly:

Pad integer part with leading zero to group 3: 011 011 → 3 3 = 33₈ ✔.

Fraction part .01101 → group from left: 011 010 (add trailing zero to make group of 3) → 3 2 = .32₈

Result: 33.32₈.

📐 5) Binary → hexadecimal : 101101.1011₂

✅ hexadecimal = 2D.B₁₆

Integer part 10 1101 → pad left: 0010 1101 → 2 D (since 1101₂ = D).

Fraction part .1011 → group 4: 1011 = B.

So 2D.B₁₆.

📐 6) Hex → binary & octal : 3A.4C₁₆

✅ binary = 00111010.01001100₂ ; octal = 72.23₈

3→0011, A→1010, 4→0100, C→1100 ⇒ 00111010.01001100₂ → strip leading: 111010.010011₂

Octal: group binary integer 111 010 = 7 2 → 72₈ ; fraction .010 011 (pad right .010011 → .010 011 = 2 3) ⇒ 72.23₈.

(decimal optional: 3×16+10 + 4/16+12/256 = 58.296875₁₀).

📐 7) Octal → hex through binary: 57.24₈

✅ hex = 2F.5₁₆

Each octal to 3 bits: 5→101, 7→111, 2→010, 4→100 → binary: 101111.010100₂

Group 4 bits for hex: integer part 10 1111 → 0010 1111 = 2F ; fraction .0101 00 (pad right .0101 0000) → .0101 = 5, rest 0000 → .5

Hence 2F.5₁₆.

📐 8) Identify non‑terminating fraction: 0.2₁₀ to binary

✅ binary = 0.0011̅0̅0̅1̅1̅... (repeating)

0.2×2=0.4 →0; 0.4×2=0.8→0; 0.8×2=1.6→1; 0.6×2=1.2→1; 0.2×2=0.4→0 (cycle).

Pattern 0011 repeats ⇒ 0.001100110011...₂

✨ Training summary: For fractions – multiply for decimal→any base; group bits for binary→oct/hex; use binary bridge for oct↔hex.

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! (Fraction‑training included)

ACADEZI 2026 – Chapter 3

📄 MIDTERM EXAMINATION

Chapter 3 – Number Systems

📌 QUICK ANSWER KEY (MCQ + T/F)

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