πŸ“„ MIDTERM EXAMINATION

Chapter 3 – Number Systems

⏱️ 90 min | total 50 marks + training
⚑TRAINING CONVERTING NUMBERS WITH FRACTIONS (BIN/OCT/HEX/DEC)
πŸ” step‑by‑step methods Β· radix point Β· group fractions
πŸ“ 1) Decimal β†’ binary (with fractional part): 13.6875₁₀
βœ… binary = 1101.1011β‚‚
Integer part 13: 13/2=6 r1; 6/2=3 r0; 3/2=1 r1; 1/2=0 r1 β†’ read up: 1101β‚‚.
Fraction 0.6875: Γ—2 = 1.375 β†’ int 1; 0.375Γ—2 = 0.75 β†’ 0; 0.75Γ—2 = 1.5 β†’ 1; 0.5Γ—2 = 1.0 β†’ 1. stop. Read down: .1011β‚‚
Result: 1101.1011β‚‚.
πŸ“ 2) Binary β†’ decimal : 101.1101β‚‚
βœ… decimal = 5.8125₁₀
Integer part 101β‚‚ = 1Γ—4 + 0Γ—2 + 1Γ—1 = 5.
Fraction .1101β‚‚ = 1Γ—2⁻¹ + 1Γ—2⁻² + 0Γ—2⁻³ + 1Γ—2⁻⁴ = 0.5 + 0.25 + 0 + 0.0625 = 0.8125.
Total = 5.8125₁₀.
πŸ“ 3) Decimal β†’ octal (with fraction): 425.3125₁₀
βœ… octal = 651.24β‚ˆ
Integer 425: 425/8=53 r1; 53/8=6 r5; 6/8=0 r6 β†’ 651β‚ˆ.
Fraction 0.3125: Γ—8 = 2.5 β†’ int 2; 0.5Γ—8 = 4.0 β†’ int 4. β†’ .24β‚ˆ
Join: 651.24β‚ˆ.
πŸ“ 4) Binary β†’ octal (group fractional bits): 11011.01101β‚‚
βœ… octal = 33.32β‚ˆ
Integer part (from right) 11 011 β†’ 3 and 3? Wait: 11011β‚‚ = 011 011 β†’ 33β‚ˆ? Let's do properly:
Pad integer part with leading zero to group 3: 011 011 β†’ 3 3 = 33β‚ˆ βœ”.
Fraction part .01101 β†’ group from left: 011 010 (add trailing zero to make group of 3) β†’ 3 2 = .32β‚ˆ
Result: 33.32β‚ˆ.
πŸ“ 5) Binary β†’ hexadecimal : 101101.1011β‚‚
βœ… hexadecimal = 2D.B₁₆
Integer part 10 1101 β†’ pad left: 0010 1101 β†’ 2 D (since 1101β‚‚ = D).
Fraction part .1011 β†’ group 4: 1011 = B.
So 2D.B₁₆.
πŸ“ 6) Hex β†’ binary & octal : 3A.4C₁₆
βœ… binary = 00111010.01001100β‚‚ ; octal = 72.23β‚ˆ
3→0011, A→1010, 4→0100, C→1100 ⇒ 00111010.01001100₂ → strip leading: 111010.010011₂
Octal: group binary integer 111 010 = 7 2 β†’ 72β‚ˆ ; fraction .010 011 (pad right .010011 β†’ .010 011 = 2 3) β‡’ 72.23β‚ˆ.
(decimal optional: 3Γ—16+10 + 4/16+12/256 = 58.296875₁₀).
πŸ“ 7) Octal β†’ hex through binary: 57.24β‚ˆ
βœ… hex = 2F.5₁₆
Each octal to 3 bits: 5β†’101, 7β†’111, 2β†’010, 4β†’100 β†’ binary: 101111.010100β‚‚
Group 4 bits for hex: integer part 10 1111 β†’ 0010 1111 = 2F ; fraction .0101 00 (pad right .0101 0000) β†’ .0101 = 5, rest 0000 β†’ .5
Hence 2F.5₁₆.
πŸ“ 8) Identify non‑terminating fraction: 0.2₁₀ to binary
βœ… binary = 0.0011Μ…0Μ…0Μ…1Μ…1Μ…... (repeating)
0.2Γ—2=0.4 β†’0; 0.4Γ—2=0.8β†’0; 0.8Γ—2=1.6β†’1; 0.6Γ—2=1.2β†’1; 0.2Γ—2=0.4β†’0 (cycle).
Pattern 0011 repeats β‡’ 0.001100110011...β‚‚

✨ Training summary: For fractions – multiply for decimalβ†’any base; group bits for binaryβ†’oct/hex; use binary bridge for oct↔hex.

A MULTIPLE CHOICE (10 Γ— 1 = 10)
1️⃣ Which of the following is a non‑positional number system?
a) Binary  b) Decimal  c) Roman  d) Hexadecimal
βœ… 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
βœ… Answer: b) 13
Hex digits: A=10, B=11, C=12, D=13, E=14, F=15.
3️⃣ What is the result of (1011)β‚‚ + (101)β‚‚ in binary?
a) 10000₂  b) 1000₂  c) 1110₂  d) 1100β‚‚
βœ… Answer: a) 10000β‚‚
1011β‚‚ = 11₁₀ ; 101β‚‚ = 5₁₀ ; 11+5=16₁₀ = 10000β‚‚.
4️⃣ How many bits are required to represent the decimal number 200 in binary?
a) 7  b) 8  c) 9  d) 10
βœ… Answer: b) 8
k = ⌈logβ‚‚200βŒ‰ = ⌈7.64βŒ‰ = 8. 200₁₀ = 11001000β‚‚ (8 bits).
5️⃣ The octal number 47β‚ˆ is equivalent to:
a) 39₁₀  b) 40₁₀  c) 41₁₀  d) 42₁₀
βœ… Answer: a) 39₁₀
4Γ—8 + 7 = 32+7 = 39.
6️⃣ Binary equivalent of hexadecimal 2F₁₆ is:
a) 101111₂  b) 101111₂  c) 101111₂  d) All the same (trick)
βœ… Answer: 101111β‚‚
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
βœ… Answer: b) 50
5 is in tens place β†’ 5Γ—10ΒΉ = 50.
8️⃣ Maximum decimal value using 4 binary digits is:
a) 8  b) 15  c) 16  d) 31
βœ… Answer: b) 15
2β΄βˆ’1 = 15. 1111β‚‚ = 15₁₀.
9️⃣ Convert 0.5₁₀ to binary:
a) 0.1₂  b) 0.01₂  c) 0.5₂  d) 0.101β‚‚
βœ… Answer: a) 0.1β‚‚
0.5Γ—2 = 1.0 β†’ integer part 1 β†’ first fractional bit = 1, remainder 0 β†’ 0.1β‚‚.
πŸ”Ÿ Binary 1101.101β‚‚ has how many bits in fractional part?
a) 2  b) 3  c) 4  d) 5
βœ… Answer: b) 3
Digits after radix point: 101 β†’ three bits.
B TRUE OR FALSE (5 Γ— 1 = 5)
1️⃣ In a positional system, value depends only on face value, not position.
βœ… Answer: FALSE
Positional means position determines weight (e.g., 5 in 50 vs 5).
2️⃣ Hexadecimal system uses 15 symbols.
βœ… Answer: FALSE
16 symbols: 0‑9 and A‑F.
3️⃣ The octal number 78β‚ˆ is valid.
βœ… Answer: FALSE
Octal digits only 0‑7, digit 8 not allowed.
4️⃣ 2β΅βˆ’1 is the maximum value representable with 5 bits.
βœ… Answer: TRUE
Max 5‑bit number = 11111β‚‚ = 31 = 2β΅βˆ’1.
5️⃣ Binary 1010β‚‚ equals decimal 10.
βœ… Answer: TRUE
1Γ—8 + 0Γ—4 + 1Γ—2 + 0Γ—1 = 10.
C SHORT ANSWER (any five, each 3 marks)
C1) Differentiate positional vs non‑positional with one example each.
βœ… Solution
Positional: value depends on position (decimal 345 = 3Γ—100+4Γ—10+5). Non‑positional: roman XIV = 10+5βˆ’1 (order independent).
C2) Convert (1011.011)β‚‚ and (2A.F)₁₆ to decimal.
βœ… Solution
(1011.011)β‚‚ = 8+0+2+1+0+0.25+0.125 = 11.375₁₀.
(2A.F)₁₆ = 2Γ—16 + AΓ—1 + FΓ—16⁻¹ = 32+10+15/16 = 42.9375₁₀.
C3) Convert 156₁₀ to binary, octal, hexadecimal.
βœ… Solution
156/2=78r0 β†’ /2=39r0 β†’ /2=19r1 β†’ /2=9r1 β†’ /2=4r1 β†’ /2=2r0 β†’ /2=1r0 β†’ /2=0r1 β‡’ 10011100β‚‚.
Octal: 156/8=19r4, 19/8=2r3, 2/8=0r2 β‡’ 234β‚ˆ.
Hex: 156/16=9r12(C), 9/16=0r9 β‡’ 9C₁₆.
C4) Convert 0.6875₁₀ to binary.
βœ… Solution
0.6875Γ—2 = 1.375 β†’ 1; 0.375Γ—2 = 0.75β†’0; 0.75Γ—2=1.5β†’1; 0.5Γ—2=1.0β†’1 β‡’ 0.1011β‚‚.
C5) Convert 110110.101β‚‚ to octal.
βœ… Solution
Group binary: 110 110 . 101 β†’ 6,6,5 β‡’ 66.5β‚ˆ.
C6) Convert 3B.4₁₆ to binary.
βœ… Solution
3→0011, B→1011, 4→0100 ⇒ 00111011.0100₂ = 111011.01₂.
D LONG ANSWER (any four, 5 marks each)
D1) a) (567.24)β‚ˆ to decimal b) (ABCD)₁₆ to decimal c) (110110.011)β‚‚ to decimal d) (10101.11)β‚‚ to hexadecimal
βœ… Step‑by‑step
a) 5Γ—64=320, 6Γ—8=48, 7Γ—1=7, 2Γ—0.125=0.25, 4Γ—0.015625=0.0625 β†’ 375.3125₁₀.
b) AΓ—4096=40960, BΓ—256=2816, CΓ—16=192, DΓ—1=13 β†’ 43981₁₀.
c) 32+16+0+4+2+0+0+0.25+0.125 = 54.375₁₀.
d) Group 4 bits: 0001 0101 . 1100 β†’ 1 5 . C β‡’ 15.C₁₆.
D2) Convert 345.625₁₀ to binary, octal, hex and verify equivalence.
βœ… Solution
Binary: 345β†’101011001, 0.625β†’0.101 β†’ 101011001.101β‚‚.
Octal: group binary 101 011 001 . 101 β†’ 531.5β‚ˆ.
Hex: 0001 0101 1001 . 1010 β†’ 159.A₁₆.
Verify: 159.A = 1Γ—256+5Γ—16+9+10/16 = 345.625 βœ“.
D3) Find number of digits needed for 500₁₀ in binary, octal, hex and verify by conversion.
βœ… Solution
Binary: ⌈logβ‚‚500βŒ‰ = ⌈8.97βŒ‰ = 9 bits. 500₁₀ = 111110100β‚‚ (9 bits).
Octal: ⌈logβ‚ˆ500βŒ‰ = ⌈2.99βŒ‰ = 3 digits. 500/8=62r4,62/8=7r6,7/8=0r7 β†’ 764β‚ˆ.
Hex: ⌈log₁₆500βŒ‰ = ⌈2.24βŒ‰ = 3 digits. 500/16=31r4,31/16=1rF,1/16=0r1 β†’ 1F4₁₆.
D4) a) 2F.8₁₆ β†’ binary, octal. b) 567β‚ˆ β†’ binary, hex.
βœ… Solution
a) 2F.8₁₆ = 0010 1111 . 1000β‚‚ = 101111.1β‚‚. Octal: group 101 111 . 100 β†’ 57.4β‚ˆ.
b) 567β‚ˆ = 101 110 111β‚‚ = 1 0111 0111 (group4) β†’ 0001 0111 0111 = 177₁₆.
D5) Largest 4‑digit octal (7777β‚ˆ) vs largest 3‑digit hex (FFF₁₆). Which is larger? Prove by decimal conversion.
βœ… Solution
7777β‚ˆ = 7Γ—512 +7Γ—64+7Γ—8+7 = 3584+448+56+7 = 4095₁₀.
FFF₁₆ = 15Γ—256+15Γ—16+15 = 3840+240+15 = 4095₁₀.
They are equal! 7777β‚ˆ = 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.
βœ… Base‑7 solution
a) symbols {0,1,2,3,4,5,6}.
b) 345₇ = 3Γ—49 + 4Γ—7 + 5 = 147+28+5 = 180₁₀.
c) 256/7=36r4, 36/7=5r1, 5/7=0r5 β†’ 514₇.
d) 56.4₇ = 5Γ—7+6 + 4/7 = 35+6+0.571428 = 41.571428₁₀.

πŸ“Œ 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.

πŸ“˜πŸš€ End of midterm – good luck! (Fraction‑training included)

ACADEZI 2026 – Chapter 3