Binary System — Set 6

Correct Answer: B. 5 bits

• **5 bits** = the minimum number of binary bits needed to represent decimal 16. Since 16 = 2⁴, it requires a '1' in the 5th position (weight 16) with all lower bits as 0: binary 10000 — a 5-digit number. • **Why not 4?** — with 4 bits the maximum value is 1111 = 15; the number 16 cannot fit, because 16 = 2⁴ needs the 2⁴ position which is the 5th bit. • General rule: to represent n in binary you need ⌊log₂(n)⌋+1 bits. For n=16: log₂(16)=4, so 4+1=5 bits are needed. • Option A (4 bits) is wrong because 4 bits max out at 15 (1111), one less than 16; Option C (8 bits) is wrong because 8 bits can represent up to 255 — far more than needed, wasting 3 bits; Option D (16 bits) is wrong because 16 bits are needed to represent 65,535, not a small number like 16.

Correct Answer: C. Digital Computer

• **Digital Computer** = the type of equipment where binary numbers are primarily used. All modern digital computers — from smartphones to supercomputers — process, store, and transmit every piece of information as binary (0s and 1s) at the hardware level. • **Why digital?** — transistors in CPUs can reliably switch between two voltage states billions of times per second, making binary the only practical encoding for high-speed electronic computation. • Even data that appears analog (music, video) is converted to binary sequences via Analog-to-Digital Converters (ADCs) before a digital computer can process it. • Option A (Analog Radio) is wrong because analog radios use continuous waveforms, not discrete binary states; Option B (Mechanical Clock) is wrong because mechanical clocks use gear ratios and springs, not electronic binary signals; Option D (Steam Engine) is wrong because steam engines are mechanical thermodynamic devices with no connection to binary or digital logic.

Correct Answer: B. OR Gate

• **OR Gate** = the logic gate that outputs 1 when either one or both of its inputs are 1. Its truth table: (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→1 — only an all-zero input produces 0. • **Bit-setting use** — OR is the standard operation for forcing a specific bit to 1 without disturbing others: ORing any register with a mask that has a 1 in the target position sets that bit unconditionally. • The '+' symbol in Boolean algebra represents OR (A+B = A OR B), and it behaves like arithmetic addition except that 1+1=1 (not 2) in Boolean logic. • Option A (AND Gate) is wrong because AND requires ALL inputs to be 1 to output 1; Option C (NOT Gate) is wrong because NOT inverts a single input, not a conjunction of two inputs; Option D (NOR Gate) is wrong because NOR is NOT-OR — it outputs 1 only when both inputs are 0, the opposite of OR.

Correct Answer: B. 1

• **1** = the result of binary addition 1 + 0. In binary arithmetic, 1+0 = 1 with no carry, identical to the same operation in decimal — adding a number to zero always returns the original number. • **Addition table** — binary addition has four cases: 0+0=0, 0+1=1, 1+0=1, 1+1=10 (carry). The case 1+0=1 is the trivial no-carry case. • This is implemented in a half-adder circuit using an XOR gate for the sum bit (XOR(1,0)=1) and an AND gate for the carry bit (AND(1,0)=0) — confirming sum=1, carry=0. • Option A (0) is wrong because 1+0=1 not 0; only 0+0=0; Option C (10) is wrong because binary 10 = 2, which would require 1+1; Option D (11) is wrong because binary 11 = 3, which would require even larger inputs.

Correct Answer: B. 2

• **2** = the decimal equivalent of binary 10. Calculation: 10 → (1×2¹)+(0×2⁰) = 2+0 = 2. The '1' in the twos-place contributes 2 and the '0' in the ones-place contributes 0. • **Counting in binary** — the binary sequence 0, 1, 10, 11, 100 corresponds to decimal 0, 1, 2, 3, 4. Binary '10' is the first number requiring two digits, just as decimal '10' is the first two-digit decimal number. • Binary 10 is also where the first carry occurs in binary addition (1+1=10), making it fundamental to understanding binary arithmetic. • Option A (1) is wrong because binary 1 = decimal 1; Option C (3) is wrong because binary 11 = 2+1 = 3; Option D (10) is wrong because decimal 10 is binary 1010, a four-bit number.

Correct Answer: C. Decimal

• **Decimal** = the base-10 number system that humans use in everyday life, using digits 0–9. It is called 'decimal' from the Latin 'decimus' (tenth) because it is based on the number 10, likely derived from humans having 10 fingers. • **Positional values** — each position in a decimal number represents a power of 10: ones (10⁰), tens (10¹), hundreds (10²), thousands (10³), and so on. • Computers internally use binary but display results in decimal for human readability, with conversion happening automatically in the CPU and output formatting software. • Option A (Binary) is wrong because binary is base 2 using only 0 and 1; Option B (Octal) is wrong because octal is base 8 using digits 0–7; Option D (Hexadecimal) is wrong because hexadecimal is base 16 using digits 0–9 and A–F.

Correct Answer: B. LSB

• **LSB (Least Significant Bit)** = the bit on the far right side of a binary number, carrying the smallest positional weight of 2⁰ = 1. Changing the LSB alters the number's value by the minimum possible amount — just ±1. • **Even/odd indicator** — the LSB instantly reveals parity: if the LSB is 0, the number is even; if 1, it is odd. This check is used in low-level programming for performance-critical parity tests. • In audio processing, LSB manipulation (dithering) adds tiny noise to the least significant bit to reduce quantization distortion in digital recordings. • Option A (MSB) is wrong because the MSB is the far LEFT bit, carrying the highest weight; Option C (Parity Bit) is wrong because the parity bit is an appended error-check bit, not a positional bit within the number itself; Option D (Sign Bit) is wrong because the sign bit is the MSB in signed integers, the leftmost bit.

Correct Answer: B. 100

• **100** = the binary equivalent of decimal 4. Calculation: 4 = 1×2² + 0×2¹ + 0×2⁰, so only the third bit from the right (weight 4) is set to 1, yielding pattern 1-0-0. • **Power-of-2 signature** — 4 is a pure power of 2 (2²), so its binary form is a 1 followed by exactly two zeros. All pure powers of 2 follow this pattern: 2=10, 4=100, 8=1000. • Verification: 100 → (1×4)+(0×2)+(0×1) = 4. Correct. • Option A (11) is wrong because binary 11 = 2+1 = 3; Option C (101) is wrong because binary 101 = 4+1 = 5; Option D (110) is wrong because binary 110 = 4+2 = 6.

Correct Answer: C. XOR Gate

• **XOR Gate** = Exclusive OR — outputs 1 only when its two inputs are different (one 0 and one 1). Its truth table: (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→0. When inputs are the same, XOR always returns 0. • **Adder circuit** — XOR is the core of a binary half-adder: XOR(A,B) gives the sum bit while AND(A,B) gives the carry, together implementing single-bit addition in hardware. • XOR's self-inverse property (A XOR A = 0) makes it ideal for simple encryption, checksums, and RAID-5 parity calculations where reversibility is needed. • Option A (AND Gate) is wrong because AND outputs 1 only when both inputs are 1 (same value), not when they differ; Option B (OR Gate) is wrong because OR outputs 1 when either or both inputs are 1, including when they are the same; Option D (NOT Gate) is wrong because NOT is a single-input gate that simply inverts one bit.

Correct Answer: D. 8192

• **8192** = the total number of bits in one Kilobyte (KB). Calculation: 1 KB = 1024 bytes (since 1 KB = 2¹⁰ bytes in binary), and each byte = 8 bits, so 1024 × 8 = 8192 bits. • **Binary vs. decimal KB** — in computing, 1 KB traditionally = 1024 bytes (2¹⁰), not 1000 bytes. The IEC standard calls 1024 bytes a 'Kibibyte (KiB)' to avoid confusion, but 'KB = 1024 bytes' remains common in practice. • Knowing that 1 KB = 8192 bits matters for network engineers who measure speed in bits per second (bps) — a 1 KB file takes 8192 bits to transmit. • Option A (1000) is wrong because 1000 bits would be if KB meant exactly 1000 bytes and each byte was 1 bit — both assumptions are wrong; Option B (1024) is wrong because 1024 is the number of bytes in 1 KB, not the number of bits; Option C (8000) is wrong because 8000 = 1000×8, using the incorrect decimal definition of KB.