Binary System — Set 4

Correct Answer: D. 101

• **101** = the binary representation of decimal 5. Calculation: 5 = 4+1 = 1×2² + 0×2¹ + 1×2⁰, so bits at position 3 and position 1 (from right) are set, giving pattern 1-0-1. • **Positional check** — binary positions from right: 1, 2, 4. Turn on positions 4 and 1 (4+1=5), turn off position 2, yielding 101. • Verification: 101 → (1×4)+(0×2)+(1×1) = 4+0+1 = 5. Correct. • Option A (100) is wrong because binary 100 = 4; Option B (111) is wrong because binary 111 = 4+2+1 = 7; Option C (110) is wrong because binary 110 = 4+2 = 6.

Correct Answer: C. NOT

• **NOT Gate** = the logic gate universally called an 'Inverter' because it reverses (inverts) its single input: NOT 0 = 1, NOT 1 = 0. Its circuit symbol is a triangle with a small bubble at the output, representing the inversion. • **Single input** — unlike AND, OR, XOR (all two-input gates), the NOT gate accepts exactly one input, making it the simplest and most fundamental building block in digital logic. • Inverters are the foundation of flip-flops (memory cells), oscillators, and all complementary logic — every CMOS chip contains thousands of them. • Option A (OR) is wrong because OR is a two-input gate that outputs 1 when any input is 1; Option B (AND) is wrong because AND outputs 1 only when all inputs are 1; Option D (XOR) is wrong because XOR outputs 1 when inputs differ, not when a single input is simply inverted.

Correct Answer: D. 10

• **10** = the binary representation of decimal 2. Calculation: 2 = 1×2¹ + 0×2⁰, so a '1' in the twos-place and a '0' in the ones-place gives the pattern 1-0. • **First carry point** — 2 is the smallest number that cannot be expressed with a single binary digit; writing it requires two digits and represents the first 'carry' in binary counting (0, 1, 10, 11, 100…). • Verification: 10 → (1×2)+(0×1) = 2. Correct. • Option A (1) is wrong because binary 1 = decimal 1; Option B (100) is wrong because binary 100 = decimal 4; Option C (11) is wrong because binary 11 = 2+1 = decimal 3.

Correct Answer: C. 8

• **8** = the number of different combinations possible with 3 bits. With each bit independently 0 or 1, the total states = 2³ = 8, covering 000, 001, 010, 011, 100, 101, 110, 111 (decimal 0 through 7). • **General rule** — n bits produce 2ⁿ unique combinations: 1 bit→2, 2 bits→4, 3 bits→8, 4 bits→16, and so on. Each extra bit doubles the capacity. • 3-bit numbers are used in octal notation (where one octal digit spans 0–7 = 3 binary bits) and in early microcontrollers with 3-bit port registers. • Option A (3) is wrong because 3 is the number of bits, not the count of combinations; Option B (6) is wrong because 6 = 2×3, a linear calculation that does not apply; Option D (9) is wrong because 9 = 3², also incorrect — combinations grow as powers of 2, not squares of the bit count.

Correct Answer: D. Binary 0

• **Binary 0** = what a low voltage signal typically represents in digital electronics. In standard positive logic convention (used by TTL and CMOS circuits), a voltage near 0 V (ground) is defined as logic 0 (off/false). • **Voltage thresholds** — in 3.3 V CMOS, any voltage below ~1.0 V is read as logic 0, and any voltage above ~2.0 V is read as logic 1, with a 'forbidden zone' between to prevent ambiguity. • This two-voltage convention is exactly why computers use binary — electronics naturally produce two stable voltage states (low and high), mapping perfectly to 0 and 1. • Option A (Binary 1) is wrong because binary 1 is represented by a HIGH voltage, not a low one; Option B (Error) is wrong because a low voltage is a valid logic state, not an error signal; Option C (Overflow) is wrong because overflow is an arithmetic condition unrelated to individual voltage levels.

Correct Answer: C. 111

• **111** = the binary representation of decimal 7. Calculation: 7 = 4+2+1 = 1×2² + 1×2¹ + 1×2⁰, so all three bit positions are set to 1, giving pattern 1-1-1. • **Maximum 3-bit value** — with 3 bits all set to 1, the value is the maximum representable by a 3-bit unsigned integer: 2³-1 = 7. • Verification: 111 → (1×4)+(1×2)+(1×1) = 4+2+1 = 7. Correct. • Option A (101) is wrong because binary 101 = 4+1 = 5; Option B (110) is wrong because binary 110 = 4+2 = 6; Option D (1000) is wrong because binary 1000 = 8, requiring a 4th bit.

Correct Answer: B. OR

• **OR** = the Boolean operation represented by the '+' symbol in Boolean algebra. The expression A + B means 'A OR B,' and it evaluates to 1 whenever at least one of A or B is 1, just like the OR gate's truth table. • **Notation convention** — Boolean algebra uses '+' for OR and '·' (or juxtaposition) for AND, analogous to addition and multiplication in regular algebra, though the rules differ for values greater than 1. • The OR operation is used in bit-setting: to force bit position i to 1 in a register, you OR the register with a mask that has a 1 only in position i. • Option A (AND) is wrong because AND is represented by '·' or simple juxtaposition (AB), not '+'; Option C (XOR) is wrong because XOR is represented by '⊕', a circled plus, to distinguish it from ordinary OR; Option D (NOT) is wrong because NOT is represented by an overbar (Ā) or a leading apostrophe (A'), never by '+'.

Correct Answer: D. Octal

• **Octal** = the base-8 number system that uses digits 0–7, where each single octal digit corresponds to exactly three binary bits (since 2³ = 8). • **Three-bit grouping** — to convert binary to octal, group bits in threes from the right: binary 110 101 = octal 65. This compact notation was common in older minicomputer and Unix file permission systems. • Unix/Linux file permissions (e.g., chmod 755) use octal: 7 = 111 (rwx), 5 = 101 (r-x), making three-bit groupings directly visible in the permission code. • Option A (Binary) is wrong because binary is base 2, using only 0 and 1; Option B (Decimal) is wrong because decimal is base 10 with digits 0–9; Option C (Hexadecimal) is wrong because hexadecimal is base 16 and groups binary in fours, not threes.

Correct Answer: B. 110

• **110** = the binary representation of decimal 6. Calculation: 6 = 4+2 = 1×2² + 1×2¹ + 0×2⁰, so positions for 4 and 2 are set to 1 and the ones-place is 0, giving pattern 1-1-0. • **Even number indicator** — since the ones-place (LSB) is 0, the number 110 is immediately identifiable as even; any binary number ending in 0 is even. • Verification: 110 → (1×4)+(1×2)+(0×1) = 4+2+0 = 6. Correct. • Option A (100) is wrong because binary 100 = 4; Option C (101) is wrong because binary 101 = 4+1 = 5; Option D (111) is wrong because binary 111 = 4+2+1 = 7.

Correct Answer: B. 0

• **0** = the result of the binary AND operation on bits 1 and 0. The AND gate outputs 1 only when ALL inputs are 1; since one input is 0, the output is 0. • **AND truth table** — (0,0)→0, (0,1)→0, (1,0)→0, (1,1)→1. Only the (1,1) combination yields 1; any input of 0 forces the output to 0. • AND is used for bit-clearing (masking): ANDing a value with a mask that has 0 in a specific position forces that bit to 0 without affecting others. • Option A (1) is wrong because 1 requires both inputs to be 1; here one input is 0; Option C (10) is wrong because 10 is binary for decimal 2, not a valid single-bit AND result; Option D (11) is wrong because 11 is binary for decimal 3, far outside the range of a single-bit operation.