Memory Units — Set 3

Correct Answer: A. 2

• **Computer memory uses multiples of 2** because digital circuits are binary — each transistor can be either ON (1) or OFF (0), so all memory capacity naturally grows as powers of 2: 2^10 = 1024, 2^20 = 1048576, and so on. • **This is why storage units jump by ×1024, not ×1000** — 1 KB = 1024 bytes, 1 MB = 1024 KB, because doubling a binary counter gives 2^10 = 1024, not a round decimal number like 1000. • Binary counting also explains why old RAM came in sizes like 256 MB, 512 MB, 1 GB, 2 GB — all powers of 2, never arbitrary numbers like 300 MB. • 💡 Option B (10) is wrong because base-10 is the decimal system used by humans, not by binary electronic circuits; Option C (100) is wrong because no binary unit corresponds to a power of 100; Option D (1000) is wrong because memory addressing uses base-2, not base-10.

Correct Answer: D. 1,474,560 bytes

• **1,474,560 bytes** is the true capacity of a 1.44 MB floppy disk, calculated as 1440 KB × 1024 bytes/KB — the '1.44 MB' label actually meant 1440 Kilobytes, mixing decimal and binary in an unusual way. • **This was a deliberate marketing choice** — floppy disk manufacturers called 1440 KB a '1.44 MB' disk by dividing 1,474,560 by 1,000,000 (not 1,048,576), giving the rounder-sounding 1.44 figure. • This non-standard labeling is a famous example of why the IEC later formalized KiB, MiB, and GiB to distinguish binary from decimal measurements. • 💡 Option A (1,440,000 bytes) is wrong because it uses 1000 as the KB multiplier, not 1024; Option B (1,512,000 bytes) is wrong because no standard conversion produces this number; Option C (1,048,576 bytes) is wrong because that equals exactly 1 MiB, which is slightly less than 1440 KiB.

Correct Answer: A. Bits per second

• **Bits per second (bps)** is the standard metric for data transfer speed because network signals transmit individual 0s and 1s across the medium, and bps directly counts how many of those binary digits pass per second. • **Real-world speeds use larger prefixes** — home broadband is rated in Mbps (Megabits per second), while fiber optic backbone links reach Gbps or even Tbps, all using bits, not bytes, as the base unit. • To convert bps to bytes-per-second, divide by 8 — a 100 Mbps connection downloads files at a maximum of 12.5 MB/s. • 💡 Option B (Bytes per minute) is wrong because network speed is standardly measured per second, not per minute, and uses bits not bytes; Option C (Pages per second) is wrong because pages have no fixed byte size and are not a data-rate unit; Option D (Megabytes per hour) is wrong because this is an informal unit, not the industry standard for connection speed.

Correct Answer: C. 16

• **2 Bytes = 16 bits**, because each byte contains 8 bits, so multiplying 2 × 8 gives 16 — a direct application of the fixed 8-bit definition of a byte. • **16 bits was the word size of early personal computers** — the Intel 8086 processor (used in original IBM PCs) operated on 16-bit words, meaning it processed 2 bytes at a time and could represent integers from 0 to 65,535. • In networking, the IPv4 header checksum field is 16 bits (2 bytes), and TCP/UDP port numbers are also stored in 16-bit fields. • 💡 Option A (8) is wrong because 8 bits equals only 1 byte, not 2; Option B (12) is wrong because 12 bits is not a standard grouping and doesn't result from 2 × 8; Option D (32) is wrong because 32 bits equals 4 bytes, not 2.

Correct Answer: D. 0.5 PB

• **0.5 PB** is the largest among the four options — 0.5 PB = 512 TB, which dwarfs 1 TB, 100 GB, and 10,000 MB when all are converted to a common unit. • **Converting all options to GB for comparison**: 100 GB = 100 GB; 1 TB = 1024 GB; 10,000 MB ≈ 9.77 GB; 0.5 PB = 524,288 GB — the Petabyte-scale value is hundreds of thousands of times larger. • This type of cross-unit comparison is commonly tested to ensure understanding of the relative magnitudes across the memory hierarchy. • 💡 Option A (100 GB) is wrong because it equals about 0.1 TB, far smaller than 0.5 PB; Option B (1 TB) is wrong because 1 TB = 1/512 of 0.5 PB; Option C (10,000 MB) is wrong because it equals approximately 9.77 GB, the smallest of all four options.

Correct Answer: D. Binary units

• **MebiByte (MiB) = 2^20 = 1,048,576 bytes** exactly, introduced by the IEC in 1998 to create an unambiguous binary unit — while 'Megabyte' had been used loosely for both 1,000,000 and 1,048,576 bytes depending on context. • **The confusion arose because disk manufacturers used decimal (1 MB = 1,000,000 bytes) while OS developers used binary (1 MB = 1,048,576 bytes)** — this made a '500 GB' hard drive appear smaller in Windows, which was reporting in MiB but calling it MB. • The IEC binary prefixes — KiB, MiB, GiB, TiB — each mean exactly a power of 1024, cleanly separating them from the SI decimal prefixes KB, MB, GB, TB. • 💡 Option A (Fractional units) is wrong because MiB has no relationship to fractions; Option B (Octal units) is wrong because octal is base-8, unrelated to the MiB/MB distinction; Option C (Hexadecimal units) is wrong because hex is base-16 and was not the source of this naming confusion.

Correct Answer: A. 700 MB

• **Standard CD-ROM = 700 MB** (approximately 703 MB or 80 minutes of audio), a capacity defined by the Orange Book and Red Book standards that governed compact disc manufacturing worldwide. • **700 MB was revolutionary in the 1990s** — earlier floppy disks held only 1.44 MB, so a single CD could replace over 480 floppy disks, enabling software like Windows 95 to ship on a single disc. • CDs store data as microscopic pits and lands on a spiral track read by a 780 nm infrared laser spinning at 200–500 RPM. • 💡 Option B (4.7 GB) is wrong because that is the capacity of a standard single-layer DVD, not a CD; Option C (1.44 MB) is wrong because that is the capacity of a 3.5-inch floppy disk; Option D (25 GB) is wrong because that is the capacity of a Blu-ray disc.

Correct Answer: C. 64

• **A 64-bit processor's word contains 64 binary digits**, because the word size is defined as the number of bits the CPU's registers, ALU, and data bus can handle in a single operation. • **64-bit processors can address 2^64 memory locations** — theoretically 16 exabytes of RAM — far beyond what 32-bit CPUs could address (only 4 GB), enabling modern servers to use terabytes of RAM. • All modern operating systems (Windows 10/11, macOS, Linux) are 64-bit, taking full advantage of the wider word size for faster integer arithmetic and larger memory access. • 💡 Option A (8) is wrong because 8 bits is the size of a byte, not a 64-bit word; Option B (32) is wrong because 32 bits is the word size of older x86 processors, not 64-bit CPUs; Option D (128) is wrong because 128-bit processors are not standard commercial CPUs.

Correct Answer: C. Terabyte

• **Terabyte (TB) = 1024 Gigabytes**, sitting one level above GB in the binary memory hierarchy and representing approximately 1,099,511,627,776 bytes (about one trillion bytes). • **TB is the standard unit for modern mass storage** — consumer hard drives, NAS devices, and backup systems are routinely sold in 1 TB, 2 TB, 4 TB, and 8 TB capacities, making it the most relevant large storage unit today. • 1 TB can store roughly 200,000 songs, 500 hours of HD video, or about 250,000 photos at 4 MB each. • 💡 Option A (Megabyte) is wrong because 1 MB = 1/1024 GB, making it smaller than GB, not larger; Option B (Petabyte) is wrong because 1 PB = 1024 TB, making it 1024 times larger than TB, not GB; Option D (Exabyte) is wrong because 1 EB = 1024 PB, far above the GB level.

Correct Answer: A. 64 Bytes

• **512 bits ÷ 8 bits/byte = 64 bytes** — a straightforward conversion using the fixed definition that every byte contains exactly 8 bits. • **64 bytes is the standard cache line size in modern Intel and AMD CPUs** — when the processor fetches data from RAM, it always retrieves 64 bytes at once (512 bits), even if only 1 byte is needed, to exploit spatial locality. • Understanding bit-to-byte conversion is foundational for network programming, where packet headers and protocol fields are specified in exact bit counts. • 💡 Option B (32 Bytes) is wrong because 32 × 8 = 256 bits, not 512; Option C (128 Bytes) is wrong because 128 × 8 = 1024 bits, which is double the given value; Option D (256 Bytes) is wrong because 256 × 8 = 2048 bits, four times the given value.