Question 1

DOS memory management (CONFIG.SYS/AUTOEXEC) Which directive enables the use of Upper Memory Blocks (UMBs) so that device drivers and TSRs can be loaded high?

Accepted Answer

Introduction / Context: In MS-DOS memory management, conserving conventional memory (the first 640 KB) is crucial. Loading device drivers and TSRs into Upper Memory Blocks (UMBs) frees conventional memory for applications, improving compatibility and performance of legacy software. Given Data / Assumptions: CONFIG.SYS and AUTOEXEC.BAT are used with HIMEM.SYS and EMM386.EXE (or equivalent) to manage memory. Question asks which setting allows UMB usage so items can be loaded high. Concept / Approach: Two things are needed: 1) enable UMBs, and 2) actually load modules into them. The directive DOS=UMB tells DOS to use Upper Memory Blocks. Once UMBs exist, you can use devicehigh=... in CONFIG.SYS and loadhigh (LH) ... in AUTOEXEC.BAT to place drivers and TSRs into UMBs. The enabling step is DOS=UMB. Step-by-Step Solution: Ensure HIMEM.SYS and EMM386.EXE (with the RAM switch) are loaded to create UMBs.Add the directive: DOS=HIGH,UMB to move DOS to HMA and allow UMB usage. The key part for UMB availability is “UMB”.Then use devicehigh= and loadhigh to place specific drivers/TSRs into UMBs. Verification / Alternative check: Microsoft DOS manuals show DOS=UMB as the switch that permits DOS to manage UMBs, while DOS=HIGH moves DOS into the High Memory Area (HMA). Why Other Options Are Wrong: DOS=High only moves DOS to HMA, not enabling UMBs by itself. devicehigh= and loadhigh= are commands used after UMBs are enabled; they are not the enabling directive. “None of the above” is incorrect because DOS=UMB is correct. Common Pitfalls: Using devicehigh/loadhigh without first enabling UMBs; forgetting EMM386 RAM switch; expecting DOS=HIGH alone to create UMBs. Final Answer: DOS=UMB.

Question 2

Field testing a fuse with a multimeter On a 2 A circuit, what resistance reading (Ohms) indicates a good fuse when measured out of circuit?

Accepted Answer

Introduction / Context: Technicians routinely check fuses during no-power or intermittent faults. A fuse is simply a low-resistance conductor that opens under overcurrent. Understanding expected resistance helps avoid unnecessary part swaps. Given Data / Assumptions: The fuse is removed from the circuit to avoid parallel paths. Digital multimeter is functioning and zeroed (or using relative mode). Fuse rating is 2 A, but rating does not materially change its intact resistance being near zero. Concept / Approach: A good fuse presents very low resistance (ideally approaching 0 Ohms). In practice, contact resistance and meter lead resistance may show a small value, but conceptually a “good” reading is near zero. An open fuse will read infinite or very high resistance. Step-by-Step Solution: Set meter to low-Ohms or continuity mode.Measure across fuse terminals out of circuit.Interpret near-zero as good; open/infinite as blown. Verification / Alternative check: Continuity beepers confirm a low resistance path; visual inspection sometimes shows a broken element, but measurement is more reliable. Why Other Options Are Wrong: 0.2 Ohm may occur due to lead resistance, but the best textbook answer is 0.0 to represent a closed conductor. 5.0 and 2.0 Ohms are far too high for a fuse and indicate excessive drop or a measurement error. Common Pitfalls: Measuring in-circuit (parallel paths give false readings); not zeroing lead resistance; assuming a visually intact fuse is necessarily good. Final Answer: 0.0.

Question 3

Installing an IDE/PATA boot hard drive Choose the correct sequence to prepare a new drive so it will boot DOS/Windows on a classic PC.

Accepted Answer

Introduction / Context: Proper IDE/PATA setup ensures the BIOS recognizes the drive and the OS can boot. The classic DOS/Windows installation flow requires correct hardware configuration followed by partitioning and formatting steps that place system files on the boot volume. Given Data / Assumptions: Legacy IDE hardware with Master/Slave/Cable-Select jumpers. BIOS supports auto-detection of drive geometry. We need a bootable primary partition with system files. Concept / Approach: The correct order is: configure jumpers; ensure BIOS detects the drive; create partitions with FDISK (or similar); perform a high-level format that writes a filesystem and transfers system files (format /s in DOS). Low-level formats are factory operations on modern drives and are not part of user setup. Step-by-Step Solution: Set drive jumpers to Master/Slave/CS as appropriate.Enter BIOS setup and auto-detect or manually set parameters.Create a primary active partition (e.g., with FDISK).High-level format the partition and transfer system files (format /s). Verification / Alternative check: Classic installation guides and OEM manuals list this sequence; deviations often result in non-bootable systems. Why Other Options Are Wrong: Options including low-level format are outdated for user procedures. Partitioning must precede formatting. A “regular format” without /s may omit boot files on pure DOS setups. Common Pitfalls: Forgetting to set the partition active; mixing Master/Slave jumpers; assuming BIOS sees the disk without detection; skipping the system file transfer step on bare DOS installs. Final Answer: jumper properly, auto-detect (or configure manually) in CMOS, partition, high-level format with /s switch.

Question 4

Intel 8088 architecture basics Pick the correct combination for the Intel 8088 CPU: internal word size and external data bus width.

Accepted Answer

Introduction / Context: The Intel 8088 powered the original IBM PC. Technicians studying PC history or supporting embedded systems based on 8088 derivatives should know its internal and external bus characteristics. Given Data / Assumptions: 8088 is closely related to the 8086, but with a narrower external data bus. Word size refers to the CPU's internal register width and instruction set data size. Concept / Approach: The 8088 has a 16-bit internal architecture (registers and ALU) but an 8-bit external data bus, enabling cheaper system designs using 8-bit components while retaining 16-bit internal processing—one reason IBM chose it for the first PC. Step-by-Step Solution: Identify internal size: 16-bit (like 8086).Identify external bus: 8-bit (unlike the 8086's 16-bit bus).Select “16 bit word size, 8 bit data path.” Verification / Alternative check: Intel datasheets and PC technical references confirm 8088's 8-bit external data bus and 16-bit internal registers (AX, BX, etc.). Why Other Options Are Wrong: Options B and E assign an 8-bit word size, which is incorrect. Option C describes the 8086, not the 8088. Option D assigns a 4-bit word size, which is unrelated to 8088. Common Pitfalls: Confusing 8088 with 8086; equating external bus width with CPU word size; overlooking the cost/design motivations for using the 8088 in early PCs. Final Answer: 16 bit word size, 8 bit data path.

Question 5

Electrostatic Discharge (ESD) Risk Conditions  Under which environmental conditions are electronics most at risk of electrostatic discharge when handling components?

Accepted Answer

Introduction / Context: Electrostatic discharge (ESD) can silently damage sensitive semiconductor devices such as CPUs, memory, and logic ICs. Understanding when ESD risk peaks helps technicians choose proper handling procedures and protective equipment. Given Data / Assumptions: The question asks for environmental conditions that make ESD most likely. Humidity and temperature both influence charge accumulation and retention on people and materials. We assume normal indoor work with common clothing and flooring. Concept / Approach: Dry air is a poor conductor and allows charges to accumulate on surfaces without bleeding off. Cooler air typically holds less moisture, increasing dryness. Therefore, cold and dry conditions yield the highest surface resistances and the greatest buildup of static charges, raising the probability and severity of ESD events when a charged person touches a grounded device. Step-by-Step Solution: Identify the key factor: relative humidity strongly affects static dissipation.Relate humidity to temperature: colder air generally means lower absolute humidity indoors, especially in heated winter environments.Infer that cold + dry leads to higher static buildup on people and tools.Select the option that matches these conditions. Verification / Alternative check: ESD control handbooks recommend keeping relative humidity between about 40% and 60% and using humidification in winter months. Facilities often see more ESD incidents during winter when air is cold outside and dry indoors after heating. Why Other Options Are Wrong: Cold and wet: Moisture provides surface conduction that dissipates static charge. Hot and wet: High humidity reduces static buildup. Hot and dry: Risky, but typically less common than wintertime cold-and-dry indoor conditions where humidity is very low. None of the above: Incorrect because one option is clearly correct. Common Pitfalls: Assuming temperature alone determines ESD risk; ignoring humidity control; neglecting ESD straps or mats because components “seem fine.” Latent ESD damage can shorten component life without immediate failure. Final Answer: Cold and dry

Question 6

PC Firmware Terminology  In the term “ROM BIOS,” what does the acronym BIOS correctly stand for?

Accepted Answer

Introduction / Context: The Basic Input Output System (BIOS) is foundational firmware that initializes hardware and starts the operating system. Knowing the correct expansion of BIOS helps distinguish it from OS software and other firmware terms like UEFI and ROM. Given Data / Assumptions: BIOS resides in non-volatile memory (historically ROM, now flash). It provides low-level routines for device initialization and bootstrapping. The question asks only for the correct expansion of the acronym. Concept / Approach: BIOS stands for Basic Input Output System. It exposes basic I/O services and firmware interfaces that allow a PC to perform POST, discover bootable devices, and hand off control to a bootloader or OS kernel. Step-by-Step Solution: Identify the context: ROM BIOS relates to firmware, not an operating system.Recall the canonical definition: Basic Input Output System.Match with the provided options and select the exact phrase. Verification / Alternative check: PC hardware documentation, service manuals, and vendor specs universally cite “Basic Input Output System” as the standard expansion for BIOS; UEFI is the modern replacement but the acronym remains historically accurate. Why Other Options Are Wrong: Basic Intuitive Output Set / Basic Input Organizational System / Basic Industry Operating System: These are not industry terms and incorrectly describe either scope or function. None of the above: Incorrect because a correct choice is present. Common Pitfalls: Confusing BIOS with the operating system; assuming BIOS and UEFI are identical (they are different firmware models). Final Answer: Basic Input Output System

Question 7

DNS Client Configuration  When you add a DNS server's IP address to a client's DNS search order, what does this configuration do?

Accepted Answer

Introduction / Context: Domain Name System (DNS) configuration on a client determines how hostnames are resolved to IP addresses. The DNS “search order” or “server list” directs the client to one or more DNS servers for queries when accessing network and internet resources. Given Data / Assumptions: A client device can be configured with multiple DNS server IPs. Servers are queried in order or based on reachability and caching rules. The item in question is the effect on the client, not the DNS server. Concept / Approach: Supplying a DNS server IP tells the client where to send name-resolution requests. It does not change the client’s own IP or restrict activity beyond how names are resolved. If the first server fails, the client may try the next server in the list. Step-by-Step Solution: Recognize that name resolution is a client-initiated query to a DNS server.Adding the server IP gives the client a destination for DNS queries.Therefore, the configuration informs the local system of the DNS server’s location.Select the option that precisely states this behavior. Verification / Alternative check: Network tools (nslookup, dig) show queries being sent to configured DNS servers. Packet captures confirm UDP/TCP traffic to port 53 at the listed DNS server IP. Why Other Options Are Wrong: Restricts browsing: DNS choice doesn’t inherently restrict web access. Overrides local IP: DNS settings don’t change client addressing. Tells the DNS server where the client is located: Not part of static client configuration. None of the above: Incorrect because a correct statement is present. Common Pitfalls: Assuming DNS changes affect routing; misunderstanding search suffixes versus server list; forgetting fallback to secondary DNS when primary is unreachable. Final Answer: It tells the local system where the DNS Server is located

Question 8

Troubleshooting “Fixed Disk Error” at Boot  Which of the following is the most likely cause of a BIOS “fixed disk error” during startup on a legacy PC?

Accepted Answer

Introduction / Context: Old BIOS firmware relied on manual CMOS setup for fixed disk (hard drive) geometry and mode. If these settings were wrong or the battery failed, boot could halt with a “fixed disk error.” Understanding this helps isolate boot failures on legacy systems. Given Data / Assumptions: Legacy BIOS with manual drive configuration (CHS/LBA, drive type). “Fixed disk” refers to hard drives, not optical media. Symptoms occur during POST or initial boot handoff. Concept / Approach: Incorrect CMOS settings for the hard disk (geometry, translation, enabled/disabled, master/slave) commonly trigger “fixed disk error.” A dead CMOS battery can also revert settings. CD-ROM presence is unrelated to the hard drive error, and RAM timing issues usually produce memory errors, not “fixed disk” errors. Step-by-Step Solution: Enter BIOS Setup and verify the hard disk is detected with correct parameters.If not auto-detecting, manually set the correct CHS or LBA mode as required.Replace/charge CMOS battery if settings are lost after power-off.Check cables and jumpers only if settings are correct but error persists. Verification / Alternative check: After fixing CMOS parameters, the drive should pass POST and allow the OS loader to proceed. Drive detection screens will list correct model and capacity. Why Other Options Are Wrong: CD-ROM not installed: Optical drives do not cause “fixed disk” errors. Incorrect RAM settings: Typically yield memory test failures or system instability, not a fixed-disk-specific message. All of the above: Incorrect because not all listed items can cause this specific error. Common Pitfalls: Replacing the HDD before verifying CMOS; overlooking cable orientation; forgetting to save BIOS changes; ignoring failing CMOS batteries that lose settings between boots. Final Answer: Incorrect CMOS settings

Question 9

PC Power Supply Output Type  The regulated output rails delivered by a computer power supply to the motherboard and peripherals are provided as which type of current?

Accepted Answer

Introduction / Context: Switch-mode power supplies (SMPS) in PCs convert AC mains into stable DC voltages for logic and storage devices. Knowing that the outputs are DC is essential for correct measurement, troubleshooting, and safety practices. Given Data / Assumptions: Typical ATX/ITX PC power supplies provide +12 V, +5 V, +3.3 V DC rails. Mains input is AC, but the output to the system is DC. Terminology like “trickle” refers to standby DC power (+5 VSB). Concept / Approach: The SMPS rectifies and filters incoming AC, then uses high-frequency switching and regulation to produce stable DC outputs. Digital electronics require DC for proper biasing and operation. Measurements of rails should be made on DC voltage scale with a multimeter. Step-by-Step Solution: Identify the function of the PC PSU: conversion from AC mains to DC rails.Recall standard output rails: +12 V, +5 V, +3.3 V, and +5 V standby.Choose “DC” as the correct current type delivered to components.Confirm by reading PSU label or motherboard manual. Verification / Alternative check: PSU specification labels, ATX standards, and motherboard manuals explicitly list DC outputs. Oscilloscope and DMM measurements confirm DC with small ripple components within spec. Why Other Options Are Wrong: AC: Present at the input, not supplied to the logic rails. Amperage / resistive / trickle: Not types of current; “trickle” is a standby DC rail concept, not the main output type. Common Pitfalls: Measuring rails on AC scale; confusing ripple with AC output; overlooking the standby rail behavior when the system is “off.” Final Answer: DC

Question 10

Memory Map Basics (DOS/Legacy PCs)  Where are Upper Memory Blocks (UMBs) located in the classic PC memory map?

Accepted Answer

Introduction / Context: Legacy DOS systems divided the first megabyte of address space into conventional memory, the upper memory area, and reserved regions for BIOS and device ROMs. Understanding where Upper Memory Blocks (UMBs) reside helps when optimizing memory with CONFIG.SYS and EMM utilities. Given Data / Assumptions: Conventional memory spans 0–640 KB. The Upper Memory Area (UMA) spans 640 KB to 1 MB. UMBs are blocks carved from reserved address space in the UMA when possible. Concept / Approach: UMBs are allocated within the reserved upper memory area (640 KB–1 MB) by mapping unused ROM shadow or adapter space for loading drivers and TSRs. They are not part of conventional memory (first 640 KB) and are distinct from extended memory (above 1 MB) and expanded memory (paged EMS windows). Step-by-Step Solution: Recall the memory map: 0–640 KB conventional; 640 KB–1 MB upper/reserved; above 1 MB extended.Identify that UMBs originate in the UMA reserved space.Choose “Reserved Memory” to reflect that UMBs are carved from the UMA.Note EMS uses a 64 KB window in UMA but is conceptually different from UMBs. Verification / Alternative check: DOS memory tools (MEM /C, EMM386, HIMEM) and classic references show UMBs appearing within the upper/ reserved region, not in conventional or extended memory. Why Other Options Are Wrong: Conventional Memory: Below 640 KB; UMBs are above this. Extended Memory: Above 1 MB; not the UMA. Expanded Memory: EMS is paging-based and different, though it can use a window in the UMA. None of the above: Incorrect because “Reserved Memory” is correct in this context. Common Pitfalls: Confusing EMS and UMBs; assuming any free RAM is “conventional”; overlooking chipset/BIOS shadowing that limits available UMB space. Final Answer: Reserved Memory

Question 11

Electronics Recycling Priorities  Among the listed PC-related items, which is typically the easiest to recycle responsibly through established programs?

Accepted Answer

Introduction / Context: Electronics recycling varies by component type. Some parts contain hazardous materials or complex mixtures, while others have mature take-back programs. Technicians should route parts through appropriate channels to minimize environmental impact. Given Data / Assumptions: Common office e-waste includes printer consumables and PC parts. Original equipment manufacturers frequently operate return or refill programs. Question asks for the easiest item to recycle via established programs. Concept / Approach: Toner cartridges have widespread OEM and third-party collection programs, prepaid return labels, and refurbish/refill pipelines. By contrast, motherboards require specialized processing to recover metals; CMOS batteries are hazardous and need specific handling; CRTs have leaded glass and are costly to process. Step-by-Step Solution: List item categories and known recycling ecosystems.Identify toner cartridges as having the most accessible, user-friendly programs.Select “Toner cartridges” as the easiest component to recycle.Recommend using OEM mail-back or local collection points. Verification / Alternative check: Most major printer brands provide free mail-in recycling; many office supply retailers accept used cartridges at the counter, confirming accessibility and scale of these programs. Why Other Options Are Wrong: Motherboards: Require specialized smelting and regulated downstream. CMOS batteries: Hazardous waste; must be handled separately. CRTs: Difficult due to leaded glass; many regions restrict disposal and require fees. Common Pitfalls: Discarding cartridges in general waste; mixing batteries with regular recyclables; ignoring data-bearing parts that need secure destruction before recycling. Final Answer: Toner cartridges

Question 12

IBM Micro Channel Architecture (MCA)  Which bus width combinations were specified by IBM’s Micro Channel Architecture for desktop systems?

Accepted Answer

Introduction / Context: IBM introduced Micro Channel Architecture (MCA) as a successor to ISA in PS/2 systems. Recognizing MCA's bus widths helps place it historically between ISA/EISA and later PCI standards in terms of performance and capability. Given Data / Assumptions: MCA targeted higher throughput and improved configuration (software setup). It was used mainly in IBM PS/2 lines, not broadly adopted industry-wide. Question asks for supported data path widths. Concept / Approach: MCA provided 16-bit and 32-bit bus implementations, improving bandwidth over 8/16-bit ISA and providing architectural features like bus mastering and better arbitration. It predated PCI but did not attain broad third-party support. Step-by-Step Solution: Recall ISA: 8/16-bit; EISA: 32-bit; MCA: 16/32-bit proprietary.Eliminate incorrect width pairs that don’t match historical specs.Select “16 bit & 32 bit.”Note MCA's role in late 1980s/early 1990s PCs. Verification / Alternative check: IBM technical references and period documentation specify 16-bit and 32-bit MCA slots and adapters, confirming the correct combination. Why Other Options Are Wrong: 8/16-bit: Matches ISA, not MCA. 16/24-bit: Not a standard pairing. 32/64-bit: 64-bit arrived later in other architectures, not in MCA desktops. Common Pitfalls: Confusing MCA with EISA or early PCI; assuming MCA reached industry-wide usage similar to PCI. Final Answer: 16 bit & 32 bit

Question 13

Parallel Port Data Transfer Concept  A standard PC parallel port (Centronics/IEEE 1284 in basic mode) communicates to a printer in what granularity per transfer?

Accepted Answer

Introduction / Context: Legacy printer ports are called “parallel” because they transmit multiple bits simultaneously. Understanding how much data moves per strobe clarifies cabling, timing, and why parallel ports historically offered higher throughput than serial ports of the same era. Given Data / Assumptions: Classic parallel port has 8 data lines (D0–D7) for output. Handshake lines synchronize byte transfers. Enhanced modes exist (EPP/ECP), but base concept is still byte granularity. Concept / Approach: With 8 dedicated data lines, one strobe transfers 8 bits simultaneously, which is exactly one byte. Control and status lines manage flow control and acknowledgments. Even advanced modes conceptually operate around byte streams, though they may burst more efficiently. Step-by-Step Solution: Identify physical data width: 8 data pins in the basic parallel interface.Associate 8 bits with one byte.Conclude the transfer granularity per strobe is one byte.Select the option stating “one byte at a time.” Verification / Alternative check: Parallel port timing diagrams show strobe and acknowledge around an 8-bit data value on D0–D7. Printer and PC technical manuals reinforce this byte-wide transfer model. Why Other Options Are Wrong: One bit at a time: That describes serial links, not parallel. 8 bytes at a time / 16,555 bytes at a time: Not aligned with the physical data pins or protocol handshakes. None of the above: Incorrect because one choice is accurate. Common Pitfalls: Assuming “parallel” means arbitrary multi-byte chunks; overlooking handshake overhead that affects effective throughput versus theoretical width. Final Answer: one byte at a time

Question 14

Dial-Up Troubleshooting Logic  Your modem does not respond when attempting to dial out. You try another modem on another computer and the result is the same (no dialing). What is the most likely root cause?

Accepted Answer

Introduction / Context: Effective troubleshooting isolates variables. When multiple devices and systems show the same failure, the shared dependency is suspect. With dial-up modems, the public switched telephone network (PSTN) line is a common external dependency beyond the PC and modem hardware. Given Data / Assumptions: Two different modems and two different computers were tested. Both attempts failed to dial out in the same manner. The telephone line is common between tests (same service or jack). Concept / Approach: If different hardware and software platforms reproduce the same symptom, the problem likely resides in the external line: no dial tone, line dead, incorrect cabling, or service outage. This aligns with the principle of controlling variables to locate the failing link in the chain. Step-by-Step Solution: Note commonality: the phone line is shared between tests while hardware/software changed.Infer that the phone line is the most probable failure point.Confirm by plugging a regular handset into the line to check for dial tone.If no tone, contact the telecom provider or repair inside wiring. Verification / Alternative check: Line testers or a simple analog phone can validate dial tone, polarity, and noise. Modem logs often show “NO DIALTONE” when the line is dead or miswired (e.g., plugged into PHONE rather than LINE jack). Why Other Options Are Wrong: Defective modem / defective computer: Unlikely since different units show the same issue. Software bug: Cross-system repetition reduces probability. Virus: Not a typical cause of identical modem failures across two systems in basic dialing tests. Common Pitfalls: Ignoring the physical line; misconnecting the wall cable to the PHONE jack; not checking for PBX or tone/pulse settings; overlooking line filters or splitters causing issues. Final Answer: bad phone line

Question 15

PC hardware upgrade planning:  When planning an upgrade for a specific motherboard, which key information must you identify first to ensure compatibility of CPU, memory, and expansion options?

Accepted Answer

Introduction / Context: Selecting parts that truly fit a given PC depends on the exact motherboard you are targeting. Chipset, socket type, BIOS support, memory standards, and I/O headers all hinge on the board’s precise identity. Knowing only the RAM or storage tells you little about CPU sockets or firmware support, so the first step is always to identify the motherboard’s make and model. Given Data / Assumptions: You plan to upgrade CPU, RAM, or storage on an existing PC. Compatibility depends on socket, chipset, form factor, memory standard, and BIOS. The question asks what you must know first for a particular motherboard. Concept / Approach: Motherboard identity (brand and model) uniquely determines chipset, CPU socket, supported RAM type and speed, M.2/SATA layouts, and maximum capacities. From the board’s support page you can read CPU QVL lists, RAM QVL lists, BIOS updates, and slot bandwidth details. Without this, you risk purchasing parts that do not work or that operate at degraded performance. Step-by-Step Solution: Find the model silkscreen on the board or query it via system info tools.Check the vendor’s support page for CPU list, memory QVL, and BIOS versions.Match your desired upgrade (CPU/RAM/SSD/GPU) to the published compatibility.Update BIOS if required and proceed with the upgrade. Verification / Alternative check: Cross-check with third-party lists and forums, but rely on official support pages for final confirmation. If in doubt, test with known-good components before purchase. Why Other Options Are Wrong: Memory / DIMM: helpful, but RAM type depends on the board’s model and chipset. Hard drive: storage is broadly compatible; not the primary determinant. None of the above: incorrect because make & model is essential. Common Pitfalls: Assuming any CPU fits any LGA/AM socket revision; ignoring BIOS microcode; mixing DDR generations; overlooking power delivery limits. Final Answer: make & model

Question 16

PC boot sequence fundamentals:  During a normal PC cold boot, which firmware or component becomes active first to begin hardware initialization and system checks?

Accepted Answer

Introduction / Context: Understanding the boot sequence helps you troubleshoot no-POST, no-boot, and device-detection issues. The very first code executed by the CPU after reset is firmware stored in non-volatile memory. Historically this is referred to as the ROM BIOS (or UEFI firmware in modern systems). Given Data / Assumptions: Cold boot scenario on an IBM-PC–compatible system. Firmware is resident in a ROM/flash device on the motherboard. CMOS holds configuration data; it is not executable code. Concept / Approach: On power-up the CPU fetches its first instructions from a fixed reset vector mapped to the system firmware. This firmware (ROM BIOS/UEFI) initializes the chipset, memory controller, timers, and buses, then reads configuration parameters stored in CMOS/NVRAM. Storage devices and boot sectors are queried later during the handoff to the OS loader. Step-by-Step Solution: CPU resets and fetches instructions from the firmware address space.ROM BIOS/UEFI begins POST, configuring memory and base devices.CMOS configuration is read by the firmware to apply user settings.After device discovery, firmware selects a boot device and loads the OS bootloader. Verification / Alternative check: Motherboard manuals confirm that CMOS is only a data store for settings, not executable code. Flash updates change the ROM content used at this first stage. Why Other Options Are Wrong: CMOS: a configuration store, not “active” code. Hard disk information: accessed later by firmware after POST. RAM BIOS / None of the above: non-standard terms or incorrect. Common Pitfalls: Confusing “CMOS setup” with the actual firmware; assuming disks initialize before firmware; overlooking the impact of corrupted firmware. Final Answer: ROM BIOS

Question 17

Clock-doubled CPUs:  A PC with an Intel 486DX2 processor runs internally at 50 MHz. What speed does its external bus (front-side bus) operate at?

Accepted Answer

Introduction / Context: Many classic processors (e.g., 486DX2) used internal multipliers to run the CPU core faster than the external front-side bus. Recognizing the relationship between internal clock and bus clock is essential when evaluating memory throughput and expansion card timing. Given Data / Assumptions: CPU is a 486DX2, which uses a 2x internal clock multiplier. Internal core frequency is 50 MHz. We are asked for the external bus frequency. Concept / Approach: The 486DX2 multiplies the external bus frequency by 2 to obtain the internal CPU frequency. Therefore, bus_clock * 2 = internal_clock. Rearranging gives bus_clock = internal_clock / 2. With a 50 MHz core, the bus runs at 25 MHz. Step-by-Step Solution: Let internal_clock = 50 MHz.For DX2, internal_clock = 2 * bus_clock.Compute bus_clock = 50 / 2 = 25 MHz.Answer is 25 MHz. Verification / Alternative check: Historical specs list 486DX2-50 as 25 MHz FSB with 2x multiplier. Similarly, a 486DX2-66 uses 33 MHz FSB with 2x multiplier to reach 66 MHz internal. Why Other Options Are Wrong: 10 MHz and 100 MHz: do not satisfy the 2x relation for a 50 MHz core. 50 MHz: that would be a DX (1x) or DX4 at different settings, not DX2. None of the above: incorrect because 25 MHz fits perfectly. Common Pitfalls: Confusing DX2 (2x) with DX4 (3x) parts; forgetting that memory and ISA bus derived timings follow the external bus, not the internal multiplier. Final Answer: 25Mhz

Question 18

Dot-matrix printer diagnostics:  On an impact (dot-matrix) printer, what common cause leads to a print pattern that alternates between light and dark (banding) across the page?

Accepted Answer

Introduction / Context: Dot-matrix printers deposit ink by striking a fabric ribbon against paper. Consistency of ribbon movement is critical to maintain uniform ink density. “Light–dark” banding is often a ribbon feed issue rather than a head or paper motion problem. Given Data / Assumptions: Symptoms: alternating light and dark horizontal bands across printed output. Printer type: impact/dot-matrix with fabric ribbon cartridge or spool. Mechanisms involved: ribbon feed, head impact, paper tractor feed. Concept / Approach: If the ribbon does not advance smoothly, the same section of ribbon is reused too frequently and becomes depleted (light), while a suddenly advanced segment appears darker. This creates a visible periodic variation in density independent of paper motion. Paper feed issues usually cause skewing, misregistration, or stretching patterns rather than neat alternating density bands. Step-by-Step Solution: Inspect ribbon cartridge for jams, stiffness, or worn gears.Verify that the ribbon is correctly threaded and tensioned.Replace the ribbon if frayed, dry, or unevenly inked.Print a test pattern to confirm uniform density. Verification / Alternative check: Swapping in a known-good ribbon that prints uniformly confirms the diagnosis. If banding persists, check head solenoids and platen pressure, but ribbon feed remains the prime suspect. Why Other Options Are Wrong: Overheating print head: would cause dropout or stalling, not periodic dark/light bands. Misaligned head: leads to positional errors, not density oscillation. Erratic paper advancement: causes spacing errors or skew, not alternating ink density. Common Pitfalls: Overlooking worn ribbon drive gears; attempting platen pressure adjustments before verifying ribbon condition; using expired ribbons. Final Answer: Erratic ribbon advancement

Question 19

Hayes AT command set basics:  Which standard AT command is used to hang up (disconnect) a modem connection?

Accepted Answer

Introduction / Context: Classic dial-up and many embedded modems use the Hayes-compatible AT command set for control. Technicians often need to drop an existing call from a terminal session or script; the correct command prevents confusion and unintended resets. Given Data / Assumptions: Modem is Hayes-compatible and currently connected or in command mode. Goal is to terminate the connection and go on-hook. Familiar AT mnemonics: D for dial, H for hook, Z for reset. Concept / Approach: The hook control command is “H”. With the AT prefix, “ATH” instructs the modem to go on-hook (disconnect). If the modem is in data mode, you may need to escape to command mode (for example, +++ with guard times) before issuing ATH. Step-by-Step Solution: Ensure modem is in command mode (escape if necessary).Type ATH and press Enter.Wait for “OK” response confirming the line is dropped.Optionally issue ATZ to reset and reload defaults. Verification / Alternative check: Vendor manuals and terminal tests confirm ATH drops carrier reliably. Why Other Options Are Wrong: ATD: dials; it is the opposite of hang-up. ATZ: resets the modem, not the canonical hang-up. ATR: not the hang-up command; function varies by device. None of the above: incorrect because ATH is correct. Common Pitfalls: Forgetting the escape sequence; confusing reset with hang-up; scripts that issue ATZ unnecessarily and alter profiles. Final Answer: ATH

Question 20

Disk geometry fundamentals:  In traditional PC storage formats, what is the standard size of a sector on a magnetic hard disk?

Accepted Answer

Introduction / Context: Understanding sector size helps with low-level disk utilities, partition alignment, and data recovery. For decades, the de facto standard sector size on PC-class hard disks has been 512 bytes, even as logical block size abstraction evolved. Given Data / Assumptions: Conventional MBR-era drives used 512-byte sectors. Advanced Format drives may expose 4K physical sectors but still emulate 512-byte logical blocks. The question asks the standard, widely taught sector size. Concept / Approach: Sector size is the minimum addressable logical block on legacy disks. While modern Advanced Format introduces 4096-byte physical sectors, logical 512-byte addressing remains common for compatibility. Therefore, the standard sector size answer in entry-level PC tech contexts is 512 bytes. Step-by-Step Solution: Recall legacy PC conventions: sector = 512 bytes.Note that 256 and 1024 bytes are not standard legacy sector sizes for PCs.512 Kb is far larger than a single sector.Select 512 bytes as the correct value. Verification / Alternative check: Disk utilities (fdisk-like tools) and OS APIs traditionally report 512-byte logical sector sizes unless specifically configured otherwise. Why Other Options Are Wrong: 256 bytes / 1024 bytes: not the established PC sector standard. 512 Kb: six orders of magnitude larger than legacy sector size. None of the above: incorrect since 512 bytes is right. Common Pitfalls: Confusing physical 4K sectors with logical emulation; mixing filesystem cluster sizes (which are larger) with disk sector size. Final Answer: 512 bytes

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