Computer Science Previous Year Questions (page 10)

Explanation

Automatically giving the boy an airplane, a toy assumed to be "for boys", reflects a gender stereotype.

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Bloom's taxonomy describes three learning domains: cognitive, affective and psychomotor. Spiritual is not one of them.

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Family type (nuclear), urbanization and religious institutions are socio-cultural influences on a child. Examinations are part of the school system, not a socio-cultural factor.

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The tribal child is excluded from peer groups and his interests are ignored by the teacher, which pushes him to the edge of the group. This is marginalization.

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Evaluation means judging or critiquing, so verbs such as criticise, judge and appraise belong to it. Define and identify are knowledge-level verbs.

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Education policies framed by the government and authorities belong to the political dimension of the socio-cultural environment.

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Later childhood (about 6-12 years) is called the "gang age", because children like to form peer groups.

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Moral ideas such as honesty, empathy and idealism develop strongly in adolescence, along with abstract thinking.

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Adolescents tend to daydream and fantasise and to live in an imaginary, idealised world, so this stage is described as the child becoming "unreal".

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The program counter always holds the address of the next instruction to be fetched; it is incremented (or loaded with a branch target) after each fetch.

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An interface unit sits between the CPU and a peripheral and resolves differences in speed, data format, signal values and timing between them.

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On a data input command the interface takes the data received from the peripheral into its buffer register and places it from that register onto the data bus, so the CPU can read it.

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When BG = 1 the CPU has given up the buses and the DMA controller is the bus master, so its RD and WR lines become outputs that control memory reads and writes.

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In powers of 1024: yottabyte (2⁸⁰) > zettabyte (2⁷⁰) > exabyte (2⁶⁰) > petabyte (2⁵⁰) > gigabyte (2³⁰) > megabyte (2²⁰).

Explanation

Required capacity is 128K × 8 bits. Each chip is 16K × 1 bit, so (128K/16K) × (8/1) = 8 × 8 = 64 chips.

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The minuend is the number from which another is subtracted. Here (011100)₂ is subtracted from (1011100)₂, so (1011100)₂ is the minuend and (011100)₂ is the subtrahend.

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The control unit directs and coordinates every other part of the computer by sending control signals, just as the nervous system coordinates the body.

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An assembler translates assembly-language mnemonics into machine code. Compilers and interpreters translate high-level languages, and a linker joins object files.

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n bits can form 2ⁿ distinct bit patterns, so an n-bit opcode can encode 2ⁿ different operations.

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(10110)₂ = 22 and (10011)₂ = 19, so the product is 22 × 19 = 418. 418 = 256 + 128 + 32 + 2 = (110100010)₂.

Explanation

(3564)₈ = 1908 and (323232)₄ = 3822. The sum is 5730, and 5730 = 1×4096 + 6×256 + 6×16 + 2, which is (1662)₁₆.

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+28 in 16 bits is 0000000000011100. Inverting the bits gives 1111111111100011, and adding 1 gives 1111111111100100.

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For base 7, the 6's complement (diminished radix complement) is found by subtracting each digit from 6: 6666 − 1543 = (5123)₇.

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The OS keeps each process's state, program counter, registers, scheduling and memory information in its Process Control Block (PCB).

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Throughput is the number of processes completed per unit time. Turnaround time is the time to complete one process, and response time is the time to first response.

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Memory allocation and managing resources for efficient utilization are both core functions of an operating system.

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The four Coffman conditions are mutual exclusion (non-sharable resources), hold and wait, NO preemption and circular wait. Preemption itself prevents deadlock, so it is not a condition.

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Shortest Seek Time First is disk scheduling, Shortest Job First is CPU scheduling, Least Recently Used is page replacement, and the Banker's algorithm is deadlock avoidance.

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Semaphores can be implemented in software (e.g., with OS/kernel code or algorithms like Peterson's) and with hardware support such as test-and-set or atomic instructions.

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Best fit: 10 KB→12 KB (2 KB left), 2 KB→the 2 KB hole, 14 KB→15 KB, so all three fit. First, next and worst fit put 10 KB into the 15 KB block, leaving no block for 14 KB.

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FIFO with 3 frames: faults on 3, 4, 2; 3 hits; then faults on 9 (out 3), 1 (out 4), 6 (out 2), 3 (out 9), 4 (out 1), 9 (out 6). Total = 9 faults.

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A process goes from running to waiting only when it waits for I/O or for an event. A lower-priority arrival never affects the running process (a higher-priority arrival may preempt it to ready).

Explanation

SSTF from 50: 34 (16), 20 (14), 19 (1), 15 (4), 10 (5), 7 (3), 6 (1), 4 (2), 2 (2), 83 (81). The total movement is 129 cylinders, so it takes 129 seconds.

Explanation

By the consensus theorem, BC is redundant: BC = BC(A + A') = ABC + A'BC, which is absorbed by AB and A'C. So AB + A'C + BC = AB + A'C.

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NAND (and NOR) are universal gates because any Boolean function, including NOT, AND and OR, can be built using only NAND gates.

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Each of the n variables can be 0 or 1, giving 2ⁿ input combinations, so the truth table has 2ⁿ rows.

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In BCD each decimal digit is coded separately in 4 bits: 4 = 0100, 5 = 0101, 8 = 1000, 9 = 1001, giving 0100 0101 1000 1001.

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DCTL stands for Direct-Coupled Transistor Logic, an early logic family in which transistors are connected directly without coupling resistors.

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A POS expression is a product of maxterms (OR terms), and an SOP expression is a sum of minterms (AND terms).

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A decimal-to-BCD encoder has 10 inputs (one for each digit 0-9) and 4 outputs (the 4-bit BCD code).

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A demultiplexer takes one input and routes it to one of several outputs according to select lines. A multiplexer does the opposite.

Explanation

The 1-cells are A'B'C'D' and ABC'D' (CD=00), A'BC'D and AB'C'D (CD=01), and ABCD (CD=11). Adding these five minterms gives option (B).

Explanation

25 = (11001)₂. For 0.625: ×2 = 1.25 (1), 0.25×2 = 0.5 (0), 0.5×2 = 1.0 (1), giving .101. So 25.625 = (11001.101)₂.

Explanation

N select lines can address 2ᴺ inputs, so a multiplexer with M data inputs needs N select lines where 2ᴺ = M.

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An AND gate gives a high output only when all inputs are high, half a byte (4 bits) is a nibble, and a D flip-flop's output follows its D input on the clock edge.

Explanation

27 = (11011)₂. The Gray code keeps the MSB and XORs adjacent bits: 1, 1⊕1=0, 0⊕1=1, 1⊕0=1, 1⊕1=0, giving 10110.

Explanation

Y = (A + BC)(B + AC') = AB + AC' + BC. AB is the consensus term of BC and AC', so Y = BC + AC'. Also (A + C)(B + C') = AB + AC' + BC, the same function, so both are correct.

Explanation

Q0 toggles on every clock and Q1 toggles whenever Q0 goes 0→1 (it is clocked by Q0). From 00: Q0→1 so Q1 toggles (11); then Q0→0 (10); then Q0→1 so Q1 toggles (01). This is a down counter.

Explanation

Portability (running on different machines) is a desirable attribute of a good language, so non-portability is not one. Clarity, abstraction and ease of verification are all good attributes.