Computer Science Previous Year Questions (page 4)

Explanation

In paging, logical memory is divided into pages and physical memory into frames of the same size, so any page fits into any frame.

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a ∧ b → b ∨ c is a tautology: whenever a ∧ b is true, b is true, so b ∨ c is true. The other implications can be false, for example when b and c are false.

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A spanning tree of n vertices keeps exactly n − 1 edges. So from a graph with e edges you remove e − (n − 1) = e − n + 1 edges.

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Treat each nationality as one block: the 3 blocks can be arranged in 3! ways. Inside the blocks, Indians arrange in 8!, Americans in 4! and Englishmen in 4!. Total = 3!·8!·4!·4!.

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Choose the vowels: C(4,2) = 6. Choose the consonants: C(5,3) = 10. Arrange the 5 letters: 5! = 120. Total = 6 × 10 × 120 = 7200.

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In a JK flip-flop, J = 0 and K = 0 is the hold condition, so the output does not change at the clock edge.

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A weak entity set has no primary key of its own. It is identified through a strong (owner) entity set plus its partial key (discriminator).

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AC + ABC = AC(1 + B) = AC, so a single 2-input AND gate is enough.

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Data warehouses store historical, integrated data for analysis, so they support OLAP (Online Analytical Processing). OLTP runs on operational databases.

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Non-volatile memory, such as ROM or flash, keeps its contents when power is lost. RAM (main memory) is volatile.

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The PolyStar tool with the Polygon style, set to 7 sides in its options, draws a heptagon directly. Flash has no separate Polygon tool.

Explanation

Outstanding salary is an expense due but not yet paid, payable within the year, so it is a current liability. Machinery and furniture are fixed assets, and debtors are current assets.

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The balance sheet shows the assets, liabilities and capital of a business on a particular date, so it shows the financial position on that date.

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Wood's Despatch (1854) laid down a full plan for English education in India, so it is called the Magna Carta of English education in India.

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Tina uses the concept of addition in a real-life situation (shopping), which shows the application component of the curriculum.

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Visual impairment is a sensory disability, not a learning disability. Dysgraphia (writing), dyslexia (reading) and dyscalculia (mathematics) are learning disabilities.

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Sarva Shiksha Abhiyan was run by the Ministry of Human Resource Development (now the Ministry of Education).

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Dysgraphia is a learning disability that affects writing: poor handwriting, spelling and difficulty putting thoughts on paper.

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The 86th Amendment Act, 2002 added Article 21A, making free and compulsory education for children aged 6 to 14 a fundamental right.

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Children with dyslexia struggle with reading text. Audio books and recorded material help them learn by listening while they follow the words.

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Putting the curriculum into practice through timetabled lesson plans, teaching strategies and suitable media is curriculum implementation.

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Following Wood's Despatch (1854), universities were set up in 1857 at Calcutta, Bombay and Madras.

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Inclusive teaching means recognising individual differences and adapting or personalising learning materials. It does not mean ignoring, isolating or pitying children with special needs.

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Classic ASP uses VBScript as its default scripting language. JScript can be chosen with a page directive.

Explanation

Bits per frame = 800 × 600 × 8 = 3,840,000. At 40 frames/s this is 3,840,000 × 40 = 153,600,000 bits/s = 153.6 Mbps.

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Quick sort and merge sort are divide-and-conquer algorithms and are normally written recursively. Bubble sort and insertion sort are simple iterative algorithms.

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In a star topology every device has its own point-to-point link to a central hub or controller. In a mesh, every device links directly to every other device.

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IEEE 754 uses 32 bits for single precision, 64 bits for double precision and 80 bits for the x87 extended-precision format.

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Corrected: the book's key says (c). The additional information is carried by the exception object that is thrown.

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Tree A is strictly binary because every node has 0 or 2 children. Tree B is not, because L and R each have only one child. Tree C is a full tree of height 2 with every level filled, so it is complete.

Explanation

DMA (Direct Memory Access) lets a peripheral transfer data directly to or from main memory without the CPU handling each word. The CPU only starts the transfer and is told when it ends.

Explanation

Rows 1 to i−1 hold 1 + 2 + … + (i−1) = i(i−1)/2 elements. A[i, j] is the j-th element of row i, so its location is i(i−1)/2 + j. Check: A[1,1] gives 0 + 1 = 1.

Explanation

In C++ an else pairs with the nearest preceding unmatched if (the dangling-else rule), whatever the indentation. So the else belongs to if#2.

Explanation

The OSI model has 7 layers: Physical, Data Link, Network, Transport, Session, Presentation and Application.

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Bit synchronization is done by the Physical layer, not the Data Link layer. Data Link handles framing, error control and flow control. The other pairs are correct.

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Each character needs 1 start + 7 data + 1 parity + 1 stop = 10 bits. So 19,200 bps ÷ 10 = 1920 characters per second.

Explanation

SSL encrypts data and is traditionally placed at the Presentation layer in OSI terms, because encryption and decryption are presentation-layer jobs. It sits between the application and transport layers.

Explanation

In Java, String is a class (a reference type), not a primitive type. boolean, byte and double are primitives.

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new String[3] creates an array of references, and each element starts with the default value null. So printing var1[1] prints null.

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An EJB (EJB 2.x) bean class comes with two interfaces, the home interface and the remote (or component) interface, plus one XML deployment descriptor (ejb-jar.xml).

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Only 90% of the code was tested, and the success probability is 0.9. So reliability is at most 0.9 × 0.9 = 0.81.

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The parentheses bind * to p first, so p is a pointer, and that pointer points to an array of 5 ints. Without parentheses, int *p[5] would be an array of 5 pointers to int.

Explanation

Since 1~2 and 2~3, all of 1, 2 and 3 fall into one equivalence class. That needs all 3 × 3 = 9 ordered pairs. R already has 2, so 9 − 2 = 7 pairs must be added.

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x²/(1+x²) is 0 at x = 0 and approaches 1 as x grows, but never reaches 1 because x² < 1 + x². So the range is [0, 1).

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Answer as per the book's key (c) 2¹⁶ (the original exam key).

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In ETL, the staging area is the intermediate storage where extracted data is cleaned and transformed before it is loaded into the data warehouse.

Explanation

Seek time is the time to move the read/write head to the correct track. Rotational latency is the wait for the sector to rotate under the head, and access time is the two combined.

Explanation

A 4-bit parallel (ripple-carry) adder uses one half adder for the least significant bit, which has no carry-in, and three full adders for the other three bits.

Explanation

10Base5 is the original Thick Ethernet (thicknet): 10 Mbps, baseband, 500 m segments on thick coaxial cable. 10Base2 is Thin Ethernet.

Explanation

Cache memory is a small, very fast memory between the CPU and main memory. It holds frequently used data and instructions, so programs run faster.