If '->' denotes increasing order of intensity, then [walk -> jog -> sprint] is analogous to [bothered -> ________ -> daunted]. Which option fits the blank?

C. Answer: (C) fazed bothered < fazed < daunted is increasing intensity of being disturbed/discouraged. "Fazed" (disconcerted) is the correct word (note: "phased", "phrased", "fused" are unrelated homophones/words).

Two wizards mix four elements (water, air, fire, earth) in all possible orders, working independently. After trying all orders, the spell fails. How many attempts does each wizard make independently?

A. Answer: (A) 24 The number of orders (permutations) of 4 distinct elements is 4!=24. Each wizard independently tries all 24 orders.

Of 10,000 students, 1,500 like neither core nor other branches. Students liking core branches =\tfrac14 of those liking other branches. 500 like both. The number who like their core branches is

A. Answer: (A) 1,800 Students liking core or other =10000-1500=8500. Let core =c, other =o, with c=o/4 and both =500. |c\cup o|=c+o-500=8500\Rightarrow c+o=9000. Substitute c=o/4: \tfrac{o}{4}+o=9000\Rightarrow o=7200,\ c=1800.

For positive non-zero x,y, if \ln\!\left(\dfrac{x+y}{2}\right)=\dfrac12[\ln x+\ln y], then \dfrac{x}{y}+\dfrac{y}{x} is

C. Answer: (C) 2 RHS =\ln\sqrt{xy}, so \dfrac{x+y}{2}=\sqrt{xy} — the arithmetic mean equals the geometric mean, which happens only when x=y. Then \dfrac{x}{y}+\dfrac{y}{x}=1+1=2.

In the sequence 6, 9, 14, x, 30, 41, a possible value of x is

B. Answer: (B) 21 The successive differences are consecutive odd numbers: 9-6=3, 14-9=5, then 7,9,11. So x=14+7=21, and indeed 21+9=30, 30+11=41.

Sequence the sentences into a coherent passage. P: This fortuitous geological event generated a colossal amount of energy and heat that resulted in the rocks rising to an average height of 4 km across the contact zone. Q: Thus, the geophysicists tend to think of the Himalayas as an active geological event rather than as a static geological feature. R: The natural process of the cooling of this massive edifice absorbed large quantities of atmospheric carbon dioxide, altering the earth's atmospher

C. Answer: (C) SPRQ S introduces the event (collision), P describes its immediate effect (energy, rocks rising), R the subsequent cooling and CO2 absorption, and Q the concluding inference ("Thus..."). Order: S -> P -> R -> Q.

A person sold two different items at the same price, making 10% profit on one and 10% loss on the other. Overall, the person made

C. Answer: (C) 1% loss Selling two items at the same price with equal +x\% and -x\% always gives a net loss of \dfrac{x^2}{100}\%. For x=10: loss =\dfrac{100}{100}=1\%.

Pie charts give shares of power generation. 2007: Coal 35, Gas 25, Wind 5, Hydro 30, Solar 5 (%). 2023: Coal 20, Gas 15, Wind 10, Hydro 35, Solar 20 (%). Renewables = Hydro+Solar+Wind. Total generation unchanged. The percentage increase in the renewable share is

D. Answer: (D) 62.5% Renewable share: 2007 =30+5+5=40\%; 2023 =35+20+10=65\%. Percentage increase =\dfrac{65-40}{40}\times100=\dfrac{25}{40}\times100=62.5\%.

A cube is cut into 8 pieces of equal size and shape; each cut is straight and goes fully through. The minimum number of cuts required is

A. Answer: (A) 3 One straight cut through each of the three axes halves the cube each time: 2\times2\times2=8 equal pieces. So 3 cuts suffice (and no fewer can double the count three times).

In the 4x4 array shown, each cell of the first three rows is a cross (X) or a number. The number in a cell = count of its immediate neighbours (left, right, top, bottom, diagonals) that do NOT contain a cross. The last row has no crosses. The sum of the four numbers to be filled in the last row is

A. Answer: (A) 11 For each cell in the last (4th) row, count its neighbours (in the 3rd and 4th rows) that are NOT crosses. Using the given 3rd-row pattern (3, X, 6, X) and that the whole 4th row is cross-free, the four counts sum to 11.

A 4 MHz processor's DMA controller transfers 8 bytes in 1 cycle via cycle stealing. If 1% of processor cycles are used for DMA, the data transfer rate (bits/sec) is

C. Answer: (C) 25,60,000 Cycles/sec =4\times10^6. DMA uses 1\%: 0.01\times4\times10^6=40000 cycles/sec. Each DMA cycle moves 8 bytes =64 bits. Rate =40000\times64=2{,}560{,}000 bits/sec.

p: Fail grade can be given. q: Student scores more than 50%. Statement: "Fail grade cannot be given when student scores more than 50% marks." Correct representation?

A. Answer: (A) q\to\neg p "When the student scores more than 50% (q), a fail grade cannot be given (\neg p)" is exactly q\to\neg p.

Parameters are evaluated right to left. int g(int p){printf("%d",p);return p;} int h(int q){printf("%d",q);return q;} void f(int x,int y){ g(x); h(y); } int main(){ f(g(10),h(20)); } Correct output?

A. Answer: (A) 20101020 f(g(10), h(20)): arguments evaluated right-to-left, so h(20) prints 20, then g(10) prints 10 -> "2010". Then inside f, g(x) prints 10, h(y) prints 20 -> "1020". Total: "2010"+"1020" = "20101020".

IEEE-754 single precision (1 sign, 8 exponent, 23 mantissa). Choose the largest number: (A) 0 | 0111 1111 | 1111...111 (B) 0 | 1111 1110 | 1111...111 (C) 0 | 1111 1111 | 1111...111 (D) 0 | 0111 1111 | 0000...000

B. Answer: (B) Exponent 11111111 (all ones) is reserved for infinity/NaN, so (C) is not a finite number. Among the valid finite numbers, the largest exponent is 11111110 (254) in (B), with a full mantissa -> this is the largest representable finite value (\approx 3.4\times10^{38}). (A) and (D) have exponent 127 (much smaller).

T(0)=1, T(1)=2, T(n)=5T(n-1)-6T(n-2) for n\ge2. Which is TRUE?

A. Answer: (A) T(n)=\Theta(2^n) Characteristic equation x^2-5x+6=0\Rightarrow x=2,3. General solution T(n)=A\cdot2^n+B\cdot3^n. Using T(0)=1=A+B and T(1)=2=2A+3B: solving gives B=0,\ A=1, so T(n)=2^n=\Theta(2^n). (The 3^n term vanishes.)

f is continuous with f(x)=1-f(2-x). The value of \int_0^2 f(x)\,dx is

B. Answer: (B) 1 Let I=\int_0^2 f(x)\,dx. Substituting x\to2-x: I=\int_0^2 f(2-x)\,dx=\int_0^2 (1-f(x))\,dx=2-I. So 2I=2\Rightarrow I=1.

A is the adjacency matrix of a simple undirected graph G, and A is its own inverse (A^2=I). Which is always TRUE?

B. Answer: (B) G is a perfect matching A^2=I means (A^2)_{ii}=1 for all i and 0 off-diagonal. (A^2)_{ii}=\deg(i), so every vertex has degree 1. A simple graph in which every vertex has degree exactly 1 is a perfect matching (a disjoint union of edges), and one checks A^2=I holds. So G must be a perfect matching.

Six unbiased dice are rolled. The probability of getting all distinct numbers (1..6) is

B. Answer: (B) \dfrac{5}{324} Favourable outcomes =6! (all permutations). Total =6^6. P=\frac{6!}{6^6}=\frac{720}{46656}=\frac{5}{324}.

Once a transaction is reported complete, its effect persists even if the system crashes before changes reach disk. This property is

A. Answer: (A) durability Durability guarantees that the effects of a committed transaction survive subsequent failures (via logging/recovery).

In the ER model with owner and weak entity sets, which statement is TRUE?

A. Answer: (A) A weak entity has no key of its own and is existence-dependent on its owner, so it MUST participate totally in the identifying relationship. The owner need not have total participation.

Match Set X to Set Y: P. Lexical Analyzer, Q. Syntax Analyzer, R. Intermediate Code Generator, S. Code Optimizer 1. Abstract Syntax Tree, 2. Token, 3. Parse Tree, 4. Constant Folding.

B. Answer: (B) P-2; Q-3; R-1; S-4 Lexical analyzer produces Tokens (2); syntax analyzer produces the Parse Tree (3); intermediate code generation works from the Abstract Syntax Tree (1); code optimization includes Constant Folding (4).

Which regular expression is equivalent to the language accepted by the DFA shown (states over \{0,1\})?

A. Answer: (A) 0^*1(0+10^*1)^* From the DFA, strings reach the accepting state by first taking any number of 0's then a 1 (0^*1), after which the self/return transitions allow repeating patterns of a single 0 or a 1 followed by 0^*1: (0+10^*1)^*. This exactly describes the accepted language.

X has a TCP connection to Y through IP router R; Ethernet switch S is the first switch on the path X-R. For a packet from X to Y when it leaves X, which statements are TRUE?

B. Answer: (B) The destination IP is the final destination's IP (Y) end-to-end -> (B) true, (A) false. The destination MAC is the next-hop L2 device, which is the first-hop router R (not the switch S, which forwards transparently by MAC, and not Y which is remote). So (C) and (D) are false.

Which tasks is/are the responsibility of the MMU in paging-based memory management?

B, C, D. Answer: (B), (C) and (D) The MMU (hardware) translates virtual->physical addresses using the page table (B), and raises faults/traps on invalid translations (page fault) (C) and on permission violations like writing a read-only page (D). Allocating a new page table (A) is done by the OS, not the MMU.

Which events ALWAYS trigger a context switch moving process P to a non-running state?

A, B. Answer: (A) and (B) - (A) A blocking syscall puts P into the blocked state and yields the CPU -> always a context switch. - (B) A major page fault (data in swap) blocks P while the page is fetched -> always a context switch. - (C) A disk interrupt for another process does not necessarily move P out of running. - (D) A timer interrupt does not always preempt P (only if the OS decides to reschedule).

Which file organization(s) is/are I/O efficient for the scan operation?

A, B. Answer: (A) and (B) Scanning reads all records sequentially. A Sorted file and a Heap file store records contiguously, so a full scan reads pages sequentially and is I/O efficient. Unclustered indexes (tree or hash) point to records scattered across pages, so scanning through them causes random I/O and is inefficient.

Which statements about the 2PL protocol is/are TRUE?

A, C, D. Answer: (A), (C) and (D) 2PL guarantees conflict-serializable schedules (A). Its defining rule is a growing phase then a shrinking phase: once any lock is released, no new lock may be acquired (C). 2PL can still deadlock (D). (B) is false — that describes a non-2PL lock-just-around-each-operation discipline, which violates the two-phase rule.

Which statements about IPv4 fragmentation is/are TRUE?

B, C. Answer: (B) and (C) In IPv4, any router may fragment a datagram if it exceeds the next link's MTU (B). Reassembly is done only at the final destination (C), never at intermediate routers (D). So (A) is false (not only the source) and (D) is false.

Which statements is/are FALSE?

B, D. Answer: (B) and (D) - (B) is FALSE: L-attributed definitions are specifically designed so that all attributes CAN be evaluated in a single depth-first (left-to-right) traversal. - (D) is FALSE: not every L-attributed SDD on an LR grammar can be evaluated bottom-up (inherited attributes can be problematic); only the S-attributed subset is always bottom-up evaluable. - (A) and (C) are TRUE statements.

For a Boolean variable x, which statements is/are FALSE?

B, C. Answer: (B) and (C) - (B) FALSE: x+1=1 (not x). - (C) FALSE: x\cdot x=x (idempotent), not 0. - (A) TRUE (x\cdot1=x); (D) TRUE (complement law).

Instructions: I1: DIV R3,R1,R2; I2: SUB R5,R3,R4; I3: ADD R3,R5,R6; I4: MUL R7,R3,R8. Which dependencies hold?

A. Answer: (A) I1: DIV R3,... I2: SUB R5,R3,... I3: ADD R3,... I4: MUL R7,R3,... - (A) I1 writes R3, I2 reads R3 -> Read-After-Write on R3 between I1 and I2. TRUE. - (B) A WAR needs an earlier READ then a later WRITE; I1 does not read R3, so there is no WAR between I1 and I3 (the actual WAR is between I2 and I3). FALSE. - (C) I2 reads R3 and I3 writes R3 - that is WAR, not RAW (RAW needs a write then a read, but I2 does not write R3). FALSE. - (D) I3 writes R3 and I4 only reads R3 - no WAW between

Which IP header field(s) is/are ALWAYS modified by any router before forwarding?

C, D. Answer: (C) and (D) Every router decrements the TTL (C); since a field changed, it must recompute the Header Checksum (D). Source IP and Protocol are not changed by an ordinary router.

int fX(char *a){ char *b=a; while(*b) b++; return b-a; } Which statements is/are TRUE?

A, B, D. Answer: (A), (B) and (D) fX computes the string length by advancing to the terminating '\0'. It works for any valid, null-terminated string: a string literal argument (A), a char array (B), and a char pointer to a literal (D) all give length 4. The code compiles fine, so (C) is false.

Partial order P=\{(x,x)\}\cup\{(1,2),(3,2),(3,4)\} on \{1,2,3,4\}. The number of total orders on \{1,2,3,4\} containing P is __________

5. Answer: 5 We need linear extensions respecting 1<2, 3<2, 3<4. Enumerating permutations consistent with these constraints: 3 1 2 4, 3 1 4 2, 1 3 2 4, 1 3 4 2, 3 4 1 2 There are exactly 5 such total orders.

Distance of array A = minimum number of elements to replace so A becomes non-decreasing. The distance of [2,5,3,1,4,2,6] is __________

3. Answer: 3 Elements that can stay form a longest non-decreasing subsequence (LNDS); the rest must be replaced. For [2,5,3,1,4,2,6], an LNDS is 2,3,4,6 (length 4). So replacements =7-4=3.

double a[2]={20.0,25.0},*p,*q; p=a; q=p+1; printf("%d,%d",(int)(q-p),(int)(*q-*p)); Output?

B. Answer: (B) 1,5 q-p is pointer difference in units of elements =1. *q-*p = 25.0-20.0 = 5.0, cast to int =5. Output: "1,5".

Processes A(0,10), B(2,6), C(4,3), D(6,7) (arrival, burst). Give the average waiting times for preemptive SRTF and non-preemptive SJF.

B. Answer: (B) SRTF=6, NP-SJF=7.5 SRTF timeline: A runs 0-2; B(6) avg =24/4=6. NP-SJF: A 0-10; then ready {B,C,D}, shortest C 10-13; B 13-19; D 19-26. Waiting: A=0,B=11,C=6,D=13 -> avg =30/4=7.5.

Which CIDR prefix exactly represents the range 10.12.2.0 to 10.12.3.255?

A. Answer: (A) 10.12.2.0/23 The range spans 2\times256=512 addresses =2^9, so the host part is 9 bits and the prefix is 32-9=23. The block 10.12.2.0/23 covers 10.12.2.0–10.12.3.255 exactly (the 3rd octet varies over 2 and 3).

A BST T is built by inserting all elements of a set V in some unknown order. Which can be determined from V alone?

A. Answer: (A) The inorder traversal of any BST is the sorted order of its keys, which is determined by the set V regardless of insertion order. The root, preorder, and postorder all depend on the (unknown) insertion sequence.

For S\to AaAb\mid BbBa, A\to cS\mid\epsilon, B\to dS\mid\epsilon, fill the partially-filled LL(1) table cells (1),(2),(3),(4). Which combination is correct?

A. Answer: (A) FIRST(AaAb)=\{c,a\} (A can be \epsilon) and FIRST(BbBa)=\{d,b\}. So under c: S\to AaAb (cell 1); under d: S\to BbBa (cell 2). For A: FOLLOW(A) includes a and b, so A\to\epsilon appears under b (cell 3). For B: B\to\epsilon appears under a (cell 4). This matches (A).

The 5-state NFA M (with \epsilon-transitions) shown accepts which language?

B. Answer: (B) 0^*+(1+0(00)^*)(11)^* Tracing the NFA: from the start there are \epsilon-moves to a branch counting 0's (loop of two 0's, with accept allowing 0^*-like behaviour) and a branch for a 1 followed by pairs of 1's. Combining the accepting paths yields 0^*+(1+0(00)^*)(11)^*, matching option (B).

A C snippet computes the longest subarray with at most two distinct integers. len1 = longest suffix of equal values, len2 = longest suffix with <=2 distinct values. Fill (P) and (Q):

B. Answer: (B) (P) 1 (Q) len1+1 When X[i] equals the second-most-recent distinct value (the else if branch), the run of equal values restarts at this element, so len1 = 1 (P). When X[i] is a brand-new third value (the else branch), the at-most-two-distinct window can only extend the previous equal-run plus this element, so len2 = len1+1 (Q).

For x[i]=(p+r)*-s[i]+u/w, fill the missing triples (1),(3),(6). Which option is correct?

A. Answer: (A) The three-address form of x[i]=(p+r)*-s[i]+u/w uses the triples: (0) + p r ; (1) =[] s i (fetch s[i]) ; (2) uminus (1) ; (3) * (0)(2) ; (4) / u w ; (5) + (3)(4) ; (6) []= x i (address of x[i]) ; (7) = (6)(5). So the missing entries are (1) =[] s i, (3) * (0)(2), and (6) []= x i - option (A).

x,y take values in [0,1] (not necessarily independent), z=xy, means \bar x,\bar y,\bar z. Which is TRUE?

D. Answer: (D) \bar z\le\bar x Since 0\le y\le1, we have xy\le x pointwise (as x\ge0), so taking expectations \bar z=E[xy]\le E[x]=\bar x. The others fail without independence (e.g. \bar z can be >\bar x\bar y with positive correlation).

Person(pid, city). Using selection, projection, cross product, rename, the minimum number of cross-product operations to find cities where at least 3 persons reside is

B. Answer: (B) 2 To require three distinct persons in the same city, we need three copies of Person joined on equal city with distinct pids. Combining three renamed copies needs 2 cross-product (join) operations: (P_1\times P_2)\times P_3, then select equal cities and pairwise-distinct pids, then project the city.

T1: wait(s1); x=x+1; print(x); wait(s2); signal(s1). T2: wait(s1); x=x+1; print(x); signal(s2); signal(s1). s1=1, s2=0, x=0. Which outcomes are possible?

B, C. Answer: (B) and (C) - If T2 acquires s1 first: it prints 1, signals s2, signals s1; then T1 acquires s1, prints 2, wait(s2) succeeds (s2=1), signal(s1). Works -> (B) possible. - If T1 acquires s1 first: it prints 1, then wait(s2) blocks (s2=0) while still holding s1; T2 can never get s1 -> deadlock -> (C) possible. - (A) impossible (T1 first always deadlocks). (D) impossible (T2 first prints 1 then T1 prints 2 is exactly (B), not a new distinct option as worded — the valid non-deadlock case is (

B is obtained from A (n\times n real) by swapping two rows. Which statements is/are TRUE?

A, B. Answer: (A) and (B) Swapping two rows negates the determinant (A). Since |\det B|=|\det A|\ne0 when A is invertible, B is also invertible (B). A row swap generally destroys symmetry (C false) and changes which diagonal entries appear, so the trace can change (D false).

S1 (capacity 4) holds 100,200,300,400 (400 top); S2 (capacity 2) empty. Operations: PushToS2 (S1->S2), PushToS1 (S2->S1), GenerateOutput (pop S1 to output). No pop on empty / push on full. Which output sequences can be generated?

B, C, D. Answer: (B), (C) and (D) This is a stack-permutation question with the auxiliary stack S2 (capacity 2) allowing limited reordering. Simulating the allowed moves, sequences (B) 200,300,400,100, (C) 400,200,100,300, and (D) 300,200,400,100 are each reachable, while (A) 100,200,400,300 cannot be produced under the capacity constraints.

Which is/are EQUAL to 224 in radix-5?

A, B, D. Answer: (A), (B) and (D) 224_5=2\cdot25+2\cdot5+4=64_{10}. - (A) 64_{10}=64. Equal. - (B) 100_8=64_{10}. Equal. - (C) 50_{16}=80_{10}. Not equal. - (D) 121_7=1\cdot49+2\cdot7+1=64_{10}. Equal.

Given f_1=\sum(0,2,3,5,7,8,11,13), f_2=\sum(1,3,5,7,11,13,15), f_3=\sum(0,1,4,11), f_4=\sum(0,2,6,13) feeding AND (f_1,f_2) and OR (f_3,f_4), whose outputs feed an XOR to give Y. Which options are CORRECT?

C, D. Answer: (C) and (D) Let g_1=f_1\cdot f_2=\sum(3,5,7,11,13) (common minterms) and g_2=f_3+f_4=\sum(0,1,2,4,6,11,13) (union). Then Y=g_1\oplus g_2= minterms in exactly one of them. Computing the symmetric difference gives Y=\sum(0,1,2,3,4,5,6,7) — option (C). Equivalently Y=\Pi(8,9,10,11,12,13,14,15) (the complementary maxterms) — option (D). (A) and (B) are different minterm sets.

G is connected with positive integer edge weights, and every spanning tree has even weight. Which statements are TRUE for every such G?

D. Answer: (D) If all spanning trees have the same parity (even), then swapping one tree edge for a non-tree edge on its fundamental cycle must not change the parity, which forces the two swapped edges to have the same parity. Propagating this around each cycle means every cycle's edges share one parity -> (D). (A),(B),(C) are too strong: edges in different cycles may have different parities as long as within each cycle they match.

GATE CS 2024 Set 2 — All 65 Questions with Solutions

Computer Science (CS). Every question from this paper with the official answer key and a step-by-step solution. 100 marks total.

Attempt in GATE exam mode →All papers

Question 1General Aptitudemcq1 mark
If '->' denotes increasing order of intensity, then [walk -> jog -> sprint] is analogous to [bothered -> ________ -> daunted]. Which option fits the blank?
- A.
  phased
- B.
  phrased
- C.
  fazed
  ✓
- D.
  fused
Answer: C
Solution
Answer: (C) fazed bothered < fazed < daunted is increasing intensity of being disturbed/discouraged. "Fazed" (disconcerted) is the correct word (note: "phased", "phrased", "fused" are unrelated homophones/words).
Question 2General Aptitudemcq1 mark
Two wizards mix four elements (water, air, fire, earth) in all possible orders, working independently. After trying all orders, the spell fails. How many attempts does each wizard make independently?
- A.
  24
  ✓
- B.
  48
- C.
  16
- D.
  12
Answer: A
Solution
Answer: (A) 24 The number of orders (permutations) of 4 distinct elements is $4! = 24$ . Each wizard independently tries all 24 orders.
Question 3General Aptitudemcq1 mark
Of 10,000 students, 1,500 like neither core nor other branches. Students liking core branches $= \frac{1}{4}$ of those liking other branches. 500 like both. The number who like their core branches is
- A.
  1,800
  ✓
- B.
  3,500
- C.
  1,600
- D.
  1,500
Answer: A
Solution
Answer: (A) 1,800 Students liking core or other $= 10000 - 1500 = 8500$ . Let core $= c$ , other $= o$ , with $c = o /4$ and both $= 500$ . $∣ c \cup o ∣ = c + o - 500 = 8500 \Rightarrow c + o = 9000$ . Substitute $c = o /4$ : $\frac{o}{4} + o = 9000 \Rightarrow o = 7200, c = 1800.$
Question 4General Aptitudemcq1 mark
For positive non-zero $x, y$ , if $ln (\frac{x + y}{2}) = \frac{1}{2} [ln x + ln y]$ , then $\frac{x}{y} + \frac{y}{x}$ is
- A.
  1
- B.
  1/2
- C.
  2
  ✓
- D.
  4
Answer: C
Solution
Answer: (C) 2 RHS $= ln x y$ , so $\frac{x + y}{2} = x y$ — the arithmetic mean equals the geometric mean, which happens only when $x = y$ . Then $\frac{x}{y} + \frac{y}{x} = 1 + 1 = 2$ .
Question 5General Aptitudemcq1 mark
In the sequence 6, 9, 14, x, 30, 41, a possible value of $x$ is
- A.
  25
- B.
  21
  ✓
- C.
  18
- D.
  20
Answer: B
Solution
Answer: (B) 21 The successive differences are consecutive odd numbers: $9 - 6 = 3$ , $14 - 9 = 5$ , then $7, 9, 11$ . So $x = 14 + 7 = 21$ , and indeed $21 + 9 = 30$ , $30 + 11 = 41$ .
Question 6General Aptitudemcq2 marks
Sequence the sentences into a coherent passage. P: This fortuitous geological event generated a colossal amount of energy and heat that resulted in the rocks rising to an average height of 4 km across the contact zone. Q: Thus, the geophysicists tend to think of the Himalayas as an active geological event rather than as a static geological feature. R: The natural process of the cooling of this massive edifice absorbed large quantities of atmospheric carbon dioxide, altering the earth's atmosphere and making it better suited for life. S: Many millennia ago, a breakaway chunk of bedrock from the Antarctic Plate collided with the massive Eurasian Plate.
- A.
  QPSR
- B.
  QSPR
- C.
  SPRQ
  ✓
- D.
  SRPQ
Answer: C
Solution
Answer: (C) SPRQ S introduces the event (collision), P describes its immediate effect (energy, rocks rising), R the subsequent cooling and CO2 absorption, and Q the concluding inference ("Thus..."). Order: S -> P -> R -> Q.
Question 7General Aptitudemcq2 marks
A person sold two different items at the same price, making 10% profit on one and 10% loss on the other. Overall, the person made
- A.
  1% profit
- B.
  2% profit
- C.
  1% loss
  ✓
- D.
  2% loss
Answer: C
Solution
Answer: (C) 1% loss Selling two items at the same price with equal $+ x %$ and $- x %$ always gives a net loss of $\frac{x ^{2}}{100} %$ . For $x = 10$ : loss $= \frac{100}{100} = 1%$ .
Question 8General Aptitudemcq2 marks
Pie charts give shares of power generation. 2007: Coal 35, Gas 25, Wind 5, Hydro 30, Solar 5 (%). 2023: Coal 20, Gas 15, Wind 10, Hydro 35, Solar 20 (%). Renewables = Hydro+Solar+Wind. Total generation unchanged. The percentage increase in the renewable share is
- A.
  25%
- B.
  50%
- C.
  77.5%
- D.
  62.5%
  ✓
Answer: D
Solution
Answer: (D) 62.5% Renewable share: 2007 $= 30 + 5 + 5 = 40%$ ; 2023 $= 35 + 20 + 10 = 65%$ . Percentage increase $= \frac{65 - 40}{40} \times 100 = \frac{25}{40} \times 100 = 62.5%.$
Question 9General Aptitudemcq2 marks
A cube is cut into 8 pieces of equal size and shape; each cut is straight and goes fully through. The minimum number of cuts required is
- A.
  3
  ✓
- B.
  4
- C.
  7
- D.
  8
Answer: A
Solution
Answer: (A) 3 One straight cut through each of the three axes halves the cube each time: $2 \times 2 \times 2 = 8$ equal pieces. So 3 cuts suffice (and no fewer can double the count three times).
Question 10General Aptitudemcq2 marks
In the 4x4 array shown, each cell of the first three rows is a cross (X) or a number. The number in a cell = count of its immediate neighbours (left, right, top, bottom, diagonals) that do NOT contain a cross. The last row has no crosses. The sum of the four numbers to be filled in the last row is
- A.
  11
  ✓
- B.
  10
- C.
  12
- D.
  9
Answer: A
Solution
Answer: (A) 11 For each cell in the last (4th) row, count its neighbours (in the 3rd and 4th rows) that are NOT crosses. Using the given 3rd-row pattern (3, X, 6, X) and that the whole 4th row is cross-free, the four counts sum to 11.
Question 11Computer Organization & Architecturemcq1 mark
A 4 MHz processor's DMA controller transfers 8 bytes in 1 cycle via cycle stealing. If 1% of processor cycles are used for DMA, the data transfer rate (bits/sec) is
- A.
  2,56,000
- B.
  3,200
- C.
  25,60,000
  ✓
- D.
  32,000
Answer: C
Solution
Answer: (C) 25,60,000 Cycles/sec $= 4 \times 1 0^{6}$ . DMA uses $1%$ : $0.01 \times 4 \times 1 0^{6} = 40000$ cycles/sec. Each DMA cycle moves 8 bytes $= 64$ bits. Rate $= 40000 \times 64 = 2, 560, 000$ bits/sec.
Question 12Discrete Mathematicsmcq1 mark
$p$ : Fail grade can be given. $q$ : Student scores more than 50%. Statement: "Fail grade cannot be given when student scores more than 50% marks." Correct representation?
- A.
  $q \to \neg p$
  ✓
- B.
  $q \to p$
- C.
  $p \to q$
- D.
  $\neg p \to q$
Answer: A
Solution
Answer: (A) $q \to \neg p$ "When the student scores more than 50% ( $q$ ), a fail grade cannot be given ( $\neg p$ )" is exactly $q \to \neg p$ .
Question 13Programming & Data Structuresmcq1 mark
Parameters are evaluated right to left.
```
int g(int p){printf("%d",p);return p;}
int h(int q){printf("%d",q);return q;}
void f(int x,int y){ g(x); h(y); }
int main(){ f(g(10),h(20)); }
```
Correct output?
- A.
  20101020
  ✓
- B.
  10202010
- C.
  20102010
- D.
  10201020
Answer: A
Solution
Answer: (A) 20101020 f(g(10), h(20)): arguments evaluated right-to-left, so h(20) prints 20, then g(10) prints 10 -> "2010". Then inside f, g(x) prints 10, h(y) prints 20 -> "1020". Total: "2010"+"1020" = "20101020".
Question 14Computer Organization & Architecturemcq1 mark
IEEE-754 single precision (1 sign, 8 exponent, 23 mantissa). Choose the largest number:
```
(A) 0 | 0111 1111 | 1111...111
(B) 0 | 1111 1110 | 1111...111
(C) 0 | 1111 1111 | 1111...111
(D) 0 | 0111 1111 | 0000...000
```
- A.
  (A)
- B.
  (B)
  ✓
- C.
  (C)
- D.
  (D)
Answer: B
Solution
Answer: (B) Exponent $11111111$ (all ones) is reserved for infinity/NaN, so (C) is not a finite number. Among the valid finite numbers, the largest exponent is $11111110$ (254) in (B), with a full mantissa -> this is the largest representable finite value ( $\approx 3.4 \times 1 0^{38}$ ). (A) and (D) have exponent 127 (much smaller).
Question 15Algorithmsmcq1 mark
$T (0) = 1, T (1) = 2, T (n) = 5 T (n - 1) - 6 T (n - 2)$ for $n \geq 2$ . Which is TRUE?
- A.
  $T (n) = Θ (2^{n})$
  ✓
- B.
  $T (n) = Θ (n 2^{n})$
- C.
  $T (n) = Θ (3^{n})$
- D.
  $T (n) = Θ (n 3^{n})$
Answer: A
Solution
Answer: (A) $T (n) = Θ (2^{n})$ Characteristic equation $x^{2} - 5 x + 6 = 0 \Rightarrow x = 2, 3$ . General solution $T (n) = A \cdot 2^{n} + B \cdot 3^{n}$ . Using $T (0) = 1 = A + B$ and $T (1) = 2 = 2 A + 3 B$ : solving gives $B = 0, A = 1$ , so $T (n) = 2^{n} = Θ (2^{n})$ . (The $3^{n}$ term vanishes.)
Question 16Engineering Mathematicsmcq1 mark
$f$ is continuous with $f (x) = 1 - f (2 - x)$ . The value of $\int_{0}^{2} f (x) d x$ is
- A.
  0
- B.
  1
  ✓
- C.
  2
- D.
  -1
Answer: B
Solution
Answer: (B) 1 Let $I = \int_{0}^{2} f (x) d x$ . Substituting $x \to 2 - x$ : $I = \int_{0}^{2} f (2 - x) d x = \int_{0}^{2} (1 - f (x)) d x = 2 - I$ . So $2 I = 2 \Rightarrow I = 1$ .
Question 17Engineering Mathematicsmcq1 mark
$A$ is the adjacency matrix of a simple undirected graph $G$ , and $A$ is its own inverse ( $A^{2} = I$ ). Which is always TRUE?
- A.
  G is a cycle
- B.
  G is a perfect matching
  ✓
- C.
  G is a complete graph
- D.
  There is no such graph G
Answer: B
Solution
Answer: (B) G is a perfect matching $A^{2} = I$ means $(A^{2})_{ii} = 1$ for all $i$ and $0$ off-diagonal. $(A^{2})_{ii} = de g (i)$ , so every vertex has degree 1. A simple graph in which every vertex has degree exactly 1 is a perfect matching (a disjoint union of edges), and one checks $A^{2} = I$ holds. So $G$ must be a perfect matching.
Question 18Engineering Mathematicsmcq1 mark
Six unbiased dice are rolled. The probability of getting all distinct numbers (1..6) is
- A.
  $\frac{1}{324}$
- B.
  $\frac{5}{324}$
  ✓
- C.
  $\frac{7}{324}$
- D.
  $\frac{11}{324}$
Answer: B
Solution
Answer: (B) $\frac{5}{324}$ Favourable outcomes $= 6!$ (all permutations). Total $= 6^{6}$ .
$P = \frac{6 !}{6 ^{6}} = \frac{720}{46656} = \frac{5}{324} .$
Question 19Databasesmcq1 mark
Once a transaction is reported complete, its effect persists even if the system crashes before changes reach disk. This property is
- A.
  durability
  ✓
- B.
  atomicity
- C.
  consistency
- D.
  isolation
Answer: A
Solution
Answer: (A) durability Durability guarantees that the effects of a committed transaction survive subsequent failures (via logging/recovery).
Question 20Databasesmcq1 mark
In the ER model with owner and weak entity sets, which statement is TRUE?
- A.
  The weak entity set MUST have total participation in the identifying relationship
  ✓
- B.
  The owner entity set MUST have total participation in the identifying relationship
- C.
  Both weak and owner entity sets MUST have total participation
- D.
  Neither must have total participation
Answer: A
Solution
Answer: (A) A weak entity has no key of its own and is existence-dependent on its owner, so it MUST participate totally in the identifying relationship. The owner need not have total participation.
Question 21Compiler Designmcq1 mark
Match Set X to Set Y: P. Lexical Analyzer, Q. Syntax Analyzer, R. Intermediate Code Generator, S. Code Optimizer <-> 1. Abstract Syntax Tree, 2. Token, 3. Parse Tree, 4. Constant Folding.
- A.
  P-4; Q-1; R-3; S-2
- B.
  P-2; Q-3; R-1; S-4
  ✓
- C.
  P-2; Q-1; R-3; S-4
- D.
  P-4; Q-3; R-2; S-1
Answer: B
Solution
Answer: (B) P-2; Q-3; R-1; S-4 Lexical analyzer produces Tokens (2); syntax analyzer produces the Parse Tree (3); intermediate code generation works from the Abstract Syntax Tree (1); code optimization includes Constant Folding (4).
Question 22Theory of Computationmcq1 mark
Which regular expression is equivalent to the language accepted by the DFA shown (states over ${0, 1}$ )?
- A.
  $0^{*} 1 (0 + 1 0^{*} 1)^{*}$
  ✓
- B.
  $0^{*} (1 0^{*} 11)^{*} 0^{*}$
- C.
  $0^{*} 1 (01 0^{*} 1)^{*} 0^{*}$
- D.
  $0 (1 + 0^{*} 1 0^{*} 1)^{*} 0^{*}$
Answer: A
Solution
Answer: (A) $0^{*} 1 (0 + 1 0^{*} 1)^{*}$ From the DFA, strings reach the accepting state by first taking any number of 0's then a 1 ( $0^{*} 1$ ), after which the self/return transitions allow repeating patterns of a single $0$ or a $1$ followed by $0^{*} 1$ : $(0 + 1 0^{*} 1)^{*}$ . This exactly describes the accepted language.
Question 23Computer Networksmsq1 mark
X has a TCP connection to Y through IP router R; Ethernet switch S is the first switch on the path X-R. For a packet from X to Y when it leaves X, which statements are TRUE?
- A.
  Destination IP is R's IP
- B.
  Destination IP is Y's IP
  ✓
- C.
  Destination MAC is S's MAC
- D.
  Destination MAC is Y's MAC
Answer: B
Solution
Answer: (B) The destination IP is the final destination's IP (Y) end-to-end -> (B) true, (A) false. The destination MAC is the next-hop L2 device, which is the first-hop router R (not the switch S, which forwards transparently by MAC, and not Y which is remote). So (C) and (D) are false.
Question 24Operating Systemsmsq1 mark
Which tasks is/are the responsibility of the MMU in paging-based memory management?
- A.
  Allocate a new page table for a newly created process
- B.
  Translate a virtual address to a physical address using the page table
  ✓
- C.
  Raise a trap when a virtual address is not found in the page table
  ✓
- D.
  Raise a trap when writing to a read-only page
  ✓
Answer: B, C, D
Solution
Answer: (B), (C) and (D) The MMU (hardware) translates virtual->physical addresses using the page table (B), and raises faults/traps on invalid translations (page fault) (C) and on permission violations like writing a read-only page (D). Allocating a new page table (A) is done by the OS, not the MMU.
Question 25Operating Systemsmsq1 mark
Which events ALWAYS trigger a context switch moving process P to a non-running state?
- A.
  P makes a blocking disk read syscall
  ✓
- B.
  P accesses a page in swap space (page fault)
  ✓
- C.
  An interrupt from the disk delivering data for another process
- D.
  A timer interrupt
Answer: A, B
Solution
Answer: (A) and (B) - (A) A blocking syscall puts P into the blocked state and yields the CPU -> always a context switch. - (B) A major page fault (data in swap) blocks P while the page is fetched -> always a context switch. - (C) A disk interrupt for another process does not necessarily move P out of running. - (D) A timer interrupt does not always preempt P (only if the OS decides to reschedule).
Question 26Databasesmsq1 mark
Which file organization(s) is/are I/O efficient for the scan operation?
- A.
  Sorted
  ✓
- B.
  Heap
  ✓
- C.
  Unclustered tree index
- D.
  Unclustered hash index
Answer: A, B
Solution
Answer: (A) and (B) Scanning reads all records sequentially. A Sorted file and a Heap file store records contiguously, so a full scan reads pages sequentially and is I/O efficient. Unclustered indexes (tree or hash) point to records scattered across pages, so scanning through them causes random I/O and is inefficient.
Question 27Databasesmsq1 mark
Which statements about the 2PL protocol is/are TRUE?
- A.
  2PL permits only serializable schedules
  ✓
- B.
  With 2PL, a transaction locks just before every operation and releases just after
- C.
  Once a lock is released in a transaction, no more locks can be obtained in that transaction
  ✓
- D.
  A deadlock is possible with 2PL
  ✓
Answer: A, C, D
Solution
Answer: (A), (C) and (D) 2PL guarantees conflict-serializable schedules (A). Its defining rule is a growing phase then a shrinking phase: once any lock is released, no new lock may be acquired (C). 2PL can still deadlock (D). (B) is false — that describes a non-2PL lock-just-around-each-operation discipline, which violates the two-phase rule.
Question 28Computer Networksmsq1 mark
Which statements about IPv4 fragmentation is/are TRUE?
- A.
  Fragmentation is performed only at the source
- B.
  Fragmentation is performed at any router whose outgoing MTU is exceeded
  ✓
- C.
  Reassembly is performed only at the destination
  ✓
- D.
  Reassembly is performed at all intermediate routers
Answer: B, C
Solution
Answer: (B) and (C) In IPv4, any router may fragment a datagram if it exceeds the next link's MTU (B). Reassembly is done only at the final destination (C), never at intermediate routers (D). So (A) is false (not only the source) and (D) is false.
Question 29Compiler Designmsq1 mark
Which statements is/are FALSE?
- A.
  An attribute grammar is an SDD whose semantic-rule functions have no side effects
- B.
  The attributes in an L-attributed definition cannot always be evaluated in depth-first order
  ✓
- C.
  Synthesized attributes can be evaluated by a bottom-up parser as the input is parsed
- D.
  All L-attributed definitions based on an LR(1) grammar can be evaluated using bottom-up parsing
  ✓
Answer: B, D
Solution
Answer: (B) and (D) - (B) is FALSE: L-attributed definitions are specifically designed so that all attributes CAN be evaluated in a single depth-first (left-to-right) traversal. - (D) is FALSE: not every L-attributed SDD on an LR grammar can be evaluated bottom-up (inherited attributes can be problematic); only the S-attributed subset is always bottom-up evaluable. - (A) and (C) are TRUE statements.
Question 30Digital Logicmsq1 mark
For a Boolean variable $x$ , which statements is/are FALSE?
- A.
  $x \cdot 1 = x$
- B.
  $x + 1 = x$
  ✓
- C.
  $x \cdot x = 0$
  ✓
- D.
  $x + x = 1$
Answer: B, C
Solution
Answer: (B) and (C) - (B) FALSE: $x + 1 = 1$ (not $x$ ). - (C) FALSE: $x \cdot x = x$ (idempotent), not $0$ . - (A) TRUE ( $x \cdot 1 = x$ ); (D) TRUE (complement law).
Question 31Computer Organization & Architecturemsq1 mark
Instructions: I1: DIV R3,R1,R2; I2: SUB R5,R3,R4; I3: ADD R3,R5,R6; I4: MUL R7,R3,R8. Which dependencies hold?
- A.
  RAW on R3 between I1 and I2
  ✓
- B.
  WAR on R3 between I1 and I3
- C.
  RAW on R3 between I2 and I3
- D.
  WAW on R3 between I3 and I4
Answer: A
Solution
Answer: (A) I1: DIV R3,... I2: SUB R5,R3,... I3: ADD R3,... I4: MUL R7,R3,... - (A) I1 writes R3, I2 reads R3 -> Read-After-Write on R3 between I1 and I2. TRUE. - (B) A WAR needs an earlier READ then a later WRITE; I1 does not read R3, so there is no WAR between I1 and I3 (the actual WAR is between I2 and I3). FALSE. - (C) I2 reads R3 and I3 writes R3 - that is WAR, not RAW (RAW needs a write then a read, but I2 does not write R3). FALSE. - (D) I3 writes R3 and I4 only reads R3 - no WAW between them. FALSE. Only (A) holds.
Question 32Computer Networksmsq1 mark
Which IP header field(s) is/are ALWAYS modified by any router before forwarding?
- A.
  Source IP Address
- B.
  Protocol
- C.
  Time to Live (TTL)
  ✓
- D.
  Header Checksum
  ✓
Answer: C, D
Solution
Answer: (C) and (D) Every router decrements the TTL (C); since a field changed, it must recompute the Header Checksum (D). Source IP and Protocol are not changed by an ordinary router.
Question 33Programming & Data Structuresmsq1 mark
int fX(char *a){ char *b=a; while(*b) b++; return b-a; } Which statements is/are TRUE?
- A.
  fX("abcd") will always return a value
  ✓
- B.
  With char c[]="abcd" in main, fX(c) always returns a value
  ✓
- C.
  The function will not compile
- D.
  With char *c="abcd" in main, fX(c) always returns a value
  ✓
Answer: A, B, D
Solution
Answer: (A), (B) and (D) fX computes the string length by advancing to the terminating '\0'. It works for any valid, null-terminated string: a string literal argument (A), a char array (B), and a char pointer to a literal (D) all give length 4. The code compiles fine, so (C) is false.
Question 34Discrete Mathematicsnat1 mark
Partial order $P = {(x, x)} \cup {(1, 2), (3, 2), (3, 4)}$ on ${1, 2, 3, 4}$ . The number of total orders on ${1, 2, 3, 4}$ containing $P$ is __________
Answer: 5
Solution
Answer: 5 We need linear extensions respecting $1 < 2$ , $3 < 2$ , $3 < 4$ . Enumerating permutations consistent with these constraints:
```
3 1 2 4,  3 1 4 2,  1 3 2 4,  1 3 4 2,  3 4 1 2
```
There are exactly 5 such total orders.
Question 35Algorithmsnat1 mark
Distance of array $A$ = minimum number of elements to replace so $A$ becomes non-decreasing. The distance of [2,5,3,1,4,2,6] is __________
Answer: 3
Solution
Answer: 3 Elements that can stay form a longest non-decreasing subsequence (LNDS); the rest must be replaced. For [2,5,3,1,4,2,6], an LNDS is $2, 3, 4, 6$ (length 4). So replacements $= 7 - 4 = 3$ .
Question 36Programming & Data Structuresmcq2 marks
```
double a[2]={20.0,25.0},*p,*q;
p=a; q=p+1;
printf("%d,%d",(int)(q-p),(int)(*q-*p));
```
Output?
- A.
  4,8
- B.
  1,5
  ✓
- C.
  8,5
- D.
  1,8
Answer: B
Solution
Answer: (B) 1,5 q-p is pointer difference in units of elements $= 1$ . *q-*p = 25.0-20.0 = 5.0, cast to int $= 5$ . Output: "1,5".
Question 37Operating Systemsmcq2 marks
Processes A(0,10), B(2,6), C(4,3), D(6,7) (arrival, burst). Give the average waiting times for preemptive SRTF and non-preemptive SJF.
- A.
  SRTF=6, NP-SJF=7
- B.
  SRTF=6, NP-SJF=7.5
  ✓
- C.
  SRTF=7, NP-SJF=7.5
- D.
  SRTF=7, NP-SJF=8.5
Answer: B
Solution
Answer: (B) SRTF=6, NP-SJF=7.5 SRTF timeline: A runs 0-2; B(6)<A(8) runs 2-4; C(3)<B(4) runs 4-7; B 7-11; D 11-18; A 18-26. Waiting = completion-arrival-burst: A=16,B=3,C=0,D=5 -> avg $= 24/4 = 6$ . NP-SJF: A 0-10; then ready {B,C,D}, shortest C 10-13; B 13-19; D 19-26. Waiting: A=0,B=11,C=6,D=13 -> avg $= 30/4 = 7.5$ .
Question 38Computer Networksmcq2 marks
Which CIDR prefix exactly represents the range 10.12.2.0 to 10.12.3.255?
- A.
  10.12.2.0/23
  ✓
- B.
  10.12.2.0/24
- C.
  10.12.0.0/22
- D.
  10.12.2.0/22
Answer: A
Solution
Answer: (A) 10.12.2.0/23 The range spans $2 \times 256 = 512$ addresses $= 2^{9}$ , so the host part is 9 bits and the prefix is $32 - 9 = 23$ . The block $10.12.2.0/23$ covers $10.12.2.0$ – $10.12.3.255$ exactly (the 3rd octet varies over 2 and 3).
Question 39Programming & Data Structuresmcq2 marks
A BST $T$ is built by inserting all elements of a set $V$ in some unknown order. Which can be determined from $V$ alone?
- A.
  Inorder traversal of T
  ✓
- B.
  Root node of T
- C.
  Preorder traversal of T
- D.
  Postorder traversal of T
Answer: A
Solution
Answer: (A) The inorder traversal of any BST is the sorted order of its keys, which is determined by the set $V$ regardless of insertion order. The root, preorder, and postorder all depend on the (unknown) insertion sequence.
Question 40Compiler Designmcq2 marks
For $S \to A a A b ∣ B b B a$ , $A \to c S ∣ ϵ$ , $B \to d S ∣ ϵ$ , fill the partially-filled LL(1) table cells (1),(2),(3),(4). Which combination is correct?
- A.
  (1) $S \to A a A b$ (2) $S \to B b B a$ (3) $A \to ϵ$ (4) $B \to ϵ$
  ✓
- B.
  (1) $S \to B b B a$ (2) $S \to A a A b$ (3) $A \to ϵ$ (4) $B \to ϵ$
- C.
  (1) $S \to A a A b$ (2) $S \to B b B a$ (3) blank (4) blank
- D.
  (1) $S \to B b B a$ (2) $S \to A a A b$ (3) blank (4) blank
Answer: A
Solution
Answer: (A) FIRST $(A a A b) = {c, a}$ (A can be $ϵ$ ) and FIRST $(B b B a) = {d, b}$ . So under $c$ : $S \to A a A b$ (cell 1); under $d$ : $S \to B b B a$ (cell 2). For $A$ : FOLLOW $(A)$ includes $a$ and $b$ , so $A \to ϵ$ appears under $b$ (cell 3). For $B$ : $B \to ϵ$ appears under $a$ (cell 4). This matches (A).
Question 41Theory of Computationmcq2 marks
The 5-state NFA $M$ (with $ϵ$ -transitions) shown accepts which language?
- A.
  $(00)^{*} + 1 (11)^{*}$
- B.
  $0^{*} + (1 + 0 (00)^{*}) (11)^{*}$
  ✓
- C.
  $(00)^{*} + (1 + (00)^{*}) (11)^{*}$
- D.
  $0^{+} + 1 (11)^{*} + 0 (11)^{*}$
Answer: B
Solution
Answer: (B) $0^{*} + (1 + 0 (00)^{*}) (11)^{*}$ Tracing the NFA: from the start there are $ϵ$ -moves to a branch counting 0's (loop of two 0's, with accept allowing $0^{*}$ -like behaviour) and a branch for a 1 followed by pairs of 1's. Combining the accepting paths yields $0^{*} + (1 + 0 (00)^{*}) (11)^{*}$ , matching option (B).
Question 42Programming & Data Structuresmcq2 marks
A C snippet computes the longest subarray with at most two distinct integers. len1 = longest suffix of equal values, len2 = longest suffix with <=2 distinct values. Fill (P) and (Q):
- A.
  (P) len1+1 (Q) len2+1
- B.
  (P) 1 (Q) len1+1
  ✓
- C.
  (P) 1 (Q) len2+1
- D.
  (P) len2+1 (Q) len1+1
Answer: B
Solution
Answer: (B) (P) 1 (Q) len1+1 When X[i] equals the second-most-recent distinct value (the else if branch), the run of equal values restarts at this element, so len1 = 1 (P). When X[i] is a brand-new third value (the else branch), the at-most-two-distinct window can only extend the previous equal-run plus this element, so len2 = len1+1 (Q).
Question 43Compiler Designmcq2 marks
For $x [i] = (p + r) * - s [i] + u / w$ , fill the missing triples (1),(3),(6). Which option is correct?
- A.
  (1) =[] s i (3) * (0)(2) (6) []= x i
  ✓
- B.
  (1) []= s i (3) - (0)(2) (6) =[] x (5)
- C.
  (1) =[] s i (3) * (0)(2) (6) []= x (5)
- D.
  (1) []= s i (3) - (0)(2) (6) =[] x i
Answer: A
Solution
Answer: (A) The three-address form of $x [i] = (p + r) * - s [i] + u / w$ uses the triples: (0) + p r ; (1) =[] s i (fetch s[i]) ; (2) uminus (1) ; (3) * (0)(2) ; (4) / u w ; (5) + (3)(4) ; (6) []= x i (address of x[i]) ; (7) = (6)(5). So the missing entries are (1) =[] s i, (3) * (0)(2), and (6) []= x i - option (A).
Question 44Engineering Mathematicsmcq2 marks
$x, y$ take values in $[0, 1]$ (not necessarily independent), $z = x y$ , means $x ˉ, y ˉ, z ˉ$ . Which is TRUE?
- A.
  $z ˉ = x ˉ y ˉ$
- B.
  $z ˉ \leq x ˉ y ˉ$
- C.
  $z ˉ \geq x ˉ y ˉ$
- D.
  $z ˉ \leq x ˉ$
  ✓
Answer: D
Solution
Answer: (D) $z ˉ \leq x ˉ$ Since $0 \leq y \leq 1$ , we have $x y \leq x$ pointwise (as $x \geq 0$ ), so taking expectations $z ˉ = E [x y] \leq E [x] = x ˉ$ . The others fail without independence (e.g. $z ˉ$ can be $> x ˉ y ˉ$ with positive correlation).
Question 45Databasesmcq2 marks
Person(pid, city). Using selection, projection, cross product, rename, the minimum number of cross-product operations to find cities where at least 3 persons reside is
- A.
  1
- B.
  2
  ✓
- C.
  3
- D.
  4
Answer: B
Solution
Answer: (B) 2 To require three distinct persons in the same city, we need three copies of Person joined on equal city with distinct pids. Combining three renamed copies needs 2 cross-product (join) operations: $(P_{1} \times P_{2}) \times P_{3}$ , then select equal cities and pairwise-distinct pids, then project the city.
Question 46Operating Systemsmsq2 marks
T1: wait(s1); x=x+1; print(x); wait(s2); signal(s1). T2: wait(s1); x=x+1; print(x); signal(s2); signal(s1). s1=1, s2=0, x=0. Which outcomes are possible?
- A.
  T1 first prints 1, T2 next prints 2
- B.
  T2 first prints 1, T1 next prints 2
  ✓
- C.
  T1 first prints 1, T2 deadlock
  ✓
- D.
  T2 first prints 1, T1 next prints 2 (no deadlock)
Answer: B, C
Solution
Answer: (B) and (C) - If T2 acquires s1 first: it prints 1, signals s2, signals s1; then T1 acquires s1, prints 2, wait(s2) succeeds (s2=1), signal(s1). Works -> (B) possible. - If T1 acquires s1 first: it prints 1, then wait(s2) blocks (s2=0) while still holding s1; T2 can never get s1 -> deadlock -> (C) possible. - (A) impossible (T1 first always deadlocks). (D) impossible (T2 first prints 1 then T1 prints 2 is exactly (B), not a new distinct option as worded — the valid non-deadlock case is (B)).
Question 47Engineering Mathematicsmsq2 marks
$B$ is obtained from $A$ ( $n \times n$ real) by swapping two rows. Which statements is/are TRUE?
- A.
  det(B) = -det(A)
  ✓
- B.
  If A is invertible, B is invertible
  ✓
- C.
  If A is symmetric, B is symmetric
- D.
  If trace(A)=0, trace(B)=0
Answer: A, B
Solution
Answer: (A) and (B) Swapping two rows negates the determinant (A). Since $∣ det B ∣ = ∣ det A ∣ \neq = 0$ when $A$ is invertible, $B$ is also invertible (B). A row swap generally destroys symmetry (C false) and changes which diagonal entries appear, so the trace can change (D false).
Question 48Programming & Data Structuresmsq2 marks
S1 (capacity 4) holds 100,200,300,400 (400 top); S2 (capacity 2) empty. Operations: PushToS2 (S1->S2), PushToS1 (S2->S1), GenerateOutput (pop S1 to output). No pop on empty / push on full. Which output sequences can be generated?
- A.
  100, 200, 400, 300
- B.
  200, 300, 400, 100
  ✓
- C.
  400, 200, 100, 300
  ✓
- D.
  300, 200, 400, 100
  ✓
Answer: B, C, D
Solution
Answer: (B), (C) and (D) This is a stack-permutation question with the auxiliary stack S2 (capacity 2) allowing limited reordering. Simulating the allowed moves, sequences (B) 200,300,400,100, (C) 400,200,100,300, and (D) 300,200,400,100 are each reachable, while (A) 100,200,400,300 cannot be produced under the capacity constraints.
Question 49Computer Organization & Architecturemsq2 marks
Which is/are EQUAL to $224$ in radix-5?
- A.
  64 in radix-10
  ✓
- B.
  100 in radix-8
  ✓
- C.
  50 in radix-16
- D.
  121 in radix-7
  ✓
Answer: A, B, D
Solution
Answer: (A), (B) and (D) $22 4_{5} = 2 \cdot 25 + 2 \cdot 5 + 4 = 6 4_{10}$ . - (A) $6 4_{10} = 64$ . Equal. - (B) $10 0_{8} = 6 4_{10}$ . Equal. - (C) $5 0_{16} = 8 0_{10}$ . Not equal. - (D) $12 1_{7} = 1 \cdot 49 + 2 \cdot 7 + 1 = 6 4_{10}$ . Equal.
Question 50Digital Logicmsq2 marks
Given $f_{1} = \sum (0, 2, 3, 5, 7, 8, 11, 13)$ , $f_{2} = \sum (1, 3, 5, 7, 11, 13, 15)$ , $f_{3} = \sum (0, 1, 4, 11)$ , $f_{4} = \sum (0, 2, 6, 13)$ feeding AND ( $f_{1}, f_{2}$ ) and OR ( $f_{3}, f_{4}$ ), whose outputs feed an XOR to give $Y$ . Which options are CORRECT?
- A.
  $Y = \sum (0, 1, 2, 11, 13)$
- B.
  $Y = Π (3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15)$
- C.
  $Y = \sum (0, 1, 2, 3, 4, 5, 6, 7)$
  ✓
- D.
  $Y = Π (8, 9, 10, 11, 12, 13, 14, 15)$
  ✓
Answer: C, D
Solution
Answer: (C) and (D) Let $g_{1} = f_{1} \cdot f_{2} = \sum (3, 5, 7, 11, 13)$ (common minterms) and $g_{2} = f_{3} + f_{4} = \sum (0, 1, 2, 4, 6, 11, 13)$ (union). Then $Y = g_{1} \oplus g_{2} =$ minterms in exactly one of them. Computing the symmetric difference gives $Y = \sum (0, 1, 2, 3, 4, 5, 6, 7)$ — option (C). Equivalently $Y = Π (8, 9, 10, 11, 12, 13, 14, 15)$ (the complementary maxterms) — option (D). (A) and (B) are different minterm sets.
Question 51Discrete Mathematicsmsq2 marks
$G$ is connected with positive integer edge weights, and every spanning tree has even weight. Which statements are TRUE for every such $G$ ?
- A.
  All edges have even weight
- B.
  All edges have even weight OR all edges have odd weight
- C.
  In each cycle C, all edges have even weight
- D.
  In each cycle C, either all edges in C are even OR all are odd
  ✓
Answer: D
Solution
Answer: (D) If all spanning trees have the same parity (even), then swapping one tree edge for a non-tree edge on its fundamental cycle must not change the parity, which forces the two swapped edges to have the same parity. Propagating this around each cycle means every cycle's edges share one parity -> (D). (A),(B),(C) are too strong: edges in different cycles may have different parities as long as within each cycle they match.
Question 52Theory of Computationmsq2 marks
$G$ : $S \to a S$ , $S \to a S b S$ , $S \to c$ . For $w \in L (G)$ , with $n_{a}, n_{b}, n_{c}$ the counts of $a, b, c$ , which statements is/are TRUE?
- A.
  $n_{a} (w) > n_{b} (w)$
- B.
  $n_{a} (w) > n_{c} (w) - 2$
  ✓
- C.
  $n_{c} (w) = n_{b} (w) + 1$
  ✓
- D.
  $n_{c} (w) = n_{b} (w) * 2$
Answer: B, C
Solution
Answer: (B) and (C) Each use of $S \to a S b S$ adds one $a$ and one $b$ ; $S \to a S$ adds an $a$ ; and each $S$ eventually becomes a $c$ . Counting: every $b$ comes with the $a S b S$ rule, and the number of $c$ 's is always one more than the number of $b$ 's -> $n_{c} = n_{b} + 1$ (C). Since each $b$ has a matching $a$ (plus possibly extra $a$ 's), $n_{a} \geq n_{b}$ , and relating to $c$ gives $n_{a} > n_{c} - 2$ (B). (A) fails when no $a S$ rule is used and (D) is false in general.
Question 53Computer Organization & Architecturenat2 marks
Disk: 6000 RPM, avg seek 5 ms, 500 sectors/track, 512-byte sectors. A file is in 3000 sectors randomly located. Assuming average rotational latency, the total read time (seconds) is _________
Answer: 29.5 to 30.5
Solution
Answer: 30.06 Rotation time $= \frac{60}{6000} = 10$ ms; average rotational latency $= 5$ ms; read time per sector $= \frac{10}{500} = 0.02$ ms. Each randomly located sector needs seek + rotational latency + read $= 5 + 5 + 0.02 = 10.02$ ms. Total $= 3000 \times 10.02 ms = 30060 ms \approx 30.06$ s.
Question 54Computer Networksnat2 marks
A TCP connection has cwnd 12 MSS when a timeout occurs. Assuming all segments in the next two RTTs are acknowledged, the cwnd (in MSS) during the third RTT is _________
Answer: 4
Solution
Answer: 4 On timeout, ssthresh $= 12/2 = 6$ and cwnd resets to 1, entering slow start (doubling each RTT):
```
RTT1: cwnd = 1
RTT2: cwnd = 2
RTT3: cwnd = 4   (still below ssthresh=6, so still doubling)
```
So during the third RTT, cwnd $= 4$ .
Question 55Computer Networksnat2 marks
Ethernet: transmission speed $1 0^{8}$ bps, max segment length 500 m, propagation speed $2 \times 1 0^{8}$ m/s. The minimum frame size (bits) for collision detection is _________
Answer: 500
Solution
Answer: 500 Propagation time one-way $T_{p} = \frac{500}{2 \times 1 0 ^{8}} = 2.5 μ s$ . Minimum frame transmission time must be $\geq 2 T_{p}$ :
$min bits = 2 T_{p} \times bandwidth = 2 (2.5 \times 1 0^{- 6}) (1 0^{8}) = 500 bits .$
Question 56Databasesnat2 marks
A useful FD $X \to Y$ requires $X \neq = \emptyset$ , $Y \neq = \emptyset$ , $X \cap Y = \emptyset$ . For a relation R with 4 attributes, the number of possible useful FDs is ________
Answer: 50
Solution
Answer: 50 Each attribute is independently placed in $X$ , in $Y$ , or in neither $= 3$ choices, giving $3^{4} = 81$ assignments. Subtract those with empty $X$ ( $X$ gets none: each attribute in $Y$ or neither $= 2^{4} = 16$ ) and empty $Y$ ( $2^{4} = 16$ ), then add back the all-neither case counted twice ( $1$ ):
$81 - 16 - 16 + 1 = 50.$
Question 57Computer Organization & Architecturenat2 marks
A processor with 16 registers uses 32-bit instructions: opcode + addressing-mode + two register operands + 16-bit scalar. 8 addressing modes are supported. The max number of unique opcodes per addressing mode is _________
Answer: 32
Solution
Answer: 32 Register field $= lo g_{2} 16 = 4$ bits (two fields $= 8$ bits). Addressing mode $= lo g_{2} 8 = 3$ bits. Scalar $= 16$ bits. Opcode bits $= 32 - 3 - 8 - 16 = 5$ , giving $2^{5} = 32$ unique opcodes.
Question 58Computer Organization & Architecturenat2 marks
Non-pipelined: 6 cycles/instruction at 2 GHz. Pipelined (5 stages, 2 GHz, ideal 1 IPC): 20% of instructions stall 2 cycles (data), 20% stall 3 cycles (control). The speedup is _________
Answer: 2.9 to 3.1
Solution
Answer: 3.0 Pipelined effective CPI $= 1 + 0.2 (2) + 0.2 (3) = 1 + 0.4 + 0.6 = 2$ . Speedup $= \frac{CPI _{non-pipe}}{CPI _{pipe}} = \frac{6}{2} = 3.0$ (same clock).
Question 59Algorithmsnat2 marks
The number of distinct minimum-weight spanning trees of the shown weighted graph (vertices a,b,c,d,e,f,g) is _________
Answer: 9
Solution
Answer: 9 All four weight-1 edges ( $a - b, a - f, c - d, d - e$ ) are forced into every MST (no cycle among them), connecting ${a, b, f}$ and ${c, d, e}$ , with $g$ isolated. The MST (weight 8) then needs 2 weight-2 edges to join the three parts ${a, b, f}$ , ${c, d, e}$ , ${g}$ :
```
to g from {a,b,f}: a-g, b-g, f-g      (3 choices)
to join {c,d,e}: either g-c / g-d / g-e (to g side)
```
Counting the valid ways the weight-2 edges connect all three components without a cycle gives $9$ distinct MSTs.
Question 60Discrete Mathematicsnat2 marks
The chromatic number of the shown graph (8 vertices drawn as an octagon with diagonals) is _________
Answer: 2
Solution
Answer: 2 The graph is bipartite — it is the complete bipartite graph $K_{4, 4}$ (each vertex on one side joins every vertex on the other). A bipartite graph with at least one edge has chromatic number exactly 2 (it contains no odd cycle).
Question 61Computer Organization & Architecturenat2 marks
32-bit instructions, 150 distinct instructions split equally into R-type and I-type. 1 opcode bit distinguishes I/R; 50 architectural registers; equal register-field sizes. X = UNUSED bits (R-type), Y = OPCODE bits, Z = immediate bits (I-type). The value of X+2Y+Z is ________
Answer: 34
Solution
Answer: 34 Registers: $⌈ lo g_{2} 50 ⌉ = 6$ bits each. Opcode $Y = 1 + ⌈ lo g_{2} 75 ⌉ = 1 + 7 = 8$ bits (75 ops per type). R-type: $32 = Y + X + 3 \times 6 \Rightarrow X = 32 - 8 - 18 = 6$ . I-type: $32 = Y + 2 \times 6 + Z \Rightarrow Z = 32 - 8 - 12 = 12$ . $X + 2 Y + Z = 6 + 16 + 12 = 34.$
Question 62Theory of Computationnat2 marks
$L_{1}$ is given by $b^{*} a b^{*} (a b^{*} a b^{*})^{*}$ and $L_{2} = {w \in (a + b)^{*} : ∣ w ∣ \leq 4}$ . The number of strings in $L_{2}$ that are also in $L_{1}$ is __________
Answer: 15
Solution
Answer: 15 $L_{1}$ is exactly the set of strings over ${a, b}$ with an ODD number of $a$ 's. Counting strings of length $\leq 4$ with an odd number of $a$ 's:
```
length 1: a                 -> 1
length 2: ab, ba            -> 2
length 3: 1 a (C(3,1)=3) + 3 a (C(3,3)=1) -> 4
length 4: 1 a (C(4,1)=4) + 3 a (C(4,3)=4) -> 8
```
Total $= 1 + 2 + 4 + 8 = 15$ .
Question 63Discrete Mathematicsnat2 marks
$Z_{n} = {0, \dots, n - 1}$ under addition mod $n$ . The number of elements in $Z_{2} \times Z_{3} \times Z_{4}$ that are their own inverses is ________
Answer: 4
Solution
Answer: 4 An element $(a, b, c)$ is its own inverse iff $2 a \equiv 0, 2 b \equiv 0, 2 c \equiv 0$ in the respective groups. - $Z_{2}$ : $2 a \equiv 0$ for $a = 0, 1$ -> 2 solutions. - $Z_{3}$ : $2 b \equiv 0 \Rightarrow b = 0$ -> 1 solution. - $Z_{4}$ : $2 c \equiv 0 \Rightarrow c = 0, 2$ -> 2 solutions. Total $= 2 \times 1 \times 2 = 4$ .
Question 64Operating Systemsnat2 marks
32-bit system, 4 KB pages, 4-byte PTEs, 2-level page table (outer directory + inner tables), demand-paged inner tables (allocated only if they hold a valid PTE). A process accesses 2000 unique pages. X = min, Y = max number of pages across the two page-table levels. X+Y is ________
Answer: 1028
Solution
Answer: 1028 Page offset $= 12$ bits; VPN $= 20$ bits split $10 + 10$ . An inner page table holds $2^{10} = 1024$ entries $= 1$ page. Outer directory is always 1 page. - Minimum: pack 2000 valid entries into as few inner tables as possible: $⌈ 2000/1024 ⌉ = 2$ inner tables -> $X = 1 + 2 = 3$ . - Maximum: spread the 2000 pages so as many inner tables as possible each hold $\geq 1$ entry; there are only $1024$ possible inner tables, and $2000 > 1024$ , so all $1024$ are used -> $Y = 1 + 1024 = 1025$ . $X + Y = 3 + 1025 = 1028.$
Question 65Compiler Designnat2 marks
Augmented grammar: $S^{'} \to S$ ; $S \to S S ∣ A a ∣ b A c ∣ B c ∣ b B a$ ; $A \to d #$ ; $B \to@$ . With $I_{0} = CLOSURE ({S^{'} \to \cdot S})$ , the number of items in $GOTO (I_{0}, S)$ is ___________
Answer: 9
Solution
Answer: 9 $GOTO (I_{0}, S)$ moves the dot past $S$ in every $I_{0}$ item with $\cdot S$ , giving $S^{'} \to S \cdot$ and $S \to S \cdot S$ , then closure of $S \to S \cdot S$ re-adds all $S$ -, $A$ -, $B$ -productions:
```
S' -> S.          S -> S.S
S -> .SS   S -> .Aa   S -> .bAc   S -> .Bc   S -> .bBa
A -> .d#   B -> .@
```
That is $2 + 5 + 2 = 9$ items.

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