## Gate-1999

 Question 1
Suppose that the expectation of a random variable X is 5. Which of the following statements is true?
 A There is a sample point at which X has the value 5. B There is a sample point at which X has value greater than 5. C There is a sample point at which X has a value greater than or equal to 5. D None of the above
Engineering-Mathematics       Probability
Question 1 Explanation:
Expectation of discrete random variable
E(X) = x1P1 + x2P2 + ... + xnPn
In question, E(X) is given as 5.
E(X) = 5, 0≤Pi≤1
P1 + P2 + ... + Pn = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the xi value is greater than 5.
 Question 2
The number of binary relations on a set with n elements is:
 A n2 B 2n C 2n (2) D None of the above
Engineering-Mathematics       Sets-And-Relation
Question 2 Explanation:
In binary relation two elements are selected from a set. So total no. of pairs possible are
n×n = n2
Now, no. of binary relations possible is
2n2
Because each pair have two possibilities either chosen or not chosen.
 Question 3
The  number  of  binary  strings  of  n  zeroes  and  k  ones  that  no  two  ones  are adjacent is
 A n-1Ck B nCk C nCk+1 D None of the above
Engineering-Mathematics       Combinatorics
Question 3 Explanation:
Since there are n zeroes, so
XOXOXOXOXOXOX
n+1 gaps can be possible, where 1's can be placed so that no two one's are adjacent. So, no. of ways in which k 1's can be placed in n+1 gaps are,
n+1Ck
 Question 4
Consider the regular expression (0 + 1) (0 + 1)…. N times. The minimum state finite  automation  that  recognizes  the  language  represented  by  this  regular expression contains
 A n states B n + 1 states C n + 2 states D None of the above
Theory-of-Computation       Finite-Automata
Question 4 Explanation:
Let's draw both NFA and DFA and see which one requires less no. of state.
DFA:

So, DFA requires (n+2) state.
NFA:

So, NFA requires (n+1) state.
min(n+1, n+2)
= n+1
 Question 5
Context-free languages are closed under:
 A Union, intersection B Union, Kleene closure C Intersection, complement D Complement, Kleene closure
Theory-of-Computation       Context-Free-Language
Question 5 Explanation:
Context free languages are not closed under Intersection and complementation.
By checking the options only option B is correct.
 Question 6
Let LD be the set of all languages accepted by a PDA by final state and LE  the set of all languages accepted by empty stack. Which of the following is true?
 A LD = LE B LD ⊃ LE C LE = LD D None of the above
Theory-of-Computation       Push-Down-Automata
Question 6 Explanation:
For any PDA which can be accepted by final state, there is an equivalent PDA which can also be accepted by an empty stack and for any PDA which can be accepted by an empty stack, there is an equivalent PDA which can be accepted by final state.
 Question 7

 A x NAND X B x NOR x C x NAND 1 D x NOR 1
Digital-Logic-Design       Logic-Gates
Question 7 Explanation:
 Question 8

 A B C D
Digital-Logic-Design       K-Map
Question 8 Explanation:

 Question 9

 A (A) – 2 (B) – 4 (C) – 3 (D) - 1 B (A) – 1 (B) – 2 (C) – 3 (D) – 4 C (A) – 3 (B) – 2 (C) – 4 (D) - 1 D (A) – 4 (B) – 1 (C) – 2 (D) – 3
Operating-Systems       Match-the-Following
Question 9 Explanation:

⇒ Threads are handled by CPU.
⇒ Virtual address is a memory type.
⇒ File system is used to manage the disk.
⇒ Interrupt is a signal.
 Question 10
Which of the following disk scheduling strategies is likely to give the best through put?
 A Farthest cylinder next B Nearest cylinder next C First come first served D Elevator algorithm
Operating-Systems       Disk-Scheduling
Question 10 Explanation:
Farthest cylinder next - Longest job first
Nearest cylinder next - SSTF
First come first served - FCFS
Elevator - SCAN
SSTF always serves the request of nearest cylinder first. Due to which the necessary movement gets reduced.
 Question 11
System calls are usually invoked by using
 A a software interrupt B polling C an indirect jump D a privileged instruction
Operating-Systems       System-Calls
Question 11 Explanation:
Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.
 Question 12
A sorting technique is called stable if
 A it takes O (nlog n) time B it maintains the relative order of occurrence of non-distinct elements C it uses divide and conquer paradigm D it takes O(n) space
Algorithms        Sorting
Question 12 Explanation:
Sorting techniques are said to be stable if it maintains the relative order of occurence of non-distinct element.
 Question 13
Suppose  we  want  to  arrange  the  n  numbers  stored  in  any  array  such  that  all negative  values  occur  before  all  positive  ones. Minimum  number  of  exchanges required in the worst case is
 A n - 1 B n C n + 1 D None of the above
Algorithms        Time-Complexity
Question 13 Explanation:
Minimum no. of exchanges required in worst case will be when 1st half will contain all +ve nos and 2nd half will conain all -ve nos.
Now we will swap 1st no. with nth no. and then 2nd no. with (n-1)th no. and then 3rd no. with (n-2)th and so on. Like this we will have to do n/2 swaps in worst case.
 Question 14

 A 8, 9, 15, 20, 47, 4, 12, 17, 30, 40 B 8, 15, 20, 47, 4, 9, 30, 40, 12, 17 C 15, 20, 47, 4, 8, 9, 12, 30, 40, 17 D 4, 8, 9, 15, 20, 47, 12, 17, 30, 40
Algorithms        Sorting
Question 14 Explanation:
 Question 15

 A 0 B 1 C 2 D 3
Data-Structures       Graphs
Question 15 Explanation:
Here, vertex 2, 3, 5 are the articulation points. By removing these vertices then the graph will be disconnected.
Total no. of articulation points = 3
 Question 16
If n is a power of 2, then the minimum number of multiplications needed to compute a* is
 A log2 n B √n C n-1 D n
Algorithms        Time-Complexity
Question 16 Explanation:

We require 4 multiplications to calculate a16 .....(I)
→ Like that 3 multiplications requires to calculate a8 .....(II)
I, II are satisfied with the option A.
 Question 17
Which of the following is the most powerful parsing method?
 A LL (1) B Canonical LR C SLR D LALR
Compiler-Design       Parsers
Question 17 Explanation:
Canonical LR is most powerful.
LR > LALR > SLR
 Question 18
Consider the join of a relation R with a relation S. If R has m tuples and S has n tuples then the maximum and minimum sizes of the join respectively are
 A m + n and 0 B mn and 0 C m + n and |m – n| D mn and m + n
Database-Management-System       Relational-Algebra
Question 18 Explanation:
For maximum:
Suppose there is no common attribute in R and S due to which natural join will act as cross product. So then in cross product total no. of tuples will be mn.
For minimum:
Suppose there is common attribute in R and S, but none of the row of R matches with rows of S then minimum no. of tuples will be 0.
 Question 19

 A σ(A=10∨B=20) (r) B σ(A=10) (r) ∪ σ(B=20) (r) C σ(A=10) (r) ∩ σ(B=20) (r) D σ(A=10) (r) - σ(B=20) (r)
Database-Management-System       Relational-Algebra
Question 19 Explanation:
The given relational algebra expression represents tuples having A=10 and B=20 which is equivalent to
σ(A=10) (r) ∩ σ(B=20) (r)
 Question 20
Booth’s coding in 8 bits for the decimal number –57 is
 A 0 – 100 + 1000 B 0 – 100 + 100 -1 C 0 – 1 + 100 – 10 + 1 D 00 – 10 + 100 - 1
Digital-Logic-Design       Number-Systems
Question 20 Explanation:
Option-B:
 Question 21
The maximum gate delay for any output to appear in an array multiplier for multiplying two n bit number is
 A On2 B O(n) C O(log n) D O(1)
Digital-Logic-Design       Multiplexer
Question 21 Explanation:
Total no. of gates being used for 'n' bit multiplication in an array multiplier (n*n) = (2n-1)
Total delay = 1 * 2n - 1 = O(2n - 1) = n
 Question 22
The main memory of a computer has 2 cm blocks while the cache has 2 c blocks. If the cache uses the set associative mapping scheme with 2 blocks per set, then block k of the main memory maps to the set
 A (k mod m) of the cache B (k mod c) of the cache C (k mod 2c) of the cache D (k mod 2 cm) of the cache
Computer-Organization       Cache
Question 22 Explanation:
No. of sets in cache = No. of blocks in cache/No. of blocks per set
= 2c/c
= c
∴ Cache set no. to which block k of main memory maps to
= (Main memory block no.) mod (Total set in cache)
= k mod c
 Question 23
The Newton-Raphson method is to be used to find the root of the equation f(x)=0 where xo is the initial approximation and f' is the derivative of f. The  method converges
 A always B only if f is a polynomial C only if f(x0) < 0 D None of the above
Engineering-Mathematics       Newton-Raphson-Method
Question 23 Explanation:
Note: Out of syllabus.
 Question 24
Let R = (a, b, c, d, e, f) be a relation scheme with the following dependencies c → f, e → a, ec → d, a → b. Which of the following is a key for R?
 A CD B EC C AE D AC
Database-Management-System       Functional-Dependency
Question 24 Explanation:
Let's check closure for each option,
A) (CD)+ = cdf
Not a key.
B) (EC)+ = ecdabf
Yes, it is a key.
C) (AE)+ = aeb
Not a key. D) (AC)+ = abcf
Not a key.
 Question 25
Which of the following is correct?
 A B-trees are for storing data on disk and B+ trees are for main memory. B Range queries are faster on B* trees. C B-trees are for primary indexes and B* trees are for secondary indexes. D The height of a B* tree is independent of the number of records.
Database-Management-System       B+-Trees
Question 25 Explanation:
Range queries are faster on B+ trees.
 Question 26

 A B Events E1 and E2 are independent C Events E1 and E2 are not independent D
Engineering-Mathematics       Probability
Question 26 Explanation:
 Question 27
Two girls have picked 10 roses, 15 sunflowers and 15 daffodils. What is the number of ways they can divide the flowers amongst themselves?
 A 1638 B 2100 C 2640 D None of the above
Engineering-Mathematics       Combinatorics
Question 27 Explanation:
Formula for distributing n identical objects into r persons is,
n+r-1Cr-1
So for 10 roses,
10+2-1C2-1 = 11C1 = 11
For 15 sunflowers,
15+2-1C2-1 = 16C1 = 16
For 15 daffodils,
15+2-1C2-1 = 16C1 = 16
11×16×16 = 2816
 Question 28
Let L be a set with a relation R which is transitive, anti-symmetric and reflexive and for any two elements a, b ∈ L let the least upper bound lub (a,b) and the greatest lower bound glb (a,b) exist. Which of the following is/are true?
 A L is a poset B L is a Boolean algebra C -L1 is context free D -L2 is regular E Both A and C
Engineering-Mathematics       Sets-And-Relation
Question 28 Explanation:
Note: Options given are wrong.
 Question 29
If L is context free language and L2 is a regular language which of the following is/are false?
 A L1 – L2 is not context free B L1 ∩ L2 is context free C ~L1 is context free D ~L2 is regular E Both A and C
Theory-of-Computation       Identify-Class-Language
Question 29 Explanation:
(A) L2 is regular language and regular language is closed under complementation. Hence ~L2 is also regular.
So L1 - L2 = L1 ∩ (~L2)
And CFL is closed under regular intersection.
So, L1 ∩ (~L2) or L1 - L2 is CFL.
So False.
(B) As we said that CFL is closed under regular intersection.
So True.
(C) CFL is not closed under complementation.
Hence False.
(D) Regular language is closed under complementation.
Hence True.
 Question 30

Given the programming constructs (i) assignment (ii) for loops where the loop parameter cannot be changed within the loop (iii) if-then-else (iv) forward go to (v) arbitrary go to (vi) non-recursive procedure call (vii) recursive procedure/function call (viii) repeat loop, which constructs will you not include in a  programming  language  such  that  it  should  be  possible  to  program  the terminates (i.e., halting) function in the same programming language.

 A (ii), (iii), (iv) B (v), (vii), (viii) C (vi), (vii), (viii) D (iii), (vii), (viii)
Programming       Programming-Constructs
Question 30 Explanation:
Arbitrary goto, recursive call and repeat loop may enter infinite loop and the program might never terminate.
 Question 31

 A This schedule is serialized and can occur in a scheme using 2PL protocol B This schedule is serializable but cannot occur in a scheme using 2PL protocol C This schedule is not serialiable but can occur in a scheme using 2PL protocol D This schedule is not seralisable and cannot occur in a scheme using 2PL protocol.
Database-Management-System       Transactions
Question 31 Explanation:
Let's draw precedence graph,

Since cycle exist so not conflict serializable.
And we know that if the schedule is not serializable then it is not 2PL.
Hence correct option is (D).
 Question 32
Consider the schema R = (S T U V) and the dependencies S → T, T → U, U → V and V → S. Let R = (R1 and R2) be a decomposition such that R1 ∩ R2 = ∅ . The decomposition is
 A not in 2NF B in 2NF but not 3NF C in 3NF but not in 2NF D in both 2NF and 3NF
Database-Management-System       Normalization
Question 32 Explanation:
Since R1 ∩ R2 = ∅, so the decomposition is lossless join. Now since all the attributes are keys, so R1 ∩ R2 will be a key of the decomposed relation.
And since every attribute is key so the decomposed relation will be in BCNF and hence in 3NF.
 Question 33

 A A = 0, B = 0, C = 1 B A = 0, B = 1, C = 1 C A = 1, B = 0, C = 1 D A = 1, B = 1, C = 1
Database-Management-System       Logic-Gates
Question 33 Explanation:

So the above equation is satisfied if either C=0 or A=0 and B=1.
Hence, Option (B) is correct.
 Question 34
Which of the following sets of component(s) is/are sufficient to implement any arbitrary Boolean function?
 A XOR gates, NOT gates B 2 to 1 multiplexors C AND gates, XOR gates D Three-input gates that output (A⋅B) + C for the inputs A⋅B and C E Both B and C
Database-Management-System       Boolean-Functions
Question 34 Explanation:
(A) Not complete because, XOR can be used to make only NOT gate and NOT gate is already available. Hence not complete.
(B) 2 to 1 multiplexors is functionally complete.
(C) XOR gate can be used to make a NOT gate. So, (AND, NOT) is functionally complete.
(D) With given gates and inputs NOT gate cannot be derived.
Hence, not complete.
 Question 35
A  multi-user,  multi-processing  operating  system  cannot  be  implemented  on hardware that does not support
 A Address translation B DMA for disk transfer C At least two modes of CPU execution (privileged and non-privileged) D Demand paging E Both A and C
Operating-Systems       Virtual Memory
Question 35 Explanation:
Address translation and atleast two modes of CPU execution (Privileged and non-privileged) are needed to implement multiuser and multiprocessing operating system, because address translation provides memory protection which ensures that a given process does not interfere with another, and we need privileged and non-privileged instruction, so that user and OS interconnects properly.
 Question 36
Which of the following is/are advantage of virtual memory?
 A Faster access to memory on an average. B Processes can be given protected address spaces. C Linker can assign addresses independent of where the program will be loaded in physical memory. D Programs larger than the physical memory size can be run. E Both B and D
Operating-Systems       Virtual Memory
Question 36 Explanation:
A) False. Because in virtual memory concept address translation is required due to which access is slow.
B) True. Because in virtual memory concept of address translation provides protected address space so that one process do not interfere the other process.
C) False.
D) True.
 Question 37
Which  of  the  following  actions  is/are  typically  not  performed  by  the  operating system when switching context from process A to process B?
 A Saving current register values and restoring saved register values for process B. B Changing address translation tables. C Swapping out the memory image of process A to the disk. D Invalidating the translation look-aside buffer.
Operating-Systems       Context-Switching
Question 37 Explanation:
A) True.
B) True.
C) False, because swapping is done when the process is suspended and not during context switching.
D) True, Invalidation of TLB is necessary because, if the TLB is not invalidated then the new process might end up using the translation of old process. Note that Invalidation of TLB is necessary but saving and reuse of TLB is not necessary.
 Question 38

 A 0.125 0.125 B 0.25 0.25 C 0.25 0.125 D 0.125 0.25
Operating-Systems       Dynamic-Scoping
Question 38 Explanation:
Note: Out of syllabus.
 Question 39
The number of tokens in the Fortran statement DO 10 I = 1.25 is
 A 3 B 4 C 5 D None of the above
Compiler-Design       Compilers
Question 39 Explanation:
DO → 1
10 → 2
I → 3
= → 4
1.25 → 5
 Question 40
A grammar that is both left and right recursive for a non-terminal, is
 A Ambiguous B Unambiguous C Information is not sufficient to decide whether it is ambiguous or unambiguous D None of the above
Theory-of-Computation       Grammar
Question 40 Explanation:
If a grammar is both left and right recursion, then grammar may or may not be ambiguous.
 Question 41
The number of full and half-adders required to add 16-bit numbers is
Question 41 Explanation:
For Least Significant Bit we do not need a full adder since initially carry is not present.
But for rest of bits we need full address since carry from previous addition has to be included into the addition operation.
So, in total 1 half adder and 15 full adders are required.
 Question 42
Zero has two representations in
 A Sign magnitude B 1’s complement C 2’s complement D None of the above E Both A and B
Digital-Logic-Design       Number-Systems
Question 42 Explanation:
Sign magnitude:
+0 = 0000
-0 = 1000
1's complement:
+0 = 0000
-0 = 1111
 Question 43
Raid configurations of the disks are used to provide
 A Fault-tolerance B High speed C High data density D None of the above
Computer-Organization       RAID
Question 43 Explanation:
Raid is a way of storing the same data in different places on multiple hard disks to protect data in the case of drive failure.
So, we can say that RAID is used to provide fault tolerence.
 Question 44
Arrange the following configuration for CPU in decreasing order of operating speeds: Hard wired control, vertical microprogramming, horizontal microprogramming.
 A Hard wired control, vertical micro-programming, horizontal micro- programming. B Hard wired control, horizontal micro-programming, vertical micro- programming. C Horizontal micro-programming, vertical micro-programming, Hard wired control. D Vertical micro-programming, horizontal micro-programming, hard wired control.
Computer-Organization       Micro-Programming
Question 44 Explanation:
Hardwired control involves only hardware which is definitely faster than microprogramming approach.
Now between horizontal and vertical microprogramming, the signals in vertical microprogramming are in encoded form which has to be decoded first using decoder. So, horizontal microprogramming is faster than vertical microprogramming.
 Question 45
The minimum number of record movements required to merge five files A (with 10 records), B (with 20 records), C (with 15 records), D (with 5 records) and E (with 25 records) is:
 A 165 B 90 C 75 D 65
Algorithms        Greedy-Algorithm
Question 45 Explanation:
Always merge two files with minimum no. of records.
10, 20, 15, 5, 25
Merge 5 & 10:
5+10=15 movements
Now the list is
15, 20, 15, 25
Merge 15 & 15:
15+15=30 movements
Now the list is
30, 20, 25
Merge 20 & 25:
20+25=45 movements
Now the list is
30, 45
Merge 30 & 45:
30+45=75 movements
∴ Total no. of movements
= 15+30+45+75
= 165
 Question 46

 A M – W N – V O – U P - X B M – W N – U O – X P - V C M – V N – W O – X P - U D None of the above
Algorithms        Time-Complexity
Question 46 Explanation:
(M) T(n) = Sum of first n matural nos. = n(n+1)/2 = O(n2)
(N) Apply Master's theorem
T(n) = θ(n) = O(n)
(O) Apply Master's theorem
T(n) = θ(n logn) = O(n logn)
(P) Here we are adding the log of firstn natural numbers.
So,
Tn = log1 + log2 + log3 + ... + logn
= log (1×2×...n)
= log(n!)
= θ(n logn)
 Question 47
The main differences(s) between a CSIC and A RISC processor is/are that a RISC processor typically
 A has fewer instructions B has fewer addressing modes C has more registers D is easier to implement using hard-wired control logic E All the above
Computer-Organization       CISC-and-RISC
Question 47 Explanation:
All are properties of RISC processor.
 Question 48
A certain processor supports only the immediate and the direct addressing modes. Which of the following programming language features cannot be implemented on this processor?
 A Pointers B Arrays C Records D Recursive procedures with local variable E All the above
Question 48 Explanation:
A) Cannot be implemented because pointers need indirect addressing mode.
B) Cannot be implemented because arrays need Register indexing.
C) Records also needs pointers which needs indirect addressing modes, so this also cannot be implemented.
D) Recursive procedures needs stack, and so it needs stack pointers which inturn needs indirect addressing. So this also cannot be implemented.
 Question 49

 A Finds the maximum of a, b, and c B Finds the minimum of a, b and c C Finds the middle number of a, b, c D None of the above
Programming       C-Programming
Question 49 Explanation:
Try for (3,2,2), it will go for infinite loop.
 Question 50
Which of the following is/are correct?
 A An SQL query automatically eliminates duplicates B An SQL query will not work if there are no indexes on the relations C SQL permits attribute names to be repeated in the same relation D None of the above
Database-Management-System       SQL
Question 50 Explanation:
→ SQL won't remove duplicates like relational algebra projection, we have to remove it explicitly by distinct.
→ If there are no indexes on the relation SQL, then also it works.
→ SQL does not permit 2 attributes to have same name in a relation.
There are 50 questions to complete.