## ISRO-2016

 Question 1
Which of the following is true?
 A √3 + √7 = √10 B √3 + √7 ≤ √10 C √3 + √7 < √10 D √3 + √7 > √10
Engineering-Mathematics       Linear-Equation
Question 1 Explanation:
√3 + √7=1.7320508075688772935274463415059 + 2.6457513110645905905016157536393
=4.3778021186334678840290620951451
√10=3.1622776601683793319988935444327
So, √3 + √7 > √10 is true.
 Question 2
A given connected graph is a Euler Graph if and only if all vertices of are of
 A same degree B even degree C odd degree D different degree
Engineering-Mathematics       Graph-Theory
Question 2 Explanation:
A given connected graph is a Euler Graph if and only if all vertices of are of even degree.
Proof:
→ Let G(V, E) be an Euler graph. Thus G contains an Euler line Z, which is a closed walk.
→ Let this walk start and end at the vertex u ∈ V. Since each visit of Z to an intermediate vertex v of Z contributes two to the degree of v and since Z traverses each edge exactly once, d(v) is even for every such vertex.
→ Each intermediate visit to u contributes two to the degree of u, and also the initial and ﬁnal edges of Z contribute one each to the degree of u. So the degree d(u) of u is also even.
 Question 3
The maximum number of edges in a n-node undirected graph without self loops is
 A n2 B n(n-1)/2 C n-1 D n(n+1)/2
Engineering-Mathematics       Graph-Theory
Question 3 Explanation:
The set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n,2)⋅C(n,2)=n(n-1)/2.
 Question 4
The minimum number of NAND gates required to implement the Boolean function A + AB’+ AB’C is equal to
 A 0 B 1 C 4 D 7
Digital-Logic-Design       Boolean-Function
Question 4 Explanation:
A + AB’+ AB’C = A(1+B’+B’C) = A
No GATE is required to implement the function A.
 Question 5
The minimum Boolean expression for the following circuit is:
 A AB + AC + BC B A + BC C A + B D A + B + C
Digital-Logic-Design       Boolean-Expression
Question 5 Explanation:
If the switches are in parallel then use “+”, and if they are serial then use “ ”.
A(B+C) + AB + (A+B)C = AB + AC + AB + AC + BC
= (AB+AB) + (AC+AC) + BC
= AB + AC + BC
 Question 6
For a binary half-subtractor having two inputs A and B, the correct set of logical expressions for the outputs D (= A minus B) and X (=borrow) are
 A D = AB + A’B , X = A’B B D = A’B + AB’ , X = AB’ C D = A’B + AB’ , X = A’B D D = AB + A’B , X = AB’
Digital-Logic-Design       Combinational-Circuit
Question 6 Explanation:
The function table for the Half Subtractor is as follows
A-B= D= A’B + AB’
X= A’B
 Question 7
Consider the following gate network Which one of the following gates is redundant?
 A Gate No. 1 B Gate No. 2 C Gate No. 3 D Gate No. 4
Digital-Logic-Design       Logic-Gates
Question 7 Explanation:
W’ + W’Z + Z’XY = W’(1+Z) + Z’XY
= W’ + Z’XY
The term W’Z is redundant which is represented by GATE 2.
 Question 8
The dynamic hazard problem occurs in
 A combinational circuit alone B sequential circuit only C Both (a) and (b) D None of the above
Digital-Logic-Design       Hazards
Question 8 Explanation:
→ A dynamic hazard is the possibility of an output changing more than once as a result of a single input change.
→ Dynamic hazards often occur in larger logic circuits where there are different routes to the output (from the input).
→ If each route has a different delay, then it quickly becomes clear that there is the potential for changing output values that differ from the required / expected output. e.g.
→ A logic circuit is meant to change output state from 1 to 0, but instead changes from 1 to 0 then 1 and finally rests at the correct value 0. This is a dynamic hazard.
→ As a rule, dynamic hazards are more complex to resolve, but note that if all static hazards have been eliminated from a circuit, then dynamic hazards cannot occur.
 Question 9
The logic circuit given below converts a binary code y1,y2,y3 into
 A Excess-3 code B Gray code C BCD code D Hamming Code
Digital-Logic-Design       Number-Systems
Question 9 Explanation:
X1= Y1
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3
 Question 10
The circuit given below in the figure below is
 A An oscillating circuit and its output is a square wave B The one whose output remains stable in ‘1’ state C The one having output remains stable in ‘0’ state D has a single pulse of three times propagation delay
Digital-Logic-Design       Logic-Gates
Question 10 Explanation:
The square wave has alternating amplitudes(0 and 1) with duty cycle 1.
An odd number of cascaded NOT gates produce a square wave.

Note: Duty cycle= Ratio of durations in which the circuit is ON and OFF in a cycle.
 Question 11
If 12A7C16 = X8, then the value of X is
 A 224174 B 425174 C 6173 D 225174
Digital-Logic-Design       Number-Systems
Question 11 Explanation:
Given, (12A7C)16 = (0001 0010 1010 0111 1100)2
MAke blocks of 3 bits each from LSB to MSB.
(Note: In the last block append zeros (as MSBs) if number bits is not three)
(000 010 010 101 001 111 100)
Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.
= (0 2 2 5 1 7 4)8
 Question 12
The Excess-3 code is also called
 A Cyclic Redundancy Code B Weighted Code C Self-Complementing Code D Algebraic Code
Digital-Logic-Design       Number-Systems
Question 12 Explanation:
Excess-3 code is also called Self-Complementing Code. Because 1’s complement of an excess-3 number is equivalent to 9’s complement of the corresponding decimal digit.
→ In excess-3 code, each of the 4-bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.
Excess-3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.
 Question 13
The simplified SOP (Sum of Product) form the Boolean expression (P + Q’ + R’)(P + Q’ + R)(P + Q + R’)
 A (P’Q + R) B (P + Q’R’) C (P Q’ + R ) D (PQ + R)
Digital-Logic-Design       Boolean-Expression
Question 13 Explanation:
 Question 14
Which of the following binary number is the same as its 2’s complement?
 A 1010 B 0101 C 1000 D 1001
Digital-Logic-Design       Number-Systems
Question 14 Explanation:
Hint: Number of bits=4
(Decimal value of maximum 4-bit number +1 )/2= (15+1)/2=8
 Question 15
The functional difference between SR flip-flop and JK flip-flop is that
 A JK Flip-flop is faster than SR flip-flop B JK flip-flop has a feedback path C JK flip-flop accepts both inputs 1 D None of them
Digital-Logic-Design       Sequential-Circuits
Question 15 Explanation:
-> JK flip flop accepts input J=K=1. When J=K=1, the state of the flip-flop gets complimented. But it's not a valid input in SR flip-flop.
-> JK flip flop doesn’t have a feedback path.
 Question 16
Consider a non-pipelined processor with a clock rate of 2.5 gigahertz and average cycles per instruction of four. The same processor is upgraded to a pipelined processor with five stages; but due to the internal pipeline delay, the clock speed is reduced to 2 gigahertz. Assume that there are no stalls in the pipeline. The speedup achieved in this pipelined processor is
 A 3.2 B 3 C 2.2 D 2
Computer-Organization       Pipelining
Question 16 Explanation:
→ Given that the processor clock rate = 2.5 GHz, the processor takes 2.5 G cycles in one second.
→ Time taken to complete one cycle = (1 / 2.5 G) seconds
→ Since it is given that average number of cycles per instruction = 4, the time taken for completing one instruction=(4/2.5 G) = 1.6 ns
→ In the pipelined case we know in the ideal case CPI = 1, and the clock speed = 2 GHz.
→ Time taken for one instruction in the pipelined case = (1 / 2 G) = 0.5 ns
→ Speedup = 1.6/0.5 = 3.2
 Question 17
What is the output of this C code?
 A 5 5 5 B 5 5 junk C 5 junk junk D Compile time error
Programming-for-Output-Problems       Pointers
Question 17 Explanation:
If we are given p instead of *p then it prints the address of k but we are printing the value of k.
The same rule applies in **m.
Output is 5 5 5
 Question 18
Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:
 A 256 Mbyte, 19 bits B 256 Mbyte, 21 bits C 512 Mbyte, 20 bits D 64 GB, 28 bits
Computer-Organization       Secondary-Memory
Question 18 Explanation:
→ Given that the disk pack has 16 surfaces, 128 tracks per surface, 256 sectors per track and each sector size is 512 bytes.
→ So the disk pack capacity = 16*128*256*512 bytes = 256 MB
→ To specify a sector we need the information about surface number, track number and sector number within a track.
→ Surface number needs 4 bits as there are 16 surfaces(24), track number needs 7 bits as there are 128 tracks(27) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (28).
→ Total number bits needed to specify a particular sector = 4+7+8 = 19 bits.
 Question 19
Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns. If the one-page fault is generated for every 106 memory accesses, what is the effective access time for the memory?
 A 21.4 ns B 29.9 ns C 23.5 ns D 35.1 ns
Operating-Systems       Memory-Management
Question 19 Explanation:
 Question 20
Register renaming is done in pipelined processors
 A as an alternative to register allocation at compile time B for efficient access to function parameters and local variables C to handle certain kinds of hazards D as part of address translation
Computer-Organization       Pipelining
Question 20 Explanation:
→ Register renaming is used to eliminate hazards that arise due to WAR (Write After Read) and WAW(Write After Write) dependencies.
 Question 21
In which class of Flynn’s taxonomy, Von Neumann architecture belongs to?
 A SISD B SIMD C MIMD D MISD
Computer-Organization       Machine-Instructions
Question 21 Explanation:
→ SISD (single instruction stream, single data stream) is a computer architecture in which a single uni-core processor, executes a single instruction stream, to operate on data stored in a single memory. This corresponds to the von Neumann architecture.
→ SISD is one of the four main classifications as defined in Flynn's taxonomy.
→ In this system, classifications are based upon the number of concurrent instructions and data streams present in the computer architecture. According to Michael J. Flynn, SISD can have concurrent processing characteristics.
→ Pipelined processors and superscalar processors are common examples found in most modern SISD computers.
→ Instructions are sent to the control unit from the Memory Module and are decoded and sent to the processing unit which processes on the data retrieved from Memory module and sent back to it.
 Question 22
What will be the output of the following program? Assume that you are running this program in little-endian processor.
 A 1 B 320 C 64 D Compilation error
Programming-for-Output-Problems       Pointers
Question 22 Explanation:
Ptr = (char*) &a ; /* It is type casted to char* */
Printf (“%d”, *ptr) ; /* Here *ptr is referring a single character */
Printf (“%d”, *ptr) ;
Output: 64 /* @ ASCII value is 64 */
Note: Assume short a is 2 byte integer.
 Question 23
Consider the following segment of C-code: The number of comparisons made in the execution of the loop for any n > 0 is:
 A ⌊log 2n⌋*n B n C ⌊log 2n⌋ D ⌊log 2n⌋+1
Programming-for-Output-Problems       Control Flow
Question 23 Explanation:
Explanation:
Let us consider n=6, then
1<=6 (correct)
2<=6 (correct)
4<=6 (correct)
8<=6 (False)
4 comparisons required
Option A:
⌊log n⌋+1
⌊log 6⌋+1
3+1=4 (correct)
Option B:
n=6 (False)
Option C:
⌊log n⌋
⌊log 6⌋=3 (False)
Option D:
⌊log 2n⌋+1
⌊log 26⌋+1=2+1=3 (False)
 Question 24
The following postfix expression with single digit operands is evaluated using a stack:
8 2 3 ^ / 2 3 * + 5 1 * –
Note that ^ is the exponentiation operator. The top two elements of the stack after the first * is evaluated are
 A 6,1 B 5,7 C 3,2 D 1,5
Data-Structures       Queues-and-Stacks
Question 24 Explanation:
Expression: 8 2 3 ^ / 2 3 * + 5 1 * –
Rule (1): If the element is a number, Push it into stack.
Rule (2): If the element is an operator, Pop operands from the stack. Evaluate the operator operation and Push the result back into the stack.
 Question 25
The average number of key comparisons required for a successful search for sequential search on items is
 A n/2 B (n-1)/2 C (n+1)/2 D None of these
Algorithms       Searching
Question 25 Explanation:
Total comparisons required
= No. of comparisons if element present in 1st position + No. of comparisons if element present in 2ndnd position + ............. + No. of comparisons if element present in nth position
= 1+2+3+ ... + n
= n(n+1)/2
Since there are n elements in the list, so average no. of comparisons
= Total comparisons/Total no. of elements
= (n(n+1)/2)/n
= n+1/2
 Question 26
A Hash Function f is defined as f(key) = key mod 7. With linear probing, while inserting the keys 37, 38, 72, 48, 98, 11, 56 into a table indexed from 0, in which location the key 11 will be stored (Count table index 0 as 0th location)?
 A 3 B 4 C 5 D 6
Data-Structures       Hashing
Question 26 Explanation:
Given data, insertion order is 37, 38, 72, 48, 98, 11, 56
Hash function = f(key) = key mod 7
 Question 27
A complete binary tree with n non-leaf nodes contains
 A log2n nodes B n+1 nodes C 2n nodes D 2n+1 nodes
Data-Structures       Binary-Trees
Question 27 Explanation:
→ A binary tree where each non-leaf node has exactly two children. The number of leaves is n+1. Total number of nodes is 2n+1.
Note: if two leaves having the same parent are removed from the tree, the number of leaves and non-leaves will decrease by 1 because the parent will be a leaf. That means the difference between them is constant. And for a tree having just one node that difference is 1.
 Question 28
Algorithm design technique used in quicksort algorithm is?
 A Dynamic programming B Backtracking C Divide and conquer D Greedy method
Algorithms       Sorting
Question 28 Explanation:
Divide and conquer sorting algorithms:
1. Quick sort
2. Merge sort
3. Heap sort
****Binary search also follows the divide and conquer.
 Question 29
A FSM(finite state machine) can be considered to be a turing machine of finite tape length
 A without rewinding capability and unidirectional tape movement B rewinding capability and unidirectional tape movement C without rewinding capability and bidirectional tape movement D rewinding capability and bidirectional tape movement
Theory-of-Computation       Finite-Automata
Question 29 Explanation:
→ A FSM(finite state machine) can be considered to be a turing machine of finite tape length without rewinding capability and unidirectional tape movement.
→ The finite state machine has less computational power than some other models of computation such as the Turing machine.
→ The computational power distinction means there are computational tasks that a Turing machine can do but a FSM cannot. This is because a FSM's memory is limited by the number of states it has.
→ A finite state machine has the same computational power as a Turing machine that is restricted such that its head may only perform "read" operations, and always has to move from left to right. That is, each formal language accepted by a finite state machine is accepted by such a kind of restricted Turing machine, and vice versa.
 Question 30
Let L = {w = (0+1)* | w has even number of 1’s}, i.e. L is the set of all bit strings with even number of 1’s. Which one of the regular expression below represents L ?
 A (0*10*1)* B 0*(10*10*)* C 0*(10*1*)*0* D 0*1(10*1)10*
Theory-of-Computation       Regular-Expression
Question 30 Explanation:
→ The best way to find correct answer is option elimination method.
→ We will guess strings which has even number of 1’s and that is not generated by wrong options OR which generate strings which doesn’t have even number of 1’s.
Option A: (Regular expression: (0*10*1)* ) doesn’t generate string such as { 110, 1100,....}
Option C: (Regular expression: 0*(10*1*)*0* generate string such as {1, 111,....} which have odd number of 1’s.
Option D: (Regular expression: 0*1(10*1)*10* doesn’t generate strings such as { 11101, 1111101, ….}.
 Question 31
Consider the following recurrence:
T(n) = 2T(√n) + 1
T(1) = 1
Which of the following is true?
 A T(n)= O(log log n) B T(n) = O(log n) C T(n) = O(√n) D T(n)= O(n)
Algorithms       Time-Complexity
Question 31 Explanation:
 Question 32
Consider the following statements about the context-free grammar
G = {S→ SS, S→ ab, S→ ba, S→ ^}
I. G is ambiguous
II. G produces all strings with an equal number of a’s and b’s
III. G can be accepted by a deterministic PDA.
Which combination below expresses all the true statements about?
 A I only B I and III only C II and III only D I, II and III
Theory-of-Computation       Context-Free-Language
Question 32 Explanation:
Is ambiguous, as it has two parse tree for the string “abbaba”

G doesn’t produce all strings of an equal number of a’s and b’s, for ex: string “aabb” doesn’t generate by grammar G.
The language generated by G can be accepted by DPDA. We can notice that grammar G generates, a’s and b’s in pair, i.e. either “ab” or “ba”, so the strings in language are {ab, ba, abab, abba, baba, ….}
We can design the DPDA:

 Question 33
If L and L’ are recursively enumerable then L is
 A Regular B Context-free C Context Sensitive D Recursive
Theory-of-Computation       Recursive-and-Recursively-Enumerable-Language
Question 33 Explanation:
→ If L is recursive enumerable, then it implies that there exist a Turing Machine (lets say M1) which always HALT for the strings which is in L.
→ If L’ are recursively enumerable, then it implies that there exist a Turing Machine (lets say M2) which always HALT for the strings which is NOT in L(as L is complement of L’ Since we can combine both Turing machines (M1 and M2) and obtain a new Turing Machine (say M3) which always HALT for the strings if it is in L and also if it is not in L.
→ This implies that L must be recursive.
 Question 34
S → aSa ∣ bSb ∣ a ∣ b The language generated by the above grammar over the alphabet is the set of
 A all palindromes B all odd length palindromes C strings that begin and end with the same symbol D all even length palindromes
Theory-of-Computation       Languages-and-Grammars
Question 34 Explanation:
From the grammar, we can easily infer that it generates all the palindromes of odd length, for ex: strings {aba, bab, aaa, bbb, ….}
 Question 35
What is the highest type number that can be assigned to the following grammar?
S → Aa
A → Ba
B → abc
 A Type 0 B Type 1 C Type 2 D Type 3
Theory-of-Computation       Languages-and-Grammars
Question 35 Explanation:
Grammar is belongs to Type 0 ,Type 1,Type 2,Type 3 because it is regular. Option D is right answer because they are asking only highest type number.
 Question 36
The access time of the symbol table will be logarithmic if it is implemented by
 A Linear list B Search tree C Hash table D Self organization list
Compiler-Design       Symbol-Table
Question 36 Explanation:
Access time of the symbolic table will be logarithmic if it is implemented by search time.
 Question 37
Recursive descent parsing is an example of
 A Top-down parsers B Bottom-up parsers C Predictive parsers D None of the above
Compiler-Design       Parsers
Question 37 Explanation:
→ A recursive descent parser is a kind of top-down parser built from a set of mutually recursive procedures (or a non-recursive equivalent) where each such procedure implements one of the nonterminals of the grammar.
→ Thus the structure of the resulting program closely mirrors that of the grammar it recognizes.
 Question 38
A top-down parser generates
 A Rightmost Derivation B Rightmost derivation in reverse C Leftmost derivation D Leftmost derivation in reverse
Compiler-Design       Parsers
Question 38 Explanation:
→ Top-down parsing can be viewed as an attempt to find leftmost derivations of an input-stream by searching for parse-trees using a top-down expansion of the given formal grammar rules.
→ The inclusive choice is used to accommodate ambiguity by expanding all alternative right-hand-sides of grammar rules.
 Question 39
Relative mode of addressing is most relevant to writing
 A Coroutines B Position-independent code C Shareable code D Interrupt Handlers
Question 39 Explanation:
The main advantage of PC- relative addressing is that code may be position independent, i.e., it can be loaded anywhere in memory without the need to adjust any address.
 Question 40
 A If there are more than two processes competing for that resources B If there are only two processes competing for that resources C If there is a single process competing for that resources D None of these
Question 40 Explanation:
Only Single resource available never occur deadlock because it violates circular wait and hold & wait condition.
 Question 41
Peephole optimization is a form of
 A Loop optimization B Local optimization C Constant folding D Data flow analysis
Compiler-Design       Code-Optimization
Question 41 Explanation:
→ Peephole optimization technique works locally on the source code to transform it into an optimized code.
→ By locally, we mean a small portion of the code block at hand.
→ These methods can be applied on intermediate codes as well as on target codes.
 Question 42
At a particular time of computation the value of a counting semaphore is 7. Then 20 P operations and xV operations were completed on this semaphore. If the new value of semaphore is 5 ,x will be
 A 18 B 22 C 15 D 13
Operating-Systems       Process-Synchronization
Question 42 Explanation:
Here, 20P operations means 20 wait operations. It decrement value by 1 every time.
xV operations means x increments operations. It increment value by 1 every time.
→ New value of semaphore is 5 after performing xV operations
= -13 + xV
= 5
xV = 5 + 13
= 18
→ After applying 20P operations in semaphore value is = 7-20 = -13
 Question 43
A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units, then
 A Deadlock can never occur B Deadlock may occur C Deadlock has to occur D None of these
Question 43 Explanation:
If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done.
 Question 44
Determine the number of page faults when references to pages occur in the following order: 1, 2, 4, 5, 2, 1, 2, 4 Assume that the main memory can accommodate 3 pages and the main memory already has the pages 1 and 2, with page one having brought earlier than page 2. (LRU page replacement algorithm is used)
 A 3 B 5 C 4 D None of these
Operating-Systems       Page-Replacement-algorithm
Question 44 Explanation:

Here, total 6 page faults but in question, they are clearly mentioned that the main memory already has the pages 1 and 2, with page one having brought earlier than page 2. It means 6-2=4.
 Question 45
Working Set(t,k) at an instant of time t is
 A the set of future references that the OS will make B the set of future references that the OS will make in the next unit of time C the set of references with high frequency D the set of pages that have been referenced in the last time units
Operating-Systems       Working-set
Question 45 Explanation:
→ Working set defines the amount of memory that a process requires in a given time interval.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
 Question 46
A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation lookaside buffer (TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is:
 A 11 bits B 13 bits C 15 bits D 20 bits
Operating-Systems       Memory-Management
Question 46 Explanation:
Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
 Question 47
For the real time operating system, which of the following is the most suitable scheduling scheme?
 A Round robin B Preemptive C First come first serve D Random scheduling
Operating-Systems       Real-Time-Operating-System
Question 47 Explanation:
→ Preemption is the act of temporarily interrupting a task being carried out by a computer system, without requiring its cooperation, and with the intention of resuming the task at a later time. Such changes of the executed task are known as context switches.
→ It is normally carried out by a privileged task or part of the system known as a preemptive scheduler, which has the power to preempt, or interrupt, and later resume, other tasks in the system.
→ Preemptive scheduling is the most suitable scheduling scheme for real time operating systems as it can always preempt other processes based on priority.
→ In real time operating systems, we have different tasks to perform simultaneously and also there are a few tasks which need to be performed first at priority basis.
 Question 48
In which one of the page replacement policies, Belady’s anomaly may occur?
 A FIFO B Optimal C LRU D MRU
Operating-Systems       Page-Replacement-algorithm
Question 48 Explanation:
→ Belady's anomaly is the phenomenon in which increasing the number of page frames results in an increase in the number of page faults for certain memory access patterns.
→ This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
→ In FIFO, the page fault may or may not increase as the page frames increase, but in Optimal and stack-based algorithms like LRU, as the page frames increase the page fault decreases.
 Question 49
Consider the join of a relation R , with a relation S . If R has m number of tuples and S has n number of tuples then the maximum and minimum sizes of the join respectively are:
 A m + n & 0 B mn & 0 C m + n & | m – n | D mn & m + n
Database-Management-System       Relational-Algebra
Question 49 Explanation:
For maximum:
If there is common attribute in R and S, and every row of R match with every row of S then total no. of tuples will be mn.
For minimum:
If there is no common attribute between R and S or if there is common attribute but none of the row of R matches with rows of S then output tuples will be 0.
 Question 50
Let R(a, b, c) and S(d, e, f) be two relations in which d is the foreign key of S that refers to the primary key of R. Consider the following four operations R and S. I. Insert into R II. Insert into S III. Delete from R IV. Delete from S Which of the following can cause a violation of the referential integrity constraint above?
 A Both I and IV B Both II and III C All of these D None of these
Database-Management-System       Constraints
Question 50 Explanation:
 Question 51
The relation book ( title & price ) contains the titles and prices of different books. Assuming that no two books have the same price, what does the following SQL query list?
select title
from book as B
where (select count(*)
from book as T
where T.price>B.price)<5
 A Titles of the four most expensive books B Title of the fifth most inexpensive book C Title of the fifth most expensive book D Titles of the five most expensive books
Database-Management-System       SQL
Question 51 Explanation:
→ Which results titles of the five most expensive books.
→ The where clause of outer query will be true for 5 most expensive books.
 Question 52
Goals for the design of the logical scheme include
 A avoiding data inconsistency B being able to construct query easily C being able to access data efficiently D All of the above
Database-Management-System
Question 52 Explanation:
Logical Schema includes
1. Avoiding data inconsistency
2. Able to construct query easily
3. Able to access data efficiently
 Question 53
Given the relations employee (name, salary, dept-no), and department (dept-no, dept-name,address) Which of the following queries cannot be expressed using the basic relational algebra operations (σ, π, x, -, ∪, p)
 A Department address of every employee B Employees whose name is the same as their department name C The sum of all employees’ salaries D All employees of a given department
Database-Management-System       Relational-Algebra
Question 53 Explanation:
The sum of all employee salaries can’t be represented by using the given six basic algebra operation. If we want to represent sum of salaries then we need to use aggregation operator.
 Question 54
Trigger is
 A Statement that enables to start any DBMS B Statement that is executed by the user when debugging an application program C The condition that the system tests for the validity of the database user D Statement that is executed automatically by the system as a side effect of a modification of the database
Database-Management-System       Trigger
Question 54 Explanation:
→ A database trigger is procedural code that is automatically executed in response to certain events on a particular table or view in a database.
→ The trigger is mostly used for maintaining the integrity of the information on the database.
→ For example, when a new record (representing a new worker) is added to the employees table, new records should also be created in the tables of the taxes, vacations and salaries.
→Triggers can also be used to log historical data, for example to keep track of employees' previous salaries.
 Question 55
The order of a leaf node in a B+ tree is the maximum number of (value, data record pointer) pairs it can hold. Given that the block size is 1K bytes, data record pointer is 7 bytes long, the value field is 9 bytes long and a block pointer is 6 bytes long, what is the order of the leaf node?
 A 63 B 64 C 67 D 68
Database-Management-System       B-and-B+-Trees
Question 55 Explanation:
Disk Block size = 1 KBytes = 210 Bytes = 1024 Bytes
Data recorder pointer size, r = 7 bytes
Value size, v = 9 bytes
Disk Block pointer, P = 6 bytes
⇒ Order of leaf be m, then
r*m+v*m+p <= 1024
7m+9m+6 <= 1024
16m <= 1024 - 6
m <= 63
 Question 56
A clustering index is defined on the fields which are of type
 A non-key and ordering B non-key and non-ordering C key and ordering D key and non-ordering
Database-Management-System       B-and-B+-Trees
Question 56 Explanation:
Single level index
1. Primary index(Sparse): Ordered with key field
2. Clustered index(Sparse): Ordered with non key field
3. Secondary index(dense): Ordered with either key or non key field.
 Question 57
The extent to which the software can continue to operate correctly despite the introduction of invalid inputs is called as
 A Reliability B Robustness C Fault tolerance D Portability
Software-Engineering       Software-testing
Question 57 Explanation:
→ The extent to which the s/w can continue to operate correctly despite the introduction of invalid inputs is called as robustness
→ Robustness is the ability of a computer system to cope with errors during execution and cope with erroneous input
 Question 58
Which one of the following is a functional requirement
 A Maintainability B Portability C Robustness D None of the mentioned
Software-Engineering       Software-requirements
Question 58 Explanation:
Functional Requirements include:
1. Descriptions of data to be entered into the system
2. Descriptions of operations performed by each screen
3. Descriptions of work-flows performed by the system
4. Descriptions of system reports or other outputs

→ The Functional Requirements Specification is designed to be read by a general audience. Readers should understand the system, but no particular technical knowledge should be required to understand the document.

Examples:
1. Interface requirements
3. Regulatory/Compliance requirements
4. Security requirements
 Question 59
Configuration management is not concerned with
 A controlling changes to the source code B choice of a hardware configuration for an application C controlling documentation changes D maintaining versions of software
Software-Engineering       Software-configuration-management
Question 59 Explanation:
Configuration management is not concerned with choice of hardware configuration for an application.
 Question 60
A company needs to develop a strategy for software product development for which it has a choice of two programming languages L1 and L2. The number of lines of code (LOC) developed using L2 is estimated to be twice the LOC developed with Ll. The product will have to be maintained for five years. Various parameters for the company are given in the table below.

Total cost of the project includes cost of development and maintenance.
What is the LOC for L1 for which the cost of the project using L1 is equal to the cost of the project using L2?
 A 10,000 B 5,000 C 7,500 D 75,000
Software-Engineering       LOC
Question 60 Explanation:
LOC L1 = X
L2 = 2X
Total cost of project: x/1000*1000000+5+100000
=2x/10000*750000+50000*5
=100×+500000
=150×+250000
⟹50×=500000-250000
 Question 61
A company needs to develop digital signal processing software for one of its newest inventions. The software is expected to have 20000 lines of code. The company needs to determine the effort in person-months needed to develop this software using the basic COCOMO model. The multiplicative factor for this model is given as 2.2 for the software development on embedded systems, while the exponentiation factor is given as 1.50. What is the estimated effort in person-months?
 A 196.77 B 206.56 C 199.56 D 210.68
Software-Engineering       COCOMO-Model
Question 61 Explanation:
Constructive Cost Model (COCOMO) formula for effort applied is
Effort Applied (E) = ab(KLOC)bb [ person-months ]
= 2.2 x(20)1.50
= 2.2 x 89.44
= 196.77
 Question 62
In the Spiral model of software development, the primary determinant in selecting activities in each iteration is
 A Iteration size B Cost C Adopted process such as Rational Unified Process or Extreme Programming D Risk
Software-Engineering       Software-design
Question 62 Explanation:
→ In the Spiral model of software development, the primary determinant in selecting activities in each iteration is risk handling
→ The spiral model is a risk-driven software development process model. Based on the unique risk patterns of a given project, the spiral model guides a team to adopt elements of one or more process models, such as incremental, waterfall, or evolutionary prototyping.
 Question 63
Bit stuffing refers to
 A inserting a ‘0’ in user stream to differentiate it with a flag B inserting a ‘0’ in flag stream to avoid ambiguity C appending a nibble to the flag sequence D appending a nibble to the use data stream
Computer-Networks       Ethernet
Question 63 Explanation:
→ Bit stuffing is most easily described as insertion of a 0 bit after a long run of 1 bits.
→ In SDLC the transmitted bit sequence "01111110" containing six adjacent 1 bits is the Flag byte.
→ Bit stuffing ensures that this pattern can never occur in normal data, so it can be used as a marker for the beginning and end of frame without any possibility of being confused with normal data.
 Question 64
Dynamic routing protocol enable routers to
 A Dynamically discover and maintain routes B Distribute routing updates to other routers C Reach agreement with other routers about the network topology D All of the above
Computer-Networks       Routing
Question 64 Explanation:
Dynamic Routing
→ Routing protocol learns about other routers automatically.
→ Router and the other routers exchange routes, and each learns about the networks that the other is connected to.
→ When new networks are added or removed, the routers update each other.
 Question 65
In Ethernet CSMA/CD, the special bit sequence transmitted by media access management to handle collision is called
 A Preamble B Postamble C Jam D None of the above
Computer-Networks       Ethernet
Question 65 Explanation:
→ In CSMA/CD once a collision is detected, the colliding devices send a jam signal.
→ Assuming you know the collision detection method used by CSMA/CD, the jam signals generated after collision informs the devices that a collision has occurred and invokes a random backoff algorithm which forces the devices on the Ethernet not to send any data until their timer expires. (everyone has a random timer value)
→ Once a transmitting device detects collision which is by examining the data over the Ethernet and identifying that it's not the data it has send, it does receive the jamming signal otherwise collision will keep happening.
→ As the jam signal is generated after collision hence the devices ideally won’t be sending any data during that time.
 Question 66
Which network protocol allows hosts to dynamically get a unique IP number on each bootup
 A DHCP B BOOTP C RARP D ARP
Question 66 Explanation:
→ A router(or) a residential gateway can be enabled to act as a DHCP server.
→ Most residential network routers receive a globally unique IP address within the ISP network.
→ Within a local network, a DHCP server assigns a local IP address to each device connected to the network.
 Question 67
In a token ring network the transmission speed is 107 bps and the propagation speed is 200 meters/ μs. The 1-bit delay in this network is equivalent to:
 A 500 meters of cable B 200 meters of cable C 20 meters of cable D 50 meters of cable
Computer-Networks       Access-Control-Methods
Question 67 Explanation:
Tt = L / B => 1/ 107 = 0.1 microsec
Given Tp = 200 m / microsec
In, 1 microsec it covers 200m
Therefore, in 0.1 microsec it is 200 * 0.1 = 20 meters
 Question 68
The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?
 A 62 subnets and 262142 hosts B 64 subnets and 262142 hosts C 62 subnets and 1022 hosts D 64 subnets and 1024 hosts
Computer-Networks       Subnetting
Question 68 Explanation:
→ It is a class B address, so there 16-bits for NID and 16-bits for HID.
→ From HID, we took 6-bits for subnetting.
→ Then total subnets possible = ( 26 ) - 2 = 62
→ Total hosts possible for each subnet = (210) - 2 = 1022
 Question 69
The message 11001001 is to be transmitted using the CRC polynomial x3 + 1 to protect it from errors. The message that should be transmitted is:
 A 11001001000 B 11001001011 C 11001010 D 110010010011
Computer-Networks       CRC
Question 69 Explanation:
CRC polynomial = x3+1 [∵ In data 3-zero’s need to be append to data]
= 1001

∴ Data transmitted is: 11001001011
 Question 70
What is the maximum size of data that the application layer can pass on to the TCP layer below?
 A Any size B 216 bytes – size of TCP header C 216 D 1500
Computer-Networks       Application-Layer-Protocol
Question 70 Explanation:
Application Layer - Any size
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
 Question 71
Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between the transmission of two frames.
 A l = 2 B l = 3 C l = 4 D l = 5
Computer-Networks       Control-Flow-Methods
Question 71 Explanation:
→ Transmission time (Tt)=1000/106 seconds = 1 ms
→ Maximum number of frames that can be transmit to maximally pack them is
=(Tt+2Tp)/Tx
= (25+1)/1
=26 which is window size
→ Minimum sequence numbers required = 26
→ Minimum number of bits required for sequence number is 5.
 Question 72
Which of the following is TRUE only for XML but NOT HTML?
 A It is derived from SGML B It describes content and layout C It allows user defined tags D It is restricted only to be used with web browsers
Web-Technologies       HTML
Question 72 Explanation:
→ User-defined tags are only allowed in XML but not in HTML. All the other options are correct for both XML and HTML.
→ SGML stands for Standard Generalized Markup Language and it was extended to HTML and XML.
→ Both XML and HTML are used to describe the content and layout of a webpage and are used with web browsers only.
 Question 73
Consider a system with 2 level cache. Access times of Level 1 cache, Level 2 cache and main memory are 1 ns, 10 ns, and 500 ns, respectively. The hit rates of Level 1 and Level 2 caches are 0.8 and 0.9, respectively. What is the average access time of the system ignoring the search time within the cache?
 A 13 B 12.8 C 12.6 D 12.4
Computer-Organization       Cache
Question 73 Explanation:
Average access time = [H1 * T1] + [(1 - H1) * Hm * Tm]
H1 = 0.8, (1 - H1) = 0.2
H2 = 0.9, (1 - H2) = 0.1
T1 = Access time for level 1 cache = 1ns
T2 = Access time for level 2 cache = 10ns
Hm = Hit rate of main memory = 1
Tm = Access time for main memory = 500ns
Average access time = [(0.8 * 1) + (0.2 * 0.9 * 10) + (0.2)(0.1) * 1 * 500]
= 0.8 + 1.8 + 10
= 12.6ns
 Question 74
If a class C is derived from class B, which is derived from class A, all through public inheritance, then a class C member function can access
 A only protected and public data of C and B B Only protected and public data of C C all data of C and private data of A and B D public and protected data of A and B and all data of C
OOPS       Class-and-object
Question 74 Explanation:
→ It is nothing but multilevel inheritance.
→ If a class C is derived from class B, which is derived from class A, all through public inheritance, then a class C member function can access public and protected data of A and B and all data of C
 Question 75
Which one of the following is correct about the statements are given below? I.  All function calls are resolved at compile time in C language II. All function calls are resolved at compile time in C++ language
 A Only II is correct B Both I and II are correct C Only I is correct D Both I and II are incorrect
Compiler-Design       Compilers
Question 75 Explanation:
Both statements are wrong. I. Logical errors can’t solve in compile time. Like dividing by error.
II. In c++ also we have to write exceptions.
 Question 76
When a DNS server accepts and uses incorrect information from a host that has no authority giving that information, then it is called
 A DNS lookup B DNS hijacking C DNS spoofing D None of the mentioned
Computer-Networks       DNS
Question 76 Explanation:
→ DNS spoofing, also referred to as DNS cache poisoning, is a form of computer security hacking in which corrupt Domain Name System data is introduced into the DNS resolves cache, causing the name server to return an incorrect result record, e.g. an IP address.
→ This results in traffic being diverted to the attacker's computer (or any other computer).
 Question 77
A simple two-pass assembler does which of the following in the first pass:
 A Checks to see if the instructions are legal in the current assembly mode B It allocates space for the literals. C It builds the symbol table for the symbols and their values. D All of these
Compiler-Design       Assembler
Question 77 Explanation:
2-pass Assembler can perform all of above operations.
There are 77 questions to complete.