Question 1 |
Let Σ be the set of all bijections from {1, ..., 5} to {1, ..., 5}, where id denotes the identity function i.e. id(j) = j,∀j. Let º denote composition on functions. For a string x = x1 x2 ... xn ∈ Σn, n ≥ 0. Let π(x) = x1 º x2 º ... º xn.
Consider the language L = {x ∈ Σ* | π(x) = id}. The minimum number of states in any DFA accepting L is ______.
120 | |
136 | |
125 | |
132 |
Question 2 |
∀x(∃z(¬β)→∀y(α)) | |
∀x(∀z(β)→∃y(¬α)) | |
∀x(∀y(α)→∃z(¬β)) | |
∀x(∃y(¬α)→∃z(¬β)) |
∀x(∃z(¬β) → ∀y(α))
⇒ ∀x(¬∃z(¬β) ∨ ∀y(α))
⇒ ∀x(∀z(β)∨y(α))
⇒ ¬∃x¬(∀z(β)∨∀y(α))
⇒ ¬∃x(¬∀z(β)∧¬∀y(α))
A is Not equivalent to the given.
Option B:
∀x(∀z(β)→∃y(¬α))
⇒ ∀x(¬∀z(β)∨∃y(¬α))
⇒ ¬∃x¬(¬∀z(β)∨∃y(¬α))
⇒ ¬∃x(∀z(β)∨∀y(α))
B is Matching and equivalent to given.
Option C:
∀x(∀y(α)→∃z(¬β))
⇒ ∀x(¬∀y(α)∨∃z(¬β))
⇒ ¬∃x¬(¬∀y(α)∨∃z(¬β))
⇒ ¬∃x(∀y(α)∧z(β))
C is equivalent to the given.
Option D:
∀x(∃y(¬α)→∃z(¬β))
⇒ ∀x(¬∃y(¬α)∨∃z(β))
⇒ ∀x(∀y(α)∨∃z(β))
⇒ ¬∃x¬(∀y(α)∨∃z(β))
⇒ ¬∃x(¬∀y(α)∧¬∃z(β))
⇒ ¬∃x(¬∀y(α)∧∀z(¬β))
So D is Not equivalent to the given.
Question 3 |
234.25 | |
932.50 | |
287.80 | |
122.40 |
Question 4 |
Embed web objects from different sites into the same page | |
Refresh the page automatically after a specified interval | |
Automatically redirect to another page upon download | |
Display the client time as part of the page |
Question 5 |
Functional Requirements
| |
Non-Functional Requirements
| |
Goals of Implementation
| |
Algorithms for Software Implementation |
Question 6 |
/* This function computes the roots of a quadratic equation a.x^2 + b.x + c = . The function stores two real roots in *root1 and *root2 and returns the status of validity of roots. It handles four different kinds of cases. (i) When coefficient a is zero irrespective of discriminant (ii) When discreminant is positive (iii) When discriminant is zero (iv) When discriminant is negative. Only in case (ii) and (iii) the stored roots are valid. Otherwise 0 is stored in roots. The function returns 0 when the roots are valid and -1 otherwise. The function also ensures root1 >= root2 int get_QuadRoots( float a, float b, float c, float *root1, float *root2); */
A software test engineer is assigned the job of doing black box testing. He comes up with the following test cases, many of which are redundant.
T1, T2, T3, T6 | |
T1, T3, T4, T5 | |
T2, T4, T5, T6 | |
T2, T3, T4, T5 |
T1 and T2 checking same condition a = 0 hence, any one of T1 and T2 is redundant.
T3, T4: in both case discriminant (D) = b2 – 4ac = 0. Hence any one of it is
T5 : D > 0
T6 : D < 0
Question 7 |

19 | |
21 | |
20 | |
10 |
Question 8 |
P. Requirements Capture 1.Module Development and Integration Q. Design 2.Domain Analysis R. Implementation 3.Structural and Behavioral Modeling S. Maintenance 4.Performance Tuning
P-3, Q-2, R-4, S-1 | |
P-2, Q-3, R-1, S-4 | |
P-3, Q-2, R-1, S-4 | |
P-2, Q-3, R-4, S-1 |
Question 9 |
begin if (a== b) {S1; exit;} else if (c== d) {S2;] else {S3; exit;} S4; endThe test cases T1, T2, T3 and T4 given below are expressed in terms of the properties satisfied by the values of variables a, b, c and d. The exact values are not given. T1 : a, b, c and d are all equal T2 : a, b, c and d are all distinct T3 : a = b and c != d T4 : a != b and c = d Which of the test suites given below ensures coverage of statements S1, S2, S3 and S4?
T1, T2, T3
| |
T2, T4 | |
T3, T4
| |
T1, T2, T4 |
T2 covers S3
T4 covers S2, S4.
Question 10 |
I-II-III-IV-V | |
V-IV-III-II-I
| |
I-III-V-II-IV
| |
IV-II-V-III-I |
Question 11 |
〈2, 2, 3〉 and 〈2, 3, 2〉 | |
〈2, 2, 3〉 and 〈2, 2, 3〉
| |
〈2, 3, 2〉 and 〈2, 3, 2〉 | |
〈2, 3, 2〉 and 〈2, 2, 3〉 |
Question 12 |
II and III
| |
II and III
| |
I and III
| |
I, II and III
|
Question 13 |
I and II | |
II and III
| |
I and III
| |
I, II and III |
Question 14 |
True, True | |
True, False | |
False, True | |
False, False |
Question 15 |
It is derived from SGML | |
It describes content and layout | |
It allows user defined tags | |
It is restricted only to be used with web browsers
|
Question 16 |
- (O1) External inputs and outputs
- (O2) Programming language to be used for the implementation
- (O3) User interactions
- (O4) External interfaces
- (O5) Number of programmers in the software project
- (O6) Files used by the system
02, 03 | |
01, 05 | |
04, 06 | |
02, 05 |
Question 17 |
5, 5 | |
4, 5 | |
5, 4 | |
4, 4 |
Question 18 |
- (S1) The two classes should be kept independent.
- (S2) Low Precision Matrix should be derived from High Precision Matrix.
- (S3) High Precision Matrix should be derived from Low Precision Matrix.
- (S4) One class should be derived from the other; the hierarchy is immaterial.
S1 | |
S2 | |
S3 | |
S4 |
Question 19 |
S4 and S3 | |
S4 and S2 | |
S3 and S1 | |
S2 and S1 |
Question 20 |
Q1 : Select e.empId From employee e Where not exists (Select * From employee s where s.department = “5” and s.salary >=e.salary) Q2 : Select e.empId From employee e Where e.salary > Any (Select distinct salary From employee s Where s.department = “5”)
Q1 is the correct query
| |
Q2 is the correct query | |
Both Q1 and Q2 produce the same answer | |
Neither Q1 nor Q2 is the correct query |
Query 1: Results the empId's which have higher salary than anyone in the department 5.
Query 2: Results the empId's which have higher salary than atleast one employee of department 5.
Question 21 |
Iteration size | |
Cost | |
Adopted process such as Rational Unified Process or Extreme Programming | |
Risk |
Question 22 |
Expert system | |
DB repository | |
Aircraft flight controller | |
Signal processing |
Question 23 |

0 | |
1 | |
2 | |
3 |
Question 24 |

The Title and Content elements | |
The Content and TOC elements | |
The Title and TOC elements | |
The Title, Content and TOC elements
|
Question 25 |

0.45 | |
0.63 | |
0.84 | |
0.95 |

The probability or reliability that the product will work for a defined period of time without failure is given by

Question 27 |

3/4 | |
2/3 | |
1/2 | |
3/8 |
Question 28 |
Consider the CPM activity chart where an arc connecting two milestones is labeled with a task identifier and the time taken in days. For example in order to go from A to B, task T1 takes 180 days. A dashed line depicts an additional dependency that is equivalent to a zero time task.
The set of activities that lie on the critical path are
T1, T2, T8, T10 | |
T1, T3, T8, T10 | |
T1, T2, T3, T4, T5, T6, T7, T8, T9, T10 | |
T1, T4, T5, T7, T8, T10 |
Question 29 |
IF ((A > B) AND (C > D)) THEN A = A + 1 B = B + 1 ENDIFThe cyclomatic complexity of the pseudo-code is
2 | |
3 | |
4 | |
5 |
Question 30 |
awk - F ' ' ' {Print $1, $2} ' t1.txt | while read a b ; do awk -v aV = $ a - v by = $b - F ' ' aV = = $1 (print aV, bV, $2 ) ' t2.txt doneWhich one of the following strings will NOT be present in the output generated when the above script in run? (Note that the given strings may be substrings of a printed line.)
"b1 c1" | |
"b2 c3" | |
"b1 c2" | |
"b1 c3" |
Question 31 |
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at most once | |
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at most once | |
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at least once | |
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at least once |
Question 32 |
- E - N + P
- E - N + 2
- P + 2
- P + 1
1 or 3 | |
2 or 3 | |
2 or 4 | |
1 or 4 |
Question 33 |
A software program consists of two modules M1 and M2 that can fail independently, but never simultaneously. The program is considered to have failed if any of these modules fails. Both the modules are ‘repairable’ and so the program starts working again as soon as the repair is done. Assume that the mean time to failure (MTTF) of M1is T1 with a mean time to repair (MTTR) of R1. The MTTF of M2 is T2 with an MTTR of R2. What is the availability of the overall program given that the failure and repair times are all exponentially distributed random variables?
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Question 34 |

III and IV | |
I and IV | |
II and IV | |
I, II and IV |
Question 35 |
<!ELEMENT Univ (Course+, Prof+)> <!ELEMENT Course (Title, Eval*)> <!ATTLIST Course Number ID #REQUIRED Instructor IDREF #IMPLIED> <!ELEMENT Title (#PCDATA)> <!ELEMENT Eval (#PCDATA)> <!ATTLIST Eval Score CDATA #REQUIRED> <!ELEMENT Prof EMPTY> <!ATTLIST Prof Name ID #REQUIRED Teaches IDREF #IMPLIED>What is returned by the following XQuery? let $as := / /@Score for $c in /Univ/Course[Eval] let $cs := $c/Eval?@Score where min($cs) > avg($as) return $c
The professor with the lowest course evaluation | |
Professors who have all their course evaluations above the university average | |
The course with the lowest evaluation | |
Courses with all evaluations above the university average |
Question 36 |
void swap(float* A1, float* A2) { float temp; if (*A1 = = *A2) return; temp = *A1; *A1 = *A2; *A2 = temp; return; }The program volume for the above module using Halstead's method is
60 | |
63 | |
66 | |
69 |
Question 37 |
void swap(float* A1, float* A2) { float temp; if (*A1 = = *A2) return; temp = *A1; *A1 = *A2; *A2 = temp; return; }The program effort for the above module using Halstead's method is
315 | |
330 | |
393 | |
403 |
Question 38 |
P1-P2-P4, 1 day | |
P1-P3-P4, 1 day | |
P1-P3-P4, 2 days | |
P1-P2-P4, 2 days |
Question 39 |
100 and 1000 | |
100 and 1200 | |
150 and 1200 | |
200 and 2000 |
Question 40 |
6 | |
4 | |
3 | |
2 |
f(Ø)=0 and f(push(S,i) = max(f(S),0) + i;
Initially stack is empty and for empty stack 0 is given.
f(push(0,2)) = max(f(Ø),0) + 2 = max(Ø,0) + 2 = 2
f(push(2,-3)) = max(2,0) + (-3) = 2 - 3 = -1
f(push(-1,2)) = max(-1,0) + 2 = 0 + 2 = 2
f(push(2,-1)) = max(2,0)+ (-1) = 2 - 1 = 1
f(push(1,2)) = max(1,0) + 2 = 1 + 2 = 3
So, 3 will be the answer.
∴ Option C is correct.
Question 41 |
Except in case of an Operating System crash | |
Except in case of a Disk crash | |
Except in case of a power failure | |
Always, even if there is a failure of any kind |
Question 42 |
In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as
01101011 | |
011010110 | |
011101100 | |
0110101100 |
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).