All questions of Practice Test: Full Length for Computer Science Engineering (CSE) Exam
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I was ___ ___ for the bus and then I ___ sight of Craig passing by.- a)Here, waiting, cot
- b)There, waiting, caught
- c)Here, weighting, caught
- d)Hare, waiting, caught
Correct answer is option 'B'. Can you explain this answer?
I was ___ ___ for the bus and then I ___ sight of Craig passing by.
a)
Here, waiting, cot
b)
There, waiting, caught
c)
Here, weighting, caught
d)
Hare, waiting, caught
|
Rohan Shah answered |
The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'
#include<stdio.h>void function(int num){ if(num>0) { function(--num); printf("%d",num); function(--num); }}int main() { function(3); return 0;}Output of the above function is:- a)0120
- b)0123
- c)1023
- d)3210
Correct answer is option 'A'. Can you explain this answer?
#include<stdio.h>
void function(int num){
if(num>0) {
function(--num);
printf("%d",num);
function(--num);
}
}
int main() {
function(3);
return 0;
}
Output of the above function is:
a)
0120
b)
0123
c)
1023
d)
3210
|
|
Subhankar Shah answered |
The recursion tree for the given program is given below:
Expanding the recursion carefully and moving from top to down and left to write, we can see that it will print 0120.
What is the output of following C code?#include<stdio.h>int main( ){char s[ ] = "C programming and Data structures" ; printf ( "\n%s", &s[2] ) ;printf ( "\n%s", s ) ;printf ( "\n%s", &s ) ;printf ( "\n%c", s[2] ) ;return 0;}- a)programming and Data structuresC programming and Data structuresC programming and Data structuresp
- b)C programming and Data structuresC programming and Data structurespp
- c)programming and Data structuresC programming and Data structuresC programming and Data structuresprogramming
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
What is the output of following C code?
#include<stdio.h>
int main( )
{
char s[ ] = "C programming and Data structures" ;
printf ( "\n%s", &s[2] ) ;
printf ( "\n%s", s ) ;
printf ( "\n%s", &s ) ;
printf ( "\n%c", s[2] ) ;
return 0;
}
a)
programming and Data structures
C programming and Data structures
C programming and Data structures
p
b)
C programming and Data structures
C programming and Data structures
p
p
c)
programming and Data structures
C programming and Data structures
C programming and Data structures
programming
d)
None of the above
|
Ravi Singh answered |
Consider a relation A with n elements. What is the total number of relations which can be formed on A which are irreflexive?- a)
- b)
- c)2n
- d)
Correct answer is option 'A'. Can you explain this answer?
Consider a relation A with n elements. What is the total number of relations which can be formed on A which are irreflexive?
a)
b)
c)
2n
d)
|
|
Ravi Singh answered |
A relation 
on a set 
is irreflexive provided that no element is related to itself; in other words, 
for no 
in 
.
on a set is irreflexive provided that no element is related to itself; in other words, for no in .An irreflexive relation contains (n2 - n) elements.
Therefore, number of irreflexive relations =
There are 6 stations in a slotted LAN. Probbability of each station transmits during a contention slot is 0.8. What is the probability of only one station transmits in a given time slot? (Compute the value upto 4 decimal places)
Correct answer is '0.0015'. Can you explain this answer?
There are 6 stations in a slotted LAN. Probbability of each station transmits during a contention slot is 0.8. What is the probability of only one station transmits in a given time slot? (Compute the value upto 4 decimal places)
|
Yash Patel answered |
The probability of only one station = 
If we increase the sides of a rectangular park by 20% , then, what is the total increase in the area of that park ?- a)21 %
- b)33 %
- c)42 %
- d)44 %
Correct answer is option 'D'. Can you explain this answer?
If we increase the sides of a rectangular park by 20% , then, what is the total increase in the area of that park ?
a)
21 %
b)
33 %
c)
42 %
d)
44 %
|
|
Preethi Deshpande answered |
Let length of park = L and breadth of park = B Original Area = BL m2 New Length = 120L/100 = 6L/5 New Breadth = 120B/100 = 6B/5 New Area = 36LB/25 m2 so, Change in area = New – Original = 11LB/25 m2 % increase = Change / original * 100 = 44%
Which of the following languages are regular over ∑ = {a, b}?
L1 = { anbn | 0 < n ≤ 10}
L2 = { a2n | n ≥ 1}
L3 = { an! | n > 0}- a)L2 only.
- b)L1 only.
- c)L1 and L2.
- d)None of L1, L2 and L3.
Correct answer is option 'C'. Can you explain this answer?
Which of the following languages are regular over ∑ = {a, b}?
L1 = { anbn | 0 < n ≤ 10}
L2 = { a2n | n ≥ 1}
L3 = { an! | n > 0}
L1 = { anbn | 0 < n ≤ 10}
L2 = { a2n | n ≥ 1}
L3 = { an! | n > 0}
a)
L2 only.
b)
L1 only.
c)
L1 and L2.
d)
None of L1, L2 and L3.
|
|
Ravi Singh answered |
Any language for which we have a finite set or we can possibly have a DFA drawn for it is regular. L1: It gives a finite set generating string with equal number of a’s and equal number of b’s for range 0
Hence L2 is also regular. L3: It gives all possible strings with a’s such that number of a’s are in factorial values of 1,2,3,4 ……. i.e. {a1, a2, a6, a24 ……………}. The set is neither finite nor a DFA is possible for such set. Hence L3 is not regular.
Find the subnet address for the IP address 165.81.35.120 and subnet mask is 255.255.192.0.- a)255.255.0.0
- b)165.81.192.0
- c)165.81.0.0
- d)165.81.35.0
Correct answer is option 'C'. Can you explain this answer?
Find the subnet address for the IP address 165.81.35.120 and subnet mask is 255.255.192.0.
a)
255.255.0.0
b)
165.81.192.0
c)
165.81.0.0
d)
165.81.35.0
|
|
Ravi Singh answered |
IP address → 165.81.35.120 → 10100101 01010001 00100011 01111000
Subnet mask → 255.255.192.0 → 11111111 11111111 11000000 00000000
IP address 10100101 01010001 00100011 01111000
Subnet mask 11111111 11111111 11000000 00000000
Subnet address 10100101 01010001 00000000 00000000
Subnet address → 10100101 01010001 00000000 00000000 → 165.81.0.0
The sum of the digits in the unit place of all the 4 digit numbers formed with the help of 3, 4, 5 and 6 taken all at a time is:- a)18
- b)108
- c)432
- d)144
Correct answer is option 'B'. Can you explain this answer?
The sum of the digits in the unit place of all the 4 digit numbers formed with the help of 3, 4, 5 and 6 taken all at a time is:
a)
18
b)
108
c)
432
d)
144
|
Shivam Sharma answered |
Sum of the digits in the units place
⇒(4−1)!(3+4+5+6)
⇒3!(3+4+5+6)
⇒6X18
⇒108
Gardener said to me, “Don’t pluck the beautiful flowers.”
Select the appropriate Indirect Speech of the above.- a)Gardner forbade me to pluck the flowers.B
- b)Gardner forbade me not to pluck the flowers.
- c)Gardner told me to pluck the flowers.
- d)Gardner told me that I should not pluck the flowers.
Correct answer is option 'A'. Can you explain this answer?
Gardener said to me, “Don’t pluck the beautiful flowers.”
Select the appropriate Indirect Speech of the above.
Select the appropriate Indirect Speech of the above.
a)
Gardner forbade me to pluck the flowers.
B
b)
Gardner forbade me not to pluck the flowers.
c)
Gardner told me to pluck the flowers.
d)
Gardner told me that I should not pluck the flowers.
|
Sanvi Kapoor answered |
The appropriate speech is:
Gardner forbade me to pluck the flowers.
Consider a sorted array of n elements. Let k inversions (swapping) have been performed on the sorted array and we want to sort it again to reverse effect of inversions. If K is very small in comparison to n, then which sorting technique will prove efficient for making again the list in sorted increasing order array.- a)Selection Sort
- b)Insertion Sort
- c)Bubble Sort
- d)Quick Sort
Correct answer is option 'B'. Can you explain this answer?
Consider a sorted array of n elements. Let k inversions (swapping) have been performed on the sorted array and we want to sort it again to reverse effect of inversions. If K is very small in comparison to n, then which sorting technique will prove efficient for making again the list in sorted increasing order array.
a)
Selection Sort
b)
Insertion Sort
c)
Bubble Sort
d)
Quick Sort
|
|
Sarita Yadav answered |
Insertion sort is directly proportional to number of inversions present in the list.
There are 5 brothers in a family. All were born at a gap of 3 years. If the sum total of ages of 5 brothers is 100. What is the age of 2nd most elder brother ?- a)13
- b)15
- c)23
- d)26
Correct answer is option 'C'. Can you explain this answer?
There are 5 brothers in a family. All were born at a gap of 3 years. If the sum total of ages of 5 brothers is 100. What is the age of 2nd most elder brother ?
a)
13
b)
15
c)
23
d)
26
|
Aditya Deshmukh answered |
Let the age of youngest brother be x
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 100 => 5x = 70 => x = 14
Therefore, age of 2nd most elder brother = 14 + 9 = 23.
Apply the Dijkstra’s shortest path algorithm on the following graph.
The correct path from vertex 1 to 6 is ______.- a)1 - 2, 2 – 3, 3 – 6
- b)1 – 2, 2 – 6
- c)1 – 4, 4 – 2, 2 – 1, 3 – 6
- d)All of the above
Correct answer is option 'A'. Can you explain this answer?
Apply the Dijkstra’s shortest path algorithm on the following graph.
The correct path from vertex 1 to 6 is ______.
a)
1 - 2, 2 – 3, 3 – 6
b)
1 – 2, 2 – 6
c)
1 – 4, 4 – 2, 2 – 1, 3 – 6
d)
All of the above
|
Uday Saraswat answered |
Correct option is 'A' because dijkstra algo is used to find shortest path from source to destination and as we see in these options option A is the shortest path.
What is the number missing from the table?
- a)45
- b)48
- c)51
- d)54
Correct answer is option 'B'. Can you explain this answer?
What is the number missing from the table?
a)
45
b)
48
c)
51
d)
54
|
Shounak Sharma answered |
XY + Z = X(Y + Z)
⇒ XY + Z = XY + XZ
⇒ Z = XZ ⇒ Z - XZ = 0 ⇒ Z (1 - X) = 0
Either Z = 0 (in this case X can take any value) OR X = 1 (in this case Z can take any value).
X = 1 or Z = 0
5 × 3 + 1 = 16
16 × 3 + 1 = 49
9 × 3 + 2 = 29
29 × 3 + 2 = 89
15 × 3 + 3 = 48
48 × 3 + 3 = 147
Match the protocol with the characteristics:
1. Mails are stored on the computer client use.
2. Stateless protocol
3. Converts MAC address to IP address.
4. Used to send error messages- a)1: POP3, 2: HTTP, 3: ARP, 4: SMTP
- b)1: IMAP, 2: HTTP, 3: RARP, 4: SMTPC
- c)1: POP3, 2: HTTP, 3: RARP, 4: ICMP
- d)1: IMAP, 2: HTTP, 3: ARP, 4: ICMP
Correct answer is option 'C'. Can you explain this answer?
Match the protocol with the characteristics:
1. Mails are stored on the computer client use.
2. Stateless protocol
3. Converts MAC address to IP address.
4. Used to send error messages
1. Mails are stored on the computer client use.
2. Stateless protocol
3. Converts MAC address to IP address.
4. Used to send error messages
a)
1: POP3, 2: HTTP, 3: ARP, 4: SMTP
b)
1: IMAP, 2: HTTP, 3: RARP, 4: SMTP
C
c)
1: POP3, 2: HTTP, 3: RARP, 4: ICMP
d)
1: IMAP, 2: HTTP, 3: ARP, 4: ICMP
|
Zoya Sharma answered |
In POP3 protocol mails are stored on client’s computer while in IMAP mails are stored on the server. HTTP is a stateless protocol, RARP is used to convert MAC to IP while ARP is used to convert IP to MAC, ICMP is used for sending error messages.
What will the output of the following C code ?
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
What will the output of the following C code ?
a)
1
b)
2
c)
3
d)
4
|
Vedika Singh answered |
Consider the while loop condition => i + 1 ? --i : j++ In first iteration : i + 1 = 3 (true) So ternary operator will return --i i.e. 1, condition part is 1 means true so while condition is true. Hence printf statement will print 1 In second iteration : i + 1 =2 (true) So ternary operator will return --i i.e 0, condition part is 0 means false so while condition is false. Hence program control will come out of the while loop.
In a data link layer, bit stuffing is used in transferring data. If the sent data after bit stuffing is 001111101101011111001111 and the flag is 01111110, then what will be the data after destuffing?- a)001111101101011111001111
- b)00111111101011111001111
- c)00111110110101111101111
- d)0011111110101111101111
Correct answer is option 'D'. Can you explain this answer?
In a data link layer, bit stuffing is used in transferring data. If the sent data after bit stuffing is 001111101101011111001111 and the flag is 01111110, then what will be the data after destuffing?
a)
001111101101011111001111
b)
00111111101011111001111
c)
00111110110101111101111
d)
0011111110101111101111
|
Vikram Bansal answered |
As the flag is 01111110, we have to delete 0 after every five consecutive 1’s.
A finite automata over the input alphabet {x,y} is given below. What does the following deterministic finite automata perform?
- a)Accepts all the strings in which every y is followed by x.
- b)Accepts all the strings whose second symbol from left hand side is x.
- c)Accepts all the strings of length divisible by 3.
- d)Accepts all the strings ending with a.
Correct answer is option 'B'. Can you explain this answer?
A finite automata over the input alphabet {x,y} is given below. What does the following deterministic finite automata perform?
a)
Accepts all the strings in which every y is followed by x.
b)
Accepts all the strings whose second symbol from left hand side is x.
c)
Accepts all the strings of length divisible by 3.
d)
Accepts all the strings ending with a.
|
|
Ravi Singh answered |
The language generated by the given automata is: {xx, yx, xxx, xxy, .........}
The given automata accepts all the strings whose second symbol from left hand side is x.
Given below is a system of linear equations.5a + b + 3c = 59a + c = -8a + 2b + 5c=8This system of equations has,- a)A unique solution
- b)Infinitely many solution
- c)No solution
- d)Cannot be determined
Correct answer is option 'B'. Can you explain this answer?
Given below is a system of linear equations.
5a + b + 3c = 5
9a + c = -8
a + 2b + 5c=8
This system of equations has,
a)
A unique solution
b)
Infinitely many solution
c)
No solution
d)
Cannot be determined
|
|
Yash Patel answered |
Comparing the system of equation with the matrix equation,
A.X = B
We have,
A = 
, B = 
, B = Augmented matrix is given by,
Applying, R3→R3−(2R1−R2)
We get,
Thus, Rank(A)=Rank(A:B)=2 < />
Hence, the system of equations is consistent and has infinitely many solutions.
The following C function takes a single-linked list of integers as a parameter and rearranges the elements of the list. The function is called with the list containing the integers 1, 2, 3, 4, 5, 6, 7 in the given order. What will be the contents of the list after function completes execution?struct node { int value; struct node *next;};void rearrange(struct node *list) { struct node *p, *q; int temp; if (!list || !list -> next) return; p = list; q = list -> next; while(q) { temp = p -> value; p->value = q -> value; q->value = temp; p = q ->next; q = p? p ->next : 0; } }- a)2, 1, 4 ,3, 6, 5, 7
- b)1, 3, 2, 5, 4, 7, 6
- c)2, 3, 4, 5, 6, 7, 1
- d)1, 2 ,3, 4, 5, 6, 7
Correct answer is option 'A'. Can you explain this answer?
The following C function takes a single-linked list of integers as a parameter and rearranges the elements of the list. The function is called with the list containing the integers 1, 2, 3, 4, 5, 6, 7 in the given order. What will be the contents of the list after function completes execution?
struct node {
int value;
struct node *next;
};
void rearrange(struct node *list) {
struct node *p, *q;
int temp;
if (!list || !list -> next) return;
p = list; q = list -> next;
while(q) {
temp = p -> value; p->value = q -> value;
q->value = temp; p = q ->next;
q = p? p ->next : 0;
}
}
a)
2, 1, 4 ,3, 6, 5, 7
b)
1, 3, 2, 5, 4, 7, 6
c)
2, 3, 4, 5, 6, 7, 1
d)
1, 2 ,3, 4, 5, 6, 7
|
|
Ravi Singh answered |
The loop is interchanging the adjacent elements of the list.
But because of (p = q -> next;) after each interchange, next interchange starts from the unchanged elements.
1st iteration 1 2 3 4 5 6 7 ------> 2 1 3 4 5 6 7
2nd iteration 2 1 4 3 5 6 7
3rd iteration 2 1 4 3 6 5 7
p will point to 7, and q=p true, hence q = p->next = null.
Given a graph G (V, E) is Bipartite, what is the chromatic number of G ?- a)2
- b)3
- c)∞
- d)Can't be decided
Correct answer is option 'A'. Can you explain this answer?
Given a graph G (V, E) is Bipartite, what is the chromatic number of G ?
a)
2
b)
3
c)
∞
d)
Can't be decided
|
|
Aravind Sengupta answered |
Since the graph G is bipartite, the vertices set V can be partitioned into two disjoint sets. This shows that we can color the graph with 2 colors such that no two adjacent vertices will have same color.
Mean of 12 observation was found to be 28. Later on, it was detected that an observation 62 was misread as 26. Then the correct mean of the observations is- a)28
- b)31
- c)33
- d)36
Correct answer is option 'B'. Can you explain this answer?
Mean of 12 observation was found to be 28. Later on, it was detected that an observation 62 was misread as 26. Then the correct mean of the observations is
a)
28
b)
31
c)
33
d)
36
|
Soumya Dey answered |
Given, mean of 12 observation was found to be 28.
Mean = sum of observations/number of observations.
Sum of observations = 12 × 28 = 336
Now, it was detected that an observation 62 was misread as 26.
∴ Correct mean = 
Can you explain the answer of this question below:Let T(n) = 2T(n/4) + 100√n
The value of T(n) can be written as:
- A:
θ(√ n)
- B:
θ(√n log n)
- C:
θ(n)
- D:
θ(n log n)
The answer is b.
Let T(n) = 2T(n/4) + 100√n
The value of T(n) can be written as:
θ(√ n)
θ(√n log n)
θ(n)
θ(n log n)
|
Pranjal Sen answered |
The time complexity of the given recurrence relation can be found using the Master Theorem.
The Master Theorem states that if a recurrence relation is of the form T(n) = aT(n/b) + f(n), where a ≥ 1, b > 1, and f(n) is a polynomial function, then:
- If f(n) = O(n^(log_b(a)-ε)) for some ε > 0, then T(n) = Θ(n^(log_b(a))).
- If f(n) = Θ(n^(log_b(a))), then T(n) = Θ(n^(log_b(a)) * log(n)).
- If f(n) = Ω(n^(log_b(a)+ε)) for some ε > 0, and if a*f(n/b) <= c*f(n)="" for="" some="" constant="" c="">< 1="" and="" sufficiently="" large="" n,="" then="" t(n)="" />
In the given recurrence relation, a = 2, b = 4, and f(n) = 100, which is a constant.
log_b(a) = log_4(2) = 1/2
f(n) = O(n^(log_b(a)-ε)) for any ε > 0, since f(n) is a constant and n^(log_b(a)-ε) approaches 0 as n goes to infinity.
Therefore, by the Master Theorem, T(n) = Θ(n^(log_b(a))) = Θ(n^(1/2)).
Hence, the time complexity of the given recurrence relation is Θ(n^(1/2)).
The Master Theorem states that if a recurrence relation is of the form T(n) = aT(n/b) + f(n), where a ≥ 1, b > 1, and f(n) is a polynomial function, then:
- If f(n) = O(n^(log_b(a)-ε)) for some ε > 0, then T(n) = Θ(n^(log_b(a))).
- If f(n) = Θ(n^(log_b(a))), then T(n) = Θ(n^(log_b(a)) * log(n)).
- If f(n) = Ω(n^(log_b(a)+ε)) for some ε > 0, and if a*f(n/b) <= c*f(n)="" for="" some="" constant="" c="">< 1="" and="" sufficiently="" large="" n,="" then="" t(n)="" />
In the given recurrence relation, a = 2, b = 4, and f(n) = 100, which is a constant.
log_b(a) = log_4(2) = 1/2
f(n) = O(n^(log_b(a)-ε)) for any ε > 0, since f(n) is a constant and n^(log_b(a)-ε) approaches 0 as n goes to infinity.
Therefore, by the Master Theorem, T(n) = Θ(n^(log_b(a))) = Θ(n^(1/2)).
Hence, the time complexity of the given recurrence relation is Θ(n^(1/2)).
The minimum number of bits required to represent - 64 in 2’s complement representation is _________.
Correct answer is '7'. Can you explain this answer?
The minimum number of bits required to represent - 64 in 2’s complement representation is _________.
|
|
Yash Patel answered |
Whenever a number is in 2n form then the minimal 2’s complement representation is 1 followed by n number of zeros.
So, – 64 = – 26 = 1000000
Consider the following infix expression. While converting this infix expression to postfix expression, what will be the stack content (from bottom to top) after scanning operand 5? ((13 × 4) + 16) ÷ 4 – ((8 × 5) ÷ 10)- a)( - ( ( ×
- b)( × + ÷ -
- c)( ÷ - ( ( ×
- d)( × - ( (
Correct answer is option 'A'. Can you explain this answer?
Consider the following infix expression. While converting this infix expression to postfix expression, what will be the stack content (from bottom to top) after scanning operand 5?
((13 × 4) + 16) ÷ 4 – ((8 × 5) ÷ 10)
a)
( - ( ( ×
b)
( × + ÷ -
c)
( ÷ - ( ( ×
d)
( × - ( (
|
Nishanth Roy answered |
The expression (P ⇒ Q) ˄ (R ⇒ Q) is equivalent to- a)(P ˅ R) ˄ Q
- b)(P ˅ R) ⇒ Q
- c)P ˄ Q ˄ R
- d)P ˅ Q ˅ R
Correct answer is option 'B'. Can you explain this answer?
The expression (P ⇒ Q) ˄ (R ⇒ Q) is equivalent to
a)
(P ˅ R) ˄ Q
b)
(P ˅ R) ⇒ Q
c)
P ˄ Q ˄ R
d)
P ˅ Q ˅ R
|
Janani Joshi answered |
The correct answer is option 2 i.e. (P ˅ R) ⇒ Q.
The truth table for the given expressions is:
A digital computer has a memory unit of 256k x 16 and a cache memory of 4k words. The cache uses direct mapping with a block size of 16 words. Q. How many bits are there in index, tag, block and words fields of address format ?- a)10, 6, 8, 2
- b)12, 6, 8, 4
- c)12, 8, 6, 4
- d)10, 6, 8, 4
Correct answer is option 'B'. Can you explain this answer?
A digital computer has a memory unit of 256k x 16 and a cache memory of 4k words. The cache uses direct mapping with a block size of 16 words.
Q. How many bits are there in index, tag, block and words fields of address format ?
a)
10, 6, 8, 2
b)
12, 6, 8, 4
c)
12, 8, 6, 4
d)
10, 6, 8, 4
|
|
Rajeev Sharma answered |
Main Memory has 256k = 2^8 x 2^10 = 2^18.
i.e. We need 18 bits to address main memory.
Cache Memory has 4k = 2^2 x 2^10 = 2^12.
i.e. We need 12 bits to address cache memory.
Cache consists of Index and tag which together are used to address main memory location.
Here Index is 12 bits and tag is 6 bits. (18 – 12 = 6).
Index is divided into block part and word part. Block part is used to address blocks in cache and word part addresses individual word in a block.
Here a block size is 16 words. i.e. 2^4, we need 4 bits to address a word and 12 – 4 = 8 bits to address a block in cache memory.
Let L1 be a regular language, L2 be a deterministic context-free language and L3 a recursively enumerable but not recursive language.Which one of the following statements is false?- a)L1 ∩ L2 is a deterministic CFL
- b)L3 ∩ L1 is recursive
- c)L1. L2 is context free
- d)L1∩ L2∩ L3 is recursively enumerable
Correct answer is option 'B'. Can you explain this answer?
Let L1 be a regular language, L2 be a deterministic context-free language and L3 a recursively enumerable but not recursive language.
Which one of the following statements is false?
a)
L1 ∩ L2 is a deterministic CFL
b)
L3 ∩ L1 is recursive
c)
L1. L2 is context free
d)
L1∩ L2∩ L3 is recursively enumerable
|
|
Anagha Chopra answered |
Explanation:
To determine which of the given statements is false, let's analyze each option one by one.
Option 'a': L1 ∩ L2 is a deterministic CFL
The intersection of a regular language (L1) and a deterministic context-free language (L2) can be either regular or context-free. Since both regular languages and deterministic context-free languages are closed under intersection, the intersection of L1 and L2 will also be either regular or deterministic context-free.
Therefore, option 'a' is true.
Option 'b': L3 ∩ L1 is recursive
The intersection of a recursively enumerable language (L3) and a regular language (L1) can be either recursively enumerable or recursive. However, the option states that L3 ∩ L1 is recursive, which may not always be true. It is possible for the intersection to be recursively enumerable but not recursive.
Therefore, option 'b' is false.
Option 'c': L1 . L2 is context free
The concatenation of a regular language (L1) and a deterministic context-free language (L2) will always result in a context-free language. Regular languages are a subset of context-free languages, and the concatenation operation preserves the context-free property.
Therefore, option 'c' is true.
Option 'd': L1 ∩ L2 ∩ L3 is recursively enumerable
The intersection of three languages, L1, L2, and L3, can be recursively enumerable. Since recursively enumerable languages are closed under intersection, the intersection of L1, L2, and L3 will also be recursively enumerable.
Therefore, option 'd' is true.
In summary, the false statement is option 'b': L3 ∩ L1 is recursive.
To determine which of the given statements is false, let's analyze each option one by one.
Option 'a': L1 ∩ L2 is a deterministic CFL
The intersection of a regular language (L1) and a deterministic context-free language (L2) can be either regular or context-free. Since both regular languages and deterministic context-free languages are closed under intersection, the intersection of L1 and L2 will also be either regular or deterministic context-free.
Therefore, option 'a' is true.
Option 'b': L3 ∩ L1 is recursive
The intersection of a recursively enumerable language (L3) and a regular language (L1) can be either recursively enumerable or recursive. However, the option states that L3 ∩ L1 is recursive, which may not always be true. It is possible for the intersection to be recursively enumerable but not recursive.
Therefore, option 'b' is false.
Option 'c': L1 . L2 is context free
The concatenation of a regular language (L1) and a deterministic context-free language (L2) will always result in a context-free language. Regular languages are a subset of context-free languages, and the concatenation operation preserves the context-free property.
Therefore, option 'c' is true.
Option 'd': L1 ∩ L2 ∩ L3 is recursively enumerable
The intersection of three languages, L1, L2, and L3, can be recursively enumerable. Since recursively enumerable languages are closed under intersection, the intersection of L1, L2, and L3 will also be recursively enumerable.
Therefore, option 'd' is true.
In summary, the false statement is option 'b': L3 ∩ L1 is recursive.
If there are 5 persons who can work individually and complete a work in 2, 4, 5, 6 and 8 hours respectively. If they work together then how much time will it take to finish the work ?- a)2.36 hours
- b)1.24 hours
- c)0.80 hours
- d)2.56 hours
Correct answer is option 'C'. Can you explain this answer?
If there are 5 persons who can work individually and complete a work in 2, 4, 5, 6 and 8 hours respectively. If they work together then how much time will it take to finish the work ?
a)
2.36 hours
b)
1.24 hours
c)
0.80 hours
d)
2.56 hours
|
|
Shounak Das answered |
In 1 hour, ½ + ¼ + ⅕ +⅙ + ⅛ = 149/120 = 1.24 is the work done. Hence, total time to complete the work would be = 1/1.24 = 0.80 hours
The output F of the 4-to-1 MUX shown in figure is
- a)
- b)x
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The output F of the 4-to-1 MUX shown in figure is
a)
b)
x
c)
d)
|
|
Sanvi Kapoor answered |
F = xy(1) + xy'(1) + x'y(0) + x'y'(0)
=xy + xy'
= x
=xy + xy'
= x
X, Y and Z are the three contestants in one km race. If X can give Y a start of 52 metres and X can also give Z a start of 83 metres, how many metres start Y can give Z?- a)33.3 m
- b)33 m
- c)32 m
- d)32.7 m
Correct answer is option 'D'. Can you explain this answer?
X, Y and Z are the three contestants in one km race. If X can give Y a start of 52 metres and X can also give Z a start of 83 metres, how many metres start Y can give Z?
a)
33.3 m
b)
33 m
c)
32 m
d)
32.7 m
|
|
Ravi Singh answered |
While X runs 1000 metre, Y runs (1000-52) = 948 metre and Z runs (1000-83) = 917 metre
i.e., when Y runs 948 metre, Z runs 917 metre
i.e., when Y runs 948 metre, Z runs 917 metre
⇒ When Y runs 1000 metre, Z runs 917/948 x 1000 = 967.30 metre
i.e., Y can give Z (1000-967.30) = 32.7 metre
Construct a B+-tree for the following set of key values:(2, 3, 5, 7, 11, 17, 19, 23, 29, 31)Assume that the tree is initially empty and values are added in ascending order. Construct B+-trees for the cases where the number of pointers is 4 and tell which element is the root node of the tree.- a)19
- b)29
- c)11
- d)17
Correct answer is option 'A'. Can you explain this answer?
Construct a B+-tree for the following set of key values:
(2, 3, 5, 7, 11, 17, 19, 23, 29, 31)
Assume that the tree is initially empty and values are added in ascending order. Construct B+-trees for the cases where the number of pointers is 4 and tell which element is the root node of the tree.
a)
19
b)
29
c)
11
d)
17
|
|
Sarita Yadav answered |
The least number of temporary variables required to create a three-address code in static single assignment form for the expression p+q*r-s/(q*r)
Correct answer is '5'. Can you explain this answer?
The least number of temporary variables required to create a three-address code in static single assignment form for the expression p+q*r-s/(q*r)
|
|
Arnab Kapoor answered |
t1= q*r
t2= p+t1
t3= q*r
t4= s/t3
t5= t2-t4
Hence, minimum 5 variables are required.
Consider the following weighted graph. Bellman-Ford algorithm is implemented on the given graph with source P. The shortest distance from source P to vertex T is ______.
Correct answer is '2'. Can you explain this answer?
Consider the following weighted graph. Bellman-Ford algorithm is implemented on the given graph with source P. The shortest distance from source P to vertex T is ______.
|
|
Ananya Kumari answered |
Consider the following weighted graph . bellman - Ford algorithm is implemented on the given on the given graph with source p. the shortest distance from source p to verte lx T is ans
2 and -1 and 2 or 1 about -1 and 1 or 2 the 3
ans 2
2 and -1 and 2 or 1 about -1 and 1 or 2 the 3
ans 2
A system has 3 resources and 5 processes competing for them. Each process can request a maximum of N instances. The largest value of N that will always avoid deadlock is _______.
Correct answer is '1'. Can you explain this answer?
A system has 3 resources and 5 processes competing for them. Each process can request a maximum of N instances. The largest value of N that will always avoid deadlock is _______.
|
Aaditya Ghosh answered |
Distribute each process to one less than maximum request resources i.e. 5(N - 1)
Total no. of resources = 5(N - 1) + 1 = 5N - 2
If total no. resources are greater than given resources then a deadlock will occur.
5N – 2 <= 3
5N <= 5
N <= 1
Direction: Study the information carefully to answer the questions that follow.In Vimla public school, students can choose among three sports Cricket, Football and Hockey. Out of 1000 students, each one is required to play at least one of the three sports. 40 students play all three sports. 380 students play only Hockey, 170 students play only Cricket, 60 students play exactly two sports Hockey and cricket and 240 students play only Football.What is the number of students who play exactly two sports?- a)155
- b)170
- c)200
- d)140
Correct answer is option 'B'. Can you explain this answer?
Direction: Study the information carefully to answer the questions that follow.
In Vimla public school, students can choose among three sports Cricket, Football and Hockey. Out of 1000 students, each one is required to play at least one of the three sports. 40 students play all three sports. 380 students play only Hockey, 170 students play only Cricket, 60 students play exactly two sports Hockey and cricket and 240 students play only Football.
What is the number of students who play exactly two sports?
a)
155
b)
170
c)
200
d)
140
|
|
Sarita Yadav answered |
It is given that the total number of students are = 1000
Only Hockey = 380
Only Cricket = 170
Only Football = 240
Hockey + Cricket + Football = 40 {all the three sports}
Now, let X be the number of students who play Cricket and Football.
And Y is the number of students who play football and hockey.
Now total number of students,
170 +60 +40 +380+ 240+ X + Y= 1000
X + Y = 110
And 60 students play only hockey and cricket, and that number we get after removing students who play all three sports from students who play hockey and cricket.
X + Y + 60 = 110+ 60 = 170
There are 170 students who play exactly two sports.
Consider the following functions:f1(a, b, c) = Ʃm(0, 1, 3, 5) + d(2, 4)f2(a, b, c) = Ʃm(1, 6) + d(2, 3, 5)How many functions are possible for f1 + f2?
Correct answer is '4'. Can you explain this answer?
Consider the following functions:
f1(a, b, c) = Ʃm(0, 1, 3, 5) + d(2, 4)
f2(a, b, c) = Ʃm(1, 6) + d(2, 3, 5)
How many functions are possible for f1 + f2?
|
Gargi Menon answered |
Therefore, the number of functions possible = 22 = 4
The 32 bit floating point representation of (-12) is ___- a)11000000010000000000000000000000
- b)11000001010000000000000000000000
- c)01000001110000000000000000000000
- d)11000000110000000000000000000000
Correct answer is option 'B'. Can you explain this answer?
The 32 bit floating point representation of (-12) is ___
a)
11000000010000000000000000000000
b)
11000001010000000000000000000000
c)
01000001110000000000000000000000
d)
11000000110000000000000000000000
|
|
Aditya Deshmukh answered |
To convert the floating point into decimal, we have 3 elements in a 32-bit floating point representation:
Sign (MSB)
Exponent (8 bits after MSB)
Mantissa (Remaining 23 bits)
Sign bit is the first bit of the binary representation. ‘1’ implies negative number and ‘0’ implies positive number. Sign bit=1
Exponent is decided by the next 8 bits of binary representation.
Hence the exponent of 2 will be 3. i.e 23=8.
127 is the unique number for 32 bit floating point representation. It is known as bias. It is determined by 2k-1-1 where ‘k’ is the number of bits in exponent field.
Thus bias = 127 for 32 bit. (28-1-1 = 128-1=127)
127+3=130 i.e. 10000010 in binary representation.
Mantissa: 12 in binary = 1100
Move the binary point so that there is only one bit from the left. Adjust the exponent of 2 so that the value does not change. This is normalizing the number.
1.100 x 23
10000000000000000000000
Thus the floating point representation of -12 is
1 10000010 10000000000000000000000
In public key and private key cryptography, host ‘A’ has PuA and PrA, host ‘B’ has PuB and PrB as public and private keys respectively. Now, if ‘A’ wants to send a message to ‘B’ securely, which key will be used by ‘A’ for encryption ?- a)PuA
- b)PuB
- c)PrA
- d)PrB
Correct answer is option 'B'. Can you explain this answer?
In public key and private key cryptography, host ‘A’ has PuA and PrA, host ‘B’ has PuB and PrB as public and private keys respectively. Now, if ‘A’ wants to send a message to ‘B’ securely, which key will be used by ‘A’ for encryption ?
a)
PuA
b)
PuB
c)
PrA
d)
PrB
|
|
Rajesh Malik answered |
If the message is encrypted using PuB, then it can be decrypted using only PrB, which will be known only to ‘B’.
A sum of Rs. 400 amounts to Rs. 480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?- a)Rs. 512
- b)520Rs.
- c)Rs.490
- d)Rs.515
Correct answer is option 'A'. Can you explain this answer?
A sum of Rs. 400 amounts to Rs. 480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?
a)
Rs. 512
b)
520Rs.
c)
Rs.490
d)
Rs.515
|
|
Sanvi Kapoor answered |
We know that, S.I = P x R x T / 100
A = S.I + P
480 = 5.1+400
⇒S.I = 480 − 400 = 80
⇒S.I = P × R × T100
⇒80 = 400 × R × 4100
⇒R = 5%
Now rate is increased by 2% So, new rate is 7% New S.I=400 × 7 × 4 / 100 = Rs.112
New Amount = S.I + P = 112 + 400 = Rs.512
The number of 4 line to-16 line decoders required to make an 8-line-to-256 line decoder is- a)12
- b)17
- c)23
- d)35
Correct answer is option 'B'. Can you explain this answer?
The number of 4 line to-16 line decoders required to make an 8-line-to-256 line decoder is
a)
12
b)
17
c)
23
d)
35
|
|
Sarita Yadav answered |
8-line to-256-line decoder using 4-line to-16-line decoders is shown below
Number of 4 x 16 MUX required
= 256 / 16 + 16/16 = 16 + 1 = 17
Find how much disk space (in surfaces) will be required to store 300,000 120-byte logical records, if the disk is fixed sector with 512 bytes/sector, with 96 sectors/track, 110 tracks per surface, and 8 usable surfaces. Ignore any file header records and track indices and assume that records cannot span two sectors.
Correct answer is '8'. Can you explain this answer?
Find how much disk space (in surfaces) will be required to store 300,000 120-byte logical records, if the disk is fixed sector with 512 bytes/sector, with 96 sectors/track, 110 tracks per surface, and 8 usable surfaces. Ignore any file header records and track indices and assume that records cannot span two sectors.
|
|
Maheshwar Majumdar answered |
Each sector can hold 4 logical records.
The required number of sectors is 303,150/4 = 75,788 sectors
= 75,788/96 tracks = 790 track
= 790/110 = 8 surfaces.
The required number of sectors is 303,150/4 = 75,788 sectors
= 75,788/96 tracks = 790 track
= 790/110 = 8 surfaces.
Consider the following regular expressions over the alphabet {0, 1}.I. 1* 0(0 + 1)*II. (0 + 1)* 01*III. 1* 0(1 + 0)* + (0 + 1)* 01*Which of the above regular expressions are equivalent?- a)I and III only
- b)I and II only
- c)II and III only
- d)I, II and III
Correct answer is option 'D'. Can you explain this answer?
Consider the following regular expressions over the alphabet {0, 1}.
I. 1* 0(0 + 1)*
II. (0 + 1)* 01*
III. 1* 0(1 + 0)* + (0 + 1)* 01*
Which of the above regular expressions are equivalent?
a)
I and III only
b)
I and II only
c)
II and III only
d)
I, II and III
|
|
Ravi Singh answered |
I and II are quite easy to understand - both denote strings containing at least one 0. III is actually the union of I and II, but III is again denoting strings containing at least one 0, as both regular expressions (i.e. I and II) are equivalent, so union won't make any difference.
Hence all 3 regular expressions are equivalent.
The language L = {WbcWR | W ∈ (a+b)*} is _____.- a)DCFL
- b)CFL but not DCFL
- c)Non-CFL
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The language L = {WbcWR | W ∈ (a+b)*} is _____.
a)
DCFL
b)
CFL but not DCFL
c)
Non-CFL
d)
None of the above
|
|
Shail Kulkarni answered |
Is the set of all strings that have the form WbcWR, where W is any string, b is a single 'b' character, c is a single 'c' character, and WR is the reverse of W.
For example, the string "abcba" would be in L because W = "a" and WR = "a", so WbcWR = "abcba". Similarly, the string "xyzbcbazyx" would also be in L because W = "xyz" and WR = "zyx", so WbcWR = "xyzbcbazyx".
However, the string "abcbazyx" would not be in L because there is no possible W such that WbcWR would result in "abcbazyx".
In summary, L is the language of all strings that can be formed by taking any string, adding a 'b' and 'c' in between, and then adding the reverse of the original string.
For example, the string "abcba" would be in L because W = "a" and WR = "a", so WbcWR = "abcba". Similarly, the string "xyzbcbazyx" would also be in L because W = "xyz" and WR = "zyx", so WbcWR = "xyzbcbazyx".
However, the string "abcbazyx" would not be in L because there is no possible W such that WbcWR would result in "abcbazyx".
In summary, L is the language of all strings that can be formed by taking any string, adding a 'b' and 'c' in between, and then adding the reverse of the original string.
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