Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Tests  >  GATE Computer Science Engineering(CSE) 2025 Mock Test Series  >  Practice Test: Computer Science Engineering (CSE) - 1 - Computer Science Engineering (CSE) MCQ

Practice Test: Computer Science Engineering (CSE) - 1 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Practice Test: Computer Science Engineering (CSE) - 1

Practice Test: Computer Science Engineering (CSE) - 1 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The Practice Test: Computer Science Engineering (CSE) - 1 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 1 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Computer Science Engineering (CSE) - 1 below.
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Practice Test: Computer Science Engineering (CSE) - 1 - Question 1

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?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 1

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

⇒ 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

Practice Test: Computer Science Engineering (CSE) - 1 - Question 2

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?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 2

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.

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Practice Test: Computer Science Engineering (CSE) - 1 - Question 3

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?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 3

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

Practice Test: Computer Science Engineering (CSE) - 1 - Question 4

If we increase the sides of a rectangular park by 20%, then, what is the total increase in the area of that park ?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 4

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%

Practice Test: Computer Science Engineering (CSE) - 1 - Question 5

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:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 5

Required sum = 3!(3 + 4 + 5 + 6) = 108 [If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]

Practice Test: Computer Science Engineering (CSE) - 1 - Question 6

Choose the one that best fits the sentence below-

The state’s duty is to _______ the safety of its citizen.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 6

The state’s duty is to ensure the safety of its citizen.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 7

Cocaine will have "pernicious" effect on your physical fitness.

Find synonym of the underlined word.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 7

Pernicious means harmful effect.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 8

Choose the one that best fits the sentence below.

The _______ of glory lead but to the ______.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 8

The path of glory lead but to the grave 

Practice Test: Computer Science Engineering (CSE) - 1 - Question 9

Gardener said to me, “Don’t pluck the beautiful flowers.”

Select the appropriate Indirect Speech of the above.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 9

The appropriate speech is:

Gardner forbade me to pluck the flowers.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 10

I haven’t had a call from the office about the meeting ________ last Monday.
Select the appropriate word to fill in the blank from the given options.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 10

I haven’t had a call from the office about the meeting since last Monday.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 11

Let T(n) = 2T(n/4) + 100√n
The value of T(n) can be written as:

Practice Test: Computer Science Engineering (CSE) - 1 - Question 12

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.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 12

Insertion sort is directly proportional to number of inversions present in the list.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 13

What will the output of the following C code?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 13

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.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 14

A balanced tree is given below. How many nodes will become unbalanced when a node is inserted as a child of node G?

Note: A node in a tree is balanced if absolute difference between its left and right subtrees is less than or equal to 1.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 14

 

In this structure, Node (C) (B) and (A) not having difference between left subtree and right subtree -1, 0, 1. Therefore 3 nodes become unbalanced.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 15

Consider the C code Below.

Q. Which of the following is the tightest upper bound on time complexity of above function.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 15

Important observation is Break statement terminates the innermost loop. So “*” is printed only n times.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 16

Consider the following statements about Bellman ford algorithm for finding shortest path in a directed connected graph G having integral edge weights.

Statement I: It will always find out negative edge weight cycle in G reachable from source.
Statement II: It will always give correct answer for the graph G.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 16

Statement 1 is true as in the nth iteration of Bellman Ford, if the length of the path of any node reachable from the source is decreased, that means we are having negative edge weight cycle in the graph. Statement 2 is false as no algorithm can give give correct answer if the Graph is having negative edge weight cycle.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 17

Consider the case: f(n) = O(g(n)). Then, following two statements are claimed to be inferred from the above case.

Statement I: 2f(n) = O(2g(n))
Statement II: 2g(n) = O(2f(n))

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 17

if f(n) = n and g(n) = 2n. then f(n) = O(g(n)) here, 2^n = O(2^(2n)) = O(4^n), but not vice versa. Hence, I is true. II is false. -------------- Now, if f(n) = 2n and g(n) = n then also f(n) = O(g(n)) because we can ignore constant but, 2^(2n) != O(2^n), hence I is false, but II is true. In both of the above cases, f(n) = O(g(n)). But both the cases are counter of each other. Hence both I and II are wrong.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 18

Let X and Y be the integers representing the number of simple graphs possible with 3 labeled vertices and 3 unlabeled vertices respectively. Let X - Y = N. Then, find the number of spanning trees possible with N labeled vertices complete graph.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 18

Number of simple graphs possible with n labeled vertices is 2^(n(n-1)/2). Number of simple graphs possible with n unlabeled vertices is n+1. Number of spanning tree possible with n vertices complete graph n^(n-2) X =8 Y = 4 X-Y=4 Therefore required answer is 4^2=16.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 19

Consider the following pseudo code.

Find the value of x, when the above code is executed in a function.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 20

Consider the following statements:

S1 : DFS of a directed graph always produces the same number of edges in the traversal, irrespective of the starting vertex.
S2 : If all of the back edges that are found while DFS traversal on directed graph are removed, the resulting graph is acyclic.
Q. Which of the following statements above are valid?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 20

Statement S1 : consider the graph

Starting with A (source vertex ) we will get 2 edges Starting with B will get only 1 edge Starting with C we will get no edge Therefore DFS on directed graph may not give same number of edges. Statement S2 : Back edges are those edges (u,v) connecting a vertex u to an ancestor u in a depth-first tree. Self-loops are considered to be back edges. Back edges describe descendant-to-ancestor relations, as they lead from “high” to “low” nodes. Suppose that there is a back edge (u, v). Then vertex v is an ancestor of vertex u in the depth-first forest. There is thus a path from v to u in G, and the back edge (u,v) completes a cycle. Removing the back edge will break the cycle. Therefore removing all the back edges will make the graph acyclic. So the statement is true.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 21

Which data structure would be the most appropriate to implement a collection of values with the following three characteristics?

i) Items are retrieved and removed from the collection in FIFO order.
ii) There is no priori limit to the number of items in the collection.
iii) The size of an item is large relative to storage required for a memory address.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 21

Head and tail pointers in singly link list will make the insertion and deletion in O(1) time complexity if we are accessing the elements in FIFO order. In doubly link list since only head pointer is given then for insertion we have to traverse the complete link list so insertion will be O(n) so not appropriate. In binary tree we have only a pointer to the root. Insertion and deletion in binary tree will be O(log n) so not appropriate. In hash table accessing the data in FIFO order will not be possible.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 22

Stack A has the entries as following sequence a, b, c (with ‘a’ on top), stack B is empty, as shown in the diagram below.

Q. An entry popped out of stack A can be printed or pushed to stack B. An entry popped out of stack B can only be printed. In this arrangement which of the following permutation of a, b, c are not possible to print?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 22

Follow these steps to print bac.
1) POP element ‘a’ from stack A, push ‘a’ to Stack B.
2) POP element ‘b’ from A, print it.
3) POP element ‘a’ from B, and print it.
4) POP element ‘c’ from A, and print it. Now, perumtation bac has been printed. Similarly, we can print bca and abc, but can’t print cab.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 23

Stack A has the entries as following sequence a, b, c (with ‘a’ on top), Stack B is empty, as shown in the diagram below.

Q. Considering the data given in the previous question, if the stack A had 4 entries, then the number of possible permutations that can be printed will be:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 23

 

permutations which start with D:
to print D first , all a,b,c must be pushed on to stack before popping D and the only arrangement possible = D C B A
permutations start with C:
you need to push a and b from stack A to stack B , now print C
content of stack B= b a (from top to bottom)
content of stack A = d
permutations possible = 3!/2! = 3 = they are c d b a, c b d a, c b a d --> 3 permutations here
permutations starting with b :
to print b first, a will be pushed onto stack B
content of stack B= a
content of stack A= c d
you can bring these out in (3! -1) as (d a c) is not possible--> 5 possible
permutations starting with a:
fix ab
a b _ _
in this a b c d or a b dc
fix ac
a c _ _
a c b d or a c d b
fix ad
a d _ _
a d c b
total 5
total = 1+3+5+5= 14

Practice Test: Computer Science Engineering (CSE) - 1 - Question 24

Given a hash table with n keys and m slots with simple uniform hashing. If collisions are resolved by chaining then what is the probability that the first slot ends up empty?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 24

Probability of one particular slot = 1/m ( because total m slots) Probability that a value should not go in one particular slot = 1 – (1/m) For n values (keys) probability = [1 – (1/m)]n

Practice Test: Computer Science Engineering (CSE) - 1 - Question 25

Consider the following instance of knapsack problem:

Q. The maximum weight of 12 is allowed in the knapsack. Find the value of maximum profit with the optimal solution of the fractional knapsack problem.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 25

Practice Test: Computer Science Engineering (CSE) - 1 - Question 26

Find the maximum value of the expression (x+y+k) where (x,y) satisfies the equation (x-2)2 + (y-3)2= 25

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 26

Since (X,Y) is a point on circle, the general form of the point is X = 2 + 5*cost, y = 3 + 5*sint We need to maximise the value of x+y+k x+y+k = 2 + 5*cost + 3 + 5*sint + k = (5+k) + 5*(cost+sint) Here, k is a constant. The maximum value of c + acost + bsint is equals to c + sqrt (a*a +b*b). Maximum value of (5+k) + 5*(cost+sint) (5+k) + 5*sqrt(2) The result is (5+k) + 5*sqrt(2).

Practice Test: Computer Science Engineering (CSE) - 1 - Question 27

Which of the following argument is invalid?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 27

“⟺” mean equal. ifA⟺B means A is equal to B.
Option (a); (p→(q⋁r) )⟺(~p⋁(q⋁r) ) ⟺((~p⋁q)⋁r) ⟺(~(p⋀~q)⋁r) [∵De morgan’s law] ⟺(p⋀~q)→r ∴(p→(q⋁r) )⟺((p⋀~q)→r)

Option (b); (p→r)⋀(q→r)⟺(~p⋁r)⋀(~q⋁r) ⟺(~p⋀~q)⋁r [∵ De morgan’s law] ⟺~(p⋁q)⋁r ⟺(p⋁q)→r ∴(p→r)⋀(q→r)⟺(p⋁q)→r

Option (c): p→(q→r)⟺~p⋁(q→r) ⟺ ~p⋁(~q⋁r) ⟺~q⋁(~p⋁r) ⟺~q⋁(p→r) ⟺q→(p→r)

Option (d): (p↔q)⟺(p→q)⋀(q→p) ⟺(~p⋁q)⋀(~q⋁p) ⟺(~p⋁q)⋀(p⋁~q) ⟺(~p⋀~q)⋁(p⋀q) ⟺ ~(p⋁q)⋁(p⋀q) (p↔q)⟺(p⋁q)⋁(p⋀q) is wrong.

Practice Test: Computer Science Engineering (CSE) - 1 - Question 28

The value of the following Integral is:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 28

Practice Test: Computer Science Engineering (CSE) - 1 - Question 29

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 29

Practice Test: Computer Science Engineering (CSE) - 1 - Question 30

Which Of the following statements are true?
S1: Every Distributive lattice has unique complement.
S2: In Complemented lattice every element has at least one complement.
S3: Boolean lattice is both complemented and distributive.
S4: In Boolean lattice every lattice has exactly one complement.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 1 - Question 30

Clearly S1 is wrong because in distributive lattice a element may have zero complement. Rest other statements are properties of the corresponding lattices.

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