Solution

Total distance =6010=600 km

New speed =600/4=150 km/hr

Increase in speed =150-60=90 km/hr

Solution

Speed in downstream =12+4=16 kmph

Solution

6:1=x:12

x=72

Solution

900 ㅡ 100

100 ㅡ ?

100 x 100 / 900=11.11%

Solution

We have, 16, 25, 36, 49, 64, 81 as perfect squares from 11 to 99.

Total numbers from 11 to 99 is 89.

Probability =6/89

Solution

Solution

The usual way to form comparatives and superlatives of adjectives is to add -er and -est: great, greater, greatest.

Solution

A pitcher is a handled container for holding and pouring liquids that usually has a lip or a spout.

Solution

Plants are green in color because they contain the green pigment chlorophyll.

Solution

The words Banquet, Carnival and Merry - making are synonyms of Feast. Whereas the word Ravenous means hunger for food. Therefore Ravenous is the odd word here.

Solution

The number of distinct BST for n nodes are given as ((2n)Cn)/(n+1)

So, for 4 distinct nodes, we can have (8C4)/5 = 14 distinct BSTs.

This is also the formula for the number of isomeric structures for n nodes - The number of distinct unlabelled trees per using nodes

Solution

I*(N-1) + 1 = L

where,

I = Number of Internal nodes.

L = Leaf Nodes.

and, N = Number of children each node can have.

=10 (9) + 1

=91

Solution

The tree is an N-ary tree. Assume it has T total nodes, which is the sum of internal nodes (I) and leaf nodes (L). A tree with T total nodes will have (T – 1) edges or branches.

In other words, since the tree is an N-ary tree, each internal node will have N branches contributing a total of N*I internal branches. Therefore we have the following relations from the above explanations,

N * I = T – 1

L + I = T

From the above two equations, we can say that L = (N – 1) * I + 1.

Solution

P: Yes, Topologically sort the vertices of G can be completed in linear time. It takes O(V+E) time.

Q: At the top level, roughly n work is done to merge all n elements. At the next level,there are two branches,each doing roughly n/2 work to merge n/2 elements. In total, roughly ‘n’ work is done on that level. This pattern continues on through to the leaves, where a constant amount of work is done on n leaves, resulting in roughly n work being done on the leaf level.

Solution

All pair shortest pair paths by repeatedly running Bellman Ford algorithm from each vertex will take O(V4)

Solution

From the recurrence relation, we note that the branching factor a=5. At level 0, there is only one problem. For every subsequent level, the number of subproblems is multiplied by a. Therefore, at the 7th level of recursion, the total number of subproblems is given by:

a^7 = 5^7 = 243.

Solution

The first move occurs when the number of elements is 2^(k-1), and the ith move occurs when the number of elements is 2^(k-i). Total number of movements is k, and total cost is ∑0_j=k-1 2j = 2k-1= Θ(2^k) = Θ(n)

Solution

Finding the correct slot in a hash table takes O(1). In the worst case, all n items will hash to the same slot. Since chaining is implemented using a heap, inserting an element into the heap and heapify() takes O(log n) time.

Solution

Solution

Do not enter a value in the foreign key field of a child table if that value does not exist in the primary key of the parent table.

Solution

If R has m tuples and S have n tuples, then the maximum size of join is: mxn

Solution

Schema I:

Registration (rollno, courses)

rollno → courses

· For the given schema Registration ‘rollno’ is a primary key.

· Left-side of the functional dependency is a superkey so, Registration is in BCNF.

Schema II:

Registrstion (rollno, courseid, email)

rollno, courseid → email

email → rollno

· From the given schema the candidate key is (rollno + courseid).

· There is no part of the key in the left hand of the FD’s so, it is in 2NF.

· In the FD email→rollno, email is non-prime attribute but rollno is a prime attribute. So, it is not a transitive dependency. No transitive dependencies so, the schema is in 3NF.

· But in the second FD email→rollno, email is not a superkey. So, it is violating BCNF.

· Hence, the schema Registration is in 3NF but not in BCNF.

Schema III:

Registration (rollno, courseid, marks, grade)

rollno, courseid → marks, grade

marks → grade

· For the schema the candidate key is (rollno + courseid).

· There are no part of the keys are determining non-prime attributes. So, the schema is in 2NF.

· In the FD marks → grade, both the attributes marks and grade are non-prime. So, it is a transitive dependency. The FD is violating 3NF.

· The schema Registration is in 2NF but not in 3NF.

Schema IV:

Registration (rollno, courseid, credit)

rollno, courseid → credit

courseid → credit

· The candidate key is (rollno + courseid).

· In the FD, courseid → credit, courseid is part of the key (prime attribute) and credit is non-prime. So, it is a partial dependency. The schema is violating 2NF.

Solution

S:

T:

∎S Natural Join T

Solution

Dimensionality reduction techniques, such as Principal Component Analysis (PCA), reduce the number of attributes while retaining the key features of the original data.

Solution

The width of each bin is calculated by dividing the range of values by the number of bins.

Width of each bin = (Xmax - Xmin) / Number of bins = (100 - 0) / 5 = 20

Solution

First, calculate the variance using the formula:

Variance = (Sum of (X - mean)^2) / N

Variance = ((10-30)^2 + (20-30)^2 + (30-30)^2 + (40-30)^2 + (50-30)^2) / 5

Variance = (400 + 100 + 0 + 100 + 400) / 5 = 200

Standard deviation is the square root of the variance:

Standard deviation = sqrt(200) ≈ 14.14

Solution

Fully Connected Architecture: Each node in one layer connects to all nodes in the next layer.

Solution

In a decision tree, the splitting process aims to maximize information gain (or equivalently, minimize entropy). As we traverse down the tree, nodes become purer (i.e., contain mostly one class), leading to lower entropy.

Solution

PDF will be the derivative of sigmoid function.

We know that the derivative of sigmoid = sigmoid(x)(1-sigmoid(x)).

At x=0, sigmoid(x)=0.5

Hence, sigmoid(x)(1-sigmoid(x))=0.5(1-0.5)=0.25.

Solution

In a negatively skewed distribution (left-skewed distribution):

• The mean is pulled to the left due to the presence of lower values (outliers).
• The mode remains at the peak of the distribution.
• The median lies between the mode and the mean.

Hence, the relationship is:

Mode > Median > Mean.

Solution

Solution

Solution

The dataset is imbalanced (90% negative, 10% positive). Accuracy is not a suitable metric in this case because it can be high even if the model predicts all points as negative, ignoring the positive points entirely.

Since predicting a positive as negative has a high cost, the metric should focus on precision and recall rather than overall accuracy.

Solution

1, 0, 3, 5, 4, 2, 6

6, 5, 4, 3, 0, 1, 2

Solution

A* guarantees an optimal solution if the heuristic

h(n) is both admissible and consistent.

Solution

Solution

The correct statements are:

• b. Complete but not optimal (it finds a solution but not necessarily the shallowest one).
• d. Less efficient than standard IDS in the worst case (due to redundant exploration and skipped depths).

Solution

Informed search algorithms like A* guarantee finding the shortest path when the heuristic function is:

• Admissible: The heuristic never overestimates the actual cost to reach the goal.
• Consistent (or Monotonic): The heuristic satisfies the triangle inequality (i.e., it does not overestimate the cost between two nodes).

Solution

This is a permutation because they are arranging the songs in order to make the program. Using the permutation formula

Using the multiplication principle

6*5*4*3*2*1 = 720

There are 720 different ways of arranging the songs to make the program.

Solution

Total Men = 300

Total Women = 1000

Good Orators among Men = 15

Good Orators among Women = 25

Total number of good orators = 15 (from men) + 25 (from women) = 40

Probability of selecting a male orator:

P(Male Orator) = Numbers of male orators / total no of orators = 15/40

Solution

= 0.85

Solution

Solution

Order is important in counting the number of ways the runners could finish 1st, 2nd, and 3rd place, so this problem can be considered a permutation probability problem.

We can use permutations to find the number of orders in which 3 of the 4 U.S. runners could finish the race. Then, we use permutations to find the number of orders in which 3 of the 8 runners could finish the race.

The probability that U.S. runners finish in 1st, 2nd, and 3rd place is 0.0714 or 7.14%.

Solution

Solution

Firstly, let us create a sample space for each event. For the event ‘A’ we have to get at least two head. Therefore, we have to include all the events that have two or more heads.

Or we can write:

A = {HHT, HTH, THH, HHH}.

This set A has 4 elements or events in it i.e. n(A) = 4

In the same way, for event B, we can write the sample as:

B = {TTT} and n(B) = 1

Again using the same logic, we can write;

C = {THT, HHH, HHT, THH} and n(C) = 4

So B & C and A & B are mutually exclusive since they have nothing in their intersection

Solution

Let T be the event of preferring tea, C be the event of preferring coffee, W be the event of being a woman, M be the event of being a man.

Given:

P(T) = 0.6

P(C) = 0.4

P(W∣T) = 0.7

P(M∣C) = 0.6

To find P(W):

P(W) = P(W∣T) ⋅P(T) + P(W∣C) ⋅P(C)

P(W) = P(W∣T) ⋅P(T) + P(W∣C) ⋅P(C)

P(W) = 0.7 × 0.6 + (1 − 0.6) × 0.4

P(W) = 0.7 × 0.6 + 0.4 × 0.4

P(W) = 0.42 + 0.16

P(W) = 0.58

Solution

By solving AB = C, we get the values of x and y. Then by substituting these values in A, we obtain A^2

Here

Solution

Solution

Solution

To find the determinant of [A], let us expand the determinant along row 1.

Solution

Adding Au1 and Au2 we get

Solution

We can get an equation for the upper left corner of the matrices because the elements located in that position have to coincide. So we have

Solution

Solution

Solution

Solution

Solution

For turning points dy/dx = 0.

Solution

Solution

Solution

Solution

The exception is not suppressed (_exit_ returns False), so the program stops after raising the ValueError. The second print is skipped.

Solution

The heapq library creates a min-heap by default, and the heap property is maintained using the first element of the tuple (-priority). The sequence of pops corresponds to the lowest (most negative) priority first: (-3, "task3"), then (-2, "task4"), then (-1, "task1"), and finally (0, "task2"). Final order: task3 task4 task1 task2.

Solution

The type function dynamically creates a class named GeneratedClass, The _init_ method (a lambda) initializes the attribute field with the value passed during instantiation, The GATE method is associated with the class as execute_method,When execute_method is called on instance_obj, it uses the attribute value ("CodeMaster") and prints.

Solution

The code generates all permutations of the list [1, 2, 3, 4, 5, 6, 7, 8, 9] and reshapes them into a 3x3 matrix. For each permutation, it computes the sum of the diagonal elements. The maximum diagonal sum for any permutation of the matrix is 21, as the highest sum is obtained from the matrix with the elements [3, 5, 7] on the diagonal.

Solution

Reference counting is Python's primary method for memory management, where each object in memory has a reference count, and when it drops to zero, the memory is immediately deallocated. However, reference cycles (where objects refer to each other, causing their reference count never to drop to zero) would lead to memory leaks if handled solely by reference counting. The cyclic garbage collector comes into play to detect and break these reference cycles, ensuring that memory can still be freed even when cyclic references exist.