Solution

Solution

Second row element = First row element - Third row element

54 - 42 = 12

76 - 46 = 30

63 - 12 = 51

Solution

Solution

• The second train has travelled 72 km more than the first train because the speed of the second train is 12 km/hr more than the first.
• Time taken by second train to cover 72 km with surplus 12 km/hr = 72/12 = 6 hours

Then the time taken by both trains before meeting is 6 hours.

• So, relative speed = 36 + 48 = 84
• Total distance covered by both trains = 84 × 6 = 504 km
• Distance between A & B = 504 km

Solution

From I, II, III, we get

K : L : M = (120000×8) : (80000×12) : (X*9)

Since M’s investment is not given, the above ratio cannot be given.

∴ Data adequate.

Solution

All the numbers in the series are alternative prime numbers. So the missing term is 23.

Solution

The words 'pain' and 'pane' 'plain' and 'plane' are homophones: they sound alike but have different meanings. Therefore you need to understand the context of the sentence to use the write word.

The above sentence is in relation with rain hitting against a window pane. Pane refers to the glass sheet of a window.

Solution

The relationship here is about the character of the person and the person.

The word tyrant means a cruel or oppressive Ruler.

Therefore the character disciplinarian best suits teacher in the options given.

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

In the passage it is given ‘Exercise regularly to improve blood circulation and to release the feel-good hormone - endorphins.’

Solution

Postorder traversal of the given binary tree will give the following sequence: 4 5 2 6 7 3 1.

Now comparing the sequence with p q – r s * + we get 1 = +, 2 = -, 3 = *, 4 = p, 5 = q, 6 = r and 7 = s.

So, option (A) is correct.

Solution

We try to keep the tree balanced and increase height as much possible in each iteration.

Max height = 3

Solution

In adjacency matrix, dimension is nn.

To find all neighbours we need O(n) operations.

∴ Tight bound = O(n2) Answer: (c)

Solution

FALSE: Multiplying all edge weights by a positive number does not change the graph’s minimum spanning tree.

TRUE: A graph can have more than one shortest path between two vertices

FALSE: A graph where all edge weights are distinct can have more than one shortest path between two vertices.

Even if no two edges have the same weight, there could be two paths with the same weight. For example, there could be two paths from s to t with lengths 3 + 5 = 8 and 2 + 6 = 8. These paths have the same length (8) even though the edges (2, 3, 5, 6) are all distinct.

FALSE: Inserting an element into a binary search tree takes O(h) time, where h is the height of the tree. If the tree is not balanced, h may be much larger than log n (as large as n − 1).

Solution

Initially table is

When we try to insert 103, it should be put at slot 3(since 103 mod 10 = 3). Since already 23 are present at slot 3 we will move to the next slot and in the next slot also 34 is present. So, we will put 103 at slot 5. Therefore the number of collisions after inserting 103 is 2.

Final hash table is

Now when we try to insert 33, collisions will happen between 33 and 23, 34, 103, 6, 7, 48, 19. So, finally it will be placed at slot 0.

Here the number of collisions after inserting 33 is 7.

Solution

TotalIn insertion sort the worst case passes are (n-1). So, asymptotically O(n).

In each pass we need to find the search time and move (or) shift the elements of an array.

= Search time + Moving or shifting the elements of an array

= O(logn)+O(n)

= O( Time complexity = Total number of passes * Each pass time.

= O(n)*O(n)

= O(n^2)

Solution

False

f(b) < f(c), but b is not ancestor of c.

Solution

Last value in post order is root.

∴ E is root.

All values before E in preorder would be in left subtree.

H, F, I will be in left subtree.

∴ Answer: 3, Option B.

Solution

A graph may have more than one minimum spanning tree .

A graph with distinct weights has a unique minimum spanning tree .

Solution

Responsibilities of transaction management are:

· Ensures that the database remains in a consistent state despite Power failures

· Ensures that the database remains in a consistent state despite Operating system crashes

· Ensure the consistency of the database during interaction among the concurrent transactions

· Ensure consistency of database by orchestrating all access requests issued by the transactions

Solution

CHECK constraints enforce domain integrity

UNIQUE constraints enforce the uniqueness of the values in a set of columns

Solution

The natural of r1 and r2 will have 7 tuples.

Solution

Both the following queries will return the employees with years of experience greater than at least one employee in the Accounts department:

a) SELECT distinct E.name FROM employee AS E, employee AS F WHERE E.years_exp > F.years_exp AND F.dept_name& = 'Accounts';

b) SELECT name FROM employee WHERE years_exp > SOME ( SELECT years_exp FROM employee WHERE dept_name = 'Accounts');

Solution

Categorical data types refer to data that can be divided into distinct categories. String values, such as names or labels, represent categorical data. Boolean represents binary values, integers and continuous data are numerical.

Solution

A concept hierarchy organizes data into increasing levels of detail, such as breaking a year down into quarters, months, and days.

Solution

Min-max normalization scales the data between a specific range, typically [0, 1]. The formula for min-max normalization is:

Normalized Value = (X - Xmin) / (Xmax - Xmin)

For the given dataset [100, 200, 300, 400, 500]:

X = 300

Xmin = 100

Xmax = 500

Applying the formula:

Normalized Value = (300 - 100) / (500 - 100) = 200 / 400 = 0.5

Solution

Highest peak curve has lowest variance

Solution

Solution

Solution

The number of distance computations is (55 C 2)=(55×54)/2=1485

Solution

All three methods, i.e., single link, complete link, and average link, can be used for finding dissimilarity between two clusters in hierarchical clustering

Solution

Models which overfit are more likely to have low bias

Models which overfit are more likely to have high variance

Solution

Batch normalization standardizes the inputs to a layer in a neural network, ensuring that they have a mean of zero and a standard deviation of one. This can help reduce the internal covariate shift, leading to faster and more stable training.

Solution

The Hill-climbing algorithm is a local search algorithm that continually moves in the direction of increasing value, i.e., uphill. It terminates when it reaches a peak where no neighboring states have a higher value. It does not look ahead beyond the immediate neighbors (option B is a property of the algorithm). It may choose randomly among the best successors if there are multiple (option C can be a property depending on the specific version of Hill-climbing). However, standard Hill-climbing does not use backtracking; once it moves to a new state, it does not go back to previous states, which is why option D is not a property of the Hill-climbing algorithm

Solution

Solution

Solution

IV.

As you can see for all x < there exists a y fails for N.

From III, second part can be proven True

(False AND True) is False

Solution

ABDEG

Solution

Given A, B are independent events.

⸫ option (c) is correct.

Solution

x: Gaussian random variable; x = 0; & variance = 2

y = max (x, 0)

Solution

Given: ‘X’ is a Random variable that can take values only +1 and -1 with equal probabilities. So the

probability distribution of x is as below.

F(x) = P (X ≤ x)

F (–1) = P (x ≤ –1) = P (x = –1) = 0.5

F (+1) = P (x ≤ 1) = P (x = –1) + P (x = 1)

= 0.5 + 0.5 = 1

∴ F (–1) = 0.5 and F (+1) = 1

Solution

Average number of cars per minute passing is λ

‘P’ is following Poisson distribution with mean = λ = 3

Solution

Given:

Solution

Solution

Given:

Σx = 6, Σy = 21, Σx^2 = 14 , Σxy = 46 , n = 3

Solution

Solution

Solution

As per Cholesky decomposition; L L^T = A

Solution

Solution

Solution

Solution

Solution

Solution

Solution

1. There is a corner at t = 0; (b) tan t has discontinuities; (c) y = 1/t has a discontinuity at t = 0.

d) It is differentiable at all points.

Solution

dy/dx at x=1 is equals to (6(1) + 5)/(3(1)2+5(1)+7) = 11/15 = 0.73

Solution

It can be solved for t =0 and t=1.

y’(0)=2

y’(1)= e^2/2

Solution

Solution

Diameter (D) = 3cm ; Height (H) = 4 cm

Error (dD or dH) = 0.02 cm

Absolute error in volume= 5.184 cm3

Solution

Solution

GATE(series, 1) returns 4.

k = 2: series[2] = 9 and series[0] = 2.

Since series[2] > series[0], we use the second condition:

GATE(series, 2) = GATE(series, 1) * 2 + series[2] = 4 * 2 + 9 = 17.

k = 3: series[3] = 4 and series[2] = 9.

Since series[3] < series[2] and series[3] % 2 == 0, we use the first condition:

GATE(series, 3) = GATE(series, 2) + 6 = 17 + 6 = 23.

k = 4: series[4] = 12 and series[2] = 4.

Since series[4] > series[2], we use the second condition:

GATE(series, 4) = GATE(series, 3) * 2 + series[4]

= 23 * 2 + 12

= 58

k = 5: series[5] = 7 and series[3] = 12

Since series[5] < series[4] and series[5] % 2 != 0, we use the third condition:

GATE(series, 5) = GATE(series, 4) - 3

= 58 - 3

= 55

Solution

The code tries to convert "NaN" to an integer, which raises a ValueError.

The except block catches this ValueError, printing "Caught ValueError".

The finally block executes after the except block, printing "Finally block executed".

Thus, the output is Caught ValueError followed by Finally block executed.

Solution

The **kwargs parameter in Python allows a function to accept an arbitrary number of keyword arguments, which are passed as a dictionary.

Each key-value pair corresponds to the keyword argument names and their values.

tester(USA = 1, China = 2, Korea = 3, New Zealand= 4) calls the function tester with the arguments USA=1, China=2, Korea=3, and New Zealand=4.

These arguments are passed to kwargs as a dictionary: {'USA': 1, 'China': 2, 'Korea': 3, 'New Zealand': 4}.

The for loop iterates over each item in the kwargs dictionary.

In each iteration, key and value hold the current dictionary key and value, respectively.

print(key, value, end=" ") prints each key and value separated by a space and keeps all outputs on the same line due to end=" ".

Solution

When the shallow_copied.value.append(4) is executed, it modifies the list that both original and shallow_copied reference since they share the same reference to the list. Thus, original.value will reflect this change, resulting in [1, 2, 3, 4] for shallow_copied.value.

The deep_copied.value.append(5) creates a completely new list, so deep_copied.value results in [1, 2, 3, 5], leaving original.value as [1, 2, 3, 4] after the shallow copy's modification.

Solution

"lambda x: (-x['age'], x['name']) "correctly sorts the list of dictionaries. The negative sign before x['age'] allows the sorted() function to sort ages in descending order while keeping names in ascending order. This effectively meets the requirement to prioritize older ages first and, for those with the same age, sorts them alphabetically by name.