Looking to make a career in one of the top IT firms in the world? Wipro offers recruitment exams for the role of Project Engineer, either in Wipro or Wipro Turbo. It's essential to note that Wipro Turbo offers a higher package than normal recruitment, so be sure to understand the difference between Wipro Turbo and Wipro TRB. Wipro TRB is the band under which all freshers are placed immediately after hiring during probation period.
To prepare for the Wipro placement exam, it's crucial to have access to the best practices and advice. You should also practice with placement exam practice questions to get a feel of the exam pattern and syllabus. Understanding the exam pattern and syllabus will help you prepare for the interview questions that will be asked during the recruitment process.
The registration process for the Wipro placement exam is easy to follow. All Wipro Talent Acquisition happens on the Wipro Synergy platform, and you can access your test results, boarding pass for an off-campus drive, or any other recruitment-related information using your candidate login details.
To crack the Wipro placement exam, you need to stay focused, stay calm, and practice consistently. It's also essential to stay up-to-date with the latest digital trends and technology, as the exam tests your knowledge in these areas.
While preparing for the Wipro placement exam, it's important to remember that cheating is not only unethical but also unnecessary. With the right preparation and practice, you can crack the exam with ease and begin your journey with one of the leading IT companies in the world.
The WIPRO Exam offers a number of benefits for those who participate, including:
To be eligible for the latest Wipro recruitment, candidates must meet the following criteria:
Please note that in some cases or specific colleges, the minimum percentage requirement may be 65%. Additionally, Wipro Turbo, a specific recruitment program, is only conducted at premium colleges such as IITs, NITs, and VIT.
The test would comprise a wide range of questions. However, we have tried to curate a list of topics from each section that are most commonly asked and have more chances of being asked. Here is the list of critical section-wise topics for the assessment test:
1. Logical Reasoning: One would need to practise the following topics for the logical reasoning section:
2. Aptitude Test: The Aptitude Test would again need a regular practice of sample questions to familiarise you with the pattern and improve your speed. It would carry questions from the following topics:
3. General English and Comprehension: This section will comprise topics that will test a candidate’s basic English language skills. These questions would be based on essential grammar topics.
4. Expected Essay Writing Topics: In this section, you will need to write an essay on a given topic. The essay topics presented are usually based on current issues or matters of socio-economic and political importance. Or it can even be based on some simple personal scenario or a situation. The word limit would be between 200 to 400 words.
Some of the probable topics could be like this:
5. Coding and Technical Section: This section will be a part of the online written test but will carry technical questions about programming language. Some of the anticipated topics would be:
The WIPRO online written test is divided into five sections, each with its own time limit and set of questions. These sections include:
The entire online test will take approximately 128 minutes to complete, and there is no negative marking for incorrect answers.
In general, getting a job is not a big thing. But getting placed in a reputed company is something beyond everything. Therefore, to grab a job you need to track the Wipro Syllabus and also the Wipro Test Patterns regularly. Our article provides you with the best PDFs and a clear-cut pattern about Wipro Syllabus. In addition to these people who want to join the top most MNCs like Wipro need to download the test papers, we have given below.
Technical Interview
Questions related to specific technical fields are asked in this round. Questions may be based on specific knowledge about the company's technical activities; understanding of the technical work required to be completed as part of the job applied for or may enquire candidates to solve actual technical problems that they would be likely to face if employed.
HR Interview
Final step to select a candidate as an employee is Interview as it helps to determine a candidate's personality. Questions can be of wide range starting from your introduction, Qualification, Experience, Industry specific experience, Courses studied, your strengths and weaknesses, salary expectations, friends, family etc.
#Popular Question 1
You are given a binary number as a string (‘S’), which is ‘N’ size. Convert this binary number into its integer decimal equivalent and then print it.
#Coding Question 1
Given a binary tree, write a function to check if it is a valid binary search tree.
#Solution 1
```sh
def isValidBST(root):
def is_valid(node, lower=float('-inf'), upper=float('inf')):
if not node:
return True
if node.val <= lower or node.val >= upper:
return False
return is_valid(node.left, lower, node.val) and is_valid(node.right, node.val, upper)
return is_valid(root)
```
#Popular Question 2
Arrange a few characters in the ‘N’ number of rows where the first pattern is a left triangle and then the second pattern is a right triangle. Finally, there will be a pattern that will be the mirror image of the combined top half. THE integer ‘N’ will denote the given number of rows. Here, the pattern for N equals 3.
#Coding Question 2
Given an array of integers, write a function to find the second largest element in the array.
#Answer 2
```sh
def second_largest(arr):
largest = max(arr[0], arr[1])
second_largest = min(arr[0], arr[1])
for i in range(2, len(arr)):
if arr[i] > largest:
second_largest = largest
largest = arr[i]
elif (arr[i] > second_largest) and (arr[i] < largest):
second_largest = arr[i]
return second_largest
```
#Popular Question 3
A person is given an array (‘arr’) of ‘N’ length. At every index, the array contains single digit elements. The person needs to return the total sum of all array elements while keeping the final sum a single digit as well. In order to return a single-digit output, the person needs to add the digits of the output till only a single digit remains. How will the person carry this out?
#Coding Question 3
How can one reverse any given number?
#Solution 3
```sh
public class Solution {
public static long reverseNumber(long n) {
// Remove all trailing zeros
while (n % 10 == 0) {
n = n / 10;
}
// Declare reverseNum and remainder and initialize them with 0
long reverseNum = 0;
long reminder = 0;
while (n > 0) {
reminder = (int) (n % 10);
reverseNum = reverseNum * 10 + reminder;
n = n / 10;
}
// Return the reverse number
return reverseNum;
}
}
```
#Coding Question 4
Let us suppose that a person is given two numbers ‘A’ and ‘B’ as two arrays of ‘N’ and ‘M’ length respectively. All the array elements individually represent a digit. How will the person find the sum of the two numbers and how can the sum be returned in the form of an array?
#Solution 4
```sh
public class Solution {
public static int[] findArraySum(int[] a, int n, int[] b, int m) {
int ans[] = new int[Math.max(n, m)];
int carry = 0;
int sum = 0;
int k = ans.length – 1;
int i = n – 1;
int j = m – 1;
while (i >= 0 || j >= 0) {
sum = 0;
// If we have some elements left in the first array, then add it to the sum.
if (i >= 0) {
sum += a[i];
i–;
}
// If we have some elements left in the second array, then add it to the sum.
if (j >= 0) {
sum += b[j];
j–;
}
sum += carry;
int lastDigit = sum % 10;
carry = sum / 10;
ans[k–] = lastDigit;
}
// If still some carry is left, then push it to the answer.
if (carry != 0) {
int[] newAns = new int[ans.length + 1];
newAns[0] = carry;
for (int p = 1; p < newAns.length; p++) {
newAns[p] = ans[p – 1];
}
return newAns;
}
return ans;
}
}
```
#Coding Question 5
Let’s assume two people are playing the game where they both get a set of ‘N’ distinct integers. They take turns to make a move where each can choose ‘X’ and ‘Y’ distinct integers from the sets in a way that the set does not end up containing the absolute difference which is X-Y. The player making the move currently then adds the integer X-Y to the set, making the size of the set grow by one. In the case where the current player cannot make a valid move, the player loses. Who will win the game, the player who goes first or the player who goes second?
#Answer 5
```sh
# Function to find the gcd of two integers.
def gcd(a, b):
if (b == 0):
return a
return gcd(b, a % b)
# Function to find the gcd of all the elements of an array.
def gcdCalc(ar, n):
ans = ar[0]
for i in range(1,n):
ans = gcd(ar[i], ans)
return ans
def gameWinner(arr, n) :
n=len(arr)
gcdVal = gcdCalc(arr, n)
maxx = -1
for i in range(n):
if (arr[i] > maxx) :
maxx = arr[i]
# Total moves performed in game.
movesLeft = maxx // gcdVal – n
ans=str()
# If total moves are odd , Alice wins.
if (movesLeft % 2 != 0) :
ans = “Alice”
else :
ans = “Bob”
return ans
```
#Popular Question 1
You are given a binary number as a string (‘S’), which is ‘N’ size. Convert this binary number into its integer decimal equivalent and then print it.
#Coding Question 1
Given a binary tree, write a function to check if it is a valid binary search tree.
#Solution 1
```sh
def isValidBST(root):
def is_valid(node, lower=float('-inf'), upper=float('inf')):
if not node:
return True
if node.val <= lower or node.val >= upper:
return False
return is_valid(node.left, lower, node.val) and is_valid(node.right, node.val, upper)
return is_valid(root)
```
#Popular Question 2
Arrange a few characters in the ‘N’ number of rows where the first pattern is a left triangle and then the second pattern is a right triangle. Finally, there will be a pattern that will be the mirror image of the combined top half. THE integer ‘N’ will denote the given number of rows. Here, the pattern for N equals 3.
#Coding Question 2
Given an array of integers, write a function to find the second largest element in the array.
#Answer 2
```sh
def second_largest(arr):
largest = max(arr[0], arr[1])
second_largest = min(arr[0], arr[1])
for i in range(2, len(arr)):
if arr[i] > largest:
second_largest = largest
largest = arr[i]
elif (arr[i] > second_largest) and (arr[i] < largest):
second_largest = arr[i]
return second_largest
```
#Popular Question 3
A person is given an array (‘arr’) of ‘N’ length. At every index, the array contains single digit elements. The person needs to return the total sum of all array elements while keeping the final sum a single digit as well. In order to return a single-digit output, the person needs to add the digits of the output till only a single digit remains. How will the person carry this out?
#Coding Question 3
How can one reverse any given number?
#Solution 3
```sh
public class Solution {
public static long reverseNumber(long n) {
// Remove all trailing zeros
while (n % 10 == 0) {
n = n / 10;
}
// Declare reverseNum and remainder and initialize them with 0
long reverseNum = 0;
long reminder = 0;
while (n > 0) {
reminder = (int) (n % 10);
reverseNum = reverseNum * 10 + reminder;
n = n / 10;
}
// Return the reverse number
return reverseNum;
}
}
```
#Coding Question 4
Let us suppose that a person is given two numbers ‘A’ and ‘B’ as two arrays of ‘N’ and ‘M’ length respectively. All the array elements individually represent a digit. How will the person find the sum of the two numbers and how can the sum be returned in the form of an array?
#Solution 4
```sh
public class Solution {
public static int[] findArraySum(int[] a, int n, int[] b, int m) {
int ans[] = new int[Math.max(n, m)];
int carry = 0;
int sum = 0;
int k = ans.length – 1;
int i = n – 1;
int j = m – 1;
while (i >= 0 || j >= 0) {
sum = 0;
// If we have some elements left in the first array, then add it to the sum.
if (i >= 0) {
sum += a[i];
i–;
}
// If we have some elements left in the second array, then add it to the sum.
if (j >= 0) {
sum += b[j];
j–;
}
sum += carry;
int lastDigit = sum % 10;
carry = sum / 10;
ans[k–] = lastDigit;
}
// If still some carry is left, then push it to the answer.
if (carry != 0) {
int[] newAns = new int[ans.length + 1];
newAns[0] = carry;
for (int p = 1; p < newAns.length; p++) {
newAns[p] = ans[p – 1];
}
return newAns;
}
return ans;
}
}
```
#Coding Question 5
Let’s assume two people are playing the game where they both get a set of ‘N’ distinct integers. They take turns to make a move where each can choose ‘X’ and ‘Y’ distinct integers from the sets in a way that the set does not end up containing the absolute difference which is X-Y. The player making the move currently then adds the integer X-Y to the set, making the size of the set grow by one. In the case where the current player cannot make a valid move, the player loses. Who will win the game, the player who goes first or the player who goes second?
#Answer 5
```sh
# Function to find the gcd of two integers.
def gcd(a, b):
if (b == 0):
return a
return gcd(b, a % b)
# Function to find the gcd of all the elements of an array.
def gcdCalc(ar, n):
ans = ar[0]
for i in range(1,n):
ans = gcd(ar[i], ans)
return ans
def gameWinner(arr, n) :
n=len(arr)
gcdVal = gcdCalc(arr, n)
maxx = -1
for i in range(n):
if (arr[i] > maxx) :
maxx = arr[i]
# Total moves performed in game.
movesLeft = maxx // gcdVal – n
ans=str()
# If total moves are odd , Alice wins.
if (movesLeft % 2 != 0) :
ans = “Alice”
else :
ans = “Bob”
return ans
```
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Daily coding challenge and the doubt session helped me in staying consistent. Also, the leadership board kept me motivated. Edyst gave me the best guiding materials for all the cohorts I joined. I like everything done by Edyst for my success.
#Popular Question 1
You are given a binary number as a string (‘S’), which is ‘N’ size. Convert this binary number into its integer decimal equivalent and then print it.
#Coding Question 1
Given a binary tree, write a function to check if it is a valid binary search tree.
#Solution 1
```sh
def isValidBST(root):
def is_valid(node, lower=float('-inf'), upper=float('inf')):
if not node:
return True
if node.val <= lower or node.val >= upper:
return False
return is_valid(node.left, lower, node.val) and is_valid(node.right, node.val, upper)
return is_valid(root)
```
#Popular Question 2
Arrange a few characters in the ‘N’ number of rows where the first pattern is a left triangle and then the second pattern is a right triangle. Finally, there will be a pattern that will be the mirror image of the combined top half. THE integer ‘N’ will denote the given number of rows. Here, the pattern for N equals 3.
#Coding Question 2
Given an array of integers, write a function to find the second largest element in the array.
#Answer 2
```sh
def second_largest(arr):
largest = max(arr[0], arr[1])
second_largest = min(arr[0], arr[1])
for i in range(2, len(arr)):
if arr[i] > largest:
second_largest = largest
largest = arr[i]
elif (arr[i] > second_largest) and (arr[i] < largest):
second_largest = arr[i]
return second_largest
```
#Popular Question 3
A person is given an array (‘arr’) of ‘N’ length. At every index, the array contains single digit elements. The person needs to return the total sum of all array elements while keeping the final sum a single digit as well. In order to return a single-digit output, the person needs to add the digits of the output till only a single digit remains. How will the person carry this out?
#Coding Question 3
How can one reverse any given number?
#Solution 3
```sh
public class Solution {
public static long reverseNumber(long n) {
// Remove all trailing zeros
while (n % 10 == 0) {
n = n / 10;
}
// Declare reverseNum and remainder and initialize them with 0
long reverseNum = 0;
long reminder = 0;
while (n > 0) {
reminder = (int) (n % 10);
reverseNum = reverseNum * 10 + reminder;
n = n / 10;
}
// Return the reverse number
return reverseNum;
}
}
```
#Coding Question 4
Let us suppose that a person is given two numbers ‘A’ and ‘B’ as two arrays of ‘N’ and ‘M’ length respectively. All the array elements individually represent a digit. How will the person find the sum of the two numbers and how can the sum be returned in the form of an array?
#Solution 4
```sh
public class Solution {
public static int[] findArraySum(int[] a, int n, int[] b, int m) {
int ans[] = new int[Math.max(n, m)];
int carry = 0;
int sum = 0;
int k = ans.length – 1;
int i = n – 1;
int j = m – 1;
while (i >= 0 || j >= 0) {
sum = 0;
// If we have some elements left in the first array, then add it to the sum.
if (i >= 0) {
sum += a[i];
i–;
}
// If we have some elements left in the second array, then add it to the sum.
if (j >= 0) {
sum += b[j];
j–;
}
sum += carry;
int lastDigit = sum % 10;
carry = sum / 10;
ans[k–] = lastDigit;
}
// If still some carry is left, then push it to the answer.
if (carry != 0) {
int[] newAns = new int[ans.length + 1];
newAns[0] = carry;
for (int p = 1; p < newAns.length; p++) {
newAns[p] = ans[p – 1];
}
return newAns;
}
return ans;
}
}
```
#Coding Question 5
Let’s assume two people are playing the game where they both get a set of ‘N’ distinct integers. They take turns to make a move where each can choose ‘X’ and ‘Y’ distinct integers from the sets in a way that the set does not end up containing the absolute difference which is X-Y. The player making the move currently then adds the integer X-Y to the set, making the size of the set grow by one. In the case where the current player cannot make a valid move, the player loses. Who will win the game, the player who goes first or the player who goes second?
#Answer 5
```sh
# Function to find the gcd of two integers.
def gcd(a, b):
if (b == 0):
return a
return gcd(b, a % b)
# Function to find the gcd of all the elements of an array.
def gcdCalc(ar, n):
ans = ar[0]
for i in range(1,n):
ans = gcd(ar[i], ans)
return ans
def gameWinner(arr, n) :
n=len(arr)
gcdVal = gcdCalc(arr, n)
maxx = -1
for i in range(n):
if (arr[i] > maxx) :
maxx = arr[i]
# Total moves performed in game.
movesLeft = maxx // gcdVal – n
ans=str()
# If total moves are odd , Alice wins.
if (movesLeft % 2 != 0) :
ans = “Alice”
else :
ans = “Bob”
return ans
```
All WIPRO Questions
All WIPRO Questions