Cognizant Placement Papers 2025

This page contains actual Cognizant placement papers from 2025 with latest questions from Online Assessment for all roles.

2025 Updates

Cognizant 2025 introduced AI/ML questions for GenC Elevate. Coding section now has partial marking. Focus on accuracy!

Cognizant 2025 assessment pattern

Role	Aptitude	Technical	Coding	Time	CTC
GenC	25 MCQs	10 MCQs	1-2	90 min	4 LPA
GenC Pro	25 MCQs	15 MCQs	2	120 min	6 LPA
GenC Elevate	20 MCQs	15 MCQs	3	150 min	9 LPA

Aptitude questions (2025 latest)

Question 1: compound interest

₹10,000 at 10% CI for 2 years. Find difference from SI.

Solution:

• CI = P(1+r)^n - P = 10000(1.1)² - 10000 = 2100
• SI = P×r×n = 10000×0.1×2 = 2000
• Difference = ₹100

Question 2: partnership

A invests 5000 for 6 months, B invests 6000 for 4 months. Profit ratio?

Solution:

• A’s investment × time = 5000 × 6 = 30000
• B’s investment × time = 6000 × 4 = 24000
• Ratio = 30000:24000 = 5:4

Question 3: probability

Two dice thrown. Probability of sum = 7?

Solution:

• Favorable: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6
• Total outcomes = 36
• P = 6/36 = 1/6

Question 4: pipes and cisterns

Pipe A fills in 20 min, B empties in 30 min. How long if both open?

Solution:

• A’s rate = 1/20, B’s rate = -1/30
• Combined = 1/20 - 1/30 = 1/60
• Time = 60 minutes

Question 5: number series

Find next: 1, 1, 2, 3, 5, 8, 13, ?

Answer: 21 (Fibonacci series)

Logical reasoning (2025)

Question 6: direction sense

A walks 10m North, turns right walks 5m, turns right walks 10m. Direction from start?

Answer: East (5m east of starting point)

Question 7: seating arrangement

5 people in a row: A is left of B, C is right of D, E is between B and D. Order?

Answer: A-B-E-D-C or similar valid arrangement

Question 8: data sufficiency

Is x > y? (I) x² > y² (II) x > 0

Answer: Both statements together are needed - C

Technical mcqs (2025)

Question 9: OOPs

Which concept achieves runtime polymorphism?

Answer: Method overriding (decided at runtime)

Question 10: DBMS

ACID stands for?

Answer: Atomicity, Consistency, Isolation, Durability

Question 11: networks

Which layer provides end-to-end delivery in OSI?

Answer: Transport Layer (Layer 4)

Question 12: data structures

Time complexity of binary search?

Answer: O(log n)

Coding problems (2025)

Problem 1: fizzbuzz

Problem

Print 1 to n. For multiples of 3 print “Fizz”, for 5 print “Buzz”, for both print “FizzBuzz”.

Python

def fizzbuzz(n):
    for i in range(1, n + 1):
        if i % 15 == 0:
            print("FizzBuzz")
        elif i % 3 == 0:
            print("Fizz")
        elif i % 5 == 0:
            print("Buzz")
        else:
            print(i)

Java

public void fizzBuzz(int n) {
    for (int i = 1; i <= n; i++) {
        if (i % 15 == 0) System.out.println("FizzBuzz");
        else if (i % 3 == 0) System.out.println("Fizz");
        else if (i % 5 == 0) System.out.println("Buzz");
        else System.out.println(i);
    }
}

Problem 2: remove duplicates from sorted array

Problem

Remove duplicates in-place from sorted array. Return new length.

Input: [1,1,2,2,3]
Output: 3 (array becomes [1,2,3,…])

Java

public int removeDuplicates(int[] nums) {
    if (nums.length == 0) return 0;
    int i = 0;
    for (int j = 1; j < nums.length; j++) {
        if (nums[j] != nums[i]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

Python

def remove_duplicates(nums):
    if not nums: return 0
    i = 0
    for j in range(1, len(nums)):
        if nums[j] != nums[i]:
            i += 1
            nums[i] = nums[j]
    return i + 1

Problem 3: valid parentheses

Problem

Check if brackets are valid.

Input: “()[]”
Output: true

Java

public boolean isValid(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        } else {
            if (stack.isEmpty()) return false;
            char top = stack.pop();
            if ((c == ')' && top != '(') ||
                (c == '}' && top != '{') ||
                (c == ']' && top != '[')) {
                return false;
            }
        }
    }
    return stack.isEmpty();
}

Python

def is_valid(s):
    stack = []
    mapping = {')': '(', '}': '{', ']': '['}
    for char in s:
        if char in mapping:
            top = stack.pop() if stack else '#'
            if mapping[char] != top:
                return False
        else:
            stack.append(char)
    return not stack

Problem 4: maximum subarray sum

Problem

Find contiguous subarray with maximum sum (Kadane’s algorithm).

Input: [-2,1,-3,4,-1,2,1,-5,4]
Output: 6 (subarray [4,-1,2,1])

Java

public int maxSubArray(int[] nums) {
    int maxSum = nums[0];
    int currentSum = nums[0];
    for (int i = 1; i < nums.length; i++) {
        currentSum = Math.max(nums[i], currentSum + nums[i]);
        maxSum = Math.max(maxSum, currentSum);
    }
    return maxSum;
}

Python

def max_subarray(nums):
    max_sum = current_sum = nums[0]
    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
    return max_sum

Problem 5: lru cache

Problem

Design LRU cache with get and put operations in O(1).

Python

from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity):
        self.cache = OrderedDict()
        self.capacity = capacity

    def get(self, key):
        if key not in self.cache:
            return -1
        self.cache.move_to_end(key)
        return self.cache[key]

    def put(self, key, value):
        if key in self.cache:
            self.cache.move_to_end(key)
        self.cache[key] = value
        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)

Key insights 2025

1. Partial Marking: Coding now has partial marks - solve test cases partially
2. AI/ML Questions: GenC Elevate includes basic AI/ML concepts
3. Sectional Cutoffs: Each section has minimum cutoff
4. Time Pressure: Practice completing tests in allocated time
5. Technical Depth: More focus on DSA for higher roles

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