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Download latest Genpact placement papers 2025 PDF with current year aptitude questions, coding problems, solutions, and updated exam patterns.
This page contains Genpact placement papers from 2025 with current year questions, solutions, and exam patterns. Practice these Genpact placement papers 2025 to understand the latest question patterns and difficulty levels.
| Section | Questions | Time | Difficulty | Focus Areas |
|---|---|---|---|---|
| Aptitude | 15-18 | 45 min | Medium | Quantitative, Logical Reasoning, Verbal |
| Technical | 5-7 | 20 min | Medium | Programming, Data Structures, DBMS |
| Coding | 1-2 | 25 min | Medium-Hard | Array, String, Basic Algorithms |
Total: 25-30 questions, 90 minutes
Platform: Genpact assessment platform
Languages Allowed: C, C++, Java, Python
Success Rate: ~25-30% cleared OA and advanced to interviews
This section contains real questions from Genpact placement papers 2025 based on candidate experiences from GeeksforGeeks, IndiaBix, and interview forums.
Solution:
Let Principal = P, Rate = R%
After 5 years: P + (P × R × 5)/100 = 9800 After 8 years: P + (P × R × 8)/100 = 12005
Subtracting: (P × R × 3)/100 = 12005 - 9800 = 2205
P × R = 73500
From first equation: P + (73500 × 5)/100 = 9800 P + 3675 = 9800 P = 6125
Rate R = 73500/6125 = 12% per annum
Answer: 12% per annum
Solution:
Let CP = 100 SP = 120 (20% profit)
New CP = 80 (20% less) New SP = 120 - 5 = 115 New Profit = 115 - 80 = 35 New Profit % = (35/80) × 100 = 43.75%
But given new profit is 25%, so:
Let actual CP = x Actual SP = 1.2x
New CP = 0.8x New SP = 1.2x - 5 New Profit = (1.2x - 5) - 0.8x = 0.4x - 5 New Profit % = ((0.4x - 5)/0.8x) × 100 = 25
(0.4x - 5)/0.8x = 0.25 0.4x - 5 = 0.2x 0.2x = 5 x = ₹25
Answer: ₹25
Solution:
Let milk = 5x, water = 3x
After adding 10 liters water: Milk = 5x, Water = 3x + 10
Ratio: 5x/(3x + 10) = 5/4 20x = 5(3x + 10) 20x = 15x + 50 5x = 50 x = 10
Quantity of milk = 5x = 5 × 10 = 50 liters
Answer: 50 liters
Solution:
Analyzing the statements:
Conclusion I: Some cats are birds
Conclusion II: All birds are cats
Answer: Only conclusion I follows
Solution:
Statement I: x² - 5x + 6 = 0 (x - 2)(x - 3) = 0 x = 2 or x = 3
Statement II: x > 0 (doesn’t help narrow down)
Combining both: x = 2 or x = 3 (both positive)
Answer: Both statements together are not sufficient to determine unique value of x
Problem Statement: Given an array, rotate it to the right by k steps.
Example:
Input: nums = [1,2,3,4,5,6,7], k = 3Output: [5,6,7,1,2,3,4]Solution (Java):
public void rotate(int[] nums, int k) { int n = nums.length; k = k % n;
reverse(nums, 0, n - 1); reverse(nums, 0, k - 1); reverse(nums, k, n - 1);}
private void reverse(int[] nums, int start, int end) { while (start < end) { int temp = nums[start]; nums[start] = nums[end]; nums[end] = temp; start++; end--; }}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Given a string, determine if it is a palindrome (ignoring case and non-alphanumeric characters).
Example:
Input: "A man, a plan, a canal: Panama"Output: trueSolution (Java):
public boolean isPalindrome(String s) { int left = 0, right = s.length() - 1;
while (left < right) { while (left < right && !Character.isLetterOrDigit(s.charAt(left))) { left++; } while (left < right && !Character.isLetterOrDigit(s.charAt(right))) { right--; }
if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) { return false; }
left++; right--; }
return true;}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Given a number n, calculate n! (factorial).
Example:
Input: 5Output: 120Solution (Java):
public long factorial(int n) { if (n <= 1) return 1;
long result = 1; for (int i = 2; i <= n; i++) { result *= i; }
return result;}
// Recursive approachpublic long factorialRecursive(int n) { if (n <= 1) return 1; return n * factorialRecursive(n - 1);}Time Complexity: O(n)
Space Complexity: O(1) for iterative, O(n) for recursive
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