Expected hiring
- Total Hires: 5500+ freshers
- Process Associate: 4950+ selections
- Associate: 400+ selections
- Locations: Pan-India
This page is a working set of Genpact placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what Genpact actually asked in the latest cycle, how hard the rounds were, and which themes (DSA, system design, aptitude, or role-specific topics) mattered most. Practice the problems below under timed conditions, then cross-check with the interview and preparation guides if you are targeting an upcoming Genpact drive.
| 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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