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Genpact Placement Papers 2025

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

Genpact Placement Papers 2025 - actual questions & solutions

Section titled “Genpact Placement Papers 2025 - actual questions & solutions”

This section contains real questions from Genpact placement papers 2025 based on candidate experiences from GeeksforGeeks, IndiaBix, and interview forums.

Q1: A sum of money amounts to ₹9800 after 5 years and ₹12005 after 8 years at the same rate of simple interest. Find the rate of interest per annum.

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

Q2: A shopkeeper sells an article at a profit of 20%. If he had bought it at 20% less and sold it for ₹5 less, he would have gained 25%. Find the cost price of the article.

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

Q3: A mixture contains milk and water in the ratio 5:3. If 10 liters of water is added, the ratio becomes 5:4. Find the quantity of milk in the mixture.

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

Q4: Statements: All cats are dogs. Some dogs are birds. Conclusions: I. Some cats are birds. II. All birds are cats.

Solution:

Analyzing the statements:

  • All cats are dogs (A → B)
  • Some dogs are birds (B → C, some)

Conclusion I: Some cats are birds

  • Since all cats are dogs, and some dogs are birds, we can say some cats are birds. Valid

Conclusion II: All birds are cats

  • This cannot be concluded from the given statements. Invalid

Answer: Only conclusion I follows

Q5: What is the value of x? Statement I: x² - 5x + 6 = 0. Statement II: x > 0.

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

Q6: Rotate an array to the right by k positions.

Problem Statement: Given an array, rotate it to the right by k steps.

Example:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [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)

Q7: Check if a string is a palindrome.

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: true

Solution (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)

Q8: Calculate factorial of a number.

Problem Statement: Given a number n, calculate n! (factorial).

Example:

Input: 5
Output: 120

Solution (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 approach
public 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

Expected hiring

  • Total Hires: 5500+ freshers
  • Process Associate: 4950+ selections
  • Associate: 400+ selections
  • Locations: Pan-India

Salary packages

  • Process Associate: ₹3.5-4.5 LPA
  • Associate: ₹6-7 LPA
  • Software Engineer: ₹8-10 LPA

Question trends

  • Aptitude: Medium (60% of questions)
  • Technical: Medium (25% of questions)
  • Coding: Medium-Hard (15% of questions)
  • Focus: Fundamentals and problem-solving
  • Aptitude Focus: Continued emphasis on quantitative ability and logical reasoning
  • Time Management: 90 minutes for 25-30 questions requires good speed and accuracy
  • Coding Basics: Focus on fundamental problems (arrays, strings, basic algorithms)
  • Technical Concepts: DBMS, OOPs, and programming fundamentals remain important
  • Problem-Solving: Emphasis on clear approach and correct implementation

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