DRDO 2024 papers
Previous year DRDO placement papers with questions and solutions
This page is a working set of DRDO placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what DRDO 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 DRDO drive.
| Section | Questions | Time | Difficulty |
|---|---|---|---|
| Technical | 40-50 | 60 min | Medium-Hard |
| Aptitude | 20-25 | 30 min | Medium |
This section contains practice questions styled on DRDO placement papers 2025 (recent-cycle pattern), with worked solutions. Use them as timed sectional drills - from student reports drives vary by college and role, so treat this as a high-signal practice bank, not an official paper dump.
Solution:
Speed = distance/time = 240/16 m/s = (240/16) × (18/5) = 54 km/h.
Answer: 54 km/h
Solution:
Old sum = 6 × 18 = 108. New sum = 108 − 12 + 24 = 120. New average = 120/6 = 20.
Answer: 20
Solution:
SI = (P × R × T)/100 = (5000 × 8 × 3)/100 = ₹1200.
Answer: ₹1200
Solution:
Combined rate = 1/10 + 1/15 = (10+15)/(10×15). Time = 10×15/(10+15) = 6 hours.
Answer: 6 hours
Solution:
Pattern: All divisible by 5 except 54
Odd one out = 54.
Answer: 54
Solution:
From the statements, pens⊆stationery and some stationery overlap useful. That does not force some pens to be useful, and ‘all stationery are pens’ reverses the first statement wrongly.
Neither conclusion follows.
Answer: Neither I nor II follows
Solution:
This is the definition of Encapsulation (often paired with abstraction in interviews).
Answer: Encapsulation
Solution:
Each step halves the search space → O(log n).
Answer: O(log n)
Problem Statement: Return the level-order traversal of a binary tree.
Example:
Input: [3,9,20,null,null,15,7]Output: [[3],[9,20],[15,7]]Solution (Java):
public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) return res; Queue<TreeNode> q = new ArrayDeque<>(); q.add(root); while (!q.isEmpty()) { int sz = q.size(); List<Integer> level = new ArrayList<>(); for (int i = 0; i < sz; i++) { TreeNode n = q.poll(); level.add(n.val); if (n.left != null) q.add(n.left); if (n.right != null) q.add(n.right); } res.add(level); } return res;}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Rotate the array to the right by k steps.
Example:
Input: [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) { k %= nums.length; reverse(nums, 0, nums.length - 1); reverse(nums, 0, k - 1); reverse(nums, k, nums.length - 1);}void reverse(int[] a, int l, int r) { while (l < r) { int t = a[l]; a[l++] = a[r]; a[r--] = t; }}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Find the longest common prefix string amongst an array of strings.
Example:
Input: ["flower","flow","flight"]Output: "fl"Solution (Java):
public String longestCommonPrefix(String[] strs) { if (strs.length == 0) return ""; String pref = strs[0]; for (int i = 1; i < strs.length; i++) { while (!strs[i].startsWith(pref)) { pref = pref.substring(0, pref.length() - 1); if (pref.isEmpty()) return ""; } } return pref;}Time Complexity: O(S)
Space Complexity: O(1)
Based on recent candidate experiences from 2025 DRDO interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
Success Tips:
For detailed interview experiences from 2025, visit DRDO Interview Experience page.
DRDO 2024 papers
Previous year DRDO placement papers with questions and solutions
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