Cadence 2025 aptitude
Quantitative, reasoning, and verbal drills with solutions
This page collects Cadence placement papers from 2025 with practice questions, worked solutions, and the exam pattern students reported that cycle. Use it when you want drive history: what the first round looked like, which topics repeated, and how to approach solutions. Work the sets below under a timer, then compare with newer material so your prep matches both established Cadence patterns and recent shifts.
Cadence 2025 aptitude
Quantitative, reasoning, and verbal drills with solutions
Cadence 2025 coding
DSA practice aligned to Cadence online assessments
Cadence interview experience
Round structure and tips from student reports
Cadence prep guide
Study plan and weekly schedule
Timed placement-style MCQs with score and explanations after you submit. Use it to check speed and accuracy before the real test.
Quantitative
The ratio of ages of A and B is 3:5. After 8 years, the ratio becomes 5:7. Find the present age of A.
Let present ages be 3x and 5x. After 8 years: - A's age: 3x + 8 - B's age: 5x + 8 Given: (3x + 8)/(5x + 8) = 5/7 7(3x + 8) = 5(5x + 8) 21x + 56 = 25x + 40 4x = 16 x = 4 Present age of A = 3x = 3 × 4 = 12 years
Quantitative
A and B finish a job in 12 and 18 days. Working together they finish in:
1/12 + 1/18 = 5/36 → 36/5 = 7.2 days.
Quantitative
Two numbers are in ratio 3:5 and their sum is 56. The larger number is:
Parts = 8. Larger = 5/8 × 56 = 35.
Quantitative
A is the father of B. B is the brother of C. How is A related to C?
Correct answer: Father
Reasoning
A walks 10m North, turns right walks 5m, turns right walks 10m. Direction from start?
Correct answer: East (5m east of starting point)
Verbal
A man sells an article at a loss of 10%. If he had sold it for ₹45 more, he would have gained 5%. Find the cost price.
Correct answer: ₹300
Reasoning
Find the next number: 2, 5, 11, 23, 47, ?
Pattern: Each number = previous × 2 + 1 2×2+1=5, 5×2+1=11, 11×2+1=23, 23×2+1=47 Next: 47×2+1 = 95
Verbal
Choose the correct form: "I _____ to the store yesterday."
"Yesterday" indicates past tense, so "went" is correct.
Verbal
Antonym of "Benevolent":
Benevolent ↔ malevolent.
Quantitative
A bank charges a processing fee of 2% on a loan amount. If the loan amount is ₹1,000,000, what is the processing fee?
Correct answer: ₹20,000
Reasoning
All pens are books. Some books are papers. Conclusion: Some pens are papers.
No definite overlap between pens and papers.
Quantitative
Find the error: "Each of the students have completed their assignment."
"Each" is singular, so "has" should be used instead of "have", and "his or her" instead of "their".
Verbal
Antonym of "Permanent":
Permanent ↔ temporary/brief.
Reasoning
Odd one out: 8, 27, 64, 100, 125
Others are perfect cubes; 100 is not.
Reasoning
Odd one out: Rose, Lily, Lotus, Leaf
Others are flowers.
Your score
0/15(0%)
| Section | What shows up | Prep focus |
|---|---|---|
| Online assessment | Coding and/or MCQ filter | Weekly timed mocks |
| Technical rounds | DSA, CS fundamentals, projects | Live problem solving |
| HR / hiring manager | Motivation and communication | Specific, evidence-based answers |
First round: Cadence Online Assessment
Skills emphasized: DSA, algorithms, C++, digital fundamentals
Languages: C++, Python
These are practice-style questions aligned to patterns students report for Cadence drives around 2025. They are not leaked live papers. Work them timed, then read the solutions only after you have an answer.
Problem: A boat’s speed in still water is 15 km/h and the stream is 3 km/h. How long to cover 36 km upstream?
Solution: Upstream speed = 15 − 3 = 12 km/h. Time = 36 / 12 = 3 hours.
Answer: 3 hours
Problem: A mixture has milk and water in the ratio 4:1. If 5 litres of water are added to 20 litres of mixture, what is the new milk:water ratio?
Solution: In 20 L: milk = 16 L, water = 4 L. After adding 5 L water: milk 16, water 9. Ratio = 16:9.
Answer: 16:9
Problem: Find compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.
Solution: Amount = 10000 × (1.1)² = 10000 × 1.21 = ₹12,100. CI = 12100 − 10000 = ₹2100. (SI for same period would be ₹2000; the extra ₹100 is interest on first-year interest.)
Answer: ₹2100
Problem: What is the angle between the hour and minute hands at 3:00?
Solution: At 3:00 the hands are exactly 90° apart (one quarter of the circle).
Answer: 90°
Problem: In how many ways can 5 different books be arranged on a shelf?
Solution: Arrangements of 5 distinct items = 5! = 120.
Answer: 120
Problem: If the sum of three consecutive integers is 72, what is the smallest of these integers?
Solution: Let the integers be x, x+1, and x+2.
x + (x+1) + (x+2) = 72 3x + 3 = 72 3x = 69 x = 23
So the integers are 23, 24, and 25.
Answer: 23
Problem: A train 150 meters long passes a pole in 15 seconds. What is its speed in km/h?
Solution: Distance = 150 m = 0.15 km. Time = 15 s = 15/3600 h = 1/240 h. Speed = 0.15 ÷ (1/240) = 0.15 × 240 = 36 km/h.
Faster check: 150/15 = 10 m/s → 10 × 18/5 = 36 km/h.
Answer: 36 km/h
Problem: If the cost price of a pen is ₹40 and it is sold at a 25% profit, what is the selling price?
Solution: Profit = 25% of 40 = ₹10. Selling price = 40 + 10 = ₹50.
Or SP = CP × 1.25 = 40 × 1.25 = ₹50.
Answer: ₹50
Problem: Given a string s, find the length of the longest substring without repeating characters. Example: ‘abcabcbb’ → 3 (‘abc’).
Approach: Sliding window with a map (or last-seen index) of characters. Expand the right pointer; when a duplicate appears inside the window, move the left pointer past the previous occurrence.
Complexity: O(n) time
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given a list of intervals [start, end], merge all overlapping intervals and return the non-overlapping set that covers the same ranges.
Approach: Sort by start time. Walk once, merging into the last interval in the result when the next start is ≤ current end; otherwise append a new interval.
Complexity: O(n log n) time from the sort
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given an integer array and an integer k, return the k most frequent elements. Order among equals can be arbitrary unless the problem says otherwise.
Approach: Count frequencies with a hash map, then use a heap of size k (or bucket sort by frequency) to extract the top k keys.
Complexity: O(n log k) with a heap
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given the head of a linked list, return true if there is a cycle and false otherwise.
Approach: Floyd’s tortoise and hare: move one pointer one step and another two steps. If they meet, a cycle exists. If the fast pointer hits null, there is no cycle.
Complexity: O(n) time, O(1) space
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given the root of a binary tree, return the level-order traversal (breadth-first) as a list of levels.
Approach: Use a queue. For each level, drain the current queue size, collect values, and enqueue children for the next level.
Complexity: O(n) time, O(n) space
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given coin denominations and an amount, return the fewest coins needed to make that amount, or -1 if it is impossible.
Approach: Unbounded knapsack DP: let dp[x] be the minimum coins for amount x. For each coin, update dp[c..amount]. Initialize dp[0] = 0 and the rest to a large sentinel.
Complexity: O(amount × coins)
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given a sorted array of distinct integers and a target, return the index of target or -1 if missing.
Approach: Maintain lo/hi. Compare mid with target and shrink the half that cannot contain it. Careful with overflow-free mid and empty arrays.
Complexity: O(log n) time
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Move all zeros in an array to the end while keeping the relative order of non-zero elements.
Approach: Two pointers: write non-zeros toward the front, then fill the remainder with zeros. Or swap zeros as you scan.
Complexity: O(n) time, O(1) space
Cadence tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Students usually say the first round is time-tight - easy marks vanish if you sit too long on one hard question. For Cadence, skim the paper in a couple of minutes, mark what you can finish cleanly, and protect accuracy. Languages people commonly use: C++, Python.
| Area | Why it matters at Cadence |
|---|---|
| DSA | What usually helps you clear the first round |
| Core CS (OOPs / DBMS / OS) | Technical interview depth |
| EDA, Computational Software awareness | Helps in managerial / HR conversations |
| Communication | Explain your approach clearly; keep a few real examples ready for HR |