Synopsys 2026 aptitude
Quantitative, reasoning, and verbal drills with solutions
This page collects Synopsys placement papers from 2026 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 Synopsys patterns and recent shifts.
Synopsys 2026 aptitude
Quantitative, reasoning, and verbal drills with solutions
Synopsys 2026 coding
DSA practice aligned to Synopsys online assessments
Synopsys interview experience
Round structure and tips from student reports
Synopsys 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
A student scored 60, 70, and 80 in three subjects. If the weights are 2, 3, and 5 respectively, find the weighted average.
Correct answer: 73
Reasoning
Find the missing number: 5, 11, 23, 47, 95, ?
Correct answer: 191
Quantitative
A sum doubles in 5 years at simple interest. The rate of interest is:
For doubling, RT = 100 → R = 100/5 = 20%.
Reasoning
A walks 9 km north, then 12 km east. Distance from start is:
Right triangle → 15 km.
Quantitative
Find the remainder when 2^31 is divided by 5.
Pattern of 2^n mod 5: 2^1=2, 2^2=4, 2^3=3, 2^4=1, 2^5=2 (cycle repeats every 4) 31 mod 4 = 3 So 2^31 mod 5 = 2^3 mod 5 = 3
Quantitative
A number is increased by 25% and then decreased by 25%. Find the net percentage change.
Let original number = 100 After 25% increase: 100 + 25 = 125 After 25% decrease: 125 - (25% of 125) = 125 - 31.25 = 93.75 Net change = 100 - 93.75 = 6.25 Net percentage change = (6.25/100) × 100 = 6.25% decrease
Verbal
Choose the correctly spelled word:
Correct spelling is Accommodation.
Reasoning
If every letter is shifted +2 in the alphabet, BOOK becomes:
Each letter +2 → DQQM.
Reasoning
Statements: All cats are animals. Some animals are dogs. Conclusions: I. Some cats are dogs. II. All dogs are cats.
Analyzing the statements: - All cats are animals (A → B) - Some animals are dogs (B → C, some) Conclusion I: Some cats are dogs - Cannot be concluded. Some animals are dogs, but cats are a subset of animals. We cannot say some cats are dogs. Invalid Conclusion
Reasoning
Find the next number: 8, 6, 9, 7, 10, ?
−2,+3 alternating → 10-2=8 → 8.
Verbal
Synonym of "Abundant":
Abundant means plentiful.
Verbal
The idiom "under the weather" means:
It means: feeling ill.
Quantitative
Price rises by 25%. By what % should consumption fall to keep expenditure the same?
Reduction = 25/(100+25)×100 = 20%.
Quantitative
The LCM of 12 and 18 is:
LCM(12,18) = 36.
Verbal
Each of the boys _____ given a prize.
"Each" takes singular verb → was.
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: Synopsys Online Test
Skills emphasized: DSA, digital design basics, C++/Python
Languages: C++, Python, SystemVerilog (role-dependent)
These are practice-style questions aligned to patterns students report for Synopsys drives around 2026. They are not leaked live papers. Work them timed, then read the solutions only after you have an answer.
Problem: Find simple interest on ₹5000 at 8% per annum for 3 years.
Solution: SI = 5000 × 8 × 3 / 100 = ₹1200.
Answer: ₹1200
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: 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
Synopsys 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
Synopsys 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
Synopsys 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)
Synopsys 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
Synopsys 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
Synopsys 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: Rotate an array to the right by k steps. Example: [1,2,3,4,5,6,7], k = 3 → [5,6,7,1,2,3,4].
Approach: Normalize k %= n. Reverse the whole array, reverse the first k elements, then reverse the rest. That yields the rotation in place.
Complexity: O(n) time, O(1) space
Synopsys 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: Find the first non-repeating character in a string and return its index, or -1 if none exists.
Approach: Count frequencies in one pass (hash map or array of 26 for lowercase). Second pass returns the first index with count 1.
Complexity: O(n) time
Synopsys 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 Synopsys, skim the paper in a couple of minutes, mark what you can finish cleanly, and protect accuracy. Languages people commonly use: C++, Python, SystemVerilog (role-dependent).
| Area | Why it matters at Synopsys |
|---|---|
| DSA | What usually helps you clear the first round |
| Core CS (OOPs / DBMS / OS) | Technical interview depth |
| EDA, Semiconductor IP awareness | Helps in managerial / HR conversations |
| Communication | Explain your approach clearly; keep a few real examples ready for HR |
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