Visa 2026 aptitude
Quantitative, reasoning, and verbal drills with solutions
This page collects Visa 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 Visa patterns and recent shifts.
Visa 2026 aptitude
Quantitative, reasoning, and verbal drills with solutions
Visa 2026 coding
DSA practice aligned to Visa online assessments
Visa interview experience
Round structure and tips from student reports
Visa 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 HCF of 36 and 48 is:
36 = 2²×3², 48 = 2⁴×3 → HCF = 2²×3 = 12.
Quantitative
If "TECHNOLOGY" is coded as "VGEPQMQNA", how is "COMPUTER" coded?
Pattern: Each letter shifted by +2 T→V, E→G, C→E, H→J, N→P, O→Q, L→N, O→Q, G→I, Y→A COMPUTER: C→E, O→Q, M→O, P→R, U→W, T→V, E→G, R→T
Quantitative
A shopkeeper marks goods 40% above cost and gives a 10% discount. Profit % is:
SP = 1.4 × 0.9 × CP = 1.26 CP → 26% profit.
Reasoning
Find next number: 2, 5, 10, 17, 26, ?
Pattern: Differences are 3, 5, 7, 9 (odd numbers) Next difference = 11 Next number = 26 + 11 = 37
Quantitative
The ratio of the ages of two persons is 4:3. After 6 years, the ratio becomes 9:7. What are their present ages?
Let present ages be 4x and 3x After 6 years: (4x + 6)/(3x + 6) = 9/7 7(4x + 6) = 9(3x + 6) 28x + 42 = 27x + 54 x = 12 Present ages: 4×12 = 48 years, 3×12 = 36 years
Verbal
Synonym of "Scarce":
Scarce means rare or insufficient.
Verbal
Identify the error type: "He don't know the answer."
He doesn't / does not - subject-verb agreement.
Quantitative
The ratio of two numbers is 3:4. If their sum is 84, find the numbers.
Let the numbers be 3x and 4x Sum = 3x + 4x = 7x = 84 x = 12 Numbers = 3(12) = 36 and 4(12) = 48
Reasoning
All roses are flowers. Some flowers fade quickly. Conclusion: All roses fade quickly.
Only some flowers.
Verbal
A tap fills a tank in 5 hours, and another tap empties it in 7 hours. How long will it take to fill the tank if both taps are open?
Filling tap fills 1/5 of tank per hour Emptying tap empties 1/7 of tank per hour Net filling = 1/5 - 1/7 = (7-5)/35 = 2/35 per hour Time to fill = 1 / (2/35) = 35/2 = 17.5 hours
Reasoning
Odd one out: Triangle, Square, Circle, Rectangle
Circle has no straight sides.
Reasoning
How many 9s are there between 1 and 100?
9,19,29,39,49,59,69,79,89,90-99 → 20 nines.
Quantitative
The sum of ages of 5 children born at intervals of 3 years is 50 years. Find the age of the youngest child.
Let youngest child's age = x Ages: x, x+3, x+6, x+9, x+12 Sum = 5x + 30 = 50 5x = 20 x = 4 years
Verbal
She is good _____ mathematics.
Good at a subject.
Reasoning
All pens are books. Some books are papers. Conclusion: Some pens are papers.
No definite overlap between pens and papers.
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: Visa Online Assessment
Skills emphasized: DSA, distributed systems, payments
Languages: Java, Python, C++
These are practice-style questions aligned to patterns students report for Visa 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 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
Visa 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
Visa 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)
Visa 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
Visa 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
Visa 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
Visa 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
Visa 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: Design a stack that supports push, pop, top, and getMin in average O(1) time.
Approach: Keep a parallel min-stack (or store pairs). When pushing, also push the new minimum. When popping, pop both stacks.
Complexity: O(1) per operation amortized
Visa 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 Visa, skim the paper in a couple of minutes, mark what you can finish cleanly, and protect accuracy. Languages people commonly use: Java, Python, C++.
| Area | Why it matters at Visa |
|---|---|
| DSA | What usually helps you clear the first round |
| Core CS (OOPs / DBMS / OS) | Technical interview depth |
| Digital Payments awareness | Helps in managerial / HR conversations |
| Communication | Explain your approach clearly; keep a few real examples ready for HR |
Mastercard · PayPal · Stripe · PhonePe · American Express