Mastercard 2026 aptitude
Quantitative, reasoning, and verbal drills with solutions
This page collects Mastercard 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 Mastercard patterns and recent shifts.
Mastercard 2026 aptitude
Quantitative, reasoning, and verbal drills with solutions
Mastercard 2026 coding
DSA practice aligned to Mastercard online assessments
Mastercard interview experience
Round structure and tips from student reports
Mastercard 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.
Reasoning
Find the next number: 6, 11, 21, 41, ?
×2−1 pattern → 81.
Quantitative
A number is first increased by 25% and then decreased by 20%. Find the net percentage change.
Let number = 100 After 25% increase: 100 + 25 = 125 After 20% decrease: 125 - 20% of 125 = 125 - 25 = 100 Net change = 0%
Quantitative
The ratio of ages of A and B is 3:5. After 10 years, the ratio becomes 5:7. Find the present age of A.
Let present ages be 3x and 5x After 10 years: (3x + 10)/(5x + 10) = 5/7 7(3x + 10) = 5(5x + 10) 21x + 70 = 25x + 50 4x = 20 x = 5 A's present age = 3 × 5 = 15 years
Reasoning
If BOOK is coded as 43 (sum of letter positions), PEN is coded as:
P+E+N = 16+5+14 = 35.
Reasoning
A is taller than B but shorter than C. D is shorter than B. Who is tallest?
C > A > B > D → C is tallest.
Quantitative
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.
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 =
Verbal
Fill in: She is good _____ mathematics.
Good at a subject.
Quantitative
A shopkeeper gives two successive discounts of 10% and 20% on an item. What is the effective discount percentage?
Let MP = ₹100 After first discount: 100 - 10% = ₹90 After second discount: 90 - 20% = ₹72 Effective discount = (100 - 72) / 100 × 100 = 28%
Reasoning
Find the next number: 5, 10, 20, 40, ?
×2 each → 80.
Verbal
Synonym of "Abundant":
Abundant means plentiful.
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
Reasoning
In a row of 40 students, A is 11th from left. Rank from right is:
From right = 40 − 11 + 1 = 30.
Verbal
Identify the correctly spelled word:
Accommodate has double "c" and double "m".
Quantitative
Find the HCF of 24 and 36.
Correct answer: 12
Verbal
A shopkeeper buys an article for ₹500 and sells it for ₹600. Find his profit percentage.
Correct answer: 20%
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: Mastercard Codility / OA
Skills emphasized: DSA, payments concepts, system design
Languages: Java, Python, C++
These are practice-style questions aligned to patterns students report for Mastercard drives around 2026. They are not leaked live papers. Work them timed, then read the solutions only after you have an answer.
Problem: Find the simple interest on ₹8000 at 10% per annum for 2 years.
Solution: SI = (P × R × T) / 100 = (8000 × 10 × 2) / 100 = ₹1600.
Answer: ₹1600
Problem: If 30% of a number is 150, what is the number?
Solution: 0.3x = 150 → x = 150 / 0.3 = 500.
Answer: 500
Problem: The average of five numbers is 20. One number 30 is replaced by 10. What is the new average?
Solution: Old sum = 5 × 20 = 100. New sum = 100 − 30 + 10 = 80. New average = 80 / 5 = 16.
Answer: 16
Problem: Find the next term: 2, 6, 12, 20, 30, ?
Solution: Pattern: 1×2, 2×3, 3×4, 4×5, 5×6, 6×7. Next = 6 × 7 = 42.
Answer: 42
Problem: What is the probability of drawing an ace from a standard 52-card deck?
Solution: There are 4 aces in 52 cards. Probability = 4/52 = 1/13.
Answer: 1/13
Problem: A number is increased by 20% and then decreased by 20%. What is the net percentage change?
Solution: Start with 100 → 120 → 96. Net change = 4% decrease. Formula: successive +a then −a gives −(a²/100)% = −4%.
Answer: 4% decrease
Problem: If the price of an item rises by 25%, by what percent should consumption fall so that expenditure stays the same?
Solution: Required reduction = r/(100+r) × 100 with r = 25. = 25/125 × 100 = 20%.
Answer: 20%
Problem: An article is marked 40% above cost and sold after a 10% discount. Find the profit percent.
Solution: SP = CP × 1.4 × 0.9 = 1.26 CP. Profit = 26%.
Answer: 26%
Problem: Write a function that returns true if n is prime and false otherwise. Handle n < 2 correctly.
Approach: Return false for n < 2. Trial-divide from 2 to floor(sqrt(n)). If any divisor divides n evenly, it is composite; otherwise prime.
Complexity: O(√n) time
Mastercard 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 string containing only ‘()[]’, decide whether the brackets are balanced and correctly nested.
Approach: Scan left to right with a stack. Push opening brackets. On a closing bracket, the stack top must be the matching opener. At the end the stack must be empty.
Complexity: O(n) time, O(n) space
Mastercard 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 array of integers and a target, return indices of two numbers that add up to the target. Assume exactly one solution and you may not use the same element twice.
Approach: Walk the array once. For each value x, check whether target − x was seen earlier in a hash map of value → index. If yes, return both indices; else store x.
Complexity: O(n) time, O(n) space
Mastercard 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 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
Mastercard 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
Mastercard 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
Mastercard 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
Mastercard 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
Mastercard 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 Mastercard, 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 Mastercard |
|---|---|
| DSA | What usually helps you clear the first round |
| Core CS (OOPs / DBMS / OS) | Technical interview depth |
| Payments Networks awareness | Helps in managerial / HR conversations |
| Communication | Explain your approach clearly; keep a few real examples ready for HR |
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