Mastercard 2025 aptitude
Quantitative, reasoning, and verbal drills with solutions
This page collects Mastercard 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 Mastercard patterns and recent shifts.
Mastercard 2025 aptitude
Quantitative, reasoning, and verbal drills with solutions
Mastercard 2025 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 2025. They are not leaked live papers. Work them timed, then read the solutions only after you have an answer.
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: An article sold at 10% loss would give 5% profit if sold for ₹60 more. Find the cost price.
Solution: 0.9P + 60 = 1.05P 60 = 0.15P P = 60 / 0.15 = ₹400.
Answer: ₹400
Problem: Eight workers finish a job in 10 days. How many days will 10 workers take for the same job (same pace)?
Solution: Total man-days = 8 × 10 = 80. Days for 10 workers = 80 / 10 = 8 days.
Answer: 8 days
Problem: A vehicle travels at 60 km/h for 2.5 hours. How far does it go?
Solution: Distance = speed × time = 60 × 2.5 = 150 km.
Answer: 150 km
Problem: If two ratios are 3:5 and 5:7, what is the compound ratio?
Solution: Compound ratio = (3/5) × (5/7) = 3/7, written as 3:7.
Answer: 3:7
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: 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.
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)
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 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
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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