Visa Placement Papers 2026 - Questions and Solutions

Last Updated: March 2026

Company Overview

Visa is a global payments technology company that connects consumers, businesses, financial institutions, and governments to fast, secure, and reliable electronic payments. Operating in over 200 countries, Visa enables digital payments across various platforms and devices.

Selection Process

Stage	Description	Duration
Online Assessment	Aptitude, Coding, Technical MCQ	120 minutes
Technical Interview 1	CS fundamentals, Coding	45 minutes
Technical Interview 2	System Design, Projects	45 minutes
HR Interview	Behavioral, Culture fit	30 minutes

Eligibility:

65% in 10th, 12th, Graduation
No active backlogs
CS/IT/MCA preferred

Exam Pattern

Section	Questions	Time	Topics
Aptitude	30	40 min	Quant, Logical, Verbal
Technical MCQ	20	20 min	CS, Networking, DBMS
Coding	3	60 min	Algorithms, DS

Aptitude Questions

1. A person travels at 40 km/hr and returns at 60 km/hr. Find average speed.

Solution: Average speed = 2xy/(x+y) = 2×40×60/(40+60) = 4800/100 = 48 km/hr

2. Find next term: 1, 1, 2, 3, 5, 8, ?

Solution: Fibonacci sequence: Each term is sum of previous two. Next = 5 + 8 = 13

3. If 12 men earn ₹240 in 4 days, how much will 8 men earn in 6 days?

Solution: 1 man in 1 day = 240/(12×4) = ₹5 8 men in 6 days = 5 × 8 × 6 = ₹240

4. Area of circle is 154 sq.cm. Find circumference.

Solution: πr² = 154 → r² = 154×7/22 = 49 → r = 7 Circumference = 2πr = 2×22/7×7 = 44 cm

5. CP of 40 articles equals SP of 32 articles. Find profit%.

Solution: 40×CP = 32×SP → SP/CP = 40/32 = 5/4 Profit = 5-4 = 1 Profit% = 1/4 × 100 = 25%

6. In what ratio should water be mixed with milk to gain 20% by selling at CP?

Solution: Let CP of 1L milk = ₹1 SP of 1L mixture = ₹1 (at CP) Gain = 20% on cost of mixture CP of mixture = 1/1.2 = 5/6 Water:Milk = (1-5/6):5/6 = 1/6:5/6 = 1:5

7. A sum doubles in 5 years at SI. How long to become 8 times?

Solution: If doubles in 5 years, rate = 20% To become 8 times: 7 times interest needed Time = 7×5 = 35 years

8. A boat goes 12 km upstream in 3 hours and 18 km downstream in 3 hours. Find stream speed.

Solution: Upstream speed = 12/3 = 4 km/hr Downstream speed = 18/3 = 6 km/hr Stream speed = (6-4)/2 = 1 km/hr

9. LCM of two numbers is 180, HCF is 15. One number is 45. Find other.

Solution: Product = LCM × HCF = 180 × 15 = 2700 Other number = 2700/45 = 60

10. How many 3-digit numbers are divisible by 7?

Solution: First = 105, Last = 994 Number of terms = (994-105)/7 + 1 = 889/7 + 1 = 127 + 1 = 128

Technical Questions

1. Difference between TCP and UDP

TCP: Connection-oriented, reliable, ordered, slower
UDP: Connectionless, unreliable, unordered, faster

2. What is REST API?

3. SQL query to find duplicate rows

SELECT column1, column2, COUNT(*)
FROM table
GROUP BY column1, column2
HAVING COUNT(*) > 1;

4. What is encapsulation?

5. Difference between process and thread

6. What is indexing in databases?

7. What is a deadlock?

8. Difference between GET and POST

GET: Parameters in URL, limited size, idempotent, for retrieval
POST: Parameters in body, larger size, not idempotent, for creation

9. What is garbage collection?

10. What is the time complexity of binary search?

Coding Questions

1. Check if a string is a valid palindrome (ignoring non-alphanumeric).

def is_palindrome(s):
    left, right = 0, len(s) - 1
    
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        
        if s[left].lower() != s[right].lower():
            return False
        
        left += 1
        right -= 1
    
    return True

# Test
print(is_palindrome("A man, a plan, a canal: Panama"))  # True

2. Merge two sorted arrays without extra space.

def merge_sorted_arrays(nums1, m, nums2, n):
    """
    nums1 has size m+n, first m elements are valid
    """
    # Start from end
    i, j, k = m - 1, n - 1, m + n - 1
    
    while j >= 0:
        if i >= 0 and nums1[i] > nums2[j]:
            nums1[k] = nums1[i]
            i -= 1
        else:
            nums1[k] = nums2[j]
            j -= 1
        k -= 1
    
    return nums1

# Test
nums1 = [1, 2, 3, 0, 0, 0]
nums2 = [2, 5, 6]
print(merge_sorted_arrays(nums1, 3, nums2, 3))  # [1, 2, 2, 3, 5, 6]

3. Find the intersection of two arrays.

def intersection(nums1, nums2):
    return list(set(nums1) & set(nums2))

# With counts
def intersection_with_counts(nums1, nums2):
    from collections import Counter
    count1 = Counter(nums1)
    count2 = Counter(nums2)
    result = []
    
    for num in count1:
        if num in count2:
            result.extend([num] * min(count1[num], count2[num]))
    
    return result

# Test
print(intersection([1, 2, 2, 1], [2, 2]))  # [2]
print(intersection_with_counts([1, 2, 2, 1], [2, 2]))  # [2, 2]

4. Move all zeros to end of array.

def move_zeros(nums):
    non_zero = 0
    
    for i in range(len(nums)):
        if nums[i] != 0:
            nums[non_zero] = nums[i]
            non_zero += 1
    
    for i in range(non_zero, len(nums)):
        nums[i] = 0
    
    return nums

# Test
print(move_zeros([0, 1, 0, 3, 12]))  # [1, 3, 12, 0, 0]

5. Find first non-repeating character.

def first_unique_char(s):
    from collections import Counter
    count = Counter(s)
    
    for i, char in enumerate(s):
        if count[char] == 1:
            return i
    
    return -1

# Test
print(first_unique_char("leetcode"))      # 0
print(first_unique_char("loveleetcode"))  # 2

Interview Tips

Focus on problem-solving approach
Practice array and string manipulation
Understand payment processing basics
Prepare for system design discussions
Research Visa's technology initiatives
Practice coding under time pressure

Best of luck with your Visa placement!