The Pseudo Code Round in Infosys placement drives (InfyTQ, HackWithInfy, or Off-Campus Hiring) tests how well you understand logic, flow, and conditions in code.

You don’t need to write code — you just need to trace what the code does.

Below, you’ll find the latest Infosys Pseudo Code questions (2025–26) taken directly from practice material and actual patterns. Each question has options and answers verified twice for accuracy.


Infosys Pseudocode Section Overview

Details Description
Total Questions 5
Time Limit 10 minutes
Question Type Output-based Logical Pseudocode
Difficulty Level Easy to Moderate
Topics Covered Loops, Conditions, Operators, Number Logic

Top Infosys Pseudocode Questions (With Answers)


Q1.

a = 5
b = 10
c = 15
if(a + b > c)
    print("A")
else if(b + c > a)
    print("B")
else
    print("C")

Options:
a) A
b) B
c) C
d) Error

Answer: b) B
Explanation:

  • (a+b > c) → 15 > 15 → False
  • (b+c > a) → 25 > 5 → True → Prints B

Q2.

a = 10
b = 20
if(a < b and b < 30)
    print("True")
else
    print("False")

Options:
a) True
b) False
c) 30
d) 10

Answer: a) True
Explanation: Both conditions are true, so output = True.


Q3.

count = 0
for i = 1 to 5
    for j = 1 to i
        count = count + 1
print(count)

Options:
a) 10
b) 15
c) 5
d) 25

Answer: b) 15
Explanation: Runs 1+2+3+4+5 = 15 times → count = 15.


Q4.

a = 0
for i = 1 to 5
    a = a + i
print(a)

Options:
a) 10
b) 15
c) 20
d) 25

Answer: b) 15
Explanation: Sum = 1+2+3+4+5 = 15.


Q5.

a = 10
b = 20
c = 30
if(a > b and a > c)
    print(a)
else if(b > a and b > c)
    print(b)
else
    print(c)

Options:
a) 10
b) 20
c) 30
d) None

Answer: c) 30
Explanation: Largest number among a,b,c → 30.


Q6.

a = 5
b = 3
for i = 1 to b
    a = a + b
print(a)

Options:
a) 8
b) 11
c) 14
d) 15

Answer: c) 14
Explanation: Adds 3 three times → 5 + 3×3 = 14.


Q7.

num = 5
fact = 1
for i = 1 to num
    fact = fact * i
print(fact)

Options:
a) 60
b) 100
c) 120
d) 24

Answer: c) 120
Explanation: 1×2×3×4×5 = 120 → factorial of 5.


Q8.

a = 4
b = 2
for i = 1 to b
    a = a * b
print(a)

Options:
a) 4
b) 8
c) 12
d) 16

Answer: d) 16
Explanation: a=4→8→16 after 2 iterations.


Q9.

a = 10
b = 5
if(a % b == 0)
    print(a / b)
else
    print(a * b)

Options:
a) 2
b) 5
c) 10
d) 50

Answer: a) 2
Explanation: 10 % 5 == 0 → prints a/b = 2.


Q10.

a = 2
b = 3
for i = 1 to 3
    a = a * b
print(a)

Options:
a) 18
b) 27
c) 36
d) 54

Answer: d) 54
Explanation: a=2×3×3×3=54.


Q11.

a = 5
b = 10
c = 0
for i = 1 to b
    c = c + a
print(c)

Options:
a) 10
b) 25
c) 50
d) 100

Answer: c) 50
Explanation: Adds 5 ten times → 50.


Q12.

sum = 0
for i = 1 to 10
    if(i % 2 == 0)
        sum = sum + i
print(sum)

Options:
a) 20
b) 25
c) 30
d) 35

Answer: c) 30
Explanation: 2+4+6+8+10 = 30.


Q13.

a = 2
b = 5
for i = 1 to b
    if(i % 2 == 0)
        a = a + i
print(a)

Options:
a) 5
b) 7
c) 8
d) 10

Answer: c) 8
Explanation: Adds even i → 2+2+4=8.


Q14.

a = 0
for i = 1 to 4
    for j = 1 to 3
        a = a + 1
print(a)

Options:
a) 9
b) 10
c) 12
d) 15

Answer: c) 12
Explanation: 4×3 = 12.


Q15.

num = 6
for i = 1 to num
    if(num % i == 0)
        print(i)

Options:
a) 1, 2, 3, 6
b) 2, 3, 4, 6
c) 1, 3, 6
d) 2, 4, 6

Answer: a) 1, 2, 3, 6
Explanation: Divisors of 6.


Q16.

for i = 1 to 5
    for j = 1 to i
        print("*")

Options:
a) *****
b) Pyramid Pattern
c) Right-Angle Triangle
d) Half Pyramid

Answer: d) Half Pyramid
Explanation: Prints * pattern of height 5.


Q17.

sum = 0
for i = 1 to 100
    sum = sum + i
print(sum)

Options:
a) 100
b) 5050
c) 5000
d) 5500

Answer: b) 5050
Explanation: Sum of 1–100 = n(n+1)/2 = 5050.


Q18.

a = 0
for i = 1 to 10
    if(i % 3 == 0)
        a = a + i
print(a)

Options:
a) 15
b) 18
c) 21
d) 24

Answer: b) 18
Explanation: 3+6+9 = 18.


Q19.

num = 153
sum = 0
temp = num
while temp > 0
    digit = temp % 10
    sum = sum + (digit * digit * digit)
    temp = temp // 10
if(sum == num)
    print("Armstrong")
else
    print("Not Armstrong")

Options:
a) Armstrong
b) Not Armstrong
c) Error
d) Infinite Loop

Answer: a) Armstrong
Explanation: 153 = 1³ + 5³ + 3³ → Armstrong number.


Q20.

num = 1234
rev = 0
while(num > 0)
    digit = num % 10
    rev = rev * 10 + digit
    num = num // 10
print(rev)

Options:
a) 1234
b) 4321
c) 3412
d) 2143

Answer: b) 4321
Explanation: Reverses digits → 4321.


Key Tips to Crack Infosys Pseudocode Section

  1. Understand flow, not syntax — It’s not about code language, it’s about logic.
  2. Practice loop tracing — Most questions use nested loops or conditions.
  3. Watch out for tricky conditionsif-else order changes the output.
  4. Run dry tests manually — simulate each line step by step.
  5. Time management: 5 questions, 10 minutes = 2 min per question.

Final Advice:

The Infosys Pseudocode Round is a scoring opportunity if you master logic-based questions.
All the 20+ questions and answers above are verified twice from the official Infosys pattern.
With regular practice, you can easily secure a 90%+ accuracy in this section.