Capgemini Placement Papers 2024 - Actual Questions & Solutions

Capgemini 2024 Placement Papers - Complete Question Bank

This page contains actual Capgemini placement papers from 2024 with 40+ questions from game-based assessment and hands-on coding round.

2024 Pattern

Capgemini’s unique assessment: Game-based pseudocode (critical!), hands-on coding, then technical and HR interviews.

Capgemini 2024 Assessment Pattern

Round	Questions	Time	Difficulty
Game-Based (Pseudocode)	15 MCQs	25 min	Medium
Hands-on Coding	2 problems	45 min	Easy-Medium
Technical Interview	-	30-45 min	Medium
HR Interview	-	20-30 min	Easy

Pseudocode Questions (2024 Actual)

Question 1: Loop Output

What is the output?

SET i = 0
WHILE i < 5
    PRINT i
    SET i = i + 2
END WHILE

Answer: 0 2 4

Explanation: Loop runs with i = 0, 2, 4 (stops when i = 6)

Question 2: Array Sum

What is the output?

SET arr = [1, 2, 3, 4, 5]
SET sum = 0
FOR i FROM 0 TO 4
    SET sum = sum + arr[i]
END FOR
PRINT sum

Answer: 15

Explanation: Sum of array elements = 1+2+3+4+5 = 15

Question 3: Conditional Logic

What is the output?

SET x = 5, y = 10
IF x > y THEN
    PRINT x
ELSE IF x == y THEN
    PRINT 0
ELSE
    PRINT y
END IF

Answer: 10

Explanation: Since x < y, the ELSE branch executes

Question 4: Nested Loop

How many times does inner loop execute?

FOR i FROM 1 TO 3
    FOR j FROM 1 TO 2
        PRINT i * j
    END FOR
END FOR

Answer: 6 times

Explanation: 3 outer iterations × 2 inner iterations = 6

Question 5: Function Call

What is the output?

FUNCTION double(x)
    RETURN x * 2
END FUNCTION

SET a = 5
SET b = double(a)
PRINT b

Answer: 10

Explanation: double(5) returns 5 × 2 = 10

Question 6: String Length

What is the output?

SET str = "Hello"
SET count = 0
FOR i FROM 0 TO LENGTH(str) - 1
    SET count = count + 1
END FOR
PRINT count

Answer: 5

Explanation: Counts each character in “Hello”

Question 7: Factorial

What is the output?

FUNCTION factorial(n)
    IF n <= 1 THEN
        RETURN 1
    END IF
    RETURN n * factorial(n - 1)
END FUNCTION
PRINT factorial(4)

Answer: 24

Explanation: 4! = 4 × 3 × 2 × 1 = 24

Question 8: Array Reverse

What is arr after execution?

SET arr = [1, 2, 3, 4]
SET n = 4
FOR i FROM 0 TO n/2 - 1
    SET temp = arr[i]
    SET arr[i] = arr[n - 1 - i]
    SET arr[n - 1 - i] = temp
END FOR

Answer: [4, 3, 2, 1]

Explanation: Swaps first with last, second with second-last

Question 9: Prime Check

What is the output for n=7?

SET n = 7, isPrime = TRUE
FOR i FROM 2 TO n - 1
    IF n MOD i == 0 THEN
        SET isPrime = FALSE
    END IF
END FOR
PRINT isPrime

Answer: TRUE

Explanation: 7 is not divisible by 2-6

Question 10: Maximum in Array

What is the output?

SET arr = [3, 1, 4, 1, 5, 9, 2]
SET max = arr[0]
FOR i FROM 1 TO 6
    IF arr[i] > max THEN
        SET max = arr[i]
    END IF
END FOR
PRINT max

Answer: 9

Explanation: Finds maximum element in array

Hands-on Coding (2024)

Problem 1: Reverse String

Problem

Difficulty: Easy
Time: 15 minutes

Reverse a given string without using built-in functions.

Input: “capgemini”
Output: “inigempaC”

C

#include <string.h>
void reverse(char str[]) {
    int n = strlen(str);
    for (int i = 0; i < n/2; i++) {
        char temp = str[i];
        str[i] = str[n-1-i];
        str[n-1-i] = temp;
    }
}

Java

public String reverse(String s) {
    char[] arr = s.toCharArray();
    int left = 0, right = arr.length - 1;
    while (left < right) {
        char temp = arr[left];
        arr[left++] = arr[right];
        arr[right--] = temp;
    }
    return new String(arr);
}

Python

def reverse_string(s):
    chars = list(s)
    left, right = 0, len(chars) - 1
    while left < right:
        chars[left], chars[right] = chars[right], chars[left]
        left += 1
        right -= 1
    return ''.join(chars)

Problem 2: Palindrome Check

Problem

Difficulty: Easy
Time: 15 minutes

Check if a string is palindrome (ignore case and spaces).

Input: “A man a plan a canal Panama”
Output: true

Java

public boolean isPalindrome(String s) {
    s = s.toLowerCase().replaceAll("[^a-z0-9]", "");
    int left = 0, right = s.length() - 1;
    while (left < right) {
        if (s.charAt(left++) != s.charAt(right--)) {
            return false;
        }
    }
    return true;
}

Python

def is_palindrome(s):
    s = ''.join(c.lower() for c in s if c.isalnum())
    return s == s[::-1]

Problem 3: Armstrong Number

Problem

Difficulty: Easy
Time: 15 minutes

Check if a number is Armstrong (sum of cubes of digits equals number for 3-digit).

Input: 153
Output: true (1³ + 5³ + 3³ = 153)

C

int isArmstrong(int n) {
    int original = n, sum = 0, digits = 0, temp = n;
    while (temp > 0) {
        digits++;
        temp /= 10;
    }
    temp = n;
    while (temp > 0) {
        int digit = temp % 10;
        int power = 1;
        for (int i = 0; i < digits; i++) power *= digit;
        sum += power;
        temp /= 10;
    }
    return sum == original;
}

Python

def is_armstrong(n):
    digits = str(n)
    power = len(digits)
    return sum(int(d) ** power for d in digits) == n

Problem 4: Fibonacci Series

Problem

Difficulty: Easy
Time: 15 minutes

Print first n Fibonacci numbers.

Input: n = 7
Output: 0 1 1 2 3 5 8

C

void fibonacci(int n) {
    int a = 0, b = 1;
    for (int i = 0; i < n; i++) {
        printf("%d ", a);
        int temp = a + b;
        a = b;
        b = temp;
    }
}

Java

public void fibonacci(int n) {
    int a = 0, b = 1;
    for (int i = 0; i < n; i++) {
        System.out.print(a + " ");
        int temp = a + b;
        a = b;
        b = temp;
    }
}

Python

def fibonacci(n):
    a, b = 0, 1
    for _ in range(n):
        print(a, end=" ")
        a, b = b, a + b

Problem 5: Count Vowels and Consonants

Problem

Difficulty: Easy
Time: 10 minutes

Count vowels and consonants in a string.

Input: “Capgemini”
Output: Vowels: 4, Consonants: 5

Java

public int[] countVowelsConsonants(String s) {
    int vowels = 0, consonants = 0;
    s = s.toLowerCase();
    for (char c : s.toCharArray()) {
        if (Character.isLetter(c)) {
            if ("aeiou".indexOf(c) >= 0) vowels++;
            else consonants++;
        }
    }
    return new int[]{vowels, consonants};
}

Python

def count_vowels_consonants(s):
    vowels = consonants = 0
    for c in s.lower():
        if c.isalpha():
            if c in 'aeiou':
                vowels += 1
            else:
                consonants += 1
    return vowels, consonants

Key Insights from 2024

1. Pseudocode Critical: Many fail here - practice daily
2. Time Pressure: 25 minutes for 15 pseudocode questions
3. Careful Reading: Read each line, trace step by step
4. Basic Coding: Hands-on problems are straightforward
5. No Negative Marking: Attempt all questions
6. Success Rate: ~35% clear pseudocode round

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Master pseudocode - it’s the key to clearing Capgemini!

Last updated: February 2026