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Access free Accenture placement papers 2024, previous year NLT questions with solutions, detailed exam pattern, interview questions, and complete preparation guide. Download Accenture 2024 placement papers PDF.
This page contains Accenture placement papers from 2024 with detailed solutions, exam patterns, and preparation insights. Use these previous year papers to understand the 2024 Accenture NLT pattern and prepare effectively for current year exams.
| Section | Questions | Time | Difficulty | Focus Areas |
|---|---|---|---|---|
| Cognitive Assessment | 20-25 | 40-45 min | Medium | Quantitative aptitude, logical reasoning, verbal ability |
| Technical Assessment | 20-25 | 40-45 min | Medium-Hard | Programming MCQs, pseudo code, output prediction |
| Coding Test | 2 | 30-35 min | Medium | Array manipulation, string processing, basic algorithms |
| Communication Assessment | 5-10 | 10-15 min | Easy-Medium | English speaking, comprehension, grammar |
Total: 60-70 questions, 120 minutes
Key Changes in 2024:
This section contains real questions from Accenture NLT 2024 based on candidate experiences and actual exam patterns. All questions include detailed solutions and explanations.
Problem: A shopkeeper marks an item 30% above cost price and gives 10% discount. Find his profit percentage.
Solution: Let CP = ₹100 Marked Price = ₹130 Discount = 10% of ₹130 = ₹13 Selling Price = ₹130 - ₹13 = ₹117 Profit = ₹117 - ₹100 = ₹17 Profit % = (17/100) × 100 = 17%
Answer: 17%
Problem: A and B can complete a work in 20 days and 30 days respectively. They work together for 6 days, then A leaves. In how many days will B complete the remaining work?
Solution: A’s 1 day work = 1/20 B’s 1 day work = 1/30 Together in 1 day = 1/20 + 1/30 = 5/60 = 1/12 Work done in 6 days = 6 × 1/12 = 1/2 Remaining work = 1 - 1/2 = 1/2 B will complete in = (1/2) / (1/30) = 15 days
Answer: 15 days
Problem: If 20% of a number is 60, what is 35% of that number?
Solution: Let the number be x 20% of x = 60 x = 60 × 100/20 = 300 35% of 300 = 35/100 × 300 = 105
Answer: 105
Problem: 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.
Solution: 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
Answer: 15 years
Problem: A sum of money becomes ₹2,400 in 2 years and ₹3,200 in 4 years at simple interest. Find the principal amount.
Solution: Interest for 2 years = 3200 - 2400 = ₹800 Interest per year = ₹400 Interest for 2 years = ₹800 Principal = 2400 - 800 = ₹1,600
Answer: ₹1,600
Problem: Find compound interest on ₹10,000 for 2 years at 10% per annum, compounded annually.
Solution: Amount = P(1 + R/100)^n Amount = 10000(1 + 10/100)^2 = 10000 × 1.1^2 = 10000 × 1.21 = ₹12,100 CI = Amount - Principal = 12100 - 10000 = ₹2,100
Answer: ₹2,100
Problem: A train 150 meters long passes a platform 200 meters long in 20 seconds. Find the speed of the train in km/hr.
Solution: Total distance = 150 + 200 = 350 meters Time = 20 seconds Speed = 350/20 = 17.5 m/s Speed in km/hr = 17.5 × 18/5 = 63 km/hr
Answer: 63 km/hr
Problem: There are two pipes in a tank. Pipe A is for filling the tank and pipe B is for emptying the tank. If A can fill the tank in 10 hours and B can empty the tank in 15 hours then find how many hours will it take to completely fill a half empty tank?
Solution: Net rate = 1/10 - 1/15 = 1/30 per hour Time to fill half tank = (1/2) / (1/30) = 15 hours
Answer: 15 hours
Problem: A code word is to consist of two English alphabets followed by two distinct numbers between 1 and 9. For example, CA23 is a code word. How many such code words are there?
Solution: Alphabets: 26 × 26 = 676 ways Numbers: 9 × 8 = 72 ways (distinct) Total: 676 × 72 = 48,672
Answer: 48,672
Problem: Two dice are thrown. What is the probability of getting a sum of 7?
Solution: Total outcomes = 6 × 6 = 36 Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 Probability = 6/36 = 1/6
Answer: 1/6
Problem: In what ratio should water be mixed with milk costing ₹12 per liter to get a mixture worth ₹8 per liter?
Solution: Using alligation: Water (₹0) Milk (₹12) \ / \ / \ / \ / \ / ₹8 Ratio = (12-8):(8-0) = 4:8 = 1:2
Answer: 1:2
Problem: The average of 5 numbers is 25. If one number is excluded, the average becomes 23. Find the excluded number.
Solution: Sum of 5 numbers = 5 × 25 = 125 Sum of 4 numbers = 4 × 23 = 92 Excluded number = 125 - 92 = 33
Answer: 33
Problem: The sum of ages of 5 children born at intervals of 3 years is 50 years. Find the age of the youngest child.
Solution: Let youngest child’s age = x Ages: x, x+3, x+6, x+9, x+12 Sum = 5x + 30 = 50 5x = 20 x = 4 years
Answer: 4 years
Problem: A and B invest ₹5,000 and ₹7,000 respectively in a business. After 6 months, A withdraws ₹2,000. If the profit at the end of the year is ₹4,500, find B’s share.
Solution: A’s investment: 5000 × 6 + 3000 × 6 = 30,000 + 18,000 = 48,000 B’s investment: 7000 × 12 = 84,000 Ratio = 48,000 : 84,000 = 4 : 7 B’s share = 7/(4+7) × 4500 = 7/11 × 4500 = ₹2,863.64
Answer: ₹2,863.64
Problem: A man can row 8 km/hr in still water. If the speed of the stream is 2 km/hr, find the time taken to row 30 km downstream.
Solution: Downstream speed = 8 + 2 = 10 km/hr Time = Distance/Speed = 30/10 = 3 hours
Answer: 3 hours
Problem: A train travels 360 km at a uniform speed. If the speed had been 5 km/h faster, it would have taken 48 minutes less for the journey. What is the speed of the train?
Solution: Let the original speed be x km/h Original time = 360/x hours New speed = (x + 5) km/h New time = 360/(x + 5) hours Time difference = 48 minutes = 48/60 = 4/5 hours
360/x - 360/(x + 5) = 4/5 360(1/x - 1/(x+5)) = 4/5 360((x+5-x)/(x(x+5))) = 4/5 360(5)/(x(x+5)) = 4/5 1800/(x(x+5)) = 4/5 9000 = 4x(x+5) 4x² + 20x - 9000 = 0 x² + 5x - 2250 = 0 (x + 50)(x - 45) = 0 x = 45 (positive value)
Answer: 45 km/h
Problem: The sum of the ages of a father and his son is 60 years. If the father’s age is four times that of his son, what are their current ages?
Solution: Let son’s age = x years Father’s age = 4x years x + 4x = 60 5x = 60 x = 12
Son’s age = 12 years Father’s age = 48 years
Answer: Son: 12 years, Father: 48 years
Problem: If 5 workers can complete a task in 20 days, how many days will it take for 8 workers to complete the same task, assuming all workers work at the same rate?
Solution: Total work = 5 workers × 20 days = 100 worker-days With 8 workers: Days required = 100 worker-days ÷ 8 workers = 12.5 days
Answer: 12.5 days
Problem: A man buys 12 liters of a liquid containing 20% substance A and the rest water. He mixes it with 10 liters of another mixture with 30% substance A. What is the percentage of water in the new mixture?
Solution: First mixture: 12L with 20% substance A Substance A = 20% of 12L = 2.4L Water = 12L - 2.4L = 9.6L
Second mixture: 10L with 30% substance A Substance A = 30% of 10L = 3L Water = 10L - 3L = 7L
Total mixture: 12L + 10L = 22L Total water = 9.6L + 7L = 16.6L Percentage of water = (16.6/22) × 100 = 75.45%
Answer: 75.45%
Problem: Two trains move in the same direction at 50 kmph and 32 kmph respectively. A man in the slower train observes that 15 seconds elapse before the faster train completely passes by him. What is the length of the faster train?
Solution: Relative speed = 50 - 32 = 18 kmph = 18 × 5/18 = 5 m/s Time = 15 seconds Length of faster train = Relative speed × Time = 5 × 15 = 75 meters
Answer: 75 meters
Problem: Find (7x + 4y) / (x - 2y) if x/2y = 3/2.
Solution: Given x/2y = 3/2 Cross multiplying: 2x = 6y, so x = 3y
Substituting x = 3y: (7x + 4y) / (x - 2y) = (7(3y) + 4y) / (3y - 2y) = (21y + 4y) / y = 25y / y = 25
Answer: 25
Problem: Find the next number: 2, 5, 11, 23, 47, ?
Solution: Pattern: Each number = previous × 2 + 1 2×2+1=5, 5×2+1=11, 11×2+1=23, 23×2+1=47 Next: 47×2+1 = 95
Answer: 95
Problem: Find the next number: 3, 8, 15, 24, 35, ?
Solution: Pattern: Differences are 5, 7, 9, 11 (odd numbers) Next difference = 13 Next number = 35 + 13 = 48
Answer: 48
Problem: Find the next letter: A, D, G, J, ?
Solution: Pattern: A(+3)→D(+3)→G(+3)→J(+3)→M Next letter = M
Answer: M
Problem: If “TECHNOLOGY” is coded as “VGEPQMQNA”, how is “COMPUTER” coded?
Solution: Pattern: Each letter shifted by +2 T→V, E→G, C→E, H→J, N→P, O→Q, L→N, O→Q, G→I, Y→A
COMPUTER: C→E, O→Q, M→O, P→R, U→W, T→V, E→G, R→T
Answer: EQORWVGT
Problem: Statements: 1. All green are blue, 2. All blue are white
Conclusions: I) Some blue are green, II) Some white are green, III) Some green are white, IV) All white are blue.
Solution: From “All green are blue”: Some blue are green (I follows) Combined: All green are white, so Some white are green (II follows) and Some green are white (III follows) IV does not follow
Answer: Only I, II, and III follow
Problem: Pointing to a man, a woman said, “His mother is the only daughter of my mother.” How is the woman related to the man?
Solution: Only daughter of woman’s mother = woman herself So woman is the mother of the man
Answer: Mother
Problem: A person walks 5 km north, then 3 km east, then 2 km south. How far and in which direction is he from the starting point?
Solution: Net north = 5 - 2 = 3 km Net east = 3 km Distance = √(3² + 3²) = √18 = 3√2 km Direction = Northeast
Answer: 3√2 km, Northeast
Problem: Five friends A, B, C, D, E are sitting in a row. A is not at either end. B is to the immediate right of A. C is at one end. D is between C and E. Who is sitting in the middle?
Solution: C is at one end. D is between C and E, so arrangement: C-D-E or E-D-C A is not at end, B is right of A If C-D-E, then A-B must be before C: A-B-C-D-E (A at end, invalid) So: E-D-C, and A-B before: A-B-E-D-C Middle position: E
Answer: E
Problem: In a class of 40 students, Ravi ranks 15th from the top. What is his rank from the bottom?
Solution: Total students = 40 Rank from top = 15 Rank from bottom = 40 - 15 + 1 = 26
Answer: 26th
Problem: Statement: All roses are flowers. Some flowers are red.
Conclusions: I) Some roses are red, II) All red things are flowers
Solution: From “All roses are flowers” and “Some flowers are red”, we cannot conclude that some roses are red (I may not follow) “All red things are flowers” is not necessarily true (II does not follow)
Answer: Neither I nor II follows
Problem: All cats are animals. Some animals are dogs. Therefore, some cats are dogs. Is the conclusion valid?
Solution: The conclusion is invalid. While all cats are animals and some animals are dogs, there is no information linking cats directly to dogs. The conclusion cannot be logically derived from the given statements.
Answer: Invalid conclusion
Problem: Pointing to a photograph, a man says, “She is the daughter of my grandfather’s only son.” How is the girl related to the man?
Solution: The man’s grandfather’s only son = the man’s father Daughter of the man’s father = the man’s sister Therefore, the girl is the man’s sister.
Answer: Sister
Problem: Find the missing number in the series: 2, 6, 12, 20, ?
Solution: Pattern: Differences are 4, 6, 8 (increasing by 2) Next difference = 10 Next number = 20 + 10 = 30
Alternative pattern: n(n+1) where n starts from 1 1×2=2, 2×3=6, 3×4=12, 4×5=20, 5×6=30
Answer: 30
Problem: If in a certain code, ‘APPLE’ is written as ‘ELPPA’, how is ‘ORANGE’ written in that code?
Solution: The code reverses the letters of the word. APPLE → ELPPA (reversed) Therefore, ORANGE → EGNARO (reversed)
Answer: EGNARO
Passage: “Artificial Intelligence has revolutionized various industries. Machine learning algorithms can now process vast amounts of data to identify patterns and make predictions. However, concerns about job displacement and ethical implications remain significant challenges.”
Question: According to the passage, what are the main concerns about AI?
Answer: Job displacement and ethical implications
Problem: Choose the correct sentence:
a) Neither the students nor the teacher were present b) Neither the students nor the teacher was present c) Neither the students or the teacher was present
Solution: With “neither…nor”, the verb agrees with the subject closer to it. “Teacher” is singular, so “was” is correct.
Answer: b) Neither the students nor the teacher was present
Problem: Find the synonym of “ABUNDANT”:
a) Scarce b) Plentiful c) Limited d) Rare
Solution: Abundant means existing in large quantities, so “Plentiful” is the synonym.
Answer: b) Plentiful
Problem: Find the antonym of “TRANSPARENT”:
a) Clear b) Opaque c) Visible d) Bright
Solution: Transparent means see-through, so “Opaque” (not see-through) is the antonym.
Answer: b) Opaque
Problem: Choose the correct sentence:
a) The data are incorrect b) The data is incorrect c) The datas are incorrect
Solution: “Data” can be singular or plural, but in modern usage, it’s often treated as singular. “The data is incorrect” is more commonly accepted.
Answer: b) The data is incorrect
Problem: Choose the synonym for “Ephemeral”:
a) Eternal b) Transient c) Permanent d) Perpetual
Solution: “Ephemeral” means lasting for a very short time, temporary. “Transient” has the same meaning.
Answer: b) Transient
Passage: “The industrial revolution led to a significant increase in urban populations as people moved to cities in search of work.”
Question: What was one major consequence of the industrial revolution according to the passage?
Solution: According to the passage, a major consequence was a significant increase in urban populations due to migration for work opportunities.
Answer: Significant increase in urban populations as people moved to cities for work
Problem: What is the output?
int main() { int x = 5; printf("%d", x++ + ++x); return 0;}Solution: Note: This exhibits undefined behavior due to sequence point violations. However, in many compilers: x++ uses 5, then x becomes 6 ++x makes x = 7, uses value 7 Output: 5 + 7 = 12
Answer: 12 (compiler-dependent, undefined behavior)
Problem: What is the output?
int main() { int i; for(i=0; i<5; i++) { if(i == 3) continue; printf("%d ", i); } return 0;}Solution: Loop executes: i=0 prints 0, i=1 prints 1, i=2 prints 2, i=3 skips (continue), i=4 prints 4
Answer: 0 1 2 4
Problem: What is the output?
int main() { int arr[] = {1, 2, 3, 4, 5}; int *p = arr; printf("%d %d", *(p+2), arr[3]); return 0;}Solution: p points to arr[0] *(p+2) = arr[2] = 3 arr[3] = 4
Answer: 3 4
Problem: What is the output?
int func(int n) { if(n <= 1) return 1; return n * func(n-1);}printf("%d", func(5));Solution: This calculates factorial: func(5) = 5 × func(4) func(4) = 4 × func(3) func(3) = 3 × func(2) func(2) = 2 × func(1) func(1) = 1 func(2) = 2 × 1 = 2 func(3) = 3 × 2 = 6 func(4) = 4 × 6 = 24 func(5) = 5 × 24 = 120
Answer: 120
Problem: What is the output?
public class Test { public static void main(String[] args) { int x = 10; System.out.println(x++ + ++x); }}Solution: x++ uses 10, then x becomes 11 ++x makes x = 12, uses value 12 Output: 10 + 12 = 22
Answer: 22
Problem: What is the output?
def func(n): if n <= 1: return 1 return n * func(n-2)
print(func(6))Solution: func(6) = 6 × func(4) func(4) = 4 × func(2) func(2) = 2 × func(0) func(0) = 1 func(2) = 2 × 1 = 2 func(4) = 4 × 2 = 8 func(6) = 6 × 8 = 48
Answer: 48
Problem: What is the output?
int main() { int arr[] = {10, 20, 30, 40, 50}; int *ptr = arr + 2; printf("%d", *(ptr+1)); return 0;}Solution: arr+2 points to arr[2] = 30 ptr = arr+2, so ptr points to arr[2] ptr+1 points to arr[3] = 40 *(ptr+1) = 40
Answer: 40
Problem: What is the output?
int main() { char str[] = "HELLO"; printf("%d", strlen(str)); return 0;}Solution: “HELLO” has 5 characters, strlen returns 5
Answer: 5
Problem: What will be the output of the following pseudo code?
Begin Integer x = 10 Integer y = 5 x = x + y y = x - y x = x - y Print x, yEndSolution: This pseudo code swaps the values of x and y without using a temporary variable: Initially: x = 10, y = 5 x = x + y = 10 + 5 = 15 y = x - y = 15 - 5 = 10 x = x - y = 15 - 10 = 5 Final: x = 5, y = 10
Answer: x = 5, y = 10
Problem: Which layer of the OSI model is responsible for end-to-end communication and error-free delivery of data?
a) Network Layer b) Transport Layer c) Session Layer d) Data Link Layer
Solution: The Transport layer (Layer 4) is responsible for end-to-end communication and error-free data delivery. It ensures that data is delivered reliably from source to destination.
Answer: b) Transport Layer
Problem: Find the second largest element in an array.
Solution (C):
int secondLargest(int arr[], int n) { int first = INT_MIN, second = INT_MIN; for (int i = 0; i < n; i++) { if (arr[i] > first) { second = first; first = arr[i]; } else if (arr[i] > second && arr[i] != first) { second = arr[i]; } } return second;}Problem: Reverse an array without using extra space.
Solution (C):
void reverseArray(int arr[], int n) { int start = 0, end = n - 1; while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; }}Problem: Check if a string is a palindrome.
Solution (C):
#include <string.h>#include <ctype.h>
int isPalindrome(char str[]) { int left = 0, right = strlen(str) - 1; while (left < right) { if (tolower(str[left]) != tolower(str[right])) return 0; left++; right--; } return 1;}Problem: Find factorial of a number using recursion.
Solution (C):
int factorial(int n) { if (n == 0 || n == 1) return 1; return n * factorial(n - 1);}Problem: Check if a number is prime.
Solution (C):
#include <math.h>
int isPrime(int n) { if (n <= 1) return 0; if (n <= 3) return 1; if (n % 2 == 0 || n % 3 == 0) return 0;
for (int i = 5; i * i <= n; i += 6) { if (n % i == 0 || n % (i + 2) == 0) return 0; } return 1;}Problem: Print first n Fibonacci numbers.
Solution (C):
void fibonacci(int n) { int a = 0, b = 1, c; if (n >= 1) printf("%d ", a); if (n >= 2) printf("%d ", b); for (int i = 3; i <= n; i++) { c = a + b; printf("%d ", c); a = b; b = c; }}Problem: Rotate an array to the right by k positions.
Solution (C):
void rotateRight(int arr[], int n, int k) { k = k % n; int temp[k]; for (int i = 0; i < k; i++) temp[i] = arr[n - k + i]; for (int i = n - 1; i >= k; i--) arr[i] = arr[i - k]; for (int i = 0; i < k; i++) arr[i] = temp[i];}Problem: Find the maximum sum of contiguous subarray (Kadane’s Algorithm).
Solution (C):
int maxSubarraySum(int arr[], int n) { int maxSoFar = arr[0], maxEndingHere = arr[0]; for (int i = 1; i < n; i++) { maxEndingHere = (maxEndingHere + arr[i] > arr[i]) ? maxEndingHere + arr[i] : arr[i]; maxSoFar = (maxEndingHere > maxSoFar) ? maxEndingHere : maxSoFar; } return maxSoFar;}Problem: Remove duplicates from sorted array.
Solution (C):
int removeDuplicates(int arr[], int n) { if (n == 0) return 0; int j = 0; for (int i = 1; i < n; i++) { if (arr[i] != arr[j]) { j++; arr[j] = arr[i]; } } return j + 1;}Problem: Find two numbers in array that add up to target sum.
Solution (C):
void twoSum(int arr[], int n, int target) { for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (arr[i] + arr[j] == target) { printf("Indices: %d, %d\n", i, j); return; } } } printf("No pair found\n");}Problem: Given an array of integers, write a function to find the maximum product of two integers in the array.
Solution (C):
#include <stdio.h>#include <limits.h>
int maxProduct(int arr[], int n) { if (n < 2) return -1;
int max1 = INT_MIN, max2 = INT_MIN; int min1 = INT_MAX, min2 = INT_MAX;
for (int i = 0; i < n; i++) { if (arr[i] > max1) { max2 = max1; max1 = arr[i]; } else if (arr[i] > max2) { max2 = arr[i]; }
if (arr[i] < min1) { min2 = min1; min1 = arr[i]; } else if (arr[i] < min2) { min2 = arr[i]; } }
int product1 = max1 * max2; int product2 = min1 * min2;
return (product1 > product2) ? product1 : product2;}Explanation: We need to consider both the product of two largest numbers and the product of two smallest numbers (in case of negative numbers).
Problem: Write a program to reverse a linked list.
Solution (C):
#include <stdio.h>#include <stdlib.h>
struct Node { int data; struct Node* next;};
struct Node* reverseList(struct Node* head) { struct Node* prev = NULL; struct Node* current = head; struct Node* next = NULL;
while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; }
return prev;}Explanation: Iteratively reverse the pointers by maintaining three pointers: prev, current, and next.
| Section | Cutoff (Approx) | Notes |
|---|---|---|
| Cognitive Assessment | 60% (12-15/20-25) | Standard cutoff |
| Technical Assessment | 60% (12-15/20-25) | Standard cutoff |
| Coding Test | 1 problem solved | Must solve at least 1 |
| Communication Assessment | 50% (3-5/5-10) | Lower cutoff |
| Overall | 60-65% | 36-45/60-70 marks |
Success Rate: Approximately 15-20% of candidates cleared NLT in 2024
Hiring Volume
Salary Packages
Process Changes
Quantitative Aptitude
Logical Reasoning
Verbal Ability
Technical Assessment
Coding Problems
Based on candidate experiences from 2024 Accenture NLT and interviews:
Common Interview Topics (2024):
Interview Process (2024):
Key Insights:
For detailed interview experiences, visit Accenture Interview Experience page.
Accenture NLT 2024 Paper 1
Complete NLT paper with solutions from December 2024
Accenture NLT 2024 Paper 2
Another 2024 NLT paper with detailed solutions from November 2024
2024 Coding Problems
All coding problems from 2024 NLT with solutions
Previous Year Papers (2020-2025)
Complete collection of Accenture previous year question papers with solutions
Based on 2024 Exam Pattern:
Practice 2024 papers to understand Accenture NLT pattern and prepare for current year exams! Download the papers above and start practicing today.