# TCS NQT 2025 Pattern Analysis | Big Change, All Shift Analysis, Latest Asked Questions, Cut Off

TCS NQT 2025 Pattern Analysis , In this blog we are going to discuss about TCS NQT 2025 Latest pattern and Important topics and asked questions.

**TCS NQT 2025 Pattern Analysis: ****Exam Check-in Process**

Step | Details |

Check-in Time | 7:30 AM (Arrived by 7:00 AM, waited 30 minutes) |

Required Documents | Admit card |

ID Proof Checks | Two ID checks |

Floor Information | Floors clearly marked (e.g., 203-400 on Floor 1), assistance available |

Validation Process | Face and thumbprint validation; provided lab and desk number |

COVID Declaration Form | Must include a photo (can be pasted at home) |

Login Assistance | Help available for login and password |

Exam Start Time | 9:10 AM |

**If you’re looking to streamline your preparation and avoid the hassle of juggling multiple resources, come join us for a comprehensive solution that brings everything you need together in one place!**

**Crack Your TCS NQT With Us**!

__TCS NQT 2025 Pattern Analysis: Topic wise Analysis__

**Section 1: Quantitative Aptitude**

Topic | Number of Questions |

Averages | 2 |

Ages | 2 |

Percentages | 2 |

Speed and Distance | 2 |

Statistics | 2 |

Number System | 1 |

Simplification | 1 |

LCM | 1 |

Time Speed Distance | 1 |

SI & CI | 1 |

Mean | 1 |

Profit & Loss | 1 |

Time and Work | 1 |

Weights | 1 |

Volume | 1 |

Ratio Partnership | 1 |

Mensuration | 1 |

TITA Question | 1 |

**Section 2: Verbal Ability**

Topic | Number of Questions |

Reading Comprehension (2 Passages) | 7-8 |

Synonyms | 1 |

Grammar-based Questions | 4-5 |

Error Correction | 2-3 |

Meaning Replacement | 2 |

Sentence Arrangement | 2-3 |

******Para jumbles will be TITA based.

**Section 3: Logical Reasoning**

Topic | Number of Questions |

Syllogism | 2-6 |

Seating Arrangement | 2-4 |

Chinese Coding | 1-3 |

Pair Identification (Vowel/Consonant Order) | 2 |

Pattern Recognition (Images) | 2 |

Coding/Decoding | 2 |

Odd Man Out | 2 |

Direction | 2 |

Data Sufficiency | 2 |

Blood Relation | 1 |

Circular Seating Arrangement (TITA) | 1 |

**Section 4: Advanced Quantitative**

Topic | Number of Questions |

TITA Questions | 5-6 |

Coding Decoding | 2 |

Eligibility | 2 |

Time Speed Distance | 1 |

Profit & Loss | 1 |

Data Interpretation | 1 |

Syllogism | 1 |

Venn Diagram | 1 |

Quadratic Equation | 1 |

**Most of the TITA questions are asked in this section.

**Section 5: Coding**

Question Number | Difficulty Level | Time Allotted |

1st Question | Easy | 35 minutes |

2nd Question | Medium | 55 minutes |

- TCS has allocated specific time limits for each coding section within the 90-minute duration this time.
- In both the questions there were 7 test cases.

**TCS NQT 2025 Cut Off**

Understanding the **cut-off criteria** is crucial to assess your readiness for the exam. The cut-off varies yearly based on factors such as the number of applicants and difficulty level. However, this is the expected cut off for TCS NQT 2024 on April 2024:

Sections | Expected Cut Off |

Numerical Ability | 8-10 Q |

Logical Reasoning | 10-12 Q |

Verbal Ability | 10-15 Q |

Advanced Quantitative + Reasoning Ability | 6-7 Q |

Advanced Coding | *** |

*****TCS NQT 2025 Role wise Cut Off:**

Role | Cut off |

Ninja | Only Foundation section cleared |

Digital | Foundation + Advanced Aptitude+ 1 coding question |

Prime | Foundation + Advanced Aptitude+ 2 coding question |

**Note:** These are approximate numbers. The actual cut-off may vary slightly based on the batch.

**TCS NQT 2025 Expected Result:**

Exam Details | Time Taken |

Written Exam to Interview | After the written exam TCS generally takes 20 days to 1 month to declare the result. (This may vary dependind upon wheather the hiring is On campus or Off campus) |

Interview to Final Selection | Following the interview TCS generally takes 15-20 days to declare the final selection result. |

**Q1. LCM of 18, 45, 408, 255.**

LCM(18, 45, 408, 255.) = 6120

Steps:

Prime factorization of the numbers:

18 = 2 * 3 * 3

45 = 3 * 3 * 5

408 = 2 * 2 * 2 * 3 * 17

255 = 3 * 5 * 17

LCM(18, 45, 408, 255.)

=2×2×2×3×3×5× 17

= 6120

**Q2. Difference between the SI and CI of 20000 with 10% rate for 3 years.**

To find the difference between Simple Interest (SI) and Compound Interest (CI) for a principal amount of ₹20,000 at an interest rate of 10% for 3 years, we need to calculate both SI and CI.

(CT-SI)_{3= }P*r^{2}*(300+r)/100^{3}

=2000*10^{2}*(300+10)/100^{3}

620

The difference between the Compound Interest (CI) and the Simple Interest (SI) is approximately ₹620.

**Q3. Sum of 2760 with 5% CI rate for 5 years.**

A = 𝑃×(1+𝑅/100)^{𝑇}

Where:

A = Final amount (sum)

P = Principal amount = ₹2,760

R = Rate of interest = 5%

T = Time = 5 years

Let’s calculate the final amount

𝐴=2760*(1+5/100)^{5}

The sum after 5 years with a compound interest rate of 5% on ₹2,760 will be approximately ₹3,522.54.

**Q4. A and B invested 20000 and 40000 . Total profit is 15000. A get his share plus salary 7000. What is the amount of A’s salary.**

Ratio of A’s investment to B’s investment =20000:40000=1:2

Calculate A’s share in the total profit: The total profit is ₹15,000, so the total ratio is 1 + 2 = 3 parts.

A’s profit share =1/3×15000=5000

Determine A’s salary: A receives ₹7,000 in total (his profit share plus salary). Therefore, his salary is:

A’s salary=7,000−5,000=2,000

Final Answer:

A’s salary is ₹2,000.

**Q5. A train travel a distance with 70kmph for 1 hour. Then travel a distance with pkm/hr for 1 and 1/2 hours. If the avg speed is 64km/hr then the value of ‘p’?**

To find the value of 𝑝, we can use the formula for average speed. The average speed is defined as:

Average Speed = Total Distance/Total Time

Step 1: Calculate the distance traveled by the train at 70 km/h for 1 hour.

Distance 1 =Speed×Time=70km/h×1h=70km

Step 2: Calculate the distance traveled by the train at

Distance 2 =p km/h×1.5h=1.5p km

Step 3: Calculate the total distance and total time.

Total Distance=Distance 1 +Distance 2 =70+1.5pkm

Total Time=1h+1.5h=2.5h

Step 4: Set up the equation using the average speed.

Given that the average speed is 64 km/h, we can set up the equation as follows:

64= (70+1.5p)/ 2.5

Step 5: Solve for

P is 60 km/h.

**Q6. Case 1 where mode is ‘M’. Median and mean is 12 and 7. Case 2 where mode is ‘N’ and Median and mean is 22 and 18. Then what is the mean of M, N?**

In Case 1 | In Case 2 |

Median = 12 Mean = 7 Mode = M Mode = 3Median-2Mean M=36-14 M=22 | Median = 22 Mean = 18 Mode = N Mode = 3Median-2Mean N=66-36 N=30 |

Mean of M and N= (22+30)/2=26

**Q7. Find the odd one**

**JDP, KFM, LIJ, MJG, NLD, ONA.**

**Q8. NPRT : GHIJ :: LXTP: ?**

**For Video solution: Click Here!**

**TCS NQT 3rd October 2024 CODING QUESTIONS **

**Q1. Find the Total minutes of exercise done and it’s average for a week**.

**Input:Day 1 exercise duration: 25Day 2 exercise duration: 26Day 3 exercise duration: 23Day 4 exercise duration: 15Day 5 exercise duration: 14Day 6 exercise duration: 38Day 7 exercise duration: 44Result: 185 26.4**

**Solution –**

**C++ code**

#include<bits/stdc++.h>

Using namespace std;

Int main() {

Int duration, sum = 0;

For(int I = 0; I < 7; i++) { Cout << “Day “ << i+1 << “ exercise duration: “; Cin >> duration;

Sum += duration;

}

Double avg = static_cast(sum) / 7;

Cout << “\nTotal minutes: “ << sum;

Cout << “\nAverage minutes per day: “ << avg;

Return 0;

}

**Java Code**

import java.util.Scanner;

Public class Main {

Public static void main(String[] args) {

Scanner scanner = new Scanner(System.in);

Int sum = 0, duration;

For (int I = 0; I < 7; i++) {

System.out.print(“Day “ + (I + 1) + “ exercise duration: “);

Duration = scanner.nextInt();

Sum += duration;

}

Double avg = (double) sum / 7;

System.out.println(“\nTotal minutes: “ + sum);

System.out.println(“Average minutes per day: “ + avg);

Scanner.close();

}

}

If the loop runs for n days instead of a fixed 7 days:

The time complexity is O(n) due to the loop running n times, while the space complexity is O(1) as only constant memory is used.

For Week : **TC is O(1)SC is O(1)**

**Q2. (print the total number of palindrome between the given range m and n,0<=m,n<=1000)For example input1 (lowest range =0 and Highest range =20)Input: 0 20Output: 11Reason: 0,1,2,3,4,5,6,7,8,9,11These numbers are palindrome**

**Solution –**

**C++ Code**

#include<bits/stdc++.h>

Using namespace std;

Bool is_palindrome(int n) {

Int original = n, reversed = 0;

While (n > 0) {

Reversed = reversed * 10 + (n % 10);

N /= 10;

}

Return original == reversed;

}

Int main() {

Int m, n, count = 0;

Cout << “Enter the range of m and n: “; Cin >> m >> n;

For (int I = m; I <= n; i++) {

If (is_palindrome(i)) count++;

}

Cout << “Number of palindromes: “ << count;

Return 0;

}

**Java Code**

import java.util.Scanner;

Public class Main {

Public static boolean isPalindrome(int n) {

Int original = n, reversed = 0;

While (n > 0) {

Reversed = reversed * 10 + (n % 10);

N /= 10;

}

Return original == reversed;

}

Public static void main(String[] args) {

Scanner scanner = new Scanner(System.in);

System.out.print(“Enter the range of m and n: “);

Int m = scanner.nextInt();

Int n = scanner.nextInt();

Int count = 0;

For (int I = m; I <= n; i++) {

If (isPalindrome(i)) {

Count++;

}

}

System.out.println(“Number of palindromes: “ + count);

Scanner.close();

}

}

The time complexity is O(k * log n), where k is the range size and log n is the number of digits in the largest number, while the space complexity is O(1) since no additional data structures are used.