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

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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*r²*(300+r)/100³

=2000*10²*(300+10)/100³

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: ?

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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: 25
Day 2 exercise duration: 26
Day 3 exercise duration: 23
Day 4 exercise duration: 15
Day 5 exercise duration: 14
Day 6 exercise duration: 38
Day 7 exercise duration: 44
Result: 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 20
Output: 11
Reason: 0,1,2,3,4,5,6,7,8,9,11
These 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.

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