TCS NQT 2025 Pattern Research | Major Change, All Shift Evaluation Recently Asked Questions, Cut Off

TCS NQT 2025 Pattern Research | Major Change, All Shift Evaluation Recently Asked Questions, Cut Off. In this blog, we will explore the latest TCS NQT 2025 pattern, as well as important themes and frequently asked questions.

TCS NQT 2025 Pattern Research: Exam Check-in Process

Section 1: Quantitative Aptitude

Also Read: The Evolution and Future of Software Development

Section 2: Verbal Ability

Section 3: Logical Reasoning

Section 4: Advanced Quantitative

Section 5: Coding

• TCS has set particular time limitations for each coding section within this 90-minute period.
• In both questions, there were seven test cases.

TCS NQT 2025 Cut-Off

Understanding the cut-off requirement is important for determining your preparation for the exam. The cut-off changes annually based on criteria such as the number of candidates and the difficulty level. However the predicted cut-off date for TCS NQT 2024 is April 2024:

TCS NQT 2025 Role wise Cut Off:

Note: These are estimated figures. The actual cut-off can vary slightly depending on the batch.

TCS NQT 2025 Expected Result:

Q1. 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. ​

Q2. 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?

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

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. 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

Q5. 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.

Q6. NPRT : GHIJ :: LXTP: ?

Ans: FLJH

Q7. 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.

Q8. Find the odd one

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

Ans: LIJ

TCS NQT 3rd October 2024 CODING QUESTIONS 

Q1. (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.

Q2. 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)

​