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TCS Coding Questions 2022 Day1 Slot 1
Coding Question 1 for 2022 (September slot)
In this article, we will discuss about the TCS Coding Question which is asked in the TCS placement test. This type of Coding Questions will help you to crack your upcoming TCS exam as well as during your inteview process.

TCS Coding Question Day 1 Slot 1 – Question 1
Problem Statement – An automobile company manufactures both a two wheeler (TW) and a four wheeler (FW). A company manager wants to make the production of both types of vehicle according to the given data below:
- 1st data, Total number of vehicle (two-wheeler + four-wheeler)=v
- 2nd data, Total number of wheels = W
The task is to find how many two-wheelers as well as four-wheelers need to manufacture as per the given data.
Example :
Input :
- 200 -> Value of V
- 540 -> Value of W
Output :
- TW =130 FW=70
Explanation:
130+70 = 200 vehicles
(70*4)+(130*2)= 540 wheels
Constraints :
- 2<=W
- W%2=0
- V<W
Print “INVALID INPUT” , if inputs did not meet the constraints.
The input format for testing
The candidate has to write the code to accept two positive numbers separated by a new line.
- First Input line – Accept value of V.
- Second Input line- Accept value for W.
The output format for testing
- Written program code should generate two outputs, each separated by a single space character(see the example)
- Additional messages in the output will result in the failure of test case
#include <bits/stdc++.h>
using namespace std;
int main ()
{
int v, w;
cin >> v >> w;
float x = ((4 * v) - w) / 2;
if ((w & 1) || w < 2 || w <= v)
{
cout << "INVALID INPUT";
return 0;
}
cout << "TW=" << x << " " << "FW=" << v - x;
}
import java.util.*;
public class Solution
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int v=sc.nextInt();
int w=sc.nextInt();
float res=((4*v)-w)/2;
if(w>=2 && (w%2==0) && v<w )
System.out.println("TW= "+(int)(res)+" FW= "+(int)(v-res));
else
System.out.println("INVALID INPUT");
}
}
v=int(input())
w=int(input())
if (w&1)==1 or w<2 or w<=v:
print("INVALID INPUT")
else:
x=((4*v) -w)//2
print("TW={0} FW={1}".format(x,v-x))
Good methods
v=int(input(“Enter no of vehicles:”))
w=int(input(“Enter no of wheels:”))
if w=w:
print(“INVALID INPUT”)
else:
tw=((4*v)-w)//2
fw=v-tw
print(f”tw={tw} fw={fw}”)
v = int(input())
w = int(input())
# x+y = v
# 2x +4y = w
if (2<w) and (w%2 == 0) and (v<w):
y = (w-v)//2
x = v – y
print(f"TW={x} FW={y}")
else:
print("INVALID INPUT")
correct c code:
#include
int main() {
int FW,TW,v,w;
printf(“Enter values for v : “);
scanf(“%d”,&v);
printf(“Enter values for w : “);
scanf(“%d”,&w);
if(2<=w && (w%2==0) && v=0){
printf(“No of TW = %d\nNo of FW = %d”,TW,FW);
}
else{
printf(“INVALID INPUT”);
}
}
else{
printf(“INVALID INPUT”);
}
}
correct code:
// Online C compiler to run C program online
#include
int main() {
int FW,TW,v,w;
printf(“Enter values for v : “);
scanf(“%d”,&v);
printf(“Enter values for w : “);
scanf(“%d”,&w);
if(2<=w && (w%2==0) && v=0){
printf(“No of TW = %d\nNo of FW = %d”,TW,FW);
}
else{
printf(“INVALID INPUT”);
}
}
else{
printf(“INVALID INPUT”);
}
}