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# Monthly Budget CodeChef Solution

## Problem – Monthly Budget CodeChef Solution

Akshat has X rupees to spend in the current month. His daily expenditure is Y rupees, i.e., he spends Y rupees each day.

Given that the current month has 30 days, find out if Akshat has enough money to meet his daily expenditures for this month.

### Input Format

• The first line will contain T – the number of test cases. Then the test cases follow.
• The first and only line of each test case contains two integers XY – the amount of money Akshat has for the current month and his daily expenditure respectively.

### Output Format

For each test case, output `YES` if Akshat has enough money to meet his daily expenditure for 3030 days of the month, otherwise output `NO`.

You may print each character of `YES` and `NO` in uppercase or lowercase (for example, `yes``yEs``Yes` will be considered identical).

• 1≤T≤100
• 1≤X,Y≤10^5

### Sample 1:

``````Input: 3
1000 10
250 50
1500 50
Output: YES
NO
YES``````

### Explanation:

Test Case 1: Akshat has 1000 rupees and he wants to spend 30×10=300 rupees in the entire month. Therefore, he has enough money for the entire month.

Test Case 2: Akshat has 250 rupees and he wants to spend 30×50=1500 rupees in the entire month. Therefore, he does not have enough money for the entire month.

Test Case 3: Akshat has 1500 rupees and he wants to spend 30×50=1500 rupees in the entire month. Therefore, he has enough money for the entire month.

### Monthly Budget CodeChef Solution in Java

``````import java.util.*;
import java.lang.*;
import java.io.*;

class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i=0;i<t;i++){
int x = sc.nextInt();
int y = sc.nextInt();
int z = y*30;
if(x>=z){
System.out.println("yes");
}else{
System.out.println("no");
}
}
}
}
``````

### Monthly Budget CodeChef Solution in C++17

``````#include <iostream>
using namespace std;

int main() {
int t;
std::cin >> t;
while(t>0){
int a,b;
std::cin >> a>>b;
if((b*30)<=a){
std::cout << "YES" << std::endl;
}
else{
std::cout << "NO" << std::endl;
}
t--;
}
return 0;
}
``````

### Monthly Budget CodeChef Solution in Pyth 3

``````a=int(input())
for i in range(a):
b,c=list(map(int,input().split()))
k=c*30
if(k<=b):
print("YES")
else:
print("NO")

``````
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