Question

Q.3.6 What does the following code output? int x = 8; int y = 2; while(x > 8) { X- y+=2; } System.out.println(y); Q.3.7 What does the following code output? System.out.println((2 + 3 - 5) * (3/3) + (5+5)); Q.3.8 What does the following code output? if((2 == 2) && (3 != 3) && ((5 + 1)==6){ System.out.println(true); else { System.out.println(false);

PLEASE SOLVE THESE QUESTIONS

Answer 1

Q-3.6: The output of the above program is 2
1. public class Main
{
2. public static void main(String[] args)
{
3. int x=8;
4. int y=2;
5. while(x>8){
6. x--; //decrement x
7. y+=2; //increment 2 }   
8. System.out.println(y); // print the value of y   
}}
In line 3 and line 4 declaring and initializing the variables x and y  i.e. x=8, y=2.
In line 5 we are checking the condition, whether x=8 is greater than 8 (i.e. 8>8) it is false so no change is made to the values of x and y( in line-6 and line-7) therefore come out of the loop and execute the statement in line-8 that is printing the value of y that is 2. Hence the output of the above program is 2.
The screenshot of the program and the output is shown below:

{ 8 1 //Question-3.6 2 public class Main 3- { 4 public static void main(String[] args) 5 6 //variable declaration and intialization 7 int x=8; int y=2; 9 10 while(x>8) // check the condition if x is greater than 8 11 - 12 X--; //decrement x y+=2; // Add 2 to the value of y and store the result in y 14 15 System.out.println(y); // print the value of y 16 } 17 } نها 13 نہا

Output:

2

Q-3.7. The output of the given question is 10.

System.out.println((2+3-5)*(3/3)+(5+5));

If we have more than one operation in a single line, then we should follow the below rules:-

The expression inside the brackets '()' is evaluated first and then follow the below table.

Operator	Associativity
++ -- !	Right to left(higher precedence)
* / %	Left to right
+ -	Left to right
> >= < <=	Left to right
== !=	Left to right
&&	Left to right
||	Left to right
= += -= *= /= %=	Right to left (least precedence)

Following the above rules the given expression is solved as follows:
(2+3-5)*(3/3)+(5+5) //Solve brackets first
= 0 * 1 + 10 // Next multiplication has higher precedence
= 0 + 10 // Finally addition is performed
= 10  
Therefore the given line prints the value 10.

Screenshot of the program along with the output is shown below:

1 //Question-3.7 2 public class Main 3- { 4. public static void main(String[] args) { 5 System.out.println((2+3-5)*(3/3)+(5+5)); 6 7 } 8

Output:

10

Q-3.8. The output of the given code is false.
1. public class Main{
2. public static void main(String[] args){
3. if((2==2)&&(3!=3)&&((5+1)==6)){
4.   System.out.println(true);
5. } else{   
6.   System.out.println(false); }
}}

In line-3 we are using the logical and operator (&&) for checking the if condition. The logical and (&&) evaluates to true only if both (or all the) operands are true, if any one of them is false, the condition becomes false.
Lets check the first operand 2==2 it's true, next check whether 3!=3 it is false, so the next operand is not checked since the if condition became false. Hence the body of the if statement(line-4) is not executed because the if condition is false. The body of the if statement (line-4) is executed only if the condition inside the brackets is true.
So come out of the loop and execute the else statement in line 6. Therefore, it prints false as the output.

Screenshot of the program along with the output is shown below:

{ 1 //Question-3.8 2 public class Main 3-{ 4 public static void main(String[] args) 5. { 6 if((2==2)&&(3!=3)&&((5+1)==6)) // if condition is true print the statement of if 7 System.out.println(true); 9 } else{ // if the condition is false print this statement 11 System.out.println(false); 12 13 } 14 } 15 8 10 }

Output:

false