Very Short Answer Questions

Java statement is:

Math.pow(a, b) + Math.cbrt(c);

(a) Constructor

(b) import

(a) Character.isLetterOrDigit()

(b) Math.random()

void check(long pno) 
{ String s = String.valueOf(pno); 
if(s.indexOf("945") >= 0) 
    System.out.println(pno); 
}                                    

void count(String s[]) 
{ 
    int i, l = s.length, c = 0; 
    for(i = 0; i < l; i++) 
    { 
        if(s[i].endsWith("KUMAR")) 
            c++; 
    }
    System.out.println(c); 
}                   

(a) "COMPASSION".substring(3, 7);

(b) "COMPASSION".substring(3);

12

1. The loop starts with k = 6.

2. The loop runs while k <= 6 * 4, i.e., k <= 24.

3. In each iteration, k increases by 6.

4. The loop checks if k % 4 == 0. If true, it breaks the loop.

Iterations:

• First: k = 6, 6 % 4 = 2 → Not 0 → Loop continues
• Second: k = 12, 12 % 4 = 0 → Condition true → Loop breaks

5. The current value of k is 12. So, the output is: 12.

char arr[]={'4', '&', 'a', 'w', 'd'}; 
int i; 
for(i=0;i<arr.length;i++) 
{ 
if(Character.isLetter(arr)) //Error 1
System.out.println(arr); // Error 2
}

1. In Character.isLetter(arr), arr is the entire array, but the method expects a single character (char), not an array.

2. In System.out.println(arr), arr prints the whole array, not the individual character.

The corrected code is as follows:

char arr[] = {'4', '&', 'a', 'w', 'd'}; 
int i; 
for(i = 0; i < arr.length; i++) 
{ 
    if(Character.isLetter(arr[i]))
        System.out.println(arr[i]);
}

(a) Implicit type casting

(b) Explicit type casting

(a) double cubeRoot = Math.cbrt(-343);

(b) int pos = s.lastIndexOf('@');

5
4

1. The variable lives is initialized to 5.
2. The statement System.out.println(lives--); first prints the current value of lives, which is 5, and then decreases the value of lives by 1 because lives-- is the postfix decrement operator.
3. After this statement, the value of lives becomes 4.
4. The next statement System.out.println(lives); prints the updated value of lives, which is 4.

k = 10;
do {
    System.out.println(k * 2);
    k--;
} while (k >= -1);
System.out.println(k * 4);

5
57
74

1. Array x = [2, 45, 7, 67, 12, 3].

2. Initial values: i = 0, t = 5.

3. Loop condition: i < 3 (loop runs for i = 0, 1, 2)

Iteration 1:

• i = 0, t = 5
• x[0] + x[5] = 2 + 3 = 5
• Print: 5
• After iteration: i = 1, t = 4

Iteration 2:

• i = 1, t = 4
• x[1] + x[4] = 45 + 12 = 57
• Print: 57
• After iteration: i = 2, t = 3

Iteration 3:

• i = 2, t = 3
• x[2] + x[3] = 7 + 67 = 74
• Print: 74
• After iteration: i = 3, t = 2

Loop ends because i is now 3 (not less than 3).

4. So the output is:

5
57
74

(a) String subjects[] = {"Maths", "Science", "English"};

(b) int marks[][] = new int[40][3]; or int marks[][] = new int[3][40];

The type of error is logical error.

The correct statement is:

double power = Math.pow(n, 2.0 / 3);

Explanation — In Java, 2/3 uses integer division, which equals 0. So the code computes Math.pow(n, 0), and any number to the power 0 is 1.0, regardless of n. Make the exponent a floating-point value—e.g., 2.0/3 or 2/3.0 or 2.0/3.0—so it evaluates to 0.666… and correctly computes $n^{2/3}$.

The corrected code is as follows:

void solve(int n) 
{ double power = Math.pow(n, 2.0 / 3);
System.out.println(power); 
}

(a) No, the program segment will not execute successfully.

(b) The error is a syntax error because the variable k is not known outside the loop. The variable k is created inside the for loop and only exists there. When the program tries to use k after the loop, it does not recognize it, so the program gives an error.

(c) To fix the error, declare k before the loop:

int k;
for (k = 1; k <= 5; k++)
System.out.println(k);
System.out.println(k);

(a) 5

(b) 2

Explanation

(a) In linear search, we check each element of the array one by one starting from the first element. We compare each element with the value we want to find, which is 56 in this case. The search will check 12, then 35, then 40, then 22, and finally 56. Since 56 is the fifth element, it takes 5 iterations to find it.

(b) If the array is sorted in descending order, it will look like this: {70, 56, 40, 35, 22, 12, 9}. Now, when we do linear search, we again start from the first element and check each element one by one. The search will check 70 first and then 56, which is the second element. So, it takes only 2 iterations to find 56 in the sorted array.

Step 1: Multiplication and Modulus have same precedence, so we evaluate left to right.

• 3 * 6 = 18
• 18 % 5 = 3 (because 18 ÷ 5 leaves remainder 3)

Expression now: 3 * 4 / 2 * 7 % 5

Step 2: Continue left to right (multiplication, division, modulus have same precedence):

• 3 * 4 = 12
• 12 / 2 = 6
• 6 * 7 = 42
• 42 % 5 = 2 (since 42 ÷ 5 leaves remainder 2)

Final result: 2

double norm() 
{ int x[][]={{1,5,6},{4,2,9},{6,1,3}}; 
int r, c, sum=0; 
for(r=0;r<3;r++) 
{ for(c=0;c<3;c++) 
   sum=sum+x[r][c] * x[r][c];
} 
return Math.sqrt(sum);
}

2
3
3

• When r = 1:

  • Inner loop runs for c from 1 to less than 1 → No iterations (because c < r is false at c=1)
  • So no output for r=1.

• When r = 2:

  • Inner loop runs for c = 1 (since 1 < 2)
  • Prints 2 once.

• When r = 3:

  • Inner loop runs for c = 1 and c = 2 (since c < 3)
  • Prints 3 twice.

The given code snippet is missing the closing curly brace } for the method encode().

The corrected code is as follows:

void encode() 
{ 
    String s= "DICe"; 
    char ch; 
    for(i=0;i<4;i++) 
    {  ch=s.charAt(i); 
        if("AEIOUaeiou".indexOf(ch)>=0) 
        System.out.println("@"); 
else 
System.out.println(ch); 
}  
}

D
@
C
@

1. void encode() — Declares a method named encode that does not return any value (void).

2. String s = "DICe"; — Inside the method, a String variable s is initialized with the value "DICe".

3. char ch; — Declares a variable ch of type char which will be used to store each character of the string during the loop.

4. for(i=0;i<4;i++) — Starts a for loop where the integer variable i begins at 0.

• The loop continues as long as i is less than 4.
• After each iteration, i is incremented by 1.

5. ch = s.charAt(i); — Inside the loop, the character at position i in the string s is extracted using charAt(i) and stored in the variable ch.

6. if("AEIOUaeiou".indexOf(ch) >= 0) — This if statement checks whether the character ch is a vowel.

• The string "AEIOUaeiou" contains all uppercase and lowercase vowels.
• The method indexOf(ch) returns the position of ch in this string if found; otherwise, it returns -1.
• If the return value is 0 or greater (meaning ch is found in the string), the condition evaluates to true.

7. System.out.println("@"); — If the condition is true (i.e., ch is a vowel), this line prints the symbol "@". 8. System.out.println(ch); — If the character is not a vowel, the original character ch is printed.

(a) The array must be sorted.

(b) Selection sort.

(a) The Scanner class method to accept a character is <Scanner object>.next().charAt(0).

(b) The jump statement which stops the execution of a construct is break.

The loop exectes 3 times.

Output

10
14
18

Explanation

The loop starts from k = 10 and runs as long as k <= 20, increasing k by 4 each time. So, the values of k will be: 10, 14, and 18. When k becomes 22, the condition k <= 20 becomes false, so the loop stops. Hence, the loop runs 3 times.