Else–if ladder in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Else–if Ladder with Multiple Conditions

What is the output of the following Java code?

Java

public class Test {
    public static void main(String[] args) {
        int score = 75;
        if (score >= 90) {
            System.out.println("Grade A");
        } else if (score >= 80) {
            System.out.println("Grade B");
        } else if (score >= 70) {
            System.out.println("Grade C");
        } else {
            System.out.println("Grade F");
        }
    }
}

A Grade C

B Grade B

C Grade A

D Grade F

💡 Hint

Check which condition the score 75 satisfies first in the else-if ladder.

The score 75 is not >= 90 or >= 80, but it is >= 70, so the else-if block for Grade C runs.

❓ Predict Output

intermediate

Output When No Conditions Match in Else–if Ladder

What will be printed when the input value is 45 in this Java program?

Java

public class Test {
    public static void main(String[] args) {
        int value = 45;
        if (value > 100) {
            System.out.println("High");
        } else if (value > 50) {
            System.out.println("Medium");
        } else if (value > 60) {
            System.out.println("Above Average");
        } else {
            System.out.println("Low");
        }
    }
}

A High

B Medium

C Low

D Above Average

💡 Hint

Check all conditions carefully and see which one matches 45.

45 is not greater than 100, 50, or 60, so the else block runs printing "Low".

🔧 Debug

advanced

Identify the Error in Else–if Ladder Syntax

What error does this Java code produce?

Java

public class Test {
    public static void main(String[] args) {
        int num = 10;
        if (num > 0) {
            System.out.println("Positive");
        } else if (num == 0) {
            System.out.println("Zero");
        } else {
            System.out.println("Negative");
        }
    }
}

A RuntimeException: Null pointer exception

B SyntaxError: Missing parentheses in else-if condition

C No error, prints "Positive"

D Logical error: prints wrong output

💡 Hint

Check the syntax of the else-if statement carefully.

In Java, else-if conditions must be enclosed in parentheses. Missing parentheses cause a syntax error.

🚀 Application

advanced

Determine the Final Value After Else–if Ladder Execution

What is the value of variable 'result' after running this Java code?

Java

public class Test {
    public static void main(String[] args) {
        int x = 15;
        String result;
        if (x < 10) {
            result = "Less";
        } else if (x < 20) {
            result = "Between";
        } else if (x < 30) {
            result = "More";
        } else {
            result = "Unknown";
        }
        System.out.println(result);
    }
}

A Between

B Less

C More

D Unknown

💡 Hint

Check which condition the value 15 satisfies first.

15 is not less than 10 but is less than 20, so 'Between' is assigned and printed.

🧠 Conceptual

expert

Understanding Else–if Ladder Execution Flow

Consider the following Java code snippet. Which statement best describes the execution flow when x = 25?

Java

int x = 25;
if (x > 30) {
    System.out.println("Above 30");
} else if (x > 20) {
    System.out.println("Above 20");
} else if (x > 10) {
    System.out.println("Above 10");
} else {
    System.out.println("10 or below");
}

A Prints "10 or below" because none of the conditions match

B Prints "Above 30" because it checks all conditions

C Prints "Above 10" because it matches the last else-if

D Prints "Above 20" and skips remaining else-if blocks

💡 Hint

Remember that else-if ladder stops checking after the first true condition.

Since x=25 is not >30 but is >20, it prints "Above 20" and skips the rest.