If statement in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Java if statement code?

Consider the following Java code snippet. What will it print when run?

Java

int x = 10;
if (x > 5) {
    System.out.println("Greater than 5");
} else {
    System.out.println("5 or less");
}

A Compilation error

B 5 or less

C Greater than 5

D No output

💡 Hint

Check the condition x > 5 and what it means for x = 10.

Since x is 10, which is greater than 5, the if condition is true, so it prints "Greater than 5".

❓ Predict Output

intermediate

What will this nested if-else print?

Look at this Java code. What is the output?

Java

int a = 3;
int b = 7;
if (a > 5) {
    System.out.println("A is big");
} else if (b > 5) {
    System.out.println("B is big");
} else {
    System.out.println("Neither is big");
}

A Neither is big

B B is big

C A is big

D Compilation error

💡 Hint

Check the values of a and b and the order of conditions.

a is 3 (not > 5), so first if is false. b is 7 (> 5), so else if is true, printing "B is big".

❓ Predict Output

advanced

What is the output of this if statement with boolean logic?

What does this Java code print?

Java

boolean flag = false;
int num = 8;
if (!flag && num < 10) {
    System.out.println("Condition met");
} else {
    System.out.println("Condition not met");
}

A No output

B Condition not met

C Compilation error

D Condition met

💡 Hint

Remember !flag means flag is false, so !flag is true.

flag is false, so !flag is true. num is 8, which is less than 10. Both conditions true, so prints "Condition met".

❓ Predict Output

advanced

What error does this Java if statement code cause?

What happens when you run this code?

Java

int x = 5;
if (x == 5) {
    System.out.println("x is five");
}

A SyntaxError: missing semicolon

B Prints: x is five

C RuntimeException

D No output

💡 Hint

Check the line endings carefully.

The first line is missing a semicolon at the end, causing a compilation syntax error.

❓ Predict Output

expert

What is the value of variable result after this if-else code?

After running this Java code, what is the value of result?

Java

int a = 4;
int b = 9;
int result;
if (a > b) {
    result = a - b;
} else if (a == b) {
    result = a + b;
} else {
    result = b - a;
}

A 5

B -5

C 13

D 0

💡 Hint

Compare a and b and see which condition is true.

a is 4, b is 9. a > b is false, a == b is false, so else runs: result = b - a = 9 - 4 = 5.

Practice

1. What does an if statement do in Java?

easy

A. It stores data in a variable.

B. It repeats code multiple times.

C. It creates a new class.

D. It runs code only if a condition is true.

Solution

1. Step 1: Understand the purpose of an if statement

   An if statement checks a condition and decides whether to run some code based on that condition.

2. Step 2: Match the description to the options

   It runs code only if a condition is true. correctly says it runs code only if the condition is true, which matches the if statement's behavior.

3. Final Answer:

   It runs code only if a condition is true. -> Option D

4. Quick Check:

   If statement = runs code if true [OK]

Hint: If checks true condition to run code [OK]

Common Mistakes:

• Confusing if with loops
• Thinking if stores data
• Mixing if with class creation

2. Which of these is the correct syntax for an if statement in Java?

easy

A. if (condition) System.out.println("Yes");

B. if condition { System.out.println("Yes"); }

C. if condition: System.out.println("Yes");

D. if (condition) then { System.out.println("Yes"); }

Solution

1. Step 1: Recall Java if statement syntax

   Java requires parentheses around the condition and braces are optional if only one statement follows.

2. Step 2: Check each option

   if (condition) System.out.println("Yes"); uses parentheses and a single statement without braces, which is valid syntax.

3. Final Answer:

   if (condition) System.out.println("Yes"); -> Option A

4. Quick Check:

   Parentheses needed around condition [OK]

Hint: Use parentheses around condition in if [OK]

Common Mistakes:

• Omitting parentheses around condition
• Using colons instead of braces
• Adding 'then' keyword

3. What will be the output of this code?

int x = 10;
if (x > 5) {
  System.out.println("Big number");
} else {
  System.out.println("Small number");
}

medium

A. Small number

B. Big number

C. No output

D. Syntax error

Solution

1. Step 1: Evaluate the condition x > 5

   Since x is 10, the condition 10 > 5 is true.

2. Step 2: Determine which block runs

   The if block runs and prints "Big number" because the condition is true.

3. Final Answer:

   Big number -> Option B

4. Quick Check:

   10 > 5 is true, prints "Big number" [OK]

Hint: Check if condition true to pick output [OK]

Common Mistakes:

• Choosing else output when condition is true
• Thinking no output if condition true
• Confusing syntax with output

4. Find the error in this code:

int num = 7;
if num > 5 {
  System.out.println("Greater");
}

medium

A. Braces not allowed in if

B. Missing semicolon after if

C. Missing parentheses around condition

D. Wrong variable name

Solution

1. Step 1: Check if statement syntax

   Java requires parentheses around the condition in an if statement.

2. Step 2: Identify the missing parentheses

   The code has if num > 5 without parentheses, which causes a syntax error.

3. Final Answer:

   Missing parentheses around condition -> Option C

4. Quick Check:

   If condition must be in parentheses [OK]

Hint: Always put condition in parentheses [OK]

Common Mistakes:

• Forgetting parentheses around condition
• Adding semicolon after if line
• Misnaming variables

5. You want to print "Even" if a number is even and "Odd" if it is odd. Which code correctly uses an if statement to do this?

int n = 4;
// Your code here

hard

A. if (n % 2 == 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

B. if (n / 2 == 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

C. if (n % 2 = 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

D. if (n % 2 != 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

Solution

1. Step 1: Understand how to check even numbers

   A number is even if dividing by 2 leaves remainder 0, so use n % 2 == 0.

2. Step 2: Check each option's condition

   if (n % 2 == 0) { System.out.println("Even"); } else { System.out.println("Odd"); } correctly uses n % 2 == 0 to check evenness and prints "Even" or "Odd" accordingly.

3. Final Answer:

   if (n % 2 == 0) { System.out.println("Even"); } else { System.out.println("Odd"); } -> Option A

4. Quick Check:

   Use modulus == 0 for even check [OK]

Hint: Use n % 2 == 0 to check even numbers [OK]

Common Mistakes:

• Using assignment = instead of comparison ==
• Checking division instead of remainder
• Reversing even and odd conditions