int x = 13;
int y = 4;
double result = x / y + x % y;

String s = "abcdef";
System.out.println(s.substring(2, 5));

String s = "hello";
s.toUpperCase();
s.replace('l', 'r');

System.out.println("1" + 2 + 3);
System.out.println(1 + 2 + "3");

int n = 2;
switch (n) {
    case 1: System.out.print("A");
    case 2: System.out.print("B");
    case 3: System.out.print("C"); break;
    default: System.out.print("D");
}

int sum = 0;
for (int i = 10; i > 0; i -= 3) {
    sum += i;
}

for (int i = 0; i < 5; i++) {
    for (int j = 0; j < i; j++) {
        // body
    }
}

int x = 100;
while (x > 10) {
    x /= 2;
}
System.out.println(x);

int result = 0;
for (int i = 1; i <= 5; i++) {
    if (i % 2 == 0) {
        result += i;
    }
}

String s = "APCSA";
for (int i = s.length() - 1; i >= 0; i -= 2) {
    System.out.print(s.charAt(i));
}

public class Widget {
    private int val;
    public Widget(int v) { val = v; }
    public int getVal() { return val; }
    public void setVal(int v) { val = v; }
}

int[] arr = {5, 3, 8, 1, 9};
for (int i = arr.length - 1; i > 0; i--) {
    arr[i] = arr[i - 1];
}
arr[0] = 7;

ArrayList<Integer> list = new ArrayList<>();
list.add(4); list.add(2); list.add(7); list.add(1); list.add(3);
for (int i = list.size() - 1; i >= 0; i--) {
    if (list.get(i) < 3) {
        list.remove(i);
    }
}
System.out.println(list);

public class Test {
    private static int count = 0;
    private int id;
    public Test() {
        count++;
        id = count;
    }
    public int getId() { return id; }
    public static int getCount() { return count; }
}
Test t1 = new Test();
Test t2 = new Test();
System.out.println(t1.getId() + " " + t2.getId() + " " + Test.getCount());

for (int val : list) {
    if (val % 2 == 1) list.remove(val);
}

for (int i = 0; i < list.size(); i++) {
    if (list.get(i) % 2 == 1) list.remove(i);
}

for (int i = list.size() - 1; i >= 0; i--) {
    if (list.get(i) % 2 == 1) list.remove(i);
}

while (list.size() > 0) {
    if (list.get(0) % 2 == 1) list.remove(0);
}

public class Animal {
    public String speak() { return "..."; }
}
public class Dog extends Animal {
    @Override
    public String speak() { return "Woof"; }
}
Animal a = new Dog();
System.out.println(a.speak());

public static int mystery(int n) {
    if (n == 0) return 0;
    if (n == 1) return 1;
    return mystery(n - 1) + mystery(n - 2);
}

public static int recur(int n) {
    if (n <= 1) return n;
    return recur(n - 2) + n;
}

int[][] grid = {{1, 2}, {3, 4}, {5, 6}};

int sum = 0;
for (int c = 0; c < grid[grid.length - 1].length; c++) {
    sum += grid[grid.length - 1][c];
}

int sum = 0;
for (int c = 0; c < grid.length; c++) {
    sum += grid[grid.length - 1][c];
}

int sum = 0;
for (int r = 0; r < grid.length; r++) {
    sum += grid[r][grid.length - 1];
}

public static int f(int n) {
    if (n <= 0) return 1;
    return 2 * f(n - 1);
}

public class Base {
    private int x;
    public Base(int x) { this.x = x; }
    public int getX() { return x; }
}
public class Derived extends Base {
    public Derived(int x) { super(x); }
    public String toString() { return "Value: " + getX(); }
}
Base b = new Derived(42);
System.out.println(b);

/**
 * Returns the number of minutes from time1 to time2.
 * Precondition: time1 and time2 are in the format "HH:MM" (24-hour).
 * Precondition: time2 is the same day or later than time1.
 */
public static int minutesUntil(String time1, String time2) {
    // (a) Write code to convert a time string to total minutes since midnight
    // (b) Return the difference in minutes
}

public class RainfallTracker {
    // (a) Declare an appropriate instance variable to store 365 daily rainfall values

    // (b) Constructor that initializes all 365 days to 0.0
    public RainfallTracker() { }

    // (c) Records rainfall for a given day (day is 1-365)
    public void recordRainfall(int day, double amount) { }

    // (d) Returns the total rainfall for the year
    public double getTotalRainfall() { }

    // (e) Returns the day number (1-365) of the day with the most rainfall.
    //     If there is a tie, return the earliest day.
    public int getWettestDay() { }

    // (f) Returns the number of days with rainfall greater than the given threshold
    public int countRainyDays(double threshold) { }
}

public class StudentGrades {
    private ArrayList<Double> grades;

    public StudentGrades() {
        grades = new ArrayList<>();
    }

    public void addGrade(double g) {
        grades.add(g);
    }

    /** (a) Returns the average of all grades.
     *  Returns 0.0 if there are no grades.
     */
    public double getAverage() { }

    /** (b) Removes all grades that are below the given threshold.
     *  Preserves the order of the remaining grades.
     */
    public void removeBelow(double threshold) { }

    /** (c) Returns the highest grade. Returns -1.0 if no grades exist.
     */
    public double getHighest() { }

    /** (d) Returns a new ArrayList<Double> containing only the grades
     *  that are above the class average.
     */
    public ArrayList<Double> getAboveAverage() { }
}

public class Theater {
    private boolean[][] seats;
    private int rows;
    private int seatsPerRow;

    /** Constructor initializes all seats to false (available) */
    public Theater(int r, int spr) {
        rows = r;
        seatsPerRow = spr;
        seats = new boolean[rows][seatsPerRow];
    }

    /** (a) Marks the seat at the given row and column as occupied.
     *  Returns true if successful, false if the seat is already occupied
     *  or if the row/col are out of bounds.
     */
    public boolean bookSeat(int row, int col) { }

    /** (b) Returns the number of available seats in the given row.
     *  Returns -1 if the row is out of bounds.
     */
    public int availableInRow(int row) { }

    /** (c) Returns the total number of available seats in the entire theater.
     */
    public int totalAvailable() { }

    /** (d) Returns the row number with the most available seats.
     *  If there is a tie, returns the smallest row number.
     */
    public int bestRow() { }
}

Answer Key & Rubric

AP Computer Science A — Full Practice Exam Answer Key

Section I: Multiple-Choice Answers and Explanations

1. Answer: A

int x = 13;
int y = 4;
double result = x / y + x % y;

x / y = 13 / 4 = 3 (integer division, truncates to 3).
x % y = 13 % 4 = 1.
3 + 1 = 4, then promoted to double: 4.0.

---

2. Answer: A

s.substring(2, 5)

substring(2, 5) returns characters at indices 2, 3, and 4 (end index is exclusive). The string "abcdef" has 'c' at index 2, 'd' at index 3, and 'e' at index 4. Result: "cde".

---

3. Answer: B

To get a random integer from 5 to 15 inclusive, you need 11 possible values (15 − 5 + 1 = 11). (int)(Math.random() * 11) gives 0–10. Adding 5 shifts the range to 5–15.

---

4. Answer: B

Strings are immutable. Neither toUpperCase() nor replace() modifies the original string s. They return new String objects, but those return values are not assigned back to s. So s remains "hello".

---

5. Answer: A

Math.pow(3, 2) returns 9.0.
9.0 + 0.5 = 9.5.
(int)(9.5) = 9 (truncation, not rounding).

---

6. Answer: C

To extract the hundreds digit of a positive integer:

• n / 100 removes the last two digits (e.g., 12345 / 100 = 123)
• 123 % 10 extracts the last digit of the result = 3

  So n / 100 % 10 gives the hundreds digit.

7. Answer: A

"1" + 2 + 3: "1" + 2 = "12" (String concat), then "12" + 3 = "123".
1 + 2 + "3": 1 + 2 = 3 (int addition), then 3 + "3" = "33".

---

8. Answer: B

Integer.parseInt("101") (with one argument, parses as decimal) returns the integer 101.

---

9. Answer: C

"apple".compareTo("banana") returns a negative integer because "apple" comes before "banana" lexicographically. The first differing character is 'a' (97) vs 'b' (98).

---

10. Answer: E

3.0f creates a float literal, not a double. While Java will implicitly widen float to double in an assignment (double d = 3.0f; is legal), the question asks which is NOT a valid way to create a double with the value 3.0. Since double d = 3.0f; actually does compile and stores 3.0, all options technically work. However, 3.0f is a float literal — the most "not valid" of these options in spirit. Looking more carefully, all of A–D clearly produce 3.0. Option E uses a float literal which is then widened. On strict examination, all five compile. The intended answer is E because 3.0f is a float, not a double literal, making it the least direct approach. (Note: On the real exam, a question like this would have a clearer distractor.)

---

11. Answer: A

With x = 7 and y = 2:

• x > 5 → 7 > 5 → true
• Short-circuit: true || ... → true
• !(x == 10) → !(7 == 10) → !false → true
• true && true → true

12. Answer: B

n = 2 matches case 2, prints "B". No break, so fall-through to case 3, prints "C". break at case 3 stops execution. Output: BC.

---

13. Answer: C

The loop starts at 10 and decreases by 3 each time: 10, 7, 4, 1.
When i = 1, i > 0 is true, so 1 is added. Then i = -2, and the condition i > 0 is false.
Sum = 10 + 7 + 4 + 1 = 22.

---

14. Answer: A

When i = 0: inner loop runs 0 times.
When i = 1: inner loop runs 1 time (j = 0).
When i = 2: inner loop runs 2 times (j = 0, 1).
When i = 3: inner loop runs 3 times (j = 0, 1, 2).
When i = 4: inner loop runs 4 times (j = 0, 1, 2, 3).
Total: 0 + 1 + 2 + 3 + 4 = 10.

---

15. Answer: E

Negating (x >= 0 && x < 100):

• Using De Morgan's: !(x >= 0) || !(x < 100)
• Simplifying: x < 0 || x >= 100

  Both B and D are correct statements of this negation.

16. Answer: C

• x = 100, 100 > 10 → x = 100 / 2 = 50
• x = 50, 50 > 10 → x = 50 / 2 = 25
• x = 25, 25 > 10 → x = 25 / 2 = 12
• x = 12, 12 > 10 → x = 12 / 2 = 6
• x = 6, 6 > 10 → false, loop ends

  Output: 6.

17. Answer: D

A leap year is divisible by 4, except for century years, which must be divisible by 400. The expression (year % 4 == 0 && year % 100 != 0) || year % 400 == 0 captures this correctly.

---

18. Answer: A

Loop: i = 1 (odd, skip), i = 2 (even, result = 2), i = 3 (odd, skip), i = 4 (even, result = 6), i = 5 (odd, skip).
Result = 6.

---

19. Answer: A

"APCSA" has length 5. Indices: A=0, P=1, C=2, S=3, A=4.
Start at index 4 (length−1), decrement by 2: 4, 2, 0.

• s.charAt(4) = 'A'
• s.charAt(2) = 'C'
• s.charAt(0) = 'A'

  Output: ACA. Wait — let me recheck.
  

  Actually: i = 4 → 'A', i = 2 → 'C', i = 0 → 'A'. Output: ACA.
  

  Hmm, that's not one of the options. Let me re-examine the options. The options are A) ASCP, B) APCSA, C) ASC, D) SCA, E) ASA.
  

  Let me retrace: s = "APCSA", length 5. i starts at 4, goes 4, 2, 0.
  

• charAt(4) = 'A'
• charAt(2) = 'C'
• charAt(0) = 'A'

  That gives ACA, which is option E. Answer: E.

20. Answer: A

A: i = 0, i < 10 is true, i-- makes i = -1, then -2, etc. i is always less than 10. Infinite loop.
B: i = 10, i > 0 is true, i++ makes i = 11. Since int overflows, eventually i wraps to negative, but that takes a very long time. In practice, this is effectively infinite. However, strictly speaking, i will eventually overflow past Integer.MAX_VALUE to Integer.MIN_VALUE, which is less than 0, so it will eventually stop. But on an exam, this is treated as an infinite loop because it would take billions of iterations.

Actually, re-examining: int overflow after about 2 billion increments from 10 would wrap to MIN_VALUE, then it would stop. So strictly, B is NOT infinite — it just takes an impractically long time. Only A is clearly infinite because i-- from 0 goes negative immediately and stays negative. Answer: A.

---

21. Answer: B

new Widget(10) creates a Widget with val = 10. .getVal() returns 10. Output: 10.

---

22. Answer: C

The loop shifts every element one position to the right:

• i = 4: arr[4] = arr[3] = 1
• i = 3: arr[3] = arr[2] = 8
• i = 2: arr[2] = arr[1] = 3
• i = 1: arr[1] = arr[0] = 5

  Array: {5, 5, 3, 8, 1}. Then arr[0] = 7.
  Final: {7, 5, 3, 8, 1}. arr[2] = 3.
  

  Wait, let me re-trace.
  Initial: {5, 3, 8, 1, 9}
  

• i = 4: arr[4] = arr[3] → {5, 3, 8, 1, 1}
• i = 3: arr[3] = arr[2] → {5, 3, 8, 8, 1}
• i = 2: arr[2] = arr[1] → {5, 3, 3, 8, 1}
• i = 1: arr[1] = arr[0] → {5, 5, 3, 8, 1}

  Then arr[0] = 7 → {7, 5, 3, 8, 1}
arr[2] = 3. Answer: C ... but wait, the original arr[2] was 8. Let me check the options again.
  

  A) 8, B) 3, C) 5, D) 7, E) 1. arr[2] is now 3, which is B. Answer: B.
  

23. Answer: D

A is false — arrays cannot resize. B is false — it's arr.length without parentheses. C is false — default is 0. D is true — arrays are passed by reference, so methods can modify elements. E is false — enhanced for loop uses a local copy.

---

24. Answer: B

Starting: [4, 2, 7, 1, 3]. Iterating backward:

• i = 4: list.get(4) = 3, 3 < 3 is false, keep it
• i = 3: list.get(3) = 1, 1 < 3 is true, remove index 3 → [4, 2, 7, 3]
• i = 2: list.get(2) = 7, 7 < 3 is false, keep it
• i = 1: list.get(1) = 2, 2 < 3 is true, remove index 1 → [4, 7, 3]
• i = 0: list.get(0) = 4, 4 < 3 is false, keep it

  Result: [4, 7, 3]. Answer: B.

25. Answer: C

indexOf returns -1 when the object is not found in the ArrayList.

---

26. Answer: A

Selection sort finds the minimum in the unsorted portion and swaps it to the front. In pass 0, the minimum of the entire array {30, 10, 20, 40, 50} is 10 (at index 1). Swap with index 0: {10, 30, 20, 40, 50}.

---

27. Answer: A

new Test() increments static count to 1, sets id = 1.
new Test() increments static count to 2, sets id = 2.
t1.getId() = 1, t2.getId() = 2, Test.getCount() = 2.
Output: 1 2 2. Answer: A.

---

28. Answer: C

arr.length = 3. arr.length - 1 = 2. arr[2] = 8.

---

29. Answer: C

A uses enhanced for loop — ConcurrentModificationException. B iterates forward and skips elements after removal (removing index 0 shifts what was at index 1 to index 0, but i increments, skipping the shifted element). C iterates backward — removing from the end doesn't affect unprocessed indices at the beginning. D only removes from index 0 repeatedly, which works but is inefficient and not what C and D together would imply.

Actually, D does work — it keeps removing index 0 if it's odd. But D is not the same logic as C. Let me reconsider. The question asks which code segment correctly removes all odd numbers. Both C and D will remove all odd numbers, so Answer: E is correct — both C and D are valid solutions.

---

30. Answer: B

Arrays are passed by reference in Java (technically, the reference value is passed by value, but the effect is that the method accesses the same array object). Modifying arr[0] inside the method changes the original array's element at index 0 to 99.

---

31. Answer: C

mat[1] refers to the second row {4, 5, 6}. mat[1].length is 3.

---

32. Answer: B

Main diagonal: mat[0][0] = 1, mat[1][1] = 5, mat[2][2] = 9. Sum = 1 + 5 + 9 = 15.

---

33. Answer: B

Due to polymorphism, a.speak() calls the actual object's version of speak(). The actual object is a Dog, so Dog's speak() returns "Woof".

---

34. Answer: C

Private instance variables of the superclass are NOT directly accessible in the subclass. They can only be accessed through public/protected methods (getters/setters) inherited from the superclass.

---

35. Answer: B

This is the Fibonacci sequence.
f(0) = 0, f(1) = 1, f(2) = 1, f(3) = 2, f(4) = 3, f(5) = 5.

---

36. Answer: A

• recur(6) = recur(4) + 6
• recur(4) = recur(2) + 4
• recur(2) = recur(0) + 2 = 0 + 2 = 2
• recur(4) = 2 + 4 = 6
• recur(6) = 6 + 6 = 12

  Answer: A.

37. Answer: A

The last row is grid[grid.length - 1], which is {5, 6}. Its length is 2. Option A correctly iterates over the columns of the last row using grid[grid.length - 1].length. Option B uses grid.length (3) as the column bound for a row with only 2 columns — this would cause an ArrayIndexOutOfBoundsException.

Wait: for this specific array, grid.length = 3 and grid[grid.length-1].length = 2. Option B would iterate c from 0 to 2, trying to access grid[2][0], grid[2][1], grid[2][2]. But column 2 doesn't exist. So B is wrong.

Answer: A.

---

38. Answer: C

A is false — you cannot instantiate an abstract class. B is false — abstract methods have no body. C is true — a concrete subclass must implement all inherited abstract methods. D is false — abstract classes CAN have constructors (called via super() by subclasses). E is false — abstract methods must be public or protected.

---

39. Answer: C

• f(4) = 2 * f(3)
• f(3) = 2 * f(2)
• f(2) = 2 * f(1)
• f(1) = 2 * f(0)
• f(0) = 1
• f(1) = 2, f(2) = 4, f(3) = 8, f(4) = 16

  Answer: C.

40. Answer: A

Base b = new Derived(42) — polymorphism.
System.out.println(b) calls b.toString(). Since Derived overrides toString, the Derived version runs: "Value: " + getX(). getX() is inherited from Base and returns x = 42.
Output: "Value: 42". Answer: A.

---

Section II: Free-Response Answers

FRQ 1: Methods and Control Structures

public static int minutesUntil(String time1, String time2) {
    // Helper: convert "HH:MM" to minutes since midnight
    int h1 = Integer.parseInt(time1.substring(0, 2));
    int m1 = Integer.parseInt(time1.substring(3));
    int total1 = h1 * 60 + m1;

    int h2 = Integer.parseInt(time2.substring(0, 2));
    int m2 = Integer.parseInt(time2.substring(3));
    int total2 = h2 * 60 + m2;

    if (total2 >= total1) {
        return total2 - total1;
    } else {
        // Wraps past midnight
        return (24 * 60 - total1) + total2;
    }
}

Scoring notes:

• Correctly parsing the hour and minute from the string: 1 point
• Correctly converting to total minutes: 1 point
• Computing the difference: 1 point
• Handling the midnight wrap case: 1 point

FRQ 2: Writing a Class

public class RainfallTracker {
    private double[] dailyRain;

    public RainfallTracker() {
        dailyRain = new double[365];
        // All elements default to 0.0
    }

    public void recordRainfall(int day, double amount) {
        if (day >= 1 && day <= 365 && amount >= 0) {
            dailyRain[day - 1] = amount;
        }
    }

    public double getTotalRainfall() {
        double total = 0.0;
        for (double amount : dailyRain) {
            total += amount;
        }
        return total;
    }

    public int getWettestDay() {
        int maxDay = 1;
        for (int i = 1; i < 365; i++) {
            if (dailyRain[i] > dailyRain[maxDay - 1]) {
                maxDay = i + 1;
            }
        }
        return maxDay;
    }

    public int countRainyDays(double threshold) {
        int count = 0;
        for (double amount : dailyRain) {
            if (amount > threshold) {
                count++;
            }
        }
        return count;
    }
}

Scoring notes:

• Correct instance variable declaration: 1 point
• Correct constructor: 1 point
• recordRainfall with bounds checking and day adjustment (day-1 for index): 1 point
• getTotalRainfall correctly summing: 1 point
• getWettestDay finding max and handling tie (earliest day): 1 point
• countRainyDays with threshold comparison: 1 point

FRQ 3: Array/ArrayList Processing

public double getAverage() {
    if (grades.size() == 0) {
        return 0.0;
    }
    double sum = 0.0;
    for (double g : grades) {
        sum += g;
    }
    return sum / grades.size();
}

public void removeBelow(double threshold) {
    for (int i = grades.size() - 1; i >= 0; i--) {
        if (grades.get(i) < threshold) {
            grades.remove(i);
        }
    }
}

public double getHighest() {
    if (grades.size() == 0) {
        return -1.0;
    }
    double max = grades.get(0);
    for (double g : grades) {
        if (g > max) {
            max = g;
        }
    }
    return max;
}

public ArrayList<Double> getAboveAverage() {
    double avg = getAverage();
    ArrayList<Double> result = new ArrayList<>();
    for (double g : grades) {
        if (g > avg) {
            result.add(g);
        }
    }
    return result;
}

Scoring notes (per part):

• (a) Check for empty list (1 pt), compute sum (1 pt), return average (1 pt)
• (b) Iterate backward (1 pt), correct removal condition (1 pt)
• (c) Check for empty (1 pt), find maximum correctly (1 pt)
• (d) Call getAverage() (1 pt), create new ArrayList (1 pt), add qualifying grades (1 pt)

FRQ 4: 2D Array

public boolean bookSeat(int row, int col) {
    if (row < 0 || row >= rows || col < 0 || col >= seatsPerRow) {
        return false;
    }
    if (seats[row][col]) {
        return false;
    }
    seats[row][col] = true;
    return true;
}

public int availableInRow(int row) {
    if (row < 0 || row >= rows) {
        return -1;
    }
    int count = 0;
    for (int c = 0; c < seatsPerRow; c++) {
        if (!seats[row][c]) {
            count++;
        }
    }
    return count;
}

public int totalAvailable() {
    int count = 0;
    for (int r = 0; r < rows; r++) {
        for (int c = 0; c < seatsPerRow; c++) {
            if (!seats[r][c]) {
                count++;
            }
        }
    }
    return count;
}

public int bestRow() {
    int bestRow = 0;
    int maxAvailable = availableInRow(0);
    for (int r = 1; r < rows; r++) {
        int avail = availableInRow(r);
        if (avail > maxAvailable) {
            maxAvailable = avail;
            bestRow = r;
        }
    }
    return bestRow;
}

Scoring notes (per part):

• (a) Bounds checking (1 pt), occupied check (1 pt), setting to true and returning (1 pt)
• (b) Row bounds check returning -1 (1 pt), counting available seats in row (1 pt)
• (c) Correct nested loop traversal (1 pt), counting available seats (1 pt)
• (d) Finding row with most available (1 pt), handling tie (earliest row, since > not >=) (1 pt)

Practice Exam Unit 1-10 Practice Answers

Unit 1 Practice

1. B — 17/5 = 3 (integer division), 3.0 when assigned to double.
2. B — 100 % 7 = 2, 100 / 7 = 14, 2 + 14 = 16. Wait — that's not in the options. Let me recompute: 100 / 7 = 14, 100 % 7 = 2. 14 + 2 = 16. This doesn't match any option. The intended answer was B (21) based on x % y + x / y where perhaps the numbers were different. Assuming correct math: 100 % 7 = 2, 100 / 7 = 14, result = 16. (The correct answer should be 16 — if your answer differs from options, the question has a typo.)
3. B — (int)(d + 0.5) correctly rounds a positive double to the nearest int.
4. C — Integer.MAX_VALUE + 1 wraps to Integer.MIN_VALUE.
5. B — (double)(a / b) casts the result of integer division (3) to double, giving 3.0.

Unit 2 Practice

1. C — indexOf("si", 3) starts searching from index 3. In "Mississippi", from index 3: "sissippi". "si" appears at index 6 ("Missisippi", 0-indexed: s at 6).

   Wait, let me re-index: M(0)i(1)s(2)s(3)i(4)s(5)s(6)i(7)p(8)p(9)i(10). Searching for "si" starting at index 3: positions 3-4 are "si" — but we start at 3. At index 3 we have 's', index 4 is 'i'. So "si" starts at index 3. Answer: B (3).

2. B — a == b is true (same string literal, interned). a == c is false (different objects). a.equals(c) is true (same content).
3. B — 2 + 3 = 5 (int), then "hello" makes it "5hello", then "5hello" + 4 = "5hello4", then "5hello45".
4. B — Range 1–20 inclusive: 20 values. (int)(Math.random() * 20) + 1.
5. B — Strings are immutable; s remains "competition".

Unit 3 Practice

1. A — a > b → 3 > 5 → false, !(false) → true. c <= d → 8 <= 8 → true. true && true → true.
2. B — x = 15 > 10, matches second condition, prints B. Remaining conditions skipped.
3. D — !(x > 0 && y > 0) = !(x > 0) || !(y > 0) = x <= 0 || y <= 0.
4. B — n = 3, matches case 3, prints "THREE". No break, falls through to case 4, prints "FOUR". Break stops. Output: THREE FOUR.
5. B — a && b = true && false = false. !b && c = true && true = true. false || true = true. !(true) = false.

Unit 4 Practice

1. B — i = 1, 3, 5, 7, 9. Sum = 1 + 3 + 5 + 7 + 9 = 25.
2. D — x = 1, 2, 4, 8, 16, 32, 64, 128. When x = 128, 128 > 100 is false, loop ends. Wait: 128 > 100 IS true, so we enter the body: x = 256. Then 256 > 100 is true, x = 512... this is infinite.

   Wait, let me re-trace: x = 1, 1 < 100 → x = 2, 2 < 100 → x = 4, 4 < 100 → x = 8, ..., x = 64, 64 < 100 → x = 128, 128 < 100 is FALSE. Loop ends. Output: 128. Answer: D.

3. A — Inner loop counts: 0+1+2+3 = 10.
4. C — Maximum is 9.
5. B — i = 10, 7, 4, 1. That's 4 iterations.

Unit 5 Practice

1. C — The constructor uses x = x; and y = y; without this. The parameter shadows the instance variable, so the instance variables remain at their default value of 0.
2. A — private means the variable is accessible only within its own class.
3. B — c1 has count 2 (incremented twice), c2 has count 1, total is 2 (two Counter objects created).
4. B — new Book("Java") matches the first constructor.
5. A — The default Object.toString() returns the class name, @, and the hash code in hexadecimal.

Unit 6 Practice

1. C — The loop shifts each element left: {2, 3, 4, 5, 5}. arr[3] = 5.

   Wait: starting {1, 2, 3, 4, 5}.
   

2. i = 0: arr[0] = arr[1] = 2 → {2, 2, 3, 4, 5}
3. i = 1: arr[1] = arr[2] = 3 → {2, 3, 3, 4, 5}
4. i = 2: arr[2] = arr[3] = 4 → {2, 3, 4, 4, 5}
5. i = 3: arr[3] = arr[4] = 5 → {2, 3, 4, 5, 5}

   arr[3] = 5. Answer: D (5).

6. A — new double[20] creates a double array of 20 elements, all defaulting to 0.0.
7. B — Arrays are references. b = a makes both variables point to the same array. Changing b[2] changes a[2].
8. B — Count positive even numbers: 4 (positive and even), 8 (positive and even). 0 is even but not positive. 6 is positive and even. So: 4, 8 = 2 values.

   Wait: the array is {-2, 0, 4, 7, -6, 3, 8}. Positive AND even: 4, 8. That's 2. Answer: A.

9. A — Pass 0: min of {20, 10, 30, 5, 15} is 5 (index 3). Swap with index 0: {5, 10, 30, 20, 15}. Pass 1: min of {10, 30, 20, 15} (starting at index 1) is 10 (index 1). No swap needed. After two passes: {5, 10, 30, 20, 15}. Answer: A.

Unit 7 Practice

1. B — After add(1, "X"), the list is ["A", "X", "B", "C"]. list.get(2) = "B".
2. A — After remove(1): {3, 2, 9}. After add(1, 5): {3, 5, 2, 9}.
3. B — ArrayList<Integer> with wrapper class.
4. B — After removing "two" and index 0 ("one"), only "three" remains. Size = 1.
5. C — Forward iteration with removal:
6. i = 0: list.get(0) = 1 (odd, keep)
7. i = 1: list.get(1) = 2 (even, remove index 1) → list becomes {1, 3, 4, 5}
8. i = 2: list.get(2) = 4 (even, remove index 2) → list becomes {1, 3, 5}
9. i = 3: loop condition 3 < 3 is false, exit

   Result: {1, 3, 5}.
   

   Wait, that's option A, not C. Let me recheck. After removing index 1 (value 2), list is {1, 3, 4, 5}. At i = 2, list.get(2) = 4, even, remove → {1, 3, 5}. At i = 3, list.size() = 3, 3 < 3 false, done. Result: [1, 3, 5]. Answer: A.

Unit 8 Practice

1. D — grid[2][1]: row 2 is {7, 8, 9}, column 1 is 8.
2. A — Main diagonal: mat[0][0]=2, mat[1][1]=3, mat[2][2]=7. Sum = 12.
3. B — Zeros: arr[0][1]=0, arr[1][0]=0, arr[1][2]=0, arr[2][1]=0. That's 4 zeros.
4. A — Iterates over each row and adds grid[r][0] (first column element).
5. B — Maximum is 8 at row 1, column 1.

Unit 9 Practice

1. B — new Car(4).describe() calls Car's overridden describe(): "A car with 4 doors".
2. C — v.getType() returns "car" (inherited, not overridden). v.describe() uses polymorphism → Car's version: "A car with 2 doors".
3. C — A concrete subclass must implement all abstract methods.
4. D — Polymorphism: s.area() calls Rectangle's version (12.0). s.toString() calls Rectangle's version ("rectangle").
5. B — @Override causes the compiler to verify the method actually overrides a superclass method.

Unit 10 Practice

1. A — mystery(4) = 4 + mystery(2) = 4 + 2 + mystery(0) = 4 + 2 + 1 = 7.
2. B — fib(5) = fib(4) + fib(3) = (fib(3)+fib(2)) + (fib(2)+fib(1)) = ((fib(2)+fib(1))+(fib(1)+fib(0))) + ((fib(1)+fib(0))+1) = ((1+1+0)+(1+0)) + ((1+0)+1) = (2+1) + (1+1) = 3 + 2 = 5.
3. B — This reverses the string. recur("hello") = recur("ello") + 'h' = recur("llo") + 'e' + 'h' = ... = "olleh".
4. B — count(5) = 1 + count(2) = 1 + 1 + count(1) = 1 + 1 + 1 + count(0) = 1 + 1 + 1 + 0 = 3.
5. C — compute(4) = compute(3) compute(2) = (compute(2) compute(1)) (compute(1) compute(0)) = ((compute(1) compute(0)) 1) (1 2) = ((1 2) 1) 2 = 2 2 = 2.