AP Computer Science A Prep, 2024 - Rob Franek 2023

Practice Test 1
Part II: Practice Test 1

Section I

The Exam

AP® Computer Science A Exam

SECTION I: Multiple-Choice Questions

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

At a Glance

Total Time

1 hour 30 minutes

Number of Questions

Percent of Total Score

50%

Writing Instrument

Pencil required

Instructions

Section I of this examination contains 40 multiple-choice questions. Fill in only the ovals for numbers 1 through 40 on your answer sheet.

Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding oval on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.

Sample Question

Chicago is a

(A) state

(B) city

(D) continent

(E) county

Sample Answer

Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.

About Guessing

Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple-choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.

Java Quick Reference

Class Constructors and Methods	Explanation
String Class
String(String str)	Constructs a new String object that represents the same sequence of characters as str
int length()	Returns the number of characters in a String object
String substring(int from, int to)	Returns the substring beginning at index from and ending at index to — 1
String substring(int from)	Returns substring(from, length())
int indexOf(String str)	Returns the index of the first occurrence of str; returns —1 if not found
boolean equals(String other)	Returns true if this is equal to other; returns false otherwise
int compareTo(String other)	Returns a value <0 if this is less than other; returns zero if this is equal to other; returns a value >0 if this is greater than other
Integer Class
Integer(int value)	Constructs a new Integer object that represents the specified int value
Integer.MIN_VALUE	The minimum value represented by an int or Integer
Integer.MAX_VALUE	The maximum value represented by an int or Integer
int intValue()	Returns the value of this Integer as an int
Double Class
Double(double value)	Constructs a new Double object that represents the specified double value
double doubleValue()	Returns the value of this Double as a double
Math Class
static int abs(int x)	Returns the absolute value of an int value
static double abs(double x)	Returns the absolute value of a double value
static double pow(double base, double exponent)	Returns the value of the first parameter raised to the power of the second parameter
static double sqrt(double x)	Returns the positive square root of a double value
static double random()	Returns a double value greater than or equal to 0.0 and less than 1.0
ArrayList Class
int size()	Returns the number of elements in the list
boolean add(E obj)	Appends obj to end of list; returns true
void add(int index, E obj)	Inserts obj at position index (0 <= index <= size), moving elements at position index and higher to the right (adds 1 to their indices) and adds 1 to size
E get(int index)	Returns the element at position index in the list
E set(int index, E obj)	Replaces the element at position index with obj; returns the element formerly at position index
E remove(int index)	Removes element from position index, moving elements at position index + 1 and higher to the left (subtracts 1 from their indices) and subtracts 1 from size; returns the element formerly at position index
Object Class
boolean equals(Object other)
String toString()

Section I

COMPUTER SCIENCE A

SECTION I

Time—1 hour and 30 minutes

Number of Questions—40

Percent of total exam grade—50%

Directions: Determine the answer to each of the following questions or incomplete statements, using the available space for any necessary scratchwork. Then decide which is the best of the choices given and fill in the corresponding oval on the answer sheet. No credit will be given for anything written in the examination booklet. Do not spend too much time on any one problem.

Notes:

· Assume that the classes listed in the Quick Reference have been imported where appropriate.

· Assume that declarations of variables and methods appear within the context of an enclosing class.

· Assume that method calls that are not prefixed with an object or class name and are not shown within a complete class definition appear within the context of an enclosing class.

· Unless otherwise noted in the question, assume that parameters in the method calls are not null and that methods are called only when their preconditions are satisfied.

1.Evaluate the following expression: 4 + 6 % 12 / 4

(A) 1

(B) 2

(D) 4.5

(E) 5

2.Which of the following expressions does NOT evaluate to 0.2?

(A) (1.0 * 2) / (1.0 * 10)

(B) 2.0 / 10

(D) (double)(2 / 10)

(E) Math.sqrt(4) / Math.sqrt(100)

3.Choose the code used to print the following:

“Friends”

(A) System.out.print(““Friends””);

(B) System.out.print(“//“Friends//””);

(D) System.out.print(“\“Friends\””);

(E) System.out.print(“\\“Friends \\””);

4.Determine the output of the following code.

String animal1 = “elephant”;

String animal2 = “lion”;

swap(animal1, animal2);

animal1.toUpperCase();

animal2.toLowerCase();

System.out.println(animal1 + “ ” + animal2);

public static void swap(String a1, String a2) {

String hold = a1;

a1 = a2;

a2 = hold;

}

(A) elephant lion

(B) ELEPHANT lion

(D) LION elephant

(E) LION ELEPHANT

Questions 5—6 refer to the Constellation class below.

public class Constellation

private String name;

private String month;

private int northernLatitude;

private int southernLatitude;

Constellation(String n, String m)

{

name = n;

month = m;

northernLatitude = 0;

southernLatitude = 0;

}

Constellation(String n, String m, int nLat, int sLat)

{

name = n;

month = m;

northernLatitude = nLat;

southernLatitude = sLat;

}

public void chgMonth(String m)

{

String month = m;

}

5.Using the Constellation class, which of the following will cause a compiler error?

(A) Constellation c1 = new Constellation(“Hercules”, “July”);

(B) Constellation c2 = new Constellation(“Pisces”, “Nov”, 90, 65);

(D) Constellation c4 = new Constellation(“Leo”, “4”, 0, 0);

(E) Constellation c5 = new Constellation(“Phoenix”, “Nov”, 32, 90);

6.A programmer has attempted to add three mutator methods to the Constellation class.

I. public void chgLatitude(String direction, int latitude)

{

if (direction.toUpperCase().equals(“N”))

northernLatitude = latitude;

else if (direction.toUpperCase().equals(“S”))

southernLatitude = latitude;

}

II. public void chgLatitude(int nLatitude, int sLatitude)

{

northernLatitude = nLatitude;

southernLatitude = sLatitude;

}

III. public void chgLatitude(double nLatitude, double sLatitude)

{

northernLatitude = (int) nLatitude;

southernLatitude = (int) sLatitude;

}

Which of the three will compile without a compiler error?

(A) I only

(B) II only

(D) I and II only

(E) I, II, and III

7.Determine the values of x and y after the following code runs.

int x = 10;

int y = 5;

if (x == 10)

{

if (y <= 5)

y++;

else if (y < 4)

x = 3;

else

y += 6;

}

if (y > 5)

{

if (x != 10)

{

x = 0;

y = 0;

}

else

x = -5;

}

(A) x = 0, y = 0

(B) x = —5, y = 6

(D) x = 3, y = 5

(E) None of the above

8.A programmer intended to write code to print three words in ascending lexicographical order. Follow the code and determine the printed output.

1 String word1 = “frog”;

2 String word2 = “dog”;

3 String word3 = “cat”;

5 if (word1.compareTo(word2) < 0)

6 if (word2.compareTo(word3) < 0)

7 System.out.println(word1 + “ ” + word2 + “ ” + word3);

8 else

9 System.out.println(word1 + “ ” + word3 + “ ” + word2);

10 else

11 if (word1.compareTo(word2) > 0)

12 if (word2.compareTo(word3) < 0)

13 System.out.println(word1 + “ ” + word2 + “ ” + word3);

14 else

15 System.out.println(word1 + “ ” + word3 + “ ” + word2);

16 else

17 if (word2.equals(word3))

18 System.out.println(“all the words are the same”);

19 else

20 System.out.println(“word1 and word2 are duplicates”);

(A) frog cat dog

(B) cat dog frog

(D) frog dog cat

(E) dog cat frog

9.Using the following variable declarations, determine which of the following would evaluate to true.

int temp = 90;

boolean cloudy = false;

I. if (temp >= 90 && !cloudy)

II. if (!(temp > 90 || cloudy))

III. if (!(temp > 90 && !cloudy))

(A) I only

(B) II only

(D) Two of the above will evaluate to true.

(E) All the above will evaluate to true.

10.Consider the following code:

1 String dog1 = new String(“Poodle”);

2 String dog2 = new String(“Beagle”);

3 dog1 = dog2;

4 String dog3 = new String(“Beagle”);

6 if (dog1 == dog2)

7 System.out.println(“dog1 and dog2 are one and the same dog”);

8 else

9 System.out.println(“dog1 and dog2 are not the same dog”);

11 if (dog1 == dog3)

12 System.out.println(“dog1 and dog3 are one and the same dog”);

13 else

14 System.out.println(“dog1 and dog3 are not the same dog”);

16 if (dog1.equals(dog3))

17 System.out.println(“dog1 and dog3 are the same breed”);

18 else

19 System.out.println(“dog1 and dog3 are not the same breed”);

Which of the following represents the output that will be produced by the code?

(A) dog1 and dog2 are one and the same dog

dog1 and dog3 are one and the same dog

dog1 and dog3 are the same breed

(B) dog1 and dog2 are one and the same dog

dog1 and dog3 are one and the same dog

dog1 and dog3 are not the same breed

dog1 and dog3 are not the same dog

dog1 and dog3 are the same breed

(D) dog1 and dog2 are one and the same dog

dog1 and dog3 are not the same dog

dog1 and dog3 are not the same breed

(E) dog1 and dog2 are not the same dog

dog1 and dog3 are not the same dog

dog1 and dog3 are the same breed

11.Choose the correct option to complete lines 3 and 4 such that str2 will contain the letters of str1 in reverse order.

1 String str1 = “banana”;

2 String str2 = “”;

3 // missing code

4 // missing code

5 {

6 str2 += str1.substring(i, i + 1);

7 i--;

8 }

(A) int i = 0;

while (i < str1.length)

(B) int i = str1.length();

while (i >= 0)

(D) int i = str1.length();

while (i > 0)

(E) int i = str1.length() - 1;

while (i > 0)

12.Consider the following code excerpt :

9 int n = // some integer greater than zero

10 int count = 0;

11 int p = 0;

12 int q = 0;

13 for (p = 1; p < n; p++)

14 for (q = 1; q <= n; q++)

15 count ++;

What will be the final value of count?

(A) nⁿ

(B) n² — 1

(D) n(n — 1)

(E) n²

13.Given the following code excerpt, determine the output.

1 int x = 0;

2 for (int j = 1; j < 4; j++)

3 {

4 if (x != 0 && j / x > 0)

5 System.out.print(j / x + “ ”);

6 else

7 System.out.print(j * x + “ ”);

8 }

(A) 0 0 0

(B) 0 0 0 0

(D) 1 0 2 0 3 0

(E) ArithmeticException: Divide by Zero

14.Consider the following code:

1 String space = “ ”;

2 String symbol = “*”;

3 int num = 5;

4 for (int i = 1; i <= num; i++)

5 {

6 System.out.print(symbol);

7 }

8 System.out.print(“\n”);

9 for (int i = 1; i <= num; i++)

10 {

11 for (int j = num - i; j > 0; j--)

12 {

13 System.out.print(space);

14 }

15 System.out.println(symbol);

16 }

17 for (int i = 1; i <= num; i++)

18 {

19 System.out.print(symbol);

20 }

Which of the following represents the output?

(A) *****

****

***

*****

(B) *****

****

***

*****

(D) *****

*****

(E) *****

***

****

*****

15.What will be printed as a result of the following code excerpt?

int sum = 0;

for (int i = 1; i < 2; i++)

for (int j = 1; j <= 3; j++)

for (int k = 1; k < 4; k++)

sum += (i * j * k);

System.out.println(sum);

(A) 18

(B) 36

(D) 60

(E) 108

16.Consider the following code:

1 int j = 0;

2 String s = “map”;

3 while (j < s.length())

4 {

5 int k = s.length();

6 while (k > j)

7 {

8 System.out.println(s.substring(j, k));

9 k--;

10 }

11 j++;

12 }

Which of the following represents the output?

(A) map

(B) map

(D) m

map

(E) p

map

17.A factorial is shown by an exclamation point(!) following a number. The factorial of 5, or 5!, is calculated by (5)(4)(3)(2)(1) = 120.

Assuming n is an integer greater than 1, choose the method that will return n!

public static int f(int n) {

int factorial = 1;

for (int i = n; i > 0; i--) {

factorial *= n;

}

return factorial;

}

II.

public static int f(int n) {

int factorial = 1;

int j = 1;

while (j <= n) {

factorial *= j;

j++;

}

return factorial;

}

III.

public static int f(int n) {

if (n == 1)

return n;

return n * f(n - 1);

}

(A) I only

(B) II only

(D) II and III only

(E) I, II, and III

Questions 18—20 refer to the code excerpt for the Tile class below:

1 public class Tile

2 {

3 private int styleNumber;

4 private String color;

5 private double width;

6 private double height;

7 private String material;

8 private double price;

9 Tile(int style, String col)

10 {

11 styleNumber = style;

12 color = col;

13 }

14 Tile(int style, String col, double w, double h, String mat, double price)

15 {

16 styleNumber = style;

17 color = col;

18 width = w;

19 height = h;

20 material = mat;

21 price = price;

22 }

23 Tile(int style, String col, String mat, double price)

24 {

25 styleNumber = style;

26 color = col;

27 material = mat;

28 price = price;

29 }

30 public void chgMaterial(String mat)

31 {

32 String material = mat;

33 }

34 public String toString()

35 {

36 return (styleNumber + “ ” + color + “ ” + width + “ ” + height + “ ” +

37 material + “ ” + price);

38 }

39 }

18.What is the output after the following client code is executed?

Tile t1 = new Tile(785, “grey”, “ceramic”, 6.95);

t1.chgMaterial(“marble”);

System.out.print(t1.toString());

(A) Tile@5ccd43c2

(B) 785 grey 0.0 0.0 marble 0.0

(D) 785 grey 0.0 0.0 ceramic 6.95

(E) 785 grey 0.0 0.0 marble 6.95

19.What is the output after the following client code is executed?

Tile t2 = new Tile(101, “blue”);

System.out.print(t2);

(A) Tile@5ccd43c2

(B) 101 blue 0.0 0.0 null 0.0

(D) NullPointerException

(E) There will be no output; the program will not compile.

20.The Tile class is going to be used for an application built for a small independent tile store. The owner wants the programmer to add a field for the number of unopened boxes of tile he has for each style of tile he has in stock and a method to change the value. What would be the proper declaration for this field?

(A) public static int inventory;

(B) private static double inventory;

(D) private int inventory;

(E) private int [] inventory;

21.Given the following code excerpt:

9 int[] nums = {11, 22, 33, 44, 55, 66 };

11 for (int i = 0; i < nums.length; i++)

12 nums[nums[i] / 11] = nums[i];

Determine the final contents of nums.

(A) 1, 2, 3, 4, 5, 6

(B) 11, 11, 33, 33, 55, 55

(D) 11, 22, 22, 33, 33, 55

(E) 11, 22, 33, 44, 55, 66

22.Given the following code excerpt:

13 int[] arr1 = {1, 2, 3, 4, 5, 6 };

14 int[] arr2 = arr1;

15 int last = arr1.length - 1;

17 for (int i = 0; i < arr1.length; i++)

18 arr2[i] = arr1[last - i];

20 for (int i = 0; i < arr1.length; i++)

21 System.out.print(arr1[i] + “ ”);

23 System.out.println(“ ”);

25 for (int i = 0; i < arr2.length; i++)

26 System.out.print(arr2[i] + “ ”);

Determine the statement below that reflects the resulting output.

(A) 1 2 3 4 5 6

1 2 3 4 5 6

(B) 1 2 3 4 5 6

6 5 4 4 5 6

(D) 6 5 4 4 5 6

1 2 3 4 5 6

(E) 6 5 4 4 5 6

6 5 4 4 5 6

23.Given the following code excerpt:

27 int[] arr3 = {1, 2, 3, 4, 5, 6 };

29 for (int element : arr3)

30 {

31 element *= 2;

32 System.out.print(element + “ ”);

33 }

34 System.out.println(“ ”);

36 for (int element : arr3)

37 System.out.print(element + “ ”);

Determine the statement below that reflects the resulting output.

(A) 1 2 3 4 5 6

1 2 3 4 5 6

(B) 2 4 6 8 10 12

1 2 3 4 5 6

2 4 6 8 10 12

(D) A compiler error will occur.

(E) A run-time exception will occur.

24.Given an array numbers containing a variety of integers and the following code excerpt:

38 int holdSmallest = Integer.MAX_VALUE;

39 int holdLargest = 0;

40 int a = 0;

41 int b = 0;

42 for (int i = 0; i < numbers.length; i++)

43 {

44 if (numbers[i] <= holdSmallest)

45 {

46 holdSmallest = numbers[i];

47 a = i;

48 }

49 if (numbers[i] >= holdLargest)

50 {

51 holdLargest = numbers[i];

52 b = i;

53 }

54 }

55 System.out.println(a + “ ” + b);

Determine the statement below that reflects the most successful outcome.

(A) The code will print the smallest and largest values in the numbers array.

(B) The code will print the locations of the smallest and largest values in the numbers array.

(D) The code will print the locations of the smallest value in the numbers array and the largest non-negative value in the numbers array.

(E) The code will print the locations of the smallest non-negative value in the numbers array and the largest value in the numbers array.

25.Choose the missing code below that will accurately find the average of the values in the sales array.

57 int i = 0;

58 int sum = 0;

59 for (int element : sales)

61 //Missing code

(A) {

sum += element;

}

double avg = (double) sum / sales.length;

(B) {

sum += sales[i];

}

double avg = (double) sum / sales.length;

sum += sales;

}

double avg = (double) sum / sales.length;

(D) {

sum += sales[element];

}

double avg = (double) sum / sales.length;

(E) {

sum += element[sales];

}

double avg = (double) sum / sales.length;

26.A programmer has written two different methods for a client program to swap the elements of one array with those of another array.

11 public static void swap1(int[] a1, int[] a2)

12 {

13 for (int i = 0; i < a1.length; i++)

14 {

15 int arrhold = a1[i];

16 a1[i] = a2[i];

17 a2[i] = arrhold;

18 }

19 }

21 public static void swap2(int[] a1, int[] a2) {

22 int [] arrhold = a1;

23 a1 = a2;

24 a2 = arrhold;

25 }

Which of the following statements best reflects the outcomes of the two methods?

(A) Both methods will swap the contents of the two arrays correctly in all cases.

(B) swap1 will swap the contents of the two arrays correctly only if both arrays have the same number of elements, whereas swap2 will work correctly for all cases.

(C) swap1 will swap the contents of the two arrays correctly only if both arrays have the same number of elements, whereas swap2 will never work correctly.

(D) swap1 will swap the contents of the two arrays correctly only if both arrays have the same number of elements or a2 has more elements, whereas swap2 will work correctly for all cases.

(E) Neither method will swap the contents of the two arrays correctly under any conditions.

27. Which code has declared and properly populated the given ArrayList?

ArrayList alist1 = new ArrayList(); alist1.add(“4.5”);

II.

ArrayList alist2 = new ArrayList(); alist1.add((int) 4.5);

III.

ArrayList alist3;

alist3 = new ArrayList();

alist3.add(4.5);

(A) I only

(B) I and II

(D) II and III

(E) I, II, and III

28.Given the following code excerpt:

ArrayList alist1 = new ArrayList();

int [] a1 = {2, 4, 6, 7, 8, 10, 11 };

for (int a : a1) {

alist1.add(a);

}

for (int i = 0; i < alist1.size(); i++) {

if (alist1.get(i) % 2 == 0){

alist1.remove(i);

}

System.out.println(alist1);

Determine the output.

(A) [4, 7, 10, 11]

(B) [2, 4, 7, 10, 11]

(D) [7, 11]

(E) An IndexOutOfBoundsException will occur.

Questions 29—30 refer to the following code excerpt.

2 ArrayList alist5 = new ArrayList();

3 int [] a1 = {21, 6, 2, 8, 1 };

4 for (int a : a1)

5 {

6 alist5.add(a);

7 }

8 for (int k = 0; k < alist5.size() - 1; k++)

9 {

10 for (int i = 0; i < alist5.size() - 2; i++)

11 {

12 if (alist5.get(i) > alist5.get(i + 1))

13 {

14 int hold = alist5.remove(i);

15 alist5.add(i + 1, hold);

16 }

17 }

18 }

19 System.out.println(alist5);

29.How many times will line 12 be executed?

(A) 6 times

(B) 12 times

(D) 16 times

(E) 20 times

30.What will be the final output after the code executes?

(A) [21, 8, 6, 2, 1]

(B) [6, 21, 2, 8, 1]

(D) [2, 6, 8, 21, 1]

(E) [1, 2, 6, 8, 21]

31.Given nums—a rectangular, but not necessarily square, two-dimensional array of integers—consider the following code intended to print the array:

4 int [][] arr2d = {{1, 2, 3, 4 }, {5, 6, 7, 8 }};

5 String s = “”;

6 for (int a = 0; a < arr2d[0].length; a++)

7 {

8 for (int b = 0; b < arr2d.length; b++)

9 {

10 s += arr2d [b][a] + “ ”;

11 }

12 s += “\n”;

13 }

14 System.out.print(s);

Determine the resulting output.

(A) 1 2 3 4

5 6 7 8

(B) 1 5 2 6

3 7 4 8

3 4

5 6

7 8

(D) 1 5

2 6

3 7

4 8

(E) 1

32. Given nums—a rectangular, two-dimensional array of integers—choose the code to print the entire array.

for (int r = 0; r < nums.length; r++)

{

for (int c = 0; c < nums[0].length; c++)

{

System.out.print(nums[r][c]);

}

System.out.print(“\n”);

}

II.

for (int [] row : nums)

{

for (int col : row)

{

System.out.print(col + “ ”);

}

System.out.println(“”);

}

III.

for (int r = 0; r < nums[0].length; r++)

{

for (int c = 0; c < nums.length; c++)

{

System.out.print(nums[r][c] + “ ”);

}

System.out.print(“\n”);

}

(A) I only

(B) I and II only

(D) II and III only

(E) I, II, and III

Questions 33—35 refer to the Percussion class and Xylophone class below.

public class Percussion {

private String name;

private double weight;

Percussion() {

}

Percussion(String n, double w)

{

name = n;

weight = w;

}

public String getName()

{

return name;

}

public double getWeight()

{

return weight;

}

public class Drums extends Percussion

{

}

public class Xylophone extends Percussion {

private int numberOfKeys;

Xylophone(String name, double weight, int numberOfKeys){

}

public int getNumKeys()

{

return numberOfKeys;

}

33.Which of the following is the most appropriate replacement for <missing code> in the Xylophone constructor?

(A) this.numberOfKeys = numberOfKeys;

super(name, weight);

(B) super(name, weight);

this.numberOfKeys = numberOfKeys;

numberOfKeys = this.numberOfKeys;

(D) this.numberOfKeys = numberOfKeys;

(E) numberOfKeys = this.numberOfKeys;

34.Assuming the above classes compile correctly, which of the following will NOT compile within a client program?

(A) Xylophone [] xylophones = new Xylophone[5];

(B) Percussion [] xylophones = new Xylophone[5];

System.out.println(x1.getNumKeys());

(D) Xylophone x1 = new Xylophone (“xylophone”, 65, 32);

System.out.println(x1.numberOfKeys);

(E) Drums [] drums;

35.A client program wishes to compare the two xylophone objects as follows:

Xylophone x2 = new Xylophone (“xylophone”, 80, 32);

Xylophone x3 = new Xylophone (“xylophone”, 65, 32);

The two objects should be considered “equally heavy” if and only if they have the same weight. Which of the following code excerpts accomplishes that task?

(A) if (x2.weight == x3.weight)

System.out.println(“equally heavy”);

else

System.out.println(“not equally heavy”);

(B) if (x2.weight() == x3.weight())

System.out.println(“equally heavy”);

else

System.out.println(“not equally heavy”);

System.out.println(“equally heavy”);

else

System.out.println(“not equally heavy”);

(D) if (x2.weight.equals(x3.weight))

System.out.println(“equally heavy”);

else

System.out.println(“not equally heavy”);

(E) The weights of the objects cannot be compared.

Questions 36—37 refer to the following classes.

public class Dog {

private int height;

private String size;

private String color;

Dog (int iheight, int iweight, String icolor)

{

height = iheight;

color = icolor;

if (iweight >= 65)

size = “large”;

else

size = “medium”;

}

public int getheight() {return height;}

public String getSize() {return size;}

public String getColor() {return color;}

public String toString() {return “ color is: ” + color;}

}

public class SportingDog extends Dog {

private String purpose;

SportingDog(int h, int w, String c)

{

super(h, w, c);

purpose = “hunting”;

}

public String getPurpose()

{

return purpose;

}

public class Retriever extends SportingDog{

private String type;

Retriever(String itype, String icolor, int iweight)

{

super(24, iweight, icolor);

type = itype;

}

public String toString() {return “ type: ” + type + super.toString();}

}

36.Which of the following declarations will NOT compile?

(A) Dog d1 = new SportingDog(30, 74, “Black”);

(B) Dog d2 = new Retriever(“Labrador”, “yellow”, 75);

(D) SportingDog d4 = new Dog(25, 80, “Red”);

(E) Retriever d5 = new Retriever(“Golden”, “Blonde”, 60);

37.What is the output after the execution of the following code in the client program:

Dog mason = new Retriever(“Labrador”, “chocolate”, 85);

System.out.println(mason.toString());

(A) type: Labrador

(B) type: Labrador, color is: chocolate, purpose: hunting

(D) type: Labrador, purpose: hunting, color is: chocolate

(E) type: Labrador, color is: chocolate

38.The following pow method was written to return b raised to the xth power where x > 0, but it does not work properly. Choose the changes that should be made to the method below so that it works properly.

1 public double pow(double b, int x)

2 {

3 if (x == 0)

4 return 1;

5 else

6 return b + pow(b, x - 1);

7 }

(A) Change lines 3 and 4 to:

3 if (x == 1)

4 return 1;

(B) Change lines 3 and 4 to:

3 if (x == 1)

4 return b;

6 return b * pow(b, x - 1);

(D) Both (A) and (C)

(E) Both (B) and (C)

39.What is output given the following code excerpt?

System.out.println(f(8765));

public static int f(int n)

{

if (n == 0)

return 0;

else

return f(n / 10) + n % 10;

}

(A) 5678

(B) 8765

(D) 26

(E) A run-time error

40.Choose the best solution to complete the missing code such that the code will implement a binary search to find the variable number in arr.

int number = <some number in arr>;

System.out.println(search(arr, 0, arr.length - 1, number));

public int search(int[] a, int first, int last, int sought) {

int mid = (first + last) / 2;

if (<missing code>) {

last = mid - 1;

return search(a, first, last, sought);

}

else if (<missing code>)) {

first = mid + 1;

return search(a, first, last, sought);

}

return mid;

}

(A) a[mid] > sought, a[mid] < sought

(B) a[mid] + 1 > sought, a[mid] < sought

(D) a[mid] + 1 > sought, a[mid] - 1 < sought

(E) a[mid] = sought, a[mid] = sought

END OF SECTION I

IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS SECTION.

DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.

Section II

COMPUTER SCIENCE A

SECTION II

Time—1 hour and 30 minutes

Number of Questions—4 Percent of Total Grade—50%

Directions: SHOW ALL YOUR WORK. REMEMBER THAT PROGRAM SEGMENTS ARE TO BE WRITTEN IN JAVA™.

Notes:

· Assume that the classes listed in the Java Quick Reference have been imported where appropriate.

· Unless otherwise noted in the question, assume that parameters in method calls are not null and that methods are called only when their preconditions are satisfied.

· In writing solutions for each question, you may use any of the accessible methods that are listed in classes defined in that question. Writing significant amounts of code that can be replaced by a call to one of these methods will not receive full credit.

FREE-RESPONSE QUESTIONS

1.This question involves the implementation of a simulation of rolling two dice. A client program will specify the number of rolls of the sample size and the number of faces on each of the two dice. A method will return the percentage of times the roll results in a double. Double in this case means when two dice match or have the same value (not a data type).

You will write two of the methods in this class.

public class DiceSimulation {

/** Sample size of simulation */

private int numSampleSize;

/** Number of faces on each die */

private int numFaces;

/** Constructs a DiceSimulation where sampleSize is the number of rolls to be simulated and

* faces is the number of faces on each die (some dice have more or fewer than 6 faces)

public DiceSimulation(int numSamples, int faces) {

numSampleSize = numSamples;

numFaces = faces;

}

/** Returns an integer from 1 to the number of faces to simulate a die roll */

public int roll() {

/* to be implemented in part (a) */

}

/** Simulates rolling two dice with the number of faces given, for the number of sample size

* rolls. Returns the percentage of matches that were rolled

* as an integer (eg. 0.50 would be 50).

public int runSimulation() {

/* to be implemented in part (b) */

}

The following table contains sample code and the expected results.

Statements and Expressions

Value Returned / Comment

DiceSimulation s1 = new

DiceSimulation(10, 6)

(no value returned) A DiceSimulation s1 is declared and instantiated.

s1.runSimulation()

10 rolls are simulated; only the percentage of matches is displayed. See further explanation below.

The 10 rolls might look like this (nothing is printed at this time)

Die1:	3	Die2:	4
Die1:	1	Die2:	5
Die1:	2	Die2:	2
Die1:	3	Die2:	4
Die1:	6	Die2:	6
Die1:	3	Die2:	4
Die1:	3	Die2:	3
Die1:	6	Die2:	4
Die1:	3	Die2:	1
Die1:	5	Die2:	5

The percentage the method would return is 40.

(a)Write the roll method to simulate the roll of one die.

/** Returns an integer from 1 to number of faces to simulate a die roll */

public int roll()

Begin your response at the top of a new page in the separate Free Response booklet and fill in the appropriate circle at the top of each page to indicate the question number. If there are multiple parts to this question, write the part letter with your response.

Class information for this question

public class DiceSimulation

private int numSampleSize;

private int numFaces;

public DiceSimulation (int numSamples, int faces)

public int roll()

public int runSimulation()

(b)Write the runSimulation method.

Class information for this question

public class DiceSimulation

private int numSampleSize;

private int numFaces;

public DiceSimulation (int numSamples, int faces)

public int roll()

public int runSimulation()

2.This question involves the implementation of a calorie counter system that is represented by the CalorieCount class. A CalorieCount object is created with 5 parameters:

· Daily calories limit—the recommended number of calories per day

· Daily calories intake—the number of calories a person has eaten in a day

· Grams of protein per day

· Grams of carbohydrate per day

· Grams of fat per day

The CalorieCount class provides a constructor and the following methods:

· addMeal—takes in calories, grams of protein, grams of carbs, and grams of fat from a meal and updates corresponding instance fields

· getProteinPercentage—returns the percent of protein in a given day (4 * grams protein / daily calorie intake)

· onTrack—returns true if the calorie intake does not exceed the daily calories limit, otherwise returns false

The following table contains sample code and the expected results.

Statements and Expressions	Value Returned (blank if no value)	Comment
CalorieCount sunday = new CalorieCount(1500);		Creates an instance with a 1500-calorie limit
sunday.addMeal(716, 38, 38, 45);		Adds 716 calories, 38 grams of protein, 38 grams of carbs, 45 grams of fat to the appropriate instance fields
sunday.addMeal(230, 16, 8, 16);		Adds 230 calories, 16 grams of protein, 8 grams of carbs, 16 grams of fat to the appropriate instance fields
sunday.addMeal(568, 38, 50, 24);		Adds 568 calories, 38 grams of protein, 50 grams of carbs, 24 grams of fat to the appropriate instance fields
onTrack()	false	Returns true if calorie intake does not exceed calorie limit
getProteinPercentage()	0.24	Multiplies grams of protein by 4 and then divides by calorie intake

Write the entire CalorieCount class. Your implementation must meet all specifications and conform to all examples.

3.This question involves the implementation of a travel planner system that is represented by the TravelPlan and Tour classes. A client will create Tour objects that will represent tours or activities of interest. Each Tour object is made up of an activity date, start time, end time, and name of the activity. The client will also create a TravelPlan object comprised of a destination and an ArrayList of Tours.

A partial declaration of the Tour class is shown below.

public class Tour {

private int actDate;

private int startTime; // times are represented in military format

private int endTime; // 1430 for 2:30 pm

private String activity;

/* Constructs a Tour

* All instance fields are initialized from parameters

Tour(int actDate, int startTime, int endTime, String activity)

{

/* implementation not shown

}

public int getActDate() {return actDate;}

public int getStartTime() {return startTime;}

public int getEndTime() {return endTime;}

public String getActivity() {return activity;}

A partial declaration of the TravelPlan class is shown below.

import java.util.ArrayList;

public class TravelPlan {

private String destination;

private ArrayList plans;

/* Constructs a TravelPlan

* All instance fields are initialized from parameters

TravelPlan(String destination)

{

/* to be implemented in part (a) */

}

/* Returns true if the timeframe overlaps with another Tour in plans;

* otherwise false

public boolean checkForConflicts(Tour t)

{

/* to be implemented in part (b) */

}

/* Calls checkForConflicts, if checkForConflicts returns false

* (the timeframe does not overlap), adds the tour to plans, returns true

* otherwise returns false

* Must call checkForConflicts for full credit

public boolean addTour(Tour t)

{

/* to be implemented in part (c) */

}

The following table contains sample code and the expected results.

Statements and Expressions	Value Returned (blank if no value)	Comment
TravelPlan p1 = new TravelPlan(“Capetown”);		Creates an instance with a destination “CapeTown” and an empty ArrayList of type Tour
Tour t1 = new Tour(1312020, 800, 1230, “Bungee jumping”);		Creates a Tour instance with date, start time, end time, and activity
Tour t2 = new Tour(1312020, 900, 1430, “Body surfing”);		Creates a Tour instance with date, start time, end time, and activity
p1.add(t1)	true	Checks for conflicts in plans; since there are none, adds the Tour object, returns true
p1.add(t2)	false	Checks for conflicts in plans; since there is a conflict, returns false
Tour t3 = new Tour(2012020, 900, 1200, “Shark cage diving”);		Creates a Tour instance with date, start time, end time, and activity
p1.add(t3)	true	Checks for conflicts in plans; since there are none, adds the Tour object, returns true

(a) Write the TravelPlan constructor. The constructor should initialize the destination and the plans ArrayList.

Class information for this question

public class Tour

private int actDate

private int startTime

private int endTime

private String activity

Tour(int actDate, int startTime, int endTime, String activity)

public int getActDate()

public int getStartTime()

public int getEndTime()

public String getActivity()

public class TravelPlan

private String destination;

private ArrayList plans;

public TravelPlan(String destination)

public boolean addTour(Tour t)

public boolean checkForConflicts(Tour t)

(b)Write the TravelPlan checkForConflicts method.

Class information for this question

public class Tour

private int actDate

private int startTime

private int endTime

private String activity

Tour(int actDate, int startTime, int endTime, String activity)

public int getActDate()

public int getStartTime()

public int getEndTime()

public String getActivity()

public class TravelPlan

private String destination;

private ArrayList plans;

public TravelPlan(String destination)

public boolean addTour(Tour t)

public boolean checkForConflicts(Tour t)

(c)Write the addTour method.

Class information for this question

public class Tour

private int actDate

private int startTime

private int endTime

private String activity

Tour(int actDate, int startTime, int endTime, String activity)

public int getActDate()

public int getStartTime()

public int getEndTime()

public String getActivity()

public class TravelPlan

private String destination;

private ArrayList plans;

public TravelPlan(String destination)

public boolean addTour(Tour t)

public boolean checkForConflicts(Tour t)

4.This question involves the implementation of a class seating chart. A SeatingChart object will represent a two-dimensional string array. The number of rows and columns for the array will be sent as parameters, as well as a one-dimensional array of type Name. You may assume there will be enough rows and columns to accommodate all the entries from the array.

The declaration of the Name class is shown.

public class Name

{

private String lastName;

private String firstName;

Name(String lName, String fName){}

public String getLastName() {return lastName;}

public String getFirstName() {return firstName;}

)

A partial declaration of the SeatingChart class is shown below.

public class SeatingChart {

private String [][] chart;

/** Constructs a SeatingChart having r rows and c columns. All elements contained in the

* names array should be placed randomly in the chart array using the format:

* lastName, firstName (e.g. Jolie, Angelina).

* Any locations not used in the chart should be

* initialized to the empty string.

SeatingChart(Name[] names, int rows, int cols){

/* to be implemented in part (a) */

}

/** Returns a string containing all elements of the chart array in row-major order.

* The method should return a string containing all the elements in the chart array.

* The method padWithSpaces should be called on each

* element of the chart before it is added to the string to ensure each name will be

* printed with the same length.

* Each row of the chart should be separated by a line break.

public String toString() {

/* to be implemented in part (b) */

}

/** Pads a string with spaces to ensure each string is exactly 35 characters long. */

private String padWithSpaces(String s) {

String str = s;

for (int a = s.length(); a < 35; a++) {

str += “ ”;

}

return str;

}

The following table contains sample code and the expected results.

Statements and Expressions

Value Returned / Comment

SeatingChart msJones = new SeatingChart(theNames, 4, 3);

(no value returned) A two-dimensional array is initialized with 4 rows and 3 columns. Every element in theNames is placed randomly in the chart in the following format: lastname, firstname (e.g., Washington, George). Empty string is placed in any unused locations.

System.out.println(msJones.toString);

Prints the names in the chart in row-major order. See example below:

Miller, Minnie

Indigo, Inde

Titon, Tim

Georgian, Greg

Fitzgerald, Fred

Banner, Boris

Robilard, Robbie

Dade, Ali

Lane, Lois

Brne, Jane

(a) Write the SeatingChart constructor.

Class information for this question

public class Name

private String lastName;

private String firstName;

Name(String lName, String fName)

public String getLastName() {return lastName;}

public String getFirstName() {return firstName;}

public class SeatingChart

private String [][] chart;

SeatingChart(Name[] names, int rows, int cols)

public String toString()

private String padWithSpaces(String s)

(b)Write the SeatingChart toString() method.

Class information for this question

public class Name

private String lastName;

private String firstName;

Name(String lName, String fName)

public String getLastName() {return lastName;}

public String getFirstName() {return firstName;)

public class SeatingChart

private String [][] chart;

SeatingChart(Name[] names, int rows, int cols)

public String toString()

private String padWithSpaces(String s)

STOP

END OF EXAM

Practice Test 1: Diagnostic Answer Key and Explanations

Click here to download a PDF of Diagnostic Answer Key Step 1.

PRACTICE TEST 1: DIAGNOSTIC ANSWER KEY

Let’s take a look at how you did on Practice Test 1. Follow the three-step process in the diagnostic answer key below and read the explanations for any questions you got wrong, or you struggled with but got correct. Once you finish working through the answer key and the explanations, go to the next chapter to make your study plan.

Check your answers.

Section I: Multiple Choice
Q #	Ans.	✔	Chapter #, Title
1	E		3, Objects & Primitive Data
2	D		3, Objects & Primitive Data 4, The Math Class
3	D		4, The String Class
4	A		4, The String Class 3, Objects & Primitive Data
5	C		4, The String Class
6	E		4, The String Class 3, Objects & Primitive Data
7	B		5, The If Statement
8	A		5, The If Statement 4, The String Class
9	E		5, The If Statement
10	C		5, The If Statement
11	C		6, The While Statement
12	D		6, The For Statement
13	A		6, The For Statement 5, The If Statement
14	C		3, Objects & Primitive Data 6, The For Statement
15	B		6, The For Statement
16	B		6, The While Statement 4, The String Class
17	D		6, The For Statement
18	C		7, Composition
19	B		7, Composition
20	D		7, Design & Structure
21	B		8, Primitives & Objects
22	E		8, Primitives & Objects
23	B		8, Primitives & Objects
24	D		8, Searches
25	A		8, Primitives & Objects
26	C		8, Primitives & Objects
27	E		9, Lists & ArrayLists
28	A		9, Lists & ArrayLists
29	B		6, Lists & ArrayLists
30	D		9, Lists & ArrayLists
31	D		10, 2D Arrays
32	B		10, 2D Arrays
33	B		11, Lists & ArrayLists
34	D		11, Lists & ArrayLists
35	C		11, Lists & ArrayLists
36	D		11, Lists & ArrayLists
37	E		11, Lists & ArrayLists
38	E		12, Recursion
39	D		12, Recursion
40	A		12, Recursively Traversing Arrays

Section II: Free-Response
Q #	Ans.	✔	Chapter #, Title
1a	See Explanation		4, The Math Class
1b	See Explanation		6, The For Statement 5, The If Statement 4, The Math Class
2	See Explanation		7, Design & Structure 3, Objects & Primitive Data
3a	See Explanation		9, Lists & ArrayLists 7, Design & Structure
3b	See Explanation		9, Lists & ArrayLists 5, The If Statement 7, Methods
3c	See Explanation		9, Lists & ArrayLists 7, Methods
4a	See Explanation		10, 2D Arrays 8, Primitives & Objects 3, Objects & Primitive Data
4b	See Explanation		10, 2D Arrays

Tally your correct answers from Step 1 by chapter. For each chapter, write the number of correct answers in the appropriate box. Then, divide your correct answers by the number of total questions (which we’ve provided) to get your percent correct.

CHAPTER 3 TEST SELF-EVALUATION

CHAPTER 4 TEST SELF-EVALUATION

CHAPTER 5 TEST SELF-EVALUATION

CHAPTER 6 TEST SELF-EVALUATION

CHAPTER 7 TEST SELF-EVALUATION

CHAPTER 8 TEST SELF-EVALUATION

CHAPTER 9 TEST SELF-EVALUATION

CHAPTER 10 TEST SELF-EVALUATION

CHAPTER 11 TEST SELF-EVALUATION

CHAPTER 12 TEST SELF-EVALUATION

Use the results above to customize your study plan. You may want to start with, or give more attention to, the chapters with the lowest percents correct.

PRACTICE TEST 1 EXPLANATIONS

Section I: Multiple-Choice Questions

1.E

Modulus division and division have the same order of precedence. Going from left to right, modulus (%) is first: 6 % 12 is 6. Division (/) is next and will be handled as integer division, since both terms of the operation are integers: 6 / 4 is 1. Finally, do the addition: 4 + 1 = 5. The correct answer is (E).

2.D

Anytime a double data type is used in an operation, the result will yield a double. In (A) (1.0 * 2), (B) (2.0), and (C) ((double) 2), the numerators are all 2.0. Choice (E) also yields 2.0, since the Math.sqrt method returns a double. Choice (D) attempts to cast to double too late. The expression inside the parentheses (2 / 10) yields 0 before it can be cast to a double.

3.D

First off, every string literal will be enclosed in quotation marks (“”). Next, to print a character that serves as a control character with specific meanings in Java, characters like \, “, or n to indicate a new line, each character will have to be preceded by its own \. Thus, to print “Friends”, each “ that’s printed will require its own \. Choices (A), (B), and (C) are missing the backslashes. Choice (E) has too many backslashes and will give a compiler error. Choice (D) is the correct answer because a backslash is used to indicate each control break.

4.A

The String class is immutable. Without additional assignment statements to change the values of animal1 and animal2, they will retain the values assigned in the first two lines.

5.C

Choices (B), (D), and (E) all pass String, String, int, int as arguments to the second Constellation constructor. Choice (A) passes two strings to the first constructor. Choice (C) is the correct answer, as a double cannot be passed to a parameter of type int because there may be a loss of precision.

6.E

Segments I and II will use an int parameter to update the instance field(s) of type int. Segment III will cast the double to int before updating the instance field of type int. There may be a loss of precision, but it would be a logic error, not a compiler error. The correct answer is (E), as all options will compile correctly.

7.B

Trace the code:

int x = 10;

int y = 5;

if (x == 10) //x is 10 so follow this branch

{

if (y <= 5) //y is 5 so follow this branch, add 1 to y, it is now 6

y++;

else if (y < 4)

x = 3;

else

y += 6;

} //the first if statement is complete

if (y > 5) //y is 6, so follow this branch

{

if (x != 10) //x is 10, so skip to the else

{

x = 0;

y = 0;

}

else //follow this branch, assign —5 to x

x = -5;

}

//Thus, x = —5 and y = 6

The correct answer is (B).

8.A

The rules of compareTo are as follows: if string1.compareTo(string2) < 0, then the strings are in lexicographical order, whereas if string1.compareTo(string2) > 0, then the strings are in reversed order.

1 String word1 = “frog”;

2 String word2 = “dog”;

3 String word3 = “cat”;

5 if (word1.compareTo(word2) < 0) //frog does not come before dog, skip to the else

6 if (word2.compareTo(word3) < 0)

7 System.out.println(word1 + “ ” + word2 + “ ” + word3);

8 else

9 System.out.println(word1 + “ ” + word3 + “ ” + word2);

10 else //skip to here

11 if (word1.compareTo(word2) > 0) //frog comes after dog, so follow this branch

12 if (word2.compareTo(word3) < 0) //dog does not precede cat, skip to the else

13 System.out.println(word1 + “ ” + word2 + “ ” + word3);

14 else

15 System.out.println(word1 + “ ” + word3 + “ ” + word2); //frog cat dog

16 else

17 if (word2.equals(word3))

18 System.out.println(“all the words are the same”);

19 else

20 System.out.println(“word1 and word2 are duplicates”);

The correct answer is (A).

9.E

The following is given: temp = 90 and cloudy = false. Segment I is evaluated as true: temp >= 90 (true) && !cloudy(true). Both sides of the && are true, so the entire condition is true. Option II is evaluated as true: De Morgan’s Law can be used to simplify the !(). The simplified version is temp <= 90 && !cloudy—which are both true, so the entire condition is true. Segment III is also evaluated as true. Again, De Morgan’s Law can be used to simplify the !(). The simplified version is temp <= 90 || cloudy. Since the temp is 90, the first condition is true. By short-circuit, the entire condition is true. The correct answer is (E).

10.C

Line 3 assigns dog2’s object reference to dog1. These two object variables are now pointing at the same object, the contents of which is “Beagle”. Thus, the result of if (dog1 == dog2) on line 6 is true. Line 4 creates another object whose contents are “Beagle”. Thus, the result of if (dog1 == dog3) on line 11 is false. The == is comparing whether the variables refer to the same object, not whether the content of the objects is the same. The result of if (dog1.equals(dog3)) on line 16 is true. The method .equals compares the contents of the two objects: they both contain “Beagle”. The correct answer is (C).

11.C

Choice (A) starts at 0 and will decrement to a negative index, causing an out of bounds exception. Choices (B) and (D) start the index at str1.length, which is out of bounds. The last character in a string should be referenced by length - 1. Choice (E) correctly starts at length - 1; however, the loop only continues while the index is greater than 0, missing the first character of str1. The correct answer is (C).

12.D

Analytical problems of this type are more easily solved by selecting a value and testing the results. In this case, substitute a small number such as 3 for n, and then trace the code. The outer loop executes from 1 to 2, which is 2 times. The inner loop will execute from 1 to 3, which is 3 times. The code inside the loops is simply counting by 1. The inner loop will be executed (2 times 3) 6 times, thereby adding 6 to count.

Now, substitute 3 for n in all the possible answers.

	Expression	Result
(A)	3³	27
(B)	3² — 1	8
(C)	(3 — 1)²	4
(D)	3(3 — 1)	6
(E)	3²	9

Thus, the answer to this problem is (D), n(n - 1). Analytically, you could have looked at the first loop processing from 1 to n — 1 and the second loop processing from 1 to n, and made the same assessment.

13.A

Choice (E) is eliminated with short-circuit. Line 4 looks to determine whether x != 0, but it IS 0, so logic immediately branches to the else statement on line 7. Variable x is initialized to 0, and j is initialized to 1, so line 7 multiplies j (1) times x (0) = 0 and prints the result. This eliminates (C) and (D). Both (A) and (B) are all zeroes, so the question becomes, how many 0s will be printed? Line 2 specifies j will start at 1 and end at 3, thus printing three 0s. The correct answer is (A).

14.C

The loop located at lines 4—7 prints symbol (*) 5 times.

Line 8 is a control break to the next line.

The loop located at lines 9—16 is executed 5 times. The loop within at lines 11—14 prints 5 — j spaces, so the first time through it will print 4 spaces, next time 3 spaces, and so on. (Note: this eliminates all answers except for (C).) After the spaces are printed on each line, a single symbol (*) is printed with println (which will then move to the next line).

The loop at 17—20 is the same as the first loop, printing symbol (*) 5 times. The correct answer is (C).

15.B

i will have only the value 1, j will range from 1 to 3, and k will range from 1 to 3. The three variables will be multiplied by each other and then added to the sum. The results will look like this:

i * j * k

1 * 1 * 1 = 1

1 * 1 * 2 = 2

1 * 1 * 3 = 3

1 * 2 * 1 = 2

1 * 2 * 2 = 4

1 * 2 * 3 = 6

1 * 3 * 1 = 3

1 * 3 * 2 = 6

1 * 3 * 3 = 9 The sum of which is 36.

The correct answer is (B).

16.B

The substring() method has two parameters. The first specifies where to start, the second how far to go (up to but NOT including).

The outer loop at lines 3—12 is controlled by j. j starts off at 0, eventually ending at 2.

The inner loop at lines 5—10 is controlled by k. k starts off at 3 and will execute as long as it is greater than j.

The first time through the outer loop the following will be printed:

s.substring(0, 3) prints map

s.substring(0, 2) prints ma

s.substring(0, 1) prints m

The second time through the outer loop the following will be printed:

s.substring(1, 3) prints ap

s.substring(1, 2) prints a

The final time through the outer loop the following will be printed:

s.substring(2, 3) prints p

The correct answer is (B).

17.D

Once again, it is helpful to choose a value for n to analyze the code. Choosing 3 for n, analyze the code.

I—Each time through the loop, factorial will be multiplied by 3. This does not follow the definition of a factorial. Eliminate (A) and (E), which include I.

II—The loop is controlled by j, which will range from 1 to n, in this case 3. Each time through the loop, factorial is multiplied by j, thereby producing a result of 1 × 2 × 3, which is correct. Eliminate (C).

III—A recursive solution that sends n (3) to the function

First pass is f(3) -> 3 * f(2)

Second pass is f(2) -> 2 * f(1)

Final pass is f(1) -> 1 3 × 2 × 1 will yield 6 as expected. Eliminate (B).

The correct answer is (D), as only II and III will work.

18.C

When a local variable is created, it is used instead of the instance variable. When the constructor is invoked, line 28 does not update the instance variable price. Without specifying this.price = price, the local parameter is assigned the same value it already holds. Thus, (D) and (E) are eliminated. Choice (A) is eliminated because the toString method has been defined in the Tile class to print the instance variables (not the object reference). The chgMaterial(mat) method at line 30 also updates a local variable rather than the instance variable, eliminating (B). The correct answer is (C).

19.B

If a print statement is passed an object, its toString() method will be invoked. This eliminates all answers except (B), which is the correct answer.

20.D

A static variable would be used for something that would belong to the entire class. Since inventory needs to exist for each style, it cannot be static, but it must be an instance of the class, eliminating (A) and (B). Choice (C) is eliminated because the keyword final is used only for constants that do not change value, but the owner has also asked for a method to change the value. Since styleNumber is an instance field, it implies that a separate instance is created for each style. Thus an array is not needed, eliminating (E). The correct answer is (D).

21.B

The array is initialized as {11, 22, 33, 44, 55, 66};

First pass: nums[nums[0] / 11] = nums[0];

nums[11 / 11] = nums[0];

nums[1] = nums[0]; The array is now: {11, 11, 33, 44, 55, 66};

Second pass: nums[nums[1] / 11] = nums[1];

nums[11 / 11] = nums[1];

nums[1] = nums[1]; The array is unchanged: {11, 11, 33, 44, 55, 66};

Third pass: nums[nums[2] / 11] = nums[2];

nums[33 / 11] = nums[2];

nums[3] = nums[2]; The array is now: {11, 11, 33, 33, 55, 66};

Fourth pass: nums[nums[3] / 11] = nums[3];

nums[33 / 11] = nums[3];

nums[3] = nums[3]; The array is unchanged: {11, 11, 33, 33, 55, 66};

Fifth pass: nums[nums[4] / 11] = nums[4];

nums[55 / 11] = nums[4];

nums[5] = nums[4]; The array is now: {11, 11, 33, 33, 55, 55};

Sixth pass: nums[nums[5] / 11] = nums[5];

nums[55 / 11] = nums[5];

nums[5] = nums[5]; The array is unchanged: {11, 11, 33, 33, 55, 55};

The correct answer is (B).

22.E

Line 14 assigns the arr1 object reference to arr2 object reference. Thus, both variables are now pointing to the exact same array in memory.

The loop at lines 17—18 is the only code that modifies the array.

both arr1 and arr2: {1, 2, 3, 4, 5, 6}; last = 5

arr2[i] = arr1[last - i];

first pass: arr2[0] = arr1[5 - 0]; {6, 2, 3, 4, 5, 6}

second pass: arr2[1] = arr1[5 - 1]; {6, 5, 3, 4, 5, 6}

third pass: arr2[2] = arr1[5 - 2]; {6, 5, 4, 4, 5, 6}

fourth pass: arr2[3] = arr1[5 - 3]; {6, 5, 4, 4, 5, 6}

fifth pass: arr2[4] = arr1[5 - 4]; {6, 5, 4, 4, 5, 6}

last pass: arr2[5] = arr1[5 - 5]; {6, 5, 4, 4, 5, 6}

The correct answer is (E).

23.B

The for loop on line 29 creates a local variable named element which will hold each value of arr3 without having to use an index. Modifying this local variable does not modify the individual contents within the array. The loop multiplies each element by 2, printing it as it does so.

2, 4, 6, 8, 10, 12

The loop at line 36 prints the contents of the array, which remain unchanged:

1, 2, 3, 4, 5, 6

The correct answer is (B).

24.D

Since index i is assigned to variables a and b, it is locations that are being printed. This eliminates (A). Scan the remaining answers and make a chart to help you understand the possibilities.

	Location of:	Location of:
(B)	Smallest integer	Largest integer
(C)	Smallest non-negative integer	Largest non-negative integer
(D)	Smallest integer	Largest non-negative integer
(E)	Smallest non-negative integer	Largest integer

The variable holdSmallest is initialized with Integer.MAX_ VALUE, which is the largest integer an int field may hold. Thus, the code will work to find the smallest number in the array even if it is a negative number. This eliminates (C) and (E). The variable holdLargest is initialized to 0, so when looking for a larger integer, it will only be replaced if it is larger than 0, or in other words, a non-negative integer. This eliminates (B). The correct answer is (D).

25.A

Choice (B) is eliminated because there is no increment to variable i. Choice (C) is eliminated because without an index, it implies the entire array (not each element) is being added to sum over and over. Choice (D) cannot use element, because it will contain the contents of a location within the array, rather than a location. Choice (E) uses the variable name of the array as the index. Choice (A) is correct because it uses the temporary variable element, which will actually hold the contents of each location within the array.

26.C

Examining the code of swap1, you can see it will work only if the arrays are the same length. There is no accommodation for one array being longer than the other. In fact, if a1 is longer, there will be an out of bounds error on the second array. This eliminates (A), (D), and (E). The code of swap2 does not work. Array variables hold a reference to the array, not to the actual elements. This eliminates (B). The correct answer is (C).

27.E

Segment I declares an ArrayList of type String and then adds “4.5”, which is a String. It is correct. Eliminate (D), which does not include I.

Segment II declares an ArrayList of type Integer and then casts 4.5 to an int before adding it to the ArrayList, which is acceptable. It is correct. Eliminate (A) and (C).

Segment III declares an ArrayList variable and then completes the declaration of the ArrayList as type double on the next line. It then adds a double to the ArrayList, which is correct. Eliminate (B).

The correct answer is (E).

28.A

The first loop loads the contents of the array into the ArrayList. The next loop begins to remove elements if those elements are even. The loop will continue to run until it reaches the size of the ArrayList. As elements of the ArrayList are removed, the size will decrease, so there is no risk of going out of bounds. Eliminate (E). However, the index i will occasionally skip elements because of the renumbering that takes place.

[2, 4, 6, 7, 8, 10, 11];

i = 0 The 2 is even, so it is removed; the array is now

[4, 6, 7, 8, 10, 11];

i = 1 Notice the 4 will now be skipped. The 6 is even, so it is removed; the array is now

[4, 7, 8, 10, 11];

i = 2 The 8 is even, so it is removed; the array is now

[4, 7, 10, 11];

i = 3 The 10 has been skipped. The 11 is odd, so the array stays the same:

[4, 7, 10, 11];

The correct answer is (A).

29.B

The size of the array is 5, so size — 1 is 4. The outer loop executes 4 times (0—3).

size — 2 is 3. The inner loop executes 3 times (0—2).

Since line 12 is executed every time the inner loop is executed, it will be executed (4)(3) = 12 times. The correct answer is (B).

30.D

The inner loop does not go far enough to process the entire array. size is 5, and size — 1 is 4, so the index can only be less than 4, stopping at index 3. The last entry in the ArrayList will never be sorted. The sort makes 4 passes through the ArrayList. The passes will look as follows:

0 [6, 21, 2, 8, 1]

1 [6, 2, 21, 8, 1]

2 [6, 2, 8, 21, 1]

3 [2, 6, 8, 21, 1]

The correct answer is (D).

31.D

The array is printed in column-major order. The outer loop runs from 0 to row length — 1 (the number of columns). The inner loop runs from 0 to the length of the array (which means the number of rows).

The original array is

1 2 3 4

5 6 7 8

The outer loop starts with column 0, prints [0] [0]: 1 [1] [0]: 5

The outer loop increments to column 1 [0] [1]: 2 [1] [1]: 6

The outer loop increments to column 2 [0] [2]: 3 [1] [2]: 7

The outer loop increments to column 3 [0] [3]: 4 [1] [3]: 8

The correct answer is (D).

32.B

Segment III will go out of bounds. The r (rows) will iterate as many times as there are columns. If there are fewer rows than columns, the index will go out of bounds. The correct answer is (B).

33.B

Since name and weight are instance variables in the Percussion class, values for those variables should be passed while calling super. The call to super must be the first line in a method. Thus, (A), (D), and (E) are eliminated. The assignment statement of numberOfKeys is reversed in (C). The local variable is being initialized by the instance field. The correct answer is (B).

34.D

The variable numberOfKeys is not visible outside the Xylophone class. Choices (A) and (B) are simply creating arrays of Xylophone objects. Choice (C) creates a xylophone object and then uses the proper accessor method to print the number of keys. Choice (E) declares a variable for an array of type Drums. Choice (D) attempts to print a private instance variable without using an accessor method. It will not comply, so the correct answer is (D).

35.C

The accessor method getWeight() will return the weight of each instance so that they can be compared. Choice (A) is incorrect because the weight field is not visible. Choice (B) is not correct because weight() is not a defined method. Choice (D) is not correct because not only is weight not visible, but .equals is not used to compare primitive types. Choice (E) is incorrect because (C) compares the fields correctly. The correct answer is (C).

36.D

Use the IS-A relationship to check the solutions:

(A)—SportingDog is a Dog (yes)

(B)—Retriever is a Dog (yes)

(C)—Retriever is a Sporting Dog (yes)

(D)—Dog is a Sporting Dog (no, the relationship is the opposite: not all dogs are sporting dogs)

(E)—Retriever is a Retriever (yes)

The correct answer is (D).

37.E

The Retriever toString() method is invoked first, returning type: Labrador + super.toString().

No toString() method is found in SportingDog, but a toString() method is found in Dog, adding color is: chocolate to the print line.

The correct answer is (E).

38.E

Try substituting numbers for the variables. Try finding 3² by making b = 3, x = 2. The solution is found by multiplying 3 × 3.

The base case will be 3 (when the exponent is 1). This should imply that the if statement at line 3 should be

if (x == 1) return b;

There is another error on line 6. Line 6 is using addition, when raising to a power is multiplying the base x times. Thus, the + sign should be changed to multiplication.

After making the changes in the code, it is advisable to test it to ensure it works:

b = 3, x = 2

1 public double pow(double b, int x)

2 {

3 if (x == 1)

4 return b;

5 else

6 return b * pow(b, x - 1);

7 }

f(3, 2)

3 * pow(3, 1)

3 * 3 = 9

The correct answer is (E).

39.D

It is best to walk the code.

System.out.println(f(8765));

public static int f(int n)

{

if (n == 0)

return 0;

else

return f(n / 10) + n % 10;

}

f(8765)

f(876) + 5

f(87) + 6

f(8) + 7

0 + 8

8 + 7 + 6 + 5 = 26

The correct answer is (D).

40.A

If sought is less than the element at index mid, the beginning of the array should be searched. The location of the middle of the array, mid — 1, should be assigned to last. If sought is greater than the element at index mid, mid + 1 should be assigned to first so that the latter half of the array can be searched. This process should be repeated until sought is found. The correct answer is (A).

Section II: Free-Response Questions

1.DiceSimulation—Canonical Solution

(a)public int roll() {

return (int)(Math.random() * numFaces + 1);

}

(b)public int runSimulation()

{

int die1 = 0;

int die2 = 0;

int countDouble = 0;

for (int i = 0; i < numSampleSize; i++) {

die1 = roll();

die2 = roll();

if (die1 == die2) {

countDouble++;

}

return (int)((1.0 * countDouble / numSampleSize) * 100);

}

DiceSimulation Rubric

Part (a)
+3	roll method
	+1	Math.random() or the Random class is used
	+1	multiplied by numFaces + 1
	+1	result of computation is cast to int appropriately and returned
Part (b)
+6	runSimulation method
	+1	local variables are declared and initialized for the two dice
	+1	roll is used to give the dice values
	+1	a loop is used to execute sample size times (no more, no less)
	+1	the values of die1 and die2 are compared with ==, doubles are counted appropriately
	+1	the percentage of doubles is calculated (avoiding integer division), multiplied by 100
	+1	percentage is returned as an int

2. CalorieCount—Canonical Solution

public class CalorieCount {

private int numCaloriesLimit;

private int numCaloriesIntake;

private int gramsProtein;

private int gramsCarbohydrate;

private int gramsFat;

public CalorieCount (int numCal) {

numCaloriesLimit = numCal;

numCaloriesIntake = 0;

gramsProtein = 0;

gramsCarbohydrate = 0;

gramsFat = 0;

}

public void addMeal(int calories, int protein, int carbs, int fat) {

numCaloriesIntake += calories;

gramsProtein += protein;

gramsCarbohydrate += carbs;

gramsFat += fat;

}

public double getProteinPercentage() {

return 4.0 * gramsProtein / numCaloriesIntake;

}

public boolean onTrack() {

return numCaloriesIntake <= numCaloriesLimit;

}

CalorieCount Rubric

+1	Declares all appropriate private instance variables
+2	Constructor
	+1	declares header: public CalorieCount (int calorieLimit)
	+1	uses parameters and appropriate values to initialize instance variables
+2	addMeal method
	+1	declares header: public void addMeal(int calories, int protein, int carbs, int fat)
	+1	updates instance variables appropriately
+2	getProteinPercentage method
	+1	declares header: public double getProteinPercentage()
	+1	calculation and return: return 4.0 * gramsProtein / numCaloriesIntake;
+2	onTrack method
	+1	declares header: public boolean onTrack()
	+1	correctly returns true or false
		e.g., return numCaloriesIntake <= numCaloriesLimit;

(a)TravelPlan(String destination){

this.destination = destination;

plans = new ArrayList();

}

(b)public boolean checkForConflicts(Tour t) {

for (int i = 0; i < plans.size(); i++)

{

if (t.getActDate() == plans.get(i).getActDate())

{

int plannedStart = plans.get(i).getStartTime();

int plannedEnd = plans.get(i).getEndTime();

if ((t.getStartTime() >= plannedStart && t.getStartTime() < plannedEnd) || (t.getEndTime() > plannedStart && t.getEndTime() < plannedEnd))

return true;

if (t.getStartTime() <= plannedStart && t.getEndTime() >= plannedEnd)

return true;

}

return false;

}

(c)public boolean String addTour(Tour t) {

if (checkForConflicts(t)) {

return false;

}

plans.add(t);

return true;

}

TravelPlan Rubric

Part (a)
+3	Constructor
	+1	constructor uses class name TravelPlan
	+1	updates destination instance field appropriately (uses this.)
	+1	creates ArrayList appropriately
Part (b)
+4	checkForConflicts method
	+1	uses a loop to traverse every item in the ArrayList (no bounds errors)
	+1	uses .get(index) to access the object in the ArrayList
	+1	uses getStartTime () and getEndTime() to access the private fields in the Tour object
	+1	uses appropriate logic to determine whether there is a time conflict on the same day; returns true if there is a conflict, false otherwise
Part (c)
+2	addTour method
	+1	calls checkForConflict method to determine whether there is a conflict (loses this point if it instead writes the logic to determine whether there is a conflict in this method), adds tour if there is no conflict
	+1	returns true if tour is added, or false if tour is not added

4. SeatingChart—Canonical Solution

(a)SeatingChart(Name[] names, int r, int c)

{

chart = new String[r][c];

for (int i = 0; i < chart.length; i++)

{

for (int j = 0; j < chart[0].length; j++)

{

chart[i][j] = “”;

}

int count = 0;

int i = (int) (Math.random() * names.length);

int row = i / c;

int col = i % c;

while (count < names.length) {

while (!chart[row][col].equals(“”))

{

i = (int) (Math.random() * names.length);

row = i / c;

col = i % c;

}

chart[row][col]= names[count].getLastName() + “, ” +

names[count].getFirstName();

count ++;

}

(b)public String toString()

{

String str = “”;

for (int a = 0; a < chart.length; a++) {

for (int b = 0; b < chart[a].length; b++) {

str += padWithSpaces(chart[a][b]);

}

str += “\n”;

}

return str;

}

SeatingChart Rubric

Part (a)
+6	Constructor
	+1	chart is initialized using rows and columns passed in parameters
	+1	random numbers are generated in the correct range
	+1	a unique random number is used each time to place the name in the 2D array (duplicate names are avoided) and all names are placed (none are skipped)
	+1	row and column in the seating chart are derived properly from the random number
	+1	the name is stored in chart as a string (last name, (comma), first name), e.g., Washington, George
	+1	any unused spaces left in the array should be initialized to the empty string (not null)
Part (b)
+3	toString method
	+1	builds a single string with all names from the 2D array, calling padWithSpaces to make all names an equal length
	+1	“\n” creates a line break after each row
	+1	returns a string

HOW TO SCORE PRACTICE TEST 1

Click here to download a PDF of How to Score Practice Test 1.

AP Computer Science A Prep, 2024 - Rob Franek 2023

Practice Test 1Part II: Practice Test 1

Practice Test 1
Part II: Practice Test 1