AP Computer Science A Prep, 2024 - Rob Franek 2023

Practice Test 3
Part VI: Practice Tests

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 of >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 the element at 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 Java 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.Consider the following methods.

public void trial()

{

int a = 10;

int b = 5;

doubleValues(a, b);

System.out.print(b);

System.out.print(a);

}

public void doubleValues(int c, int d)

{

c = c * 2;

d = d * 2;

System.out.print(c);

System.out.print(d);

}

What is printed as the result of the call trial()?

(A) 2010

(B) 2010105

(D) 20102010

(E) 20101020

2.Consider the following method.

/**

* Precondition: a > b > 0

public static int mystery(int a, int b)

{

int d = 0;

for (int c = a; c > b; c--)

{

d = d + c;

}

return d;

}

What is returned by the call mystery(x, y)?

(A) The sum of all integers greater than y but less than or equal to x

(B) The sum of all integers greater than or equal to y but less than or equal to x

(D) The sum of all integers greater than or equal to y but less than x

(E) The sum of all integers less than y but greater than or equal to x

3.Consider the following method.

public void mystery (int n)

{

int k;

for (k = 0; k < n; k++)

{

mystery(k);

System.out.print (k);

}

What is printed by the call mystery(3) ?

(A) 0123

(B) 00123

(D) 00100123

(E) 001001200100123

4.Consider an array of integers.

4 10 1 2 6 7 3 5

If selection sort is used to order the array from smallest to largest values, which of the following represents a possible state of the array at some point during the selection sort process?

(A) 1 4 10 2 3 6 7 5

(B) 1 2 4 6 10 7 3 5

(D) 4 3 1 2 6 7 10 5

(E) 5 3 7 6 2 1 10 4

5.Consider the following code segment:

int k;

int A[];

a = new int [7];

for (k = 0; k < A.length; k++)

{

A[k] = A.length — k;

}

for (k = 0; k < A.length — 1; k++)

{

A[k + 1] = A[k];

}

What values will A contain after the code segment is executed?

(A) 1 1 2 3 4 5 6

(B) 1 2 3 4 5 6 7

(D) 7 7 6 5 4 3 2

(E) 7 7 7 7 7 7 7

Questions 6—7 refer to the following two classes.

public class PostOffice

{

// constructor initializes boxes

// to length 100

public PostOffice()

{ /* implementation not shown */}

// returns the P.O. box based on the given P.O. box number

// 0 <= theBox < getNumBoxes ()

public Box getBox(int theBox)

{ /* implementation not shown */}

// returns the number of P.O. boxes

public int getNumBoxes()

{ /* implementation not shown */}

// private data members and

// other methods not shown

}

public class Box

{

// constructor

public Box()

{ /* implementation not shown */}

// returns the number of this box

public int getBoxNumber()

{ /* implementation not shown */}

// returns the number of pieces

// of mail in this box

public int getMailCount()

{ /* implementation not shown */}

// returns the given piece of mail

// 0 <= thePiece < getMailCount ()

public Mail getMail(int thePiece)

{ /* implementation not shown */}

// true if the box has been assigned

// to a customer

public boolean isAssigned()

{ /* implementation not shown */}

// true if the box contains mail

public boolean hasMail()

{ /* implementation not shown */}

// private data members and

// other methods not shown

}

public class Mail

{

// private members, constructors, and

// other methods not shown

}

6.Consider the following code segment:

PostOffice p[];

p = new PostOffice[10];

Assuming that the box has been assigned and that it has at least four pieces of mail waiting in it, what is the correct way of getting the fourth piece of mail from the 57th box of the 10th post office of p?

(A) Mail m = p[10].getBox(57).getmail(4);

(B) Mail m = p[9].getBox(56).getMail(3);

(D) Mail m = getMail(getBox(p[9], 560, 3);

(E) Mail m = new Mail(10, 57, 4);

7.Consider the incomplete function printEmptyBoxes given below. printEmptyBoxes should print the box numbers of all of the boxes that have been assigned to a customer but do not contain mail.

public void printEmptyBoxes (PostOffice[] p)

{

for (int k = 0; k < p.length - 1; k++)

{

for (int x = 0; x < p[k].getNumBoxes() - 1; x++)

{

/ * missing code * /

}

Which of the following could be used to replace / * missing code * / so that printBoxesWithoutMail works as intended?

(A) if (p[k].getBox(x).isAssigned() &&

!p[k].getBox(x).hasMail())

{

System.out.println(p[k].getBox(x).getBoxNumber());

}

(B) if (p[x].getBox(k).isAssigned() &&

!p[x].getBox(k).hasMail())

{

System.out.println(p[x].getBox(k).getBoxNumber());

}

!p[k].getBox(x).hasMail())

{

System.out.println (p[k].getBoxNumber (x));

}

(D) if (p[x].getBox(k).isAssigned() &&

!p[x].getBox (k).hasMail())

{

System.out.println(p[x].getBoxNumber(k));

}

(E) if (p[x].getBox(k).isAssigned() &&

p[x].getBox(k).getMail() == 0)

{

System.out.println(k);

}

8.Assume that a and b are boolean variables that have been initialized. Consider the following code segment.

a = a && b;

b = a || b;

Which of the following statements is always true?

I. The final value of a is the same as the initial value of a.

II. The final value of b is the same as the initial value of b.

III. The final value of a is the same as the initial value of b.

(A) I only

(B) II only

(D) I and II only

(E) II and III only

9.Consider the following code segment.

int x;

x = 53;

if (x > 10)

{

System.out.print(“A”);

}

if (x > 30)

{

System.out.print(“B”);

}

else if (x > 40)

{

System.out.print(“C”);

}

if (x > 50)

{

System.out.print (“D”);

}

if (x > 70)

{

System.out.print (“E”);

}

What is the output when the code is executed?

(A) A

(B) D

(D) ABCD

(E) ABCDE

10.Consider the following code segment:

int j;

int k;

for (j = -2; j <= 2; j = j + 2)

{

for (k = j; k < j + 3; k++)

{

System.out.print(k + “ ”);

}

What is the output when the code is executed?

(A) -2 -1 0

(B) -2 -1 0 1 2

(D) -2 0 2

(E) -2 -1 0 0 1 2 2 3 4

11.Consider the following method.

public void mystery (int count, String s)

{

if (count <= 0)

{

return;

}

if (count % 3 == 0)

{

System.out.print(s + “--” + s);

}

else if (count % 3 == 1)

{

System.out.print(s + “-” + s);

}

else

{

System.out.print(s);

}

mystery(count - 1, s);

}

What is outputted by the call mystery(5, “X”)?

(A) XX-XX--XXX-X

(B) XX-XX-XX-XX

(D) XX-XXX--XXX-XX

(E) XXXXX

Questions 12—13 refer to the following classes and method descriptions.

Class Table has a method getPrice, which takes no parameters and returns the price of the table.

Class Chair also has a method getPrice, which takes no parameters and returns the price of the chair.

Class DiningRoomSet has a constructor which is passed a Table object and an ArrayList of Chair objects. It stores these parameters in its private data fields myTable and myChairs.

Class DiningRoomSet has a method getPrice, which takes no parameters and returns the price of the dining room set. The price of a dining room set is calculated as the sum of the price of its table and all of its chairs.

12.What is the correct way to define the signature of the constructor for the DiningRoomSet class?

(A) public void DiningRoomSet(Table t, ArrayList, chairs)

(B) public DiningRoomSet(Table t, ArrayList chairs)

(D) public DiningRoomSet(Table t, ArrayList Chair Chairs)

(E) public DiningRoomSet(Table t, Chair Chairs)

13.What is the correct way to implement the getPrice method of the DiningRoomSet class?

(A) public double getPrice(Table t, ArrayList chairs)

{

return t.getPrice() + chairs.getPrice();

}

(B) public double getPrice(Table t, ArrayList chairs)

{

return myTable.getPrice() + myChairs.getPrice();

}

{

return myTable.getPrice() + myChairs.getPrice();

}

(D) public double getPrice()

{

double result = myTable.getPrice();

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

{

result += ((Chair)myChairs.get(k)).getPrice();

}

return result;

}

(E) public double getPrice()

{

double result = myTable.getPrice();

for (int k = 0; k < myChairs.length - 1; k++)

{

result += ((Chair)myChairs[k]).getPrice();

}

return result;

}

14.Consider the following output:

6 5 4 3 2 1

5 4 3 2 1

4 3 2 1

3 2 1

2 1

Which of the following code segments produces the above output when executed?

(A) for (int j = 6; j < 0; j--)

{

for (int k = j; k > 0; k--)

{

System.out.print(k + “ ”);

}

System.out.println(“ ”);

}

(B) for (int j = 6; j >= 0; j--)

{

for (int k = j; k >= 0; k--)

{

System.out.print(k + “ ”);

}

System.out.println(“ ”);

}

{

for (int k = 6 - j; k > 0; k--)

{

System.out.print(k + “ ”);

}

system.out.println(“ ”);

}

(D) for (int j = 0; j < 6; j++)

{

for (int k = 7 - j; k > 0; k--)

{

System.out.print(k + “ ”);

}

System.out.println(“ ”);

}

(E) for (int j = 0; j < 6; j++)

{

for (int k = 6 - j; k >= 0; k--)

{

System.out.print(k + “ ”);

}

System.out.println(“ ”);

}

15.Consider the following code segment.

List list = new ArrayList();

list.add(new Integer(7));

list.add(new Integer(6));

list.add(1, new Integer(5));

list.add(1, new Integer(4));

list.add(new Integer(3));

list.set(2, new Integer(2));

list.add(1, new Integer(1));

System.out.println(list);

What is printed as a result of executing this code segment?

(A) [1, 4, 2, 7, 6, 3]

(B) [7, 1, 4, 2, 6, 3]

(D) [7, 6, 2, 4, 3, 1]

(E) [7, 1, 2]

16.Consider the following declarations.

public class Animal

{

String makeSound()

{

// Implementation not shown

}

String animalType()

{

// Implementation not shown

}

public static class Dog extends Animal

{

public String makeSound(Animal a)

{

// Implementation not shown

}

Which of the following methods must be included in the declaration of the Dog class in order for the class to successfully compile?

I. public String makeSound()

II. public String animalType()

III. public String animalType(Animal b)

(A) I only

(B) II only

(D) II and III only

(E) None

17.Consider the following two classes.

public class Fish

{

public String endoskeleton = “bone”;

public void action()

{

System.out.println(“splash splash”);

}

public class Shark extends Fish

{

public void action()

{

System.out.println(“chomp chomp”);

}

public String endoskeleton = “cartilage”;

}

Which of the following is the correct output after the following code segment is executed?

Fish Bob = new Shark();

System.out.println(Bob.endoskeleton);

Bob.action();

(A) bone

chomp chomp

(B) bone

splash splash

(D) cartilage

chomp chomp

(E) cartilage

splash splash

chomp chomp

Questions 18—19 refer to the following incomplete method.

The following insertSort method sorts the values in an integer array, sort, in ascending order.

1 public static void insertSort(int[] sort)

2 {

3 for (int index = 1; index < sort.length; index++)

4 {

5 int temp = sort[index];

6 while (index > 0 && sort[index - 1] > temp)

7 {

8 / * missing code * /

9 }

10 sort[index] = temp;

11 }

12 }

18.Which of the following can be used to replace / * missing code * / so that the insertSort method will execute properly?

(A) sort[index] = sort[index - 1];

index++;

(B) sort[index - 1] = sort[index];

index--;

index++;

(D) sort[index] = sort[index - 1];

index--;

(E) sort[index] = sort[index + 1];

index--;

19.Assuming that the / * missing code * / is implemented properly, what change can be made to the code in order for the array to be sorted in descending order?

(A) Replace Line 6 with: while (index < 0 && sort[index - 1] > temp)

(B) Replace Line 6 with: while (index < 0 && sort[index - 1] < temp)

(D) Replace Line 3 with: for (int index = sort.length - 1; index > 0; index--)

(E) Replace Line 3 with: for (int index = 1; index > 0; index--)

20.Which of the following arrays would be sorted the slowest using insertion sort?

(A) [3 4 6 2 7 3 9]

(B) [3 2 5 4 6 7 9]

(D) [2 3 4 5 6 7 9]

(E) [9 3 2 4 5 7 6]

Questions 21—23 refer to the following incomplete class declaration used to represent fractions with integer numerators and denominators.

public class Fraction

{

private int numerator;

private int denominator;

public Fraction()

{

numerator = 0;

denominator = 1;

}

public Fraction(int n, int d)

{

numerator = n;

denominator = d;

}

// postcondition: returns the

// numerator

public int getNumerator()

{ /* implementation not shown */ }

// postcondition: returns the

// denominator

public int getDenominator()

{ /* implementation not shown*/ }

// postcondition: returns the greatest

// common divisor of x and y

public int gcd(int x, int y)

{ /* implementation not shown*/ }

// postcondition: returns the Fraction

// that is the result of multiplying

// this Fraction and f

public Fraction multiply(Fraction f)

{ /* implementation not shown */ }

//…other methods not shown

}

21.Consider the method multiply of the Fraction class.

// postcondition: returns the Fraction

// that is the result of multiplying

// this Fraction and f

public Fraction multiply(Fraction f)

{ / * missing code * / }

Which of the following statements can be used to replace / * missing code * / so that the multiply method is correctly implemented?

I. return Fraction(

numerator * f.getNumerator(),

denominator * f.getDenominator());

II. return new Fraction(

numerator * f.numerator,

denominator * f.denominator());

III. return new Fraction(

numerator * f.getNumerator(),

denominator * f.getDenominator());

(A) I only

(B) II only

(D) I and III only

(E) II and III only

22.Consider the use of the Fraction class to multiply the fractions and . Consider the following code:

Fraction fractionOne;

Fraction fractionTwo;

Fraction answer;

fractionOne = new Fraction(3, 4);

fractionTwo = new Fraction(7, 19);

/ * missing code * /

Which of the following could be used to replace / * missing code * / so that the answer contains the result of multiplying fractionOne by fractionTwo?

(A) answer = fractionOne * fractionTwo;

(B) answer = multiply(fractionOne, fractionTwo);

(D) answer = new Fraction(fractionOne, fractionTwo);

(E) answer = (fractionOne.getNumerator() * fractionTwo.getNumerator()) / (fractionOne.getDenominator() * fractionTwo.getDenominator());

23.The following incomplete class declaration is intended to extend the Fraction class so that fractions can be manipulated in reduced form (lowest terms).

Note that a fraction can be reduced to lowest terms by dividing both the numerator and denominator by the greatest common divisor (gcd) of the numerator and denominator.

public class ReducedFraction extends Fraction

{

private int reducedNumerator;

private int reducedDenominator;

//…constructors and other methods not shown

}

Consider the following proposed constructors for the ReducedFraction class:

I. public ReducedFraction( )

{

reducedNumerator = 0;

reducedDenominator = 1;

}

II. public ReducedFraction(int n, int d)

{

numerator = n;

denominator = d;

reducedNumerator = n / gcd(n, d);

reducedDenominator = d / gcd(n, d);

}

III. public ReducedFraction(int n, int d)

{

super(n, d);

reducedNumerator = n / gcd(n, d);

reducedDenominator = d / gcd(n, d);

}

Which of these constructor(s) would be legal for the ReducedFraction class?

(A) I only

(B) II only

(D) I and III only

(E) II and III only

24.Consider s1 and s2 defined as follows.

String s1 = new String(“hello”);

String s2 = new String(“hello”);

Which of the following is/are correct ways to see whether s1 and s2 hold identical strings?

I. if (s1 == s2)

/** s1 and s2 are identical */

II. if (s1.equals(s2))

/** s1 and s2 are identical */

III. if (s1.compareTo(s2) == 0)

/** s1 and s2 are identical */

(A) I only

(B) II only

(D) II and III only

(E) I, II, and III

25.Consider the following variable and method declarations:

String s;

String t;

public void mystery (String a, String b)

{

a = a + b;

b = b + a;

}

Assume that s has the value “Elizabeth” and t has the value “Andrew” and mystery (s, t) is called. What are the values of s and t after the call to mystery?

(A) s: Elizabeth, t: Andrew

(B) s: ElizabethAndrew, t: AndrewElizabeth

(D) s: ElizabethAndrew, t: ElizabethAndrewAndrew

(E) s: ElizabethAndrewElizabeth, t: AndrewElizabethAndrew

26.Consider the following incomplete and incorrect class and interface declarations:

public class ComparableObject

{

public int compareTo(Object o)

{

//method body not shown

}

//other methods and variables not shown

}

public class Point extends ComparableObject

{

private int x;

private int y;

public boolean compareTo(Point other)

{

return (x == other.x &&

y == other.y);

}

//…constructors and other methods

// not shown

}

For which of the following reasons, if any, is the above class structure incorrect?

I. Objects may not access private data fields of other objects in the same class.

II. The ComparableObject class requires that compareTo be passed as an Object rather than a Point.

III. The ComparableObject class requires that compareTo return an int rather than a boolean.

(A) I only

(B) III only

(D) II and III only

(E) None, the above class declarations are correct.

27.Consider the following abstraction of a for loop where <1>, <2>, <3>, and <4> represent legal code in the indicated locations:

for (<1>; <2>; <3>)

{

<4>

}

Which of the following while loops has the same functionality as the above for loop?

(A) <1>;

while (<2>)

{

<3>;

<4>

}

(B) <1>;

while (<2>)

{

<4>

<3>;

}

while (!<2>)

{

<3>;

<4>

}

(D) <1>;

while (! <2>)

{

<4>

<3>;

}

(E) <1>;

<3>;

while (<2>)

{

<4>

<3>;

}

28.Consider the following expression:

a / b + c — d % e * f

Which of the expressions given below is equivalent to the one given above?

(A) ((a / b) + (c — d)) % (e * f)

(B) ((((a / b) + c) — d) % e) * f

(D) (a / ((b + c) — d) % e) * f

(E) ((a / b) + c) — ((d % e) * f)

29.Assume that a program declares and initializes x as follows:

String[] x ;

x = new String[10] ;

initialize(x); // Fills the array x with

// valid strings each of

// length 5

Which of the following code segments correctly traverses the array and prints out the first character of all ten strings followed by the second character of all ten strings, and so on?

I. int i;

int j;

for (i = 0; i < 10; i++)

for (j = 0; j < 5; j++)

System.out.print(x[i].substring(j, j + 1));

II. int i;

int j;

for (i = 0; i < 5; i++)

for (j = 0; j < 10; j++)

System.out.print(x[j].substring(i, i + 1));

III. int i;

int j;

for (i = 0; i < 5; i++)

for (j = 0; j < 10; j++)

System.out.print(x[i].substring(j, j + 1));

(A) I only

(B) II only

(D) II and III only

(E) I, II, and III

30.Consider the following declaration and assignment statements:

int a = 7;

int b = 4;

double c;

c = a / b;

After the assignment statement is executed, what’s the value of c?

(A) 1.0

(B) 1.75

(D) An error occurs because c was not initialized.

(E) An error occurs because a and b are integers and c is a double.

31.Consider the following code segment:

int x;

x = /* initialized to an integer */

if (x % 2 == 0 && x / 3 == 1)

System.out.print(“Yes”);

For what values of x will the word “Yes” be printed when the code segment is executed?

(A) 0

(B) 4

(D) Whenever x is odd and x is divisible by 3

(E) Whenever x is even and x is divisible by 3

32.Consider the following incomplete class definition:

public class SomeClass

{

private String myName;

// postcondition: returns myName

public String getName( )

{/* implementation not shown */ }

// postcondition: myName == name

public void setName(String name)

{ /* implementation not shown */ }

//…constructors, other methods,

// and private data not shown

}

Now consider the method swap, not part of the SomeClass class.

// precondition: x and y are correctly

// constructed

// postcondition: the names of objects

// x and y are swapped

public void swap (SomeClass x, SomeClass y)

{

/ * missing code * /

}

Which of the following code segments can replace / * missing code * / so that the method swap works as intended?

I. SomeClass temp;

temp = x;

x = y;

y = temp;

II. String temp;

temp = x.myName;

x.myName = y.myName;

y.myName = temp;

III. String temp;

temp = x.getName();

x.setName(y.getName());

y.setName(temp);

(A) I only

(B) III only

(D) II and III only

(E) I, II, and III

33.A bookstore wants to store information about the different types of books it sells.

For each book, it wants to keep track of the title of the book, the author of the book, and whether the book is a work of fiction or nonfiction.

If the book is a work of fiction, then the bookstore wants to keep track of whether it is a romance novel, a mystery novel, or science fiction.

If the book is a work of nonfiction, then the bookstore wants to keep track of whether it is a biography, a cookbook, or a self-help book.

Which of the following is the best design?

(A) Use one class, Book, which has three data fields: String title, String author, and int bookType.

(B) Use four unrelated classes: Book, Title, Author, and BookType.

(D) Use a class Book which has two data fields: String title, String author, and six subclasses: RomanceNovel, Mystery, ScienceFiction, Biography, Cookbook, and SelfHelpBook.

(E) Use a class Book which has two data fields: String title, String author, and two subclasses: FictionWork and NonFictionWork. The class FictionWork has three subclasses, RomanceNovel, Mystery, and ScienceFiction. The class NonFictionWork has three subclasses: Biography, Cookbook, and SelfHelpBook.

34.Consider the following code:

public int mystery(int x)

{

if (x == 1)

return ;

else

return(2 * mystery(x - 1)) + x;

}

Which of the following can be used to replace so that mystery (4) returns 34 ?

(A) 0

(B) 1

(D) 3

(E) 4

35.Consider the following code segment:

int [ ] X;

int [ ] Y;

int k;

X = initializeX(); // returns a valid

// initialized int [ ]

Y = initializeY(); // returns a valid

// initialized int [ ]

for (k = 0;

k < X.length && X[k] == Y[k];

k++)

{

/* some code */

}

Assuming that after X and Y are initialized, X.length == Y.length, which of the following must be true after executing this code segment?

(A) k < X.length

(B) k < X.length && X[k] == Y[k]

(D) k >= X.length || X[k] == Y[k]

(E) k >= X.length || X[k] != Y[k]

36.Which of the following would NOT cause a run-time exception?

(A) Dividing an integer by zero

(B) Using an object that has been declared but not instantiated

(D) Attempting to create a substring beginning at a negative index

(E) Attempting to call a method with the wrong number of arguments

37.Assume that a and b are properly initialized variables of type Double.

Which of the following is an equivalent expression to:

a.doubleValue() != b.doubleValue()

(A) a != b

(B) a.notEquals(b)

(D) !(a.compareTo(b))

(E) a.compareTo(b) != 0

38.Which of the following would be the LEAST effective way of ensuring reliability in a program?

(A) Encapsulating functionality in a class by declaring all data fields to be public

(B) Defining and following preconditions and postconditions for every method

(D) Using descriptive variable names

(E) Indenting code in a consistent and logical manner

39.Consider a dictionary that has 1,024 pages with 50 words on each page.

In order to look up a given target word, a student is considering using one of the following three methods:

Method 1

Use a binary search technique to find the correct page (comparing the target word with the first word on a given page). When the correct page is found, use a sequential search technique to find the target word on the page.

Method 2

Use a sequential search technique to find the correct page (comparing the target word with the first word on a given page). When the correct page is found, use another sequential search technique to find the target word on the page.

Method 3

Use a sequential search technique on all of the words in the dictionary to find the target word.

Which of the following best characterizes the greatest number of words that will be examined using each method?

(A) Method 1: 10, Method 2: 50, Method 3: 1,024

(B) Method 1: 55, Method 2: 512, Method 3: 2,560

(D) Method 1: 60, Method 2: 1,074, Method 3: 1,074

(E) Method 1: 60, Method 2: 1,074, Method 3: 51,200

40.Consider the following recursive method.

public static int mystery(int m)

{

if (m == 0)

{

return 0;

}

else

{

return 4 + mystery(m — 2);

}

Assuming that j is a positive integer and that m = 2j, what value is returned as a result of the call mystery(m)?

(A) 0

(B) m

(D) j

(E) 2j

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.A day care has a program to keep track of its employees and which children they teach during the day. An Employee has a minimum and maximum age they can teach. The DayCare also has a maximum ratio that specifies the maximum number of children a single employee can teach. Below is the full DayCare class:

public class DayCare

{

private ArrayList employees;

private ArrayList children;

private int maxRatio;

public DayCare(int maxRatio)

{

employees = new ArrayList();

children = new ArrayList();

this.maxRatio = maxRatio;

}

public boolean findEmployeeForChild(Child c)

{

/* To be completed in part (a) */

}

public boolean runDayCare()

{

/* To be completed in part (b) */

}

public boolean addChild(Child c)

{

/* To be completed in part (c) */

}

(a)An Employee can only teach children between the employee’s minimum age (inclusive) and maximum age (inclusive). They can also only teach children up to the day care’s maximum ratio (inclusive). Below is the full Employee class.

public class Employee

{

/* Instance variables not shown */

public Employee(String name, String id, int min, int max)

{

/* Implementation not shown */

}

// Return the number of children currently assigned to this Employee

public int childrenAssigned()

{

/* Implementation not shown */

}

// Assign a new child to this Employee

public void assignChild(Child c)

{

/* Implementation not shown */

}

// Determine whether this Employee can teach a Child based on the child’s age

public boolean canTeach(int age)

{

/* Implementation not shown */

}

A Child has accessors to get their name and age. While the implementation of Child is not shown, you can assume the accessors are called getName and getAge.

Complete the findEmployeeForChild method below that assigns a Child to the first Employee who can teach the Child and who has not reached the maximum ratio of the DayCare.

/* Return true if an Employee was found for the Child, false otherwise */

public boolean findEmployeeForChild(Child c)

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.

(b)In order for the DayCare to run for a day, each Child must be assigned an Employee. If an Employee cannot be found for a Child, the DayCare cannot run for the day.

Complete the runDayCare method below that finds an Employee for each Child in the children ArrayList.

/* Return true if an Employee was found for each Child, false otherwise */

public boolean runDayCare()

(c)When a Child is added to the roster of the DayCare, the DayCare should first make sure there is an Employee available to teach that Child.

Complete the addChild method below that adds a Child to the children ArrayList if an Employee is available to teach that Child.

/* Return true if the Child was added to the ArrayList, false otherwise */

public boolean addChild(Child c)

2.A baseball team consists of different people including players, coaches, and people who work in the front office making trades and other transactions. The following Person class is used for all of the people who work for the team.

Each person has a name and an age.

public class Person

{

private String fullName;

private int age;

public Person(String s, int a)

{

fullName = s;

age = a;

}

// Accessors for name and age

public String getName()

{

return fullName;

}

public int getAge()

{

return age;

}

A Player has a name and age just like any person on the team, but also has a position. The position could be something like “catcher,” “left fielder,” or “infielder.” Players should also be able to change their positions using a method called changePosition. Here is an example of a Player object:

Player p = new Player(“Sammy Sosa”, 32, “right fielder”);

p.changePosition(“outfielder”);

Write the entire Player class.

3.A class is designed to store someone’s full name. You can assume the name has only one space between the first name and the last name. The class has methods to extract the first name, last name, and number of vowels in the name. You can see an example below.

String fullName = “Katherine Johnson”;

Name.getFirstName(fullName); // Returns “Katherine”

Name.getLastName(fullName); // Returns “Johnson”

Name.countVowels(fullName); // Returns 6

(a)The getFirstName method returns the first name based on a given full name. You can assume that fullName has only one space between the first name and the last name. Write the getFirstName method.

public static String getFirstName(String name)

(b)The getLastName method returns the last name based on a given full name. You can assume that fullName has only one space between the first name and the last name. Write the getLastName method.

public static String getLastName(String name)

(c)The countVowels method counts the number of vowels in the given full name. You can assume we will count only the letters a, e, i, o, and u as vowels. Write the entire countVowels method.

public static int countVowels(String name)

4.A city parking lot has a sign that keeps track of how many parking spaces are available in the lot. The class for the parking lot is detailed below.

public class ParkingLot

{

private Car[ ][ ] lot;

public ParkingLot(int rows, int cols)

{

lot = new Car[rows][cols];

}

public int openSpaces()

{

// Complete in part (a)

}

public boolean parkCar(Car newCar)

{

// Complete in part (b)

}

(a)Write the openSpaces method that returns the number of spaces available in the parking lot. If a space is empty, it will be equal to null.

/* Return the number of empty spaces in the parking lot */

public int openSpaces( )

(b)Complete the parkCar method that puts a new car in any space in the parking lot and returns true if it was able to do so. It should return false if there are no empty spaces. You should use the openSpaces method to receive full credit.

/* Return true if there is an open spot to park the newCar, false otherwise. The car should be added to the lot 2D array if there is an open spot. */

public boolean parkCar(Car newCar)

STOP

END OF EXAM

Practice Test 3: Answers and Explanations

ANSWER EXPLANATIONS

Section I: Multiple-Choice Questions

1.C

This question tests how well you understand assigning values to variables and following the steps of the code. When trial() is called, a is assigned to the integer value of 10 and b is assigned to the integer value of 5. Next, doubleValues() is called with a and b as inputs c and d. The method doubleValues() multiplies the input values c and d by 2 and prints them out. Thus, the value of c is 10 * 2 = 20 and the value of d is 5 * 2 = 10. Because System.out.print() does not print any line breaks or spaces, the resulting print is 2010. While c and d have been reassigned new values, these values exist only within the doubleValues() call, so the values of a and b are unchanged. After the doubleValues() method is completed, the trial() method then prints the values of b and then a, printing 510. Once again, no spaces or line breaks are printed, so, combined together, what is printed from calling trial() is 2010510. Therefore, the correct answer is (C).

2.A

The method mystery(int a, int b) takes as input integers a and b with the precondition that a > b > 0. Plug in x = 5 and y = 2. These values are passed as parameters to make a = 5 and b = 2. Set d = 0. Now go to the for loop, initializing c = a = 5. Since it is still the case that c > b, execute the for loop. Execute d = d + c by setting d = 0 + 5 = 5. Decrease c by 1 to get c = 4. Since c > b, execute the for loop again. Execute d = d + c by setting d = 5 + 4 = 9. Decrease c by 1 to get c = 3. Since it is still the case that c > b, execute the for loop again. Execute d = d + c by setting d = 9 + 3 = 12. Decrease c by 1 to get c = 2. Since it is no longer the case that c > b, stop executing the for loop. Return d = 12. Now go through the choices and eliminate any that don’t describe a return of 12. Choice (A) is the sum of all integers greater than y but less than or equal to x. The sum of all integers greater than 2 but less than or equal to 5 is 3 + 4 + 5 = 12, so keep (A). Choice (B) is the sum of all integers greater than or equal to y but less than or equal to x. The sum of all integers greater than or equal to 2 but less than or equal to 5 is 2 + 3 + 4 + 5 = 14, so eliminate (B). Choice (C) is the sum of all integers greater than y but less than x. The sum of all integers greater than 2 but less than 5 is 3 + 4 = 7, so eliminate (C). Choice (D) is the sum of all integers greater than or equal to y but less than x. The sum of all integers greater than or equal to 2 but less than 5 is 2 + 3 + 4 = 9, so eliminate (D). Choice (E) is the sum of all integers less than y but greater than or equal to x. However, there are no integers that are both less than 2 and greater than or equal to 5, so eliminate (E). Only one answer remains. The correct answer is (A).

3.C

The question asks for what is printed by the call mystery(3). In the call, 3 is taken as a parameter and assigned to n. The integer k is then initialized in the for loop and set equal to 0. Since k < n, execute the for loop. Call mystery(0), while keeping your place in the call of mystery(3). Again, k is initialized in the for loop and set equal to 0. In this call, since n = 0, it is not the case that k < n, so do not execute the for loop. This completes the call of mystery(0), so return to mystery(3). The next command is System.out.print(k). Since k = 0, the first character printed is 0. Look to see whether any choices can be eliminated. All choices begin with 0, so no choices can be eliminated. Continue with the method call. Since this is the last command of the for loop, increment k to get k = 1. Since it is still the case that k < n, execute the for loop again. Call mystery(1), again keeping your place in the call of mystery(3). Again, k is initialized in the for loop and set equal to 0. In this call, since n = 1, it is the case that k < n, so execute the for loop. Call mystery(k), which is mystery(0). As seen above, mystery(0) prints nothing, so execute the next line in mystery(1), which is System.out.print(k). Since k = 0, print 0. Eliminate (A), since the second character printed is not 0. Increment k to get k = 1. Since it is no longer the case that k < n, do not execute the for loop again and end this call of mystery(1). Return to the original call of mystery(3), where k = 1 and System.out.print(k) is the next statement. Print 1. All remaining choices have 1 as the third character, so do not eliminate any choices. This is the end of the for loop, so increment k to get k = 2. Since n = 3, it is still the case that k < n, so execute the for loop. Call mystery(2) while holding your place in mystery(3). Again, k is initialized in the for loop and set equal to 0. In this call, since n = 2, it is the case that k < n, so execute the for loop. Call mystery(0), which prints nothing. Execute System.out.print(k) to print 0. Eliminate (B), since the next character is not 0. Increment k to get k = 1. Since k < n, execute the for loop. Call mystery(1), which, as described above, prints 0. All remaining choices have 0 as the next character, so don’t eliminate any choices. The next command in mystery(2) is System.out. print(k), so print 1. Again, keep all the remaining choices. In mystery(2), increment k to get k = 2. Since n = 2, it is no longer the case that k < n, so stop executing the for loop. End mystery(2) and return to mystery(3), where k = 2 and the next command is System.out. print(k), so print 2. Again, keep all remaining choices. Increment k to get k = 3. Since it is no longer the case that k < n, stop executing the for loop. The method call is complete for mystery(3), so there will be no more printed characters. Eliminate (D) and (E), which have further characters printed.

In other words, the calling sequence looks like this

Function Call	Prints
mystery(3)
mystery(0)
System.out.print(0)	0
mystery(1)
mystery(0)
System.out.print(0)	0
System.out.print(1)	1
mystery(2)
mystery(0)
System.out.print(0)	0
mystery(1)
mystery(0)
System.out.print(0)	0
System.out.print(1)	1
System.out.print(2)	2

The correct answer is (C).

4.C

SelectionSort walks through the array to find the smallest element in the part of the array not yet sorted. It then swaps that smallest element with the first unsorted element.

Start with the original array of values.

4 10 1 2 6 7 3 5

The smallest element is 1. Swap that with the first unsorted element, 4. The array now looks like this.

1 10 4 2 6 7 3 5

This is not a choice, so continue. The smallest element in the unsorted part of the array (from 10 to 5) is 2. Swap that with the first unsorted element, 10.

1 2 4 10 6 7 3 5

This is still not a choice, so continue this process. The smallest element in the unsorted part of the array (from 4 to 5) is 3. Swap that with the first unsorted element, 4.

1 2 3 10 6 7 4 5

This is (C), so stop here. If you’re interested, here are the complete moves for selection sort.

1 2 3 4 6 7 10 5

1 2 3 4 5 7 10 6

1 2 3 4 5 6 10 7

1 2 3 4 5 6 7 10

The correct answer is (C).

5.E

Execute the commands. The integer k and the integer array A are initialized. The array A is given length 7. Thus, the array, A, has seven elements with indexes from 0 to 6. Execute the first for loop. Set k = 0. Since k < A.length, execute the for loop, which sets A[0] equal to A.length — k = 7 — 0 = 7. Increment k to get k = 1. Since 1 < 7, execute the for loop, which sets A[1] equal to A.length — k = 7 — 1 = 6. Increment k to get k = 2. Since 2 < 7, execute the for loop, which sets A[2] equal to A.length — k = 7 — 2 = 5. Increment k to get k = 3. Since 3 < 7, execute the for loop, which sets A[3] equal to A.length — k = 7 — 3 = 4. Increment k to get k = 4. Since 4 < 7, execute the for loop, which sets A[4] equal to A.length — k = 7 — 4 = 3. Increment k to get k = 5. Since 5 < 7, execute the for loop, which sets A[5] equal to A.length — k = 7 — 5 = 2. Increment k to get k = 6. Since 6 < 7, execute the for loop, which sets A[6] equal to A.length — k = 7 — 6 = 1. Increment k to get k = 7. Since it is not the case that 7 < 7, stop executing the for loop. Therefore, the array has the values

7 6 5 4 3 2 1

The second for loop puts the value of the kth element of the array into the (k + 1)st element. Be careful!

A quick glance might lead you to believe that each value is shifted in the array one spot to the right, giving an answer of (D). However, a step-by-step analysis demonstrates that this will not be the case. The for loop sets k = 0. Since 0 < A.length — 1, execute the for loop. The command in the for loop is A[k + 1] = A[k]. Since k = 0, set A[1] = A[0] = 7. Increment k to get k = 1. Since it is still the case that k < A.length — 1, execute the for loop. Since k = 1, set A[2] = A[1] = 7. Continue in this manner. Since A.length — 1 = 7 — 1 = 6, only execute the for loop through k = 5. Each execution of the for loop is shown below.

Assignment

A[]

A[1] = A[0]

A[2] = A[1]

A[3] = A[2]

A[4] = A[3]

A[5] = A[4]

A[6] = A[5]

7 7 5 4 3 2 1

7 7 7 4 3 2 1

7 7 7 7 3 2 1

7 7 7 7 7 2 1

7 7 7 7 7 7 1

7 7 7 7 7 7 7

The final values contained by A are 7 7 7 7 7 7 7. The correct answer is (E).

6.B

Choices (C), (D), and (E) are syntactically invalid according to the given class definitions. In (C), p is used as a PostOffice object rather than an array of PostOffice objects. Choice (D) treats getMail and getBox as static methods without invoking them from an object. Choice (E) creates a new Mail object attempting to use a Mail constructor. Even if such a constructor were available, there would be no way for the constructor to know about p, the array of PostOffices.

Choices (A) and (B) differ only in the indexes of the array and methods. There are two clues in the question that indicate that (B) is the correct answer. First, p is declared as an array of 10 Postoffices. This means that p[10] would raise an ArrayListOutOfBoundsException. Remember that p[9] actually refers to the 10th post office, since array indexes begin with 0. Second, the comments in the class definitions for the getBox and getMail methods indicate that the parameter they take is zero-based. Therefore, they should be passed an integer one less than the number of the box or piece of mail needed. For either reason, eliminate (A). The correct answer is (B).

7.A

In the method printEmptyBoxes, the loop variable k refers to the index of the post office and the loop variable x refers to the index of the box within the post office. Choice (A) is correct. It checks to see whether the box is assigned and whether it does not have mail using the appropriate methods of the Box class. It then prints out the box number of the box. Choice (B) is similar to (A) but incorrectly interchanges x and k. Eliminate (B). Choice (C) omits the call to the method getBox. Therefore, it attempts to call the getBoxNumber() method of the PostOffice object p[k]. Since PostOffice objects do not have a getBoxNumber() method, this would result in a compile error. Eliminate (C). Choice (D) is similar to (C), but interchanges x and k. Eliminate (D). Choice (E) prints the value of k, which is the index of the post office rather than the index of the box. Eliminate (E). The correct answer is (A).

8.B

There are four possibilities for the values of a and b. Either both a and b are true, both a and b are false, a is true and b is false, or a is false and b is true.

Analyze the possibilities in a table.

a initial value)	b (initial value)	a = a && b (final value)	b = a \|\| b (final value)
true	true	true	true
true	false	false	false
false	true	false	true
false	false	false	false

Remember when calculating b = a || b that a has already been modified. Therefore, use the final, rather than the initial, value of a when calculating the final value of b.

Go through each statement. Statement I says that the final value of a is equal to the initial value of a. This is not the case in the second row of the table, so Statement I is not always true. Eliminate any choice that includes it: (A) and (D). Statement II says that the final value of b is equal to the initial value of b. This is true in each row and, thus, Statement II is always true, so eliminate the choice that does not include it: (C). Statement III says that the final value of a is equal to the initial value of b. This is not the case in the third row of the table, so Statement III is not always true. Eliminate any choice that includes it: (E). The correct answer is (B).

9.C

The best way to solve this problem is to look at each if statement individually. The integer x is initialized and then set equal to 53. The next statement is an if statement with the condition (x > 10). Since 53 > 10, execute the statement, System.out.print(“A”), and print A. The next statement is an if-else statement, this one having the condition (x > 30). Since 53 > 30, execute the if statement, System.out.print(“B”), and print B. Even though 53 > 40, because the next part of the statement is an else statement, it cannot be executed if the original if statement was executed. Therefore, do not execute the else statement, and do not print C. The next statement is another if statement, this one having the condition (x > 50). Since 53 > 50, execute the statement, System.out.print(“D”), and print D. The next statement is another if statement, this one having the condition (x > 70). Since it is not the case that 53 > 70, do not execute the statement. There is no further code to execute, so the result of the printing is ABD. The correct answer is (C).

10.E

This problem tests your ability to work with nested for loops. Although it appears complicated at first glance, it can easily be solved by systematically walking through the code.

Be sure to write the values of the variables j and k down on paper; don’t try to keep track of j and k in your head. Use the empty space in the question booklet for this purpose.

Code	j	k	Output
Set j to the value —2.	—2
Test the condition: j <= 2? Yes.	—2
Set k to the value that j has.	—2	—2
Test the condition: k < j + 3? Yes.	—2	—2
Output k.	—2	—2	—2
Update k: k++	—2	—1	—2
Test the condition: k < j + 3? Yes.	—2	—1	—2
Output k.	—2	—1	—2 —1
Update k: k++	—2	0	—2 —1
Test the condition: k < j + 3? Yes.	—2	0	—2 —1
Output k.	—2	0	—2 —1 0
Update k: k++	—2	1	—2 —1 0
Test the condition: k < j + 3? No.	—2	1	—2 —1 0
Update j: j = j + 2	0	1	—2 —1 0
Test the condition: j <= 2? Yes.	0	1	—2 —1 0
Set k to the value that j has.	0	0	—2 —1 0
Test the condition: k < j + 3? Yes.	0	0	—2 —1 0
Output k.	0	0	—2 —1 0 0

At this point, stop because (E) is the only choice that starts —2 —1 0 0. However, repeating this process will result in the correct output of —2 —1 0 0 1 2 2 3 4. The correct answer is (E).

11.A

To solve this problem, first note that count % 3 is equal to the remainder when count is divided by 3. Execute the method call mystery(5, “X”), taking count = 5 and s = “X” as parameters. Because it is not the case that 5 <= 0, do not execute the first if statement.

Because 5 % 3 = 2, calling mystery(5, “X”) will execute the else statement, printing X. Then, it will call mystery(4, “X”), and return. Because 4 % 3 = 1, calling mystery(4, “X”) will execute the else if statement, printing X-X, call mystery(3, “X”), and return. At this point, (C) and (E) can be eliminated. Because 3 % 3 = 0, calling mystery(3, “X”) will execute the if statement, printing X—X, call mystery(2, “X”), and return. Note that you can stop at this point because (B) and (D) can be eliminated. If you’re interested, here is the remainder of the execution. Because 2 % 3 = 2, calling mystery(2, “X”) will execute the else statement, printing X, call mystery(1, “X”), and return. Because 1 % 3 = 1, calling mystery(1, “X”) will execute the else if statement, printing X—X, call mystery(0, “X”), and return. Finally, calling mystery(0, “X”) will simply return because count is less than or equal to zero. Putting it all together, mystery(5, “X”) prints

XX-XX—XXX-X

The correct answer is (A).

12.B

The only information that you need to solve this problem is the first sentence in the description of the constructor. Class DiningRoomSet has a constructor, which is passed a Table object and an ArrayList of Chair objects. Because you are writing a constructor, you can immediately eliminate (A) and (C), which are void methods. Constructors never return anything, so there is never a need to specify a void return. Choice (E) is incorrect because the constructor is passed a Table object and a Chair object, not a Table object and an ArrayList of Chair objects. Choice (D) is incorrect because the second parameter has two types associated with it. It should have only one type: ArrayList. Choice (B) is correct.

13.D

The best way to solve this problem is to eliminate choices. Choices (A) and (B) are incorrect because the class description states that the getPrice method of the DiningRoomSet class does not take any parameters. Choice (C) can be eliminated because the private data field myChairs is not a Chair; it is an ArrayList. Therefore, it does not have a getPrice method. This leaves (D) and (E). You need to know that ArrayLists are accessed using the get method while arrays are accessed using the [] notation. MyChairs is an ArrayList, so the correct answer is (D).

14.C

To solve this type of problem, use information about the output and the loops to eliminate incorrect choices. Then, if necessary, work through the code for any remaining choices to determine the correct answer. There are six lines of output. By examining the outer loop of each of the choices, you can eliminate (B) because it will traverse the loop seven times, and during each traversal at least one number will be printed. You can also eliminate (A) because the outer loop will never be executed. The initial value of j is 6, but the condition j < 0 causes the loop to terminate immediately. The remaining choices have the same outer loop, so turn your attention to the inner loop. Eliminate (D) because the first time through the loop, a 7 will be printed—clearly not the correct output. Finally, eliminate (E) because the condition of the inner loop, k >= 0, will cause a 0 to be printed at the end of each line. This leaves (C), which is indeed the correct answer.

15.B

This problem tests your knowledge of the methods of the ArrayList class. The following table shows the contents of list after each line of code.

Code	Contents of list	Explanation
list = new ArrayList();	[]	A newly created ArrayList is empty
list.add(new Integer(7));	[7]	Adds 7 to the end of list
list.add(new Integer(6));	[7, 6]	Adds 6 to the end of list
list.add(1, new Integer(5));	[7, 5, 6]	Inserts 5 into list as position 1, shifting elements to the right as necessary
list.add(1, new Integer(4));	[7, 4, 5, 6]	Inserts 4 into list as position 1, shifting elements to the right as necessary
list.add(new Integer(3));	[7, 4, 5, 6, 3]	Adds 3 to the end of list
list.set(2, new Integer(2));	[7, 4, 2, 6, 3]	Replaces the number at position 2 in list with 2
list.add(1, new Integer(1));	[7, 1, 4, 2, 6, 3]	Inserts 1 into list at position 1, shifting elements to the right as necessary

The next command is to print list. Therefore, the correct answer is (B).

16.E

Although the Dog class extends Animal, Java does not require Dog to have any specific methods. New methods will be unique to Dog objects (and those of its subclasses, when applicable) and methods inherited from Animal can be invoked as normal.

The correct answer is (E).

17.A

This question is testing your understanding of static and dynamic types.

Fish Bob = new Shark();

This line creates the Bob variable with the static type of Fish and the dynamic type of Shark.

System.out.println(Bob.endoskeleton);

This line prints the endoskeleton field of the variable Bob. When looking up a field, the static type is used, so “bone” is printed.

Bob.action();

This line executes the method action(). When looking up a method, the dynamic type is used, so “chomp chomp” is printed. The correct answer is (A).

18.D

The insertion sort algorithm creates a sorted array by sorting elements one at a time.

The for loop of the code shows that the elements are being sorted from left to right. To determine the position of the new element to be sorted, the value of the new element must be compared with the values of the sorted elements from right to left. This requires index— in the while loop, not index++. Choices (A) and (C) are wrong.

In order to place the new element to be sorted in its correct position, any sorted elements larger than the new element must be shifted to the right. This requires sort[index] = sort[index -1]. Choices (B) and (E) are wrong. The correct answer is (D).

19.C

In order for the array to be sorted in descending order, you will need to make a change. Plug each answer choice into the array to see which is correct. In (A) and (B), as the index will never be less than 0, the contents of the while loop will never be executed. Choice (C) is the correct answer, as it changes the condition of finding the position of the new element from being less than the compared element to greater. In (D), altering the for loop this way would lead to the elements being sorted from right to left but still in ascending order. In (E), the index is initialized at 1 and decremented while index > 0, so it will execute the for loop only once. The correct answer is (C).

20.C

The rate at which insertion sort runs depends on the number of comparisons that are made. The number of comparisons is minimized with an array with elements that are already sorted in ascending order and maximized with an array with elements that are sorted in descending order. The array in (C) is sorted in reverse order and will require the most comparisons to sort. The correct answer is (C).

21.C

All three responses look very similar, so look carefully at the difference. Statement I is missing the keyword new, which is needed to create a new object. Eliminate any choice that includes Statement I: (A) and (D). Statements II and III differ in that Statement II uses f.numerator and f.denominator while Statement III uses f.getNumerator() and f.getDenominator(). Since the integers numerator and denominator are private, while the methods getNumerator() and getDenominator() are public, the methods must be called by another object. Therefore, Statement II is not valid. Eliminate any choice that includes it: (B) and (E). The correct answer is (C).

22.C

Go through the choices one at a time. Choice (A) is incorrect because the multiplication operator, *, is not defined to work on Fraction objects. Eliminate (A). Choice (B) is incorrect because the multiply method takes only one parameter and it is not correctly invoked by a Fraction object. Eliminate (B). Choice (C) correctly calls the multiply() method as through a fraction object, taking the other fraction as a parameter. Keep (C). Choice (D) attempts to create a new Fraction object but incorrectly constructs it by passing two Fraction objects rather than two integers. Eliminate (D). Finally, while (E) calculates the value of the result of the multiplication, the “/” operator assigns integer types, which cannot be applied to answer, which is an object of type Fraction. Eliminate (E). The correct answer is (C).

23.D

Go through each statement one at a time. Constructor I is legal. This default constructor of the ReducedFraction class will automatically call the default constructor of the Fraction class and the private data of both classes will be set appropriately. Eliminate any choice that does not include Constructor I: (B), (C), and (E). Because Constructor II is not included in the remaining choices, do not worry about analyzing it. Constructor III is also legal. The call super(n, d) invokes the second constructor of the Fraction class, which sets the private data of the Fraction class appropriately. The private data of the ReducedFraction class is then set explicitly in the ReducedFraction constructor. Note that if the call super(n, d) were not present, Constructor III would still be legal. However, it would create a logical error in the code, as the default constructor of the Fraction class would be invoked, and the private data of the Fraction class would not be set appropriately. Eliminate (A), which does not include Constructor III. Only one choice remains, so there is no need to continue. However, to see why Constructor II is illegal, remember that derived classes may not access the private data of their super classes. In other words, the constructor for a ReducedFraction may not directly access numerator and denominator in the Fraction class. The correct answer is (D).

24.D

Go through each statement one at a time. Statement I is incorrect. “==” checks object references as opposed to their contents. This is an important distinction as two different objects may hold the same data. Eliminate (A), (C), and (E), which contain Statement I. Both of the remaining choices contain Statement II, so don’t worry about analyzing it. Statement III is correct. The compareTo() method of the String class returns 0 if the two String objects hold the same strings. Eliminate (B), which does not include Statement III. Only one choice remains, so there is no need to continue. However, you should see why Statement II is also correct. The equals method of the String class returns true if the two String objects hold the same strings. The correct answer is (D).

25.A

The values of s and t are not changed in mystery(). Even though s and t are both parameters of the method, only the instance variables a and b are changed. Thus, the original Strings s and t are unaffected by any action in mystery(). The correct answer is (A).

26.E

Though these two classes are related through an inheritance relationship, there is no rule in Java that requires this structure to share methods and/or data. Therefore, there are no requirements on either class. The correct answer is (E).

27.B

An example will help clarify this question.

Consider the following for loop:

for (int k = 0; k < 3; k++)

{

Systemout.println(k);

}

This prints out the integers 0 to 2, one number per line. Matching int k = 0 to <1>; k < 3 to <2>; k++ to <3> and System.out.println(k); to <4>, go through each choice one at a time and determine whether each has the same functionality.

Choice (A) initializes k and assigns it the value 0. Then it tests to determine whether k < 3. This is true, so execute the while loop. The next command is k++, so increase k by 1 to get k = 0 + 1 = 1. Now execute System.out.println(k) to print 1. However, the first number printed in the original was 0, so this is incorrect. Eliminate (A).

Choice (B) initializes k and assigns it the value 0. Then it tests to determine whether k < 3. This is true, so execute the while loop. Now execute System.out.println(k) to print 0. The next command is k++, so increase k by 1 to get k = 0 + 1 = 1. Go back to the top of the while loop. Since k = 1, it is still the case that k < 3, so execute the while loop. Now execute System.out. println(k) to print 1. The next command is k++, so increase k by 1 to get k = 1 + 1 = 2. Go back to the top of the while loop. Since k = 2, it is still the case that k < 3, so execute the while loop. Now execute System.out.println(k) to print 2. The next command is k++, so increase k by 1 to get k = 2 + 1 = 3. Go back to the top of the while loop. Since k = 3, it is no longer the case that k < 3, so stop executing the while loop. The result of the program is printing the integers 0 to 2, one number per line, so keep (B).

Choice (C) initializes k and assigns it the value 0. Then it tests the condition ! (k < 3). Since it is the case that k < 3, the statement (k < 3) has the boolean value true, making the boolean value of ! (k < 3) false. Thus the while loop is not executed. Since the while loop is not executed, nothing is printed. Eliminate (C).

Choice (D) has the same initialization statement and while loop condition as (C), so nothing is printed by this choice either. Eliminate (D).

Choice (E) initializes k and assigns it the value 0. The next command is k++, so increase k by 1 to get k = 0 + 1 = 1. Then it tests to determine whether k < 3. This is true, so execute the while loop. Execute System.out.println(k) to print 1. However, the first number printed in the original was 0, so this is incorrect. Eliminate (E).

The correct answer is (B).

28.A

This question tests your knowledge of operator precedence. In Java, multiplication, division, and modulus are performed before addition and subtraction. If more than one operator in an expression has the same precedence, the operations are performed left-to-right. Parenthesizing the expression one step at a time

a / b + c — d % e * f

→ (a / b) + c — d % e * f

→ (a / b) + c — (d % e) * f

→ (a / b) + c — ((d % e) * f)

→ ((a / b) + c) — ((d % e) * f)

The correct answer is (E).

29.B

Come up with a sample array of 10 strings with length 5. Let String[] x be {“gator”, “teeth”, “ducky”, “quack”, “doggy”, “woofs”, “kitty”, “meows”, “bears”, “growl”}. The code must print the first letter of all 10 strings, followed by the second letter of all 10 strings, and so on. Therefore, it must print

gtdqdwkmbgaeuuooieer…

Go through each segment one at a time.

Segment I initializes i and j with no values. The outer for loop sets i = 0, and the inner for loop sets j = 0. The instruction of the inner for loop is System.out.print(x[i].substring(j, j + 1). The substring(a, b) method of the String class returns a string made up of all the characters starting with the index of a through the index of b — 1. Therefore, x[i].substring(j, j + 1) returns the characters of x[i] from index j through index (j + 1) — 1. Since (j + 1) — 1 = j, it returns all the characters from indexes j through j—a single character string made up of the character at index j. Therefore, if i = 0 and j = 0, System.out.print(x[i].substring(j, j + 1) returns the character at index 0 of the string at index 0. This is the “g” from “gator”. This is what should be printed, so continue. The inner for loop increments j to get j = 1. Since j < 5, the loop is executed again, printing the character at index 1 of the string at index 0. This is the character “a” from “gator”. However, the character “t” from “teeth” should be the next character printed, so Segment I does not execute as intended. There is no need to continue with I, but note that it will eventually print all of the characters of the first string followed by all of the characters of the second string, and so on. Eliminate (A), (C), and (E), which include I.

Both remaining choices include II, so don’t worry about checking this one. Instead, analyze III.

Segment III initializes i and j with no values. The outer for loop sets i = 0, and the inner for loop sets j = 0. The instruction of the inner for loop is System.out.print(x[i].substring(j, j + 1). As discussed above, when i = 0 and j = 0, this command returns the character at index 0 of the string at index 0, which is the “g” from “gator”. This is what should be printed, so continue. The inner for loop increments j to get j = 1. Since j < 5, the loop is executed again, printing the character at index 1 of the string at index 0. This is the character “a” from “gator”. Again, the character “t” from “teeth” should be the next character printed, so Segment III does not execute as intended. There is no need to continue with Segment III, but note that it will attempt to print the first 10 characters of the first 5 strings. Since each String contains only 5 characters, an IndexOutOfBoundsException will be thrown when j = 5 and the program attempts to print the character at index 5. Eliminate (D), which includes Segment III.

Only one answer remains, so there is no need to continue. However, to see why Segment II is correct, note that it is similar to Segment I, but it reverses the roles of i and j. The command System.out.print(x[j].substring(i, i + 1)) prints the character at index j of the string at index i. The inner for loop increments the index of the String array before the outer loop increments the index of the character in each String. Therefore, Segment II correctly prints out the first character of all 10 strings followed by the second character of all 10 strings, and so on. The correct answer is (B).

30.A

Begin with (D) and (E), which discuss whether an error would occur. Because c is initialized with the command in the third line, double c;, eliminate (D). It is perfectly legal to assign a value of type int to a variable of type double in an expression, so eliminate (E). However, certain rules apply when evaluating the expression. In the expression c = a / b; a and b are both integers. Therefore, the / operator represents integer division. The result of the division is truncated by discarding the fractional component. In this example, 7 / 4 has the value 1—the result of truncating 1.75. When assigning an integer to a variable of type double, the integer value is converted to its equivalent double value. Therefore, the correct answer is (A).

31.B

For the code to print the word Yes two conditions must be true. The remainder must be zero when x is divided by 2. In other words, x must be divisible by 2: that is, even. The value after truncating the result when x is divided by 3 must be 1. In other words, 1 ≤ x / 3 < 2. The second condition is more narrow than the first. The only integers that fulfill the second condition are 3, 4, and 5. Of those, only 4 is even and therefore also fulfills the first condition. Therefore, the correct answer is (B).

32.B

Go through each statement one at a time. Segment I is incorrect because all parameters in Java are passed by value. After a method returns to its caller, the value of the caller’s parameters is not modified. The strings will not be swapped. Eliminate (A), (C), and (E), which include Segment I. Since both remaining choices include Segment III, don’t worry about it. Segment II is incorrect because myName is a private data field of the SomeClass class and may not be accessed by the swap method. Note that the question specifically states that the swap method is not a method of the SomeClass class. Eliminate (D), which includes Segment II. Only one choice remains, so there is no need to continue. To see that Segment III correctly swaps the names of the objects, note that it calls the public methods setName() and getName() rather than the private String myName. Because the references of the instance object variables are the same as the references of the class object variables, modifying the data of the instance variables also changes the data of the class variable. The correct answer is (B).

33.E

The way to approach this type of design problem is to look for HAS-A and IS-A relationships among the distinct pieces of data. A book HAS-A title and author. The title and author should be data fields of the Book class, either as Strings or as their own unrelated classes. This information is not enough to answer the question though. Looking for the IS-A relationships, a mystery novel IS-A work of fiction, which IS-A book. Therefore, it makes good design sense for these three items to be separate classes. Specifically, Mystery should be a subclass of FictionWork, which should be a subclass of Book. Similarly, RomanceNovel and ScienceFiction should be subclasses of FictionWork and Biography, Cookbook, and SelfHelpBook should be subclasses of NonFictionWork, which should be a subclass of Book. Only (E) meets all of these design criteria.

34.C

In order to solve this recursive problem, work backward from the base case to the known value of mystery(4).

Let y represent the . Note that mystery(1) = y. Now calculate mystery(2), mystery(3), and mystery(4) in terms of y.

mystery(2) = 2 * mystery(1) + 2

= 2 * y + 2

mystery(3) = 2 * mystery(2) + 3

= 2 * (2 * y + 2) + 3

= 4 * y + 4 + 3

= 4 * y + 7

mystery(4) = 2 * mystery(3) + 4

= 2 * (4 * y + 7) + 4

= 8 * y + 14 + 4

= 8 * y + 18

Because mystery(4) also equals 34, set 8 * y + 18 = 34 and solve for y.

8 * y + 18 = 34

8 * y = 16

y = 2

The correct answer is (C).

35.E

The for loop terminates when the condition is no longer true. This can happen either because k is no longer less than X.length or because X[k] does not equal Y[k]. Choice (E) states this formally. Another way to approach the problem is to use DeMorgan’s Law to negate the condition in the for loop. Recall that DeMorgan’s Law states that

!(p && q) is equivalent to !p || !q

Negating the condition in the for loop gives

!(k < X.length && X[k] == Y[k])

=> !(k < X.length) || !(X[k] == Y[k])

=> k >= X.length || X[k] != Y[k]

This method also gives (E) as the correct answer.

36.E

Choices (A), (B), (C), and (D) are examples of an ArithmeticException, a nullPointerException, an ArrayIndexOutOfBoundsException, and an IndexOutOfBoundsException, respectively. Be careful! While (E) may appear to be an example of an IllegalArgumentException, it is actually an example of an error that is caught at compile time rather than at runtime. An IllegalArgumentException occurs when a method is called with an argument that is either illegal or inappropriate—for instance, passing —1 to a method that expects to be passed only positive integers. Therefore, the correct answer is (E).

37.E

First note that a and b are of type Double, not of type double. This distinction is important. The type double is a primitive type; the type Double is a subclass of the Object class that implements the Comparable interface. A Double is an object wrapper for a double value. Go through each choice one at a time. Choice (A) is incorrect. It compares a and b directly rather than the double values inside the objects. Even if a and b held the same value, they might be different objects. Eliminate (A). Choice (B) is incorrect. The Double class does not have a notEquals method. Eliminate (B). Choice (C) is incorrect, because a.doubleValue() returns a double. Since double is a primitive type, it does not have any methods. Had this choice been !(a.equals(b)), it would have been correct.

The expression a.compareTo(b) returns a value less than zero if a.doubleValue() is less than b.doubleValue(), a value equal to zero if a.doubleValue() is equal to b.doubleValue(), and a value greater than zero if a.doubleValue() is greater than b.doubleValue(). Choice (D) is incorrect because the compareTo method returns an int rather than a boolean. Choice (E) is correct. Since compareTo() returns 0 if and only if the two objects hold the same values, it will not return 0 if the values are different. The correct answer is (E).

38.A

While “encapsulating functionality in a class” sounds like (and is) a good thing, “declaring all data fields to be public” is the exact opposite of good programming practice. Data fields to a class should be declared to be private in order to hide the underlying representation of an object. This, in turn, helps increase system reliability. Choices (B), (C), (D), and (E) all describe effective ways to ensure reliability in a program. The correct answer is (A).

39.E

Go through each method one at a time. Method 1 examines at most 10 words using a binary search technique on 1,024 pages to find the correct page. Remember, the maximum number of searches using binary search is log₂(n) where n is the number of entries. It then searches sequentially on the page to find the correct word. If the target word is the last word on the page, this method will examine all 50 words on the page. Therefore, Method 1 will examine at most 60 words. Eliminate (A), (B), and (C), which don’t have 60 for Method 1. The two remaining choices both have the same number for Method 2, so worry only about Method 3. Method 3 sequentially searches through all of the words in the dictionary. Because there are 51,200 words in the dictionary and the target word may be the last word in the dictionary, this method may have to examine all 51,200 words in order to find the target word. Eliminate (D), which does not have 51,200 for Method 3. Only one choice remains, so there is no need to continue. However, note that Method 2 first uses a sequential search technique to find the correct page. If the target word is on the last page, this method will examine the first word on all 1,024 pages. Then, as with Method 1, it may examine as many as all 50 words on the page to find the target word. Therefore, Method 2 will examine at most 1,074 words. The correct answer is (E).

40.C

Pick a positive value of j. The question says that j is a positive integer. A recursive method is easiest if it takes fewer recursions to get to the base case, so try j = 1. If j = 1, then m = 2j = 2(1) = 2. Determine the return of mystery(2). Since it is not the case that m = 0, the if condition is false, so execute the else statement, which is to return 4 + mystery(m — 2). Since m — 2 = 2 — 2 = 0, mystery(2) returns 4 + mystery(0). Determine the return of mystery(0). In mystery(0), m = 0, so the if condition is true, which causes the method to return 0. Therefore, mystery(0) returns 0, and mystery(2) returns 4 + mystery(0) = 4 + 0 = 4. Go through each choice and eliminate any that are not 4. Choice (A) is 0, so eliminate (A); (B) is m, which is 2, so eliminate (B). Choice (C) is 2m, which is 2(2) = 4, so keep (C). Choice (D) is j, which is 1, so eliminate (D). Choice (E) is 2j, which is 2(1) = 2, so eliminate (E). Only one choice remains. The correct answer is (C).

Section II: Free-Response Questions

1.DayCare—Canonical Solution

public class DayCare

{

private ArrayList employees;

private ArrayList children;

private int maxRatio;

public DayCare(int maxRatio)

{

employees = new ArrayList();

children = new ArrayList();

this.maxRatio = maxRatio;

}

(a) public boolean findEmployeeForChild(Child c)

{

for (Employee e : employees)

{

if (e.childrenAssigned() < maxRatio && e.canTeach(c.getAge()))

{

e.assignChild(c);

return true;

}

return false;

}

This can also be done with a typical for loop or a while loop and the .get method. A loop is necessary to look at each Employee in the ArrayList. You must then make sure the chosen Employee doesn’t already have the maximum number of children allowed. You must also send the age of the Child using the getAge accessor to the canTeach method to see whether the chosen Employee is eligible to teach the given Child.

If an Employee is found for the given Child, you need to assign the Child to the Employee using the assignChild method and return true.

You should not return false inside the loop, since it is possible a different Employee is eligible to teach the given Child.

(b)public boolean runDayCare()

{

for (Child c : children)

{

if (findEmployeeForChild(c) == false)

return false;

}

return true;

}

This could also be done with a typical for loop or a while loop and the .get method. A loop is needed to look at each Child in the ArrayList. You must then call the findEmployeeForChild method from part (a) to see whether there is an Employee eligible to teach the current Child. If not, you need to return false. You shouldn’t return true inside the loop, since it is possible there is a later Child who can’t be taught by any of the Employees in the ArrayList.

{

if (findEmployeeForChild(c) == true)

{

children.add(c);

return true;

}

return false;

}

The solution must call the findEmployeeForChild method from part (a) to see whether there is an Employee eligible to teach the given Child. If there is, you add the Child to the ArrayList using the add method and return true. If there isn’t, you return false.

DayCare Rubric

Part (a)

(+1)for loop correctly iterates through all elements of the employees list

(+1)if statement correctly determines whether the number of children assigned to the current employee is less than the maximum allowed

(+1)if statement correctly determines whether current employee can teach the given child based on age

(+1)The child is assigned to the employee if both conditions are met

(+1)true is returned if the child is assigned to an employee; false is returned otherwise

Part (b)

(+1)for loop correctly iterates through all elements of the children list

(+1)The findEmployeeForChild method is called correctly

Part (c)

(+1)The findEmployeeForChild method is called correctly

(+1)Children are added to the children list correctly if an employee is found

2.Person—Canonical Solution

public class Player extends Person

{

private String position;

public Player(String name, int age, String pos)

{

super(name, age);

position = pos;

}

public void changePosition(String p)

{

position = p;

}

The class header must look exactly the same as the header above. The public class Player part would be necessary for any class you are writing called Player. The extends Person part is necessary because a Player is a Person.

The position variable must be declared as private and it must be a string.

The constructor (public Player) must take three parameters in the order shown above, since the example shows the name, age, and position in that order. They can be called whatever you want, however. The name and age variables must be sent to the Person class using the super( ) call and they must be in the given order. The position variable should be set after the super( ) call.

The changePosition method should be void and should take a string parameter. The only thing it needs to do is set the class-level position variable.

Person Rubric

(+1)public class Player

(+1)extends Person

(+1)A string variable is declared as private

(+1)The constructor header is correct (public Player).

(+1)The constructor takes a string parameter, an integer parameter, and a string parameter, in that order

(+1)The constructor call uses super to initialize the name and age

(+1)The constructor initializes the class-level string variable

(+1)The header for changePosition is correct

(+1)The changePosition method correctly modifies the class-level string variable

3.fullName—Canonical Solution

(a)public static String getFirstName(String name)

{

int space = name.indexOf(“ ”);

String first = name.substring(0, space);

return first;

}

You need to use the indexOf method to find the location of the space. Once you know where the space is located, you can use the substring method to extract from the beginning of the name (index 0) up to the space. Since the second parameter of the substring method is excluded, the space will not be included when the first name is returned.

(b)public static String getLastName(String name)

{

int space = name.indexOf(“ ”);

String last = name.substring(space + 1);

return last;

}

The solution to part (b) is similar to that of part (a). You still need to find the location of the space, but the substring starts at the location of the space plus 1, which will be the first letter of the last name. You could add a second parameter of name.length(), but it isn’t required.

(c)public static int countVowels(String name)

{

int count = 0;

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

{

String letter = name.substring(i, i + 1);

if (letter.equals(“a”) || letter.equals(“e”) || letter.equals(“i”) || letter. equals(“o”) || letter.equals(“u”))

count++;

}

return count;

}

You need to create a loop to go through the entire name. You could use a for-each loop to extract characters, but characters are not part of the AP subset. If you use them, make sure you use them correctly. With a traditional for loop, you need to extract each letter using the substring method and see whether it equals one of the vowels. A count variable is increased by 1 each time a vowel is found.

fullName Rubric

Part (a)

(+1)The indexOf method is used correctly to find the first space

(+1)The substring method is used correctly to extract the first name

(+1)The first name is returned correctly

Part (b)

(+1)The indexOf method is used correctly to find the first space

(+1)The substring method is used correctly to extract the last name

(+1)The last name is returned correctly

Part (c)

(+1)A loop is used to look at each letter in the name

(+1)An if statement is used to determine whether the current letter is a vowel

(+1)The correct vowel count is returned

4.ParkingLot—Canonical Solution

(a)public int openSpaces()

{

int taken = 0;

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

{

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

{

if (lot[r][c] != null)

taken++;

}

return (lot.length * lot[0].length) - taken;

}

You need to use nested for loops to iterate through the lot 2D array. You could also use lot[0].length in the second for loop that iterates through the columns instead of lot[r].length. If you find a spot that is not equal to null (meaning there is already a car parked there), then you should increase a counter variable by 1. That variable should be subtracted from the size of the 2D array to get the final result.

(b)public boolean parkCar(Car newCar)

{

if (openSpaces() > 0)

{

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

{

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

{

if (lot[r][c] == null)

{

lot[r][c] = newCar;

return true;

}

return false;

}

You need to use nested for loops to iterate through the lot 2D array. You could also use lot[0].length in the second for loop that iterates through the columns instead of lot[r].length. If you find a spot that is null, you need to set that spot to newCar and return true, in that order. There should be a return false at the end of the method or in an else statement. On the AP exam, if a question tells you to use a method (such as openSpaces) and you don’t use it, you can lose points.

ParkingLot Rubric

Part (a)

(+1)A variable is declared to keep track of the taken parking spaces

(+1)Nested for loops are used correctly to iterate through the lot 2D array

(+1)An if statement checks whether the current spot in the array is not null

(+1)The correct number of open spaces is returned

Part (b)

(+1)The openSpaces method is called correctly to determine whether there are spaces available

(+1)Nested for loops are used correctly to iterate through the lot 2D array

(+1)An if statement checks whether the current spot in the array is null

(+1)The newCar is assigned correctly to a null element in the 2D array

(+1)The method returns true if a spot was found, and false otherwise

AP Computer Science A Prep, 2024 - Rob Franek 2023

Practice Test 3Part VI: Practice Tests

Practice Test 3
Part VI: Practice Tests