Next-Generation ACCUPLACER 2019-02-20T21:26:35+00:00

Next-Generation ACCUPLACER

Preparing to take the ACCUPLACER?

Awesome!

You’ve found the right page. We will answer every question you have and tell you exactly what you need to study to score well on the ACCUPLACER.

Quick Facts

Get the “need to know” information at a quick glance.

Overview

The ACCUPLACER is a web-based assessment that measures student readiness for college courses. This test focuses on the knowledge and skills that research shows to be essential for college readiness and success. Your ACCUPLACER test score will help colleges, including universities and technical schools, make placement decisions and determine where you will be likely to succeed. The ACCUPLACER focuses on three major areas of content knowledge: Reading, Writing, and Math.

Format

Let’s take a look at how you can expect the ACCUPLACER to be formatted:

Feeling overwhelmed by the number of questions on the ACCUPLACER? Don’t be – that’s because the ACCUPLACER is untimed!

It’s also important to know that you cannot return to previous questions on the ACCUPLACER. This is one reason why you should take advantage of the unlimited time available to you as you take the test.

The ACCUPLACER is a computer adaptive test (CAT). This means that the computer will assign the next question or question set based on the response you gave to the previous question. In other words, this test is designed to fit your ability level.

Scoring

One aspect of the ACCUPLACER (which students really appreciate) is that you cannot fail the test. This is because the ACCUPLACER is not designed to measure what you do or do not know; the test is designed to measure what you are ready to learn. Think of it this way, ACCUPLACER means “accurate placer” – in other words, it focuses on placing you where you will succeed. It does not focus on your prior learning.

Although you cannot pass or fail the ACCUPLACER, you will receive a score after taking the exam. Scores on the ACCUPLACER range from 200 – 300. You will receive your score as soon as you finish the exam. No wait, no sweat!

 

Study Time

Studying appropriately for the ACCUPLACER is going to be your best bet when it comes to putting yourself in a position for future success. Consider this – if you do not score your best on the ACCUPLACER, you may be placed in a remedial class that you do not need.

The amount of time you will need to spend studying for the ACCUPLACER exam depends upon your existing skills and how much information you have retained from previous courses. You can use 240Tutoring practice questions to determine the areas in which you need to strengthen your skills.

Remember, it’s best to spend some time studying each day instead of cramming for the exam shortly before you take it.

What test takers wish they would’ve known:

  • Read every answer choice thoroughly, even if you think that the first or second choice is the best answer. Remember that some answer choices may be “almost correct.” You wouldn’t want to accidentally skip the correct answer choice by selecting an “almost correct” answer and moving on without reading the rest of the choices.
  • When given passages to read, read over the questions first. By doing this, you’ll be on the watch for relevant information.
  • Many test-takers wish they had studied for the ACCUPLACER.

Information and screenshots obtained from the College Board ACCUPLACER website: https://accuplacer.collegeboard.org/sites/default/files/next-generation-test-specifications-manual.pdf

Reading Placement Test

The Reading Placement Test has about 20 questions.

There are four broad categories:

  • Information and Ideas
  • Rhetoric
  • Synthesis
  • Vocabulary

So, let’s talk about Information and Ideas first.

Information and Ideas

This category tests your ability to understand what a text states explicitly, as well as what it implies. You will need to be able to identify the overall message, or theme, of a text, as well as draw conclusions.

Let’s look at a concept that will more than likely appear on the test.

Identifying the Theme of a Text

Finding a central idea or theme of a text is all about getting to the point. As you read, think about the “big picture” created by the details in the text in order to determine the author’s overall message. Is the author trying to make a general statement about life, or is he/she helping you understand a major concept?

You should be able to determine central ideas and themes, as well as analyze how details in the text support these themes. Remember that a theme is rarely stated outright.

Let’s take a moment to think about Aesop’s fable, The Crow and the Pitcher:

“A thirsty crow found a pitcher with a little water at the bottom. Unfortunately, the pitcher had a narrow neck and the crow couldn’t reach the water.

However, the crow had an idea. He began dropping pebbles into the pitcher. With each pebble, the water rose higher.

Eventually, the water rose high enough that the crow was able to drink.”

So, what’s the theme of this fable? Well, Aesop wants us to understand a central message about life. Aesop uses the details in the fable to tell us that it is better to be determined and resourceful than to accept defeat. This overall message is the theme of The Crow and the Pitcher.

When determining the theme, think beyond the details and what is explicitly stated. How do the details come together to create a larger meaning?

Rhetoric

This category tests your ability to analyze both figurative and literal words and phrases and determine how they contribute to the overall mood, tone, and meaning of a text. You’ll also identify and analyze structural formats. On this section, you will need to determine an author’s point of view, purpose, and reasoning.

Here are some concepts you should know.

Text Structure

Text structure refers to the way in which the information in a text is organized. Knowing the different types of text structures can help you understand the author’s purpose, the main idea, and the relationship between ideas in a text.

Let’s take a look at some common text structures:

  • Description: In this type of text, the author describes a topic in detail. One example of a descriptive text is an encyclopedia article, regardless of the topic.
  • Sequence of Events: The author uses a chronological order to show the sequence in which events happen. An example of this type of text is a passage describing the steps to build a garden fence. Most novels and short stories tend to have a sequence of events structure, as well.
  • List: The author uses numerical or chronological order to list items or ideas. For example, most recipes begin with a list of ingredients.
  • Compare and Contrast: The author compares and/or contrasts two or more events, objects, or ideas. A comparison shows how two ideas are alike. When an author contrasts ideas, he/she is showing you how the ideas are different. A passage showing the similarities and differences between evergreen and deciduous trees is an example of a text with this structure.
  • Cause and Effect: The author introduces one or more causes and then describes the effects of the cause(s). A passage about how reading improves literacy is an example of this text structure.
  • Problem and Solution: The author describes a problem or asks a question and then gives possible answers or solutions. An example of this type of writing is a chapter in an article which explains how to treat mosquito bites.

Point of View

Point of view refers to the perspective from which a text is narrated. Together, we’ll look at the different types of perspectives a text may be written from, as well as some examples of each type:

Synthesis

Synthesis questions will test your ability to understand and describe how similar ideas are expressed by more than one text.

Here are a few examples of tasks you may be asked to perform when answering synthesis questions:

  • Compare and contrast the opinions of two authors
  • Explain how two passages are related
  • Choose a statement that two different authors would be likely to agree/disagree on
  • Determine what information is included in one passage, but not in another passage
  • Find the main idea of two passages

Vocabulary

This category tests your ability to determine the literal and figurative meanings of words and phrases in a text. While you complete vocabulary questions, you will need to use your prior knowledge of terms. You’ll also look at word parts and the context in which terms are used in order to determine their meanings.

The main concept you need to know is how to use context clues.

Using Context Clues

Context clues are hints embedded in a text that help readers determine the meaning of a word or phrase. Remember to use context clues when you encounter an unknown word.

Here are a few examples of how to use context clues to find the meaning of unknown words:

  • Definition clues: The author gives the meaning of a term outright. Consider the term “eastern white pine” in the example below.

“The eastern white pine, a type of tree with long white needles, remains green even in winter.”

Notice that you need no knowledge of eastern white pines to determine that they are a type of tree.

  • Synonym clues: The author includes a synonym to help the reader understand the meaning of a word.

“That hotel was luxurious! We were so impressed with how grand and upscale the rooms were.”

In this example, “upscale” and “grand” can help a reader determine the meaning of “luxurious.”

  • Antonym clues: The author includes an antonym to help the reader understand the meaning of a word.

“The top of the table was illuminated, but the rest of the room was dark.”

In this example, “dark” helps the reader determine the meaning of “illuminated.”

And that’s some basic info about the Reading Placement Test.

Now, let’s look at a few practice questions to see how these concepts might actually appear on the real test.

Directions for questions 1-4

Read the passage(s) below and answer the questions based on what is stated or implied in the passage(s) and in any introductory material that may be provided.

Passage 1 is by Dorothy Sayers; Passage 2 is adapted from a work by Raymond Chandler.

Passage 1

The detective story does not and cannot attain the loftiest level of literary achievement. Though it deals with the most desperate effects of rage, jealousy, and revenge, it rarely touches the heights and depths of human passion. It presents us with an accomplished fact and looks upon death with a dispassionate eye. It does not show us the inner workings of the murderer’s mind—it must not, for the identity of the criminal is hidden until the end of the book. The most successful writers are those who contrive to keep the story running from beginning to end at the same emotional level, and it is better to err in the direction of too little feeling than too much.

Passage 2

I think what was really gnawing at Dorothy Sayers in her critique of the detective story was the realization that her kind of detective story was an arid formula unable to satisfy its own implications. If the story started to be about real people, they soon had to do unreal things to conform to the artificial pattern required by the plot. When they did unreal things, they ceased to be real themselves. Sayers’ own stories show that she was annoyed by this triteness. Yet she would not give her characters their heads and let them make their own mystery.

Reading Practice Questions

Question 1

Which of the following best paraphrases the main idea in Passage 1?

  1. Detective stories have plots without direction
  2. Detective stories may someday be improved upon
  3. Detective stories don’t have relatable characters
  4. Detective stories are not great literature

Correct answer: 4. Passage 1 makes the argument that detective stories “cannot attain the loftiest level of literary achievement.” In other words, detective stories “are not great literature.” So, this choice is correct.

 

Question 2

Which of the following statements would the author of Passage 2 most likely agree with?

  1. Detective writers should hold on tightly to their plots and characters’ actions
  2. Dorothy Sayers writes formulaic and predictable detective stories
  3. Detective story writers should ensure that the characters are not too deeply developed
  4. Dorothy Sayers’ writings have inspired a new type of detective story

Correct answer: 2. The author in Passage 2 makes it clear that the reason why Dorothy Sayers (author of Passage 1) finds fault with the detective genre is because she comes to it with the assumption that it is prescribed. Thus, this choice is correct.

Question 3

Which of the following best describes the relationship between the two passages?

  1. Passage 1 focuses on the uniqueness of the genre; Passage 2 focuses on its weaknesses
  2. Passage 1 discusses the limitations of the genre; Passage 2 discusses the uniqueness of the genre
  3. Passage 1 highlights the merits of the genre; Passage 2 downplays its faults
  4. Passage 1 relates the details the particulars of the genre; Passage 2 generalizes the limitations

Correct answer: 2. This is the best answer, because the first passage focuses on the limitations of writing in the detective genre, while Passage 2 focuses on how the detective genre is unique and has merit.

Question 4

Which of the following would the author of Passage 1 most likely agree with?

  1. The characters in detective stories are well-developed and have deep inner worlds
  2. The detective genre involves a creative and unpredictable writing style
  3. The plot of a detective story is formulated
  4. Detective story writers will soon be known as world-class writers

Correct answer: 3. This is the correct answer, because Dorothy Sayers, author of Passage 1, describes the predictable, strict, and formulaic writing style that characterizes the detective genre, in her opinion.

 

Directions for questions 5-6

Read the passage(s) below and answer the questions based on what is stated or implied in the passage(s) and in any introductory material that may be provided.

While most people can name plenty of their favorite artists, ask someone what makes an artist great, and you’ll likely get a different answer from each person you ask. Try to compare the greatness of different artists and you might start an argument. That’s because feeling connected to a work of art is an incredibly personal experience. The same piece of work may affect two people in very different ways, ranging from delight to indifference to disgust. Some works of art end up in the trash, some incite riots, and some are put on the cover of magazines. Still, the art that ends up in the trash could be discovered and treasured years later, while the art on the magazine cover can end up forgotten. No matter what happens to the art, as long as it exists, it always has the potential to inspire others.

Question 5

The author’s main purpose in this selection is:

  1. to show that responses to art are subjective and personal.
  2. to show which qualities make art great.
  3. to show that people don’t care about art.
  4. to show that artistic standards don’t change.

Correct answer: 1. “Responses to art are subjective…” is correct, because the author explicitly states that responses to art are personal. In addition, the author shows that art is subjective, or influenced by personal feelings, tastes, or opinions, by stating that “the same piece of work may affect two people in very different ways.”

Question 6

According to the information presented in the selection, people disagree on the greatness of art and artists because:

  1. art has the potential to inspire others.
  2. the standards of great art haven’t changed since the Renaissance.
  3. feeling connected to art is a personal experience.
  4. art is irrelevant in society today.

Correct answer: 3. “Art is a personal experience” is correct, because the author explicitly states in the selection that “feeling connected to a work of art is an incredibly personal experience.”

Directions for questions 7-8

Read the passage(s) below and answer the questions based on what is stated or implied in the passage(s) and in any introductory material that may be provided.

1    President Kennedy was not the first to imagine sending a man to the moon. A little more than 100 years earlier, in 1865, science fiction writer Jules Verne also imagined space travel. He put his innovative thoughts in a book called From the Earth to the Moon. In it he described a lunar expedition that is so eerily close to the Apollo 11 mission that a reader would think he was predicting the future. He called his spaceship with a crew of three the Columbiad. In his book the spacecraft launches from Florida, and the United States Navy recovers it from the Pacific Ocean. In 1969, Florida was the launch site of Apollo 11. The command module was named Columbia. When the spacecraft returned to Earth, it splashed down in the Pacific, where the Navy recovered it along with its three-astronaut crew. Verne accurately delineated the future when the technology of his own time made his predictions seem highly unlikely to occur. How could he have known that his far-fetched idea was not so far-fetched after all?

2    Like Verne, other science fiction writers have accurately described inventions that are commonplace today. Many of H. G. Wells’s ideas, for example, have become a reality. Considered by many to be one of the best science fiction writers of all time, Wells wrote about lasers, wireless communication, automatic doors, and other gadgets that did not exist at the time of his writings. But today these gadgets are such an integral part of our society that we probably cannot imagine living without them. Wells also describes a journey to the moon on a spaceship made from anti gravity material. We can only speculate that these writers might have inspired those who later turned their fiction into reality.

3    In 2012, a Mars rover, developed by the National Aeronautics and Space Administration (NASA), landed on the planet Mars. No one would have been more excited to hear the news than Ray Bradbury, one of America’s greatest science fiction writers. In 1950 he wrote about travel to Mars in his book The Martian Chronicles. The book describes an expedition that lands humans on Mars. The story then tells how the people inhabit the planet and bring their families to live there. Since NASA has successfully landed a rover on Mars, Bradbury’s fantasy may yet become reality. The Mars rover, appropriately called Curiosity, is gathering information that will help NASA plan a manned mission to Mars sometime in the 2030s. Will future families travel to Mars to live there, as Bradbury imagined? If so, the world as we know it today will certainly be different.

Question 7

What is one detail that illustrates how Jules Verne’s book connects with the real Apollo 11 mission?

  1. When the spacecraft returned to earth, it landed in the Pacific Ocean
  2. President Kennedy was in charge of the space launch
  3. When they landed on Mars, it looked eerily similar to the way Verne had described it
  4. The mission’s name, Apollo 11, was taken from Verne’s book

Correct answer: 1. This question asks you to recall simple details, so the answer will be right in the passage. This is the only answer that correctly states what is implicitly written.

 

Question 8

How is the information in this selection organized?

  1. Cause/Effect
  2. Problem/Solution
  3. Compare/Contrast
  4. Spatial

Correct answer: 3. The paragraph does not discuss how fiction writing necessarily caused these modern inventions to take place; it simply focuses on describing their similarities. Likewise, the passage doesn’t describe any effects of the writings on reality; it’s simply showing us how they’re oddly similar. Therefore, “Cause/Effect” is incorrect. Also, “Problem/Solution” is incorrect since there is no problem stated to be solved. “Spatial” is incorrect, because the passage does not focus on the geography or locations. These facts are mentioned, but they’re mentioned as part of the main idea that they’re the same as in the past writings that seem to predict their realities.

Directions for questions 9-10

The following sentences have a blank indicating that something has been left out. Beneath the sentence are four words or phrases. Choose the word or phrase that, when inserted in the sentence, best fits the meaning of the sentence as a whole.

Question 9

After hearing the testimony of more than ten women, the head detective decided to keep investigating in order to determine if anyone else, in addition to the main suspect, was ____________.

  1. irreproachable
  2. culpable
  3. incognizant
  4. uninformed

Correct answer: 2. Culpable, an adjective, is defined as criminal or guilty. This is the best answer choice.

 

Question 10

The petulant worker, who was ____________ throwing tantrums and treating guests unfairly, demanded a tip for carrying the couple’s luggage.

  1. known for
  2. indebted to
  3. pleased with
  4. described as

Correct answer: 1. The phrase known for best completes the sentences.

Writing Placement Test

The Writing Placement Test has about 25 questions.

There are six broad categories:

  • Development
  • Organization
  • Effective Language Use
  • Sentence Structure
  • Conventions of Usage
  • Conventions of Punctuation

So, let’s talk about Development first.

Development

This category tests your ability to revise passages and sentences.

Let’s take a look at a concept that is more than likely to appear on the test.

Focus

A text with a clear focus does not include information that strays from the main idea. Sentences that don’t support the main idea serve no real purpose and confuse the reader. On the test, you’ll encounter focus questions which will ask you to add, revise, delete, or retain information in a passage in order to ensure that the text does not stray from its purpose or main idea.

Here are a few ways that you can revise a passage to ensure that it has a clear focus:

  • Add a topic sentence if the passage does not include one.
  • Delete details that are not relevant.
  • Change the wording of a sentence so that its relationship to the main idea is stronger.
  • Add sentences that clarify the relationship of details with the main idea.

Let’s take a look at an example passage together:

The planet Jupiter has a mass which is more than 300 times the mass of the Earth. Jupiter was named after the Roman god of the sky. In ancient Greece, this god was known as Zeus. Both gods, Zeus and Jupiter, were said to use a lightning bolt as a weapon. The planet Jupiter is the fifth planet from the Sun, and it is about 150 times farther from the sun than the Earth; therefore, the surface of Jupiter is much colder than the surface of Earth. Most images of Jupiter show the planet as being white, orange, and brown.

Take a second look at the underlined sentences. What in the world are they doing there? They certainly don’t help you learn about the planet Jupiter, which is the topic of the passage!

The author should delete these two sentences to maintain focus.

Organization

This category tests your ability to revise passages as needed so that information and ideas appear in the most logical order. You’ll answer questions about introductions, conclusions, and transitional sentences, phrases, and words.

Here is a concept you should know.

Sequencing Ideas and Events

Proficient writers are able to order their ideas logically. In a nonfiction text, writers generally introduce a main idea, give details to support and explain the main idea, and then give the reader a concluding message about the main idea. In both fiction and nonfiction texts, writers often need to explain events in the order in which they occur.

When reading a passage on the Writing Placement Test, ask yourself what the purpose of each sentence is. Is the sentence clarifying the previous sentence? Is it introducing a new idea? Is it showing the connection between two ideas? Determining the purpose of each sentence allows you to decide whether or not it is correctly placed in a passage.

When completing the test, you may be asked whether or not a sentence in a passage should be moved to a different place in the passage. Remember that you can take your time on this test, so try the sentence out in each suggested place in the passage. Ask yourself where the sentence makes the most sense. Usually, your “gut feeling” is correct. 

Effective Language Use

This category tests your ability to revise a text by replacing existing words with more precise and appropriate words. You’ll also show that you can identify words that should be deleted, because they are pointless or redundant. Furthermore, you will show that you are able to create and recognize a variety of sentence structures.

Take a look at this concept.

Sentence Structures

In order to perform well on this section, you will need to be able to identify and create sentences with a variety of structures. Let’s take a look at some different types of sentence structures, as well as some examples of each:

A simple sentence has only one clause. A clause is a phrase which contains a subject and a verb. The clause in a simple sentence is called an independent clause, because it is able to stand alone. Here are examples of simple sentences:

  • He is cooking chicken for dinner tonight.
  • This soup contains fresh vegetables, a variety of savory herbs, and oven-roasted chicken.

A compound sentence contains two or more independent clauses, joined by a conjunction. Let’s look at some examples of compound sentences. Notice that the conjunctions between clauses are underlined.

  • Rob wants to visit New York City, so he will buy a plane ticket.
  • The boy did not want to tattle on his friend, nor did he want to feel guilty about staying silent.
  • Alice definitely did not want to go to the movie theater, but she did not feel like going to a restaurant either, so she ended up staying home.

Complex sentences have at least one independent clause, as well as at least one dependent clause. Unlike independent clauses, dependent clauses cannot stand alone. A dependent clause is missing either a subject or a verb, or it contains a subject and a verb, but it is not a complete thought. In the following  examples of complex sentences, the dependent clauses are underlined:

  • Despite Ian’s doubts, he found that his new professor was very agreeable.
  • When I went to visit my cousin, who lives in Greensboro, I met some of her friends.

Compound-complex sentences have at least two independent clauses, as well as at least one dependent clause. In the following examples, the dependent clauses are underlined:

  • I want to watch television, though considering the state of the house, I need to finish cleaning the bedroom first.
  • After Sarah’s dalmation got sick, she started choosing pet food more carefully, and her dog is perfectly healthy now.

Sentence Structure

This category tests your ability to recognize and appropriately use verb tenses, pronouns, and punctuation. You will also be tested on your knowledge of sentence structures, subordinate and coordinate clauses, and modifiers.

Take a look at this concept that may come up on the test.

Modifiers

A modifier is a word, phrase, or clause that describes another word. Usually, modifiers are adjectives, like “friendly” or “personable.” Sometimes, a modifier is included in the wrong part of a sentence. Consider this example:

“The live bowl of fish looks great in her office.”

Since the fish are alive and the bowl is not, “live” is a misplaced modifier. You could revise the sentence by moving the modifier:

“The bowl of live fish looks great in her office.”

A dangling modifier is a modifier that does not have a clear subject. The word that a dangling modifier is meant to describe is missing. Consider this example:

Staring over the fields, the storm approached.”

“Staring” is a modifier, but who or what is staring? There are multiple ways to revise this sentence, and here is one example:

Uncle George was staring over the fields as the storm approached.”

Conventions of Usage

This category tests your ability to correctly use words that are commonly misused, revise expressions, and correct errors in the agreement between subjects and verbs and pronouns and antecedents.

Let’s look at a concept that is more than likely to pop up on the test.

Pronoun-Antecedent Agreement

Pronouns, such as “he,” are used to replace more specific nouns like “Barkley.” If a more specific noun, like “Barkley,” occurs first in the text, then that noun is an antecedent. Consider this information as you read the following sentence:

“If Barkley could speak, he would probably ask you to get his leash and take him for a walk in the dog park.”

In the example, “Barkley” is the antecedent and “he” and “his” are pronouns. Pronouns should agree with their antecedents in terms of number and gender.

Conventions of Punctuation

This category tests your ability to correctly use punctuation to end sentences, write lists, show possession, and separate ideas.

Here is a concept to know.

Semicolons

We tend to use semicolons (;) less often than many other types of punctuation marks; therefore, many people are confused about how to use them.

Think back to when you read about independent clauses, which are able to stand on their own. We often use conjunctions like “and” and “for” to join independent clauses. Sometimes, however, it may be appropriate to join two independent clauses using a semicolon. A semicolon divides two independent clauses, but it tells you that the information in those clauses is related.

Here are a few examples of appropriate semicolon use to separate independent clauses:

I know you don’t like going to the gym; nevertheless, it is very good for you.

We should plan to see each other soon; we haven’t spent any time together lately.

Another appropriate time to use semicolons is when separating items in a list that already contains commas. Here’s an example:

Andrea visited Lima, Peru; Santiago, Chile; and Caracas, Venezuela.

And that’s some basic info about the Writing Placement Test.

Now, let’s look at a few practice questions to see how these concepts might actually appear on the real test.

Directions for questions 1-3

Read the following early draft of an essay and then choose the best answer to the question or the best completion of the statement.

Sports are a wonderful means for mankind to exercise one of its most basic principles: competition with our fellow man. Surrounding all types of sports is the concept of sportsmanship – the respect and ethical behavior shown to all participants of a contest. The spirit of the game, in many cases, is more important than the outcome of the match; a true competitor understands this. This is why many of our most beloved athletes are not always the most talented performers—it is the players who play with the purest motive, for the sake of the team, and with respect for all opponents, who gain the respect and admiration of the fans.

There are greater lessons to be learned from sports than being well liked by fans. Sports, and by extension, the athletes who play them, extend beyond cultural differences; surely styles of play can vary between countries and regions, but in general, sports are played the same everywhere. Similarly, fans of a sport are able to appreciate incredible athletic feats or displays of true sportsmanship regardless of the player. Simply put, in a day and age when settling cultural differences is of utmost importance, sports are a reasonably viable way to bring the world closer together.

Lastly, international events such as the Olympic Games or World Cup are perfect opportunities to show the world that international cooperation and peace are possible. Sports can and should be used as instruments of change in an uncertain world. They can also be proponents of peace.

The Olympic Creed says it best: “The most important thing in the Olympic Games is not to win but to take part, just as the most important thing in life is not the triumph, but the struggle. The essential thing is not to have conquered, but to have fought well.”

Writing Practice Questions

Question 1

Which transition would best connect the mention of sportsmanship in paragraph one to the first sentence of paragraph two?

  1. In other words
  2. Therefore
  3. For example
  4. However

Correct answer: 4. “Therefore” is incorrect, because it is used when authors provide a conclusion to an argument. “In other words” is incorrect, because it suggests that the author is simply restating details in a different way. “For example” is incorrect, because it would be used to provide a piece of evidence that supports previously stated ideas.

Question 2

The passage above discusses the importance of sports. Select the best evidence from the passage to support the author’s belief that sports can connect different cultures.

  1. “Sports are a reasonably viable way to bring the world closer together”
  2. “The essential thing is not to have conquered, but to have fought well”
  3. “The spirit of the game, in many cases, is more important than the outcome of the match”
  4. “There are greater lessons to be learned from sports than being well-liked by fans”

Correct answer: 1. “Sports are a reasonably viable way to bring the world closer together” is correct. It best supports the statement, because bringing the world closer together would require the connection of cultures.

Question 3

Which of the following is the most effective way to revise and combine the following sentences from paragraph three? “Sports can and should be used as instruments of change in an uncertain world. They can also be proponents of peace.”

  1. Leave it as it is now
  2. Sports can and should be used as instruments of change in an uncertain world, and they can also be proponents of peace
  3. Sports can and should be used as instruments of change in an uncertain world; they can also be proponents of peace
  4. Sports can and should be used as instruments of change and proponents of peace in an uncertain world

Correct answer: 4. This is the correct answer.

 

Directions for questions 4-10

Choose the best answer to the question.

Question 4

It is important to brush your teeth twice a day, even if they are pinched for time in your schedule.

  1. (as it is now)
  2. you are
  3. everyone is
  4. one is

Correct answer: 2. “Your” is a second-person pronoun. To ensure the pronouns agree, the underlined words must also be in second person. “They,” “everyone,” and “one” are all third-person pronouns. “You” is the only second-person pronoun in the answer choices.

 

Question 5

In the arctic, many animals have adaptable fur, it turns white in the winter.

  1. (as it is now)
  2. fur, but
  3. fur, which
  4. fur, and

Correct answer: 3. This avoids a comma-splice error by replacing “it” with “which,” thus making the second independent clause into a dependent clause. Two independent clauses can’t be joined by a comma.

Question 6

Texas experiences very hot and humid summers; therefore, it is recommended to pack cool clothes and a small, portable fan.

  1. (as it is now)
  2. moreover
  3. in contrast
  4. however

Correct answer: 1. This is the correct answer. Answer choices 2, 3, and 4 create illogical phrases.

 

Question 7

Salt and baking powder is needed for the recipe.

  1. (as it is now)
  2. both is
  3. are
  4. was

Correct answer: 3. The subject and verb must agree in number. There are two ingredients listed (plural), so the verb used must also be plural.

Question 8

Which is the best version of this sentence?

All of the resources in the library are outdated, leaving the teachers to purchase their own materials.

  1. Leave it as it is now
  2. All of the resources, leaving the teachers to purchase their own materials, in the library are outdated.
  3. The teachers purchase their own materials, because all of the resources in the library are outdated.
  4. All of the resources in the library are outdated, because the teachers purchase their own materials.

Correct answer: 3. The sentence should be rewritten, “The teachers purchase their own materials, because all of the resources in the library are outdated.” This is the only answer choice that shows the relationship between the two ideas.

Question 9

Which is the best version of this sentence?

Carol, who recently divorced Jim, and nearly forty, was nervous to begin a new life journey.

  1. Leave it as it is now
  2. Carol was nervous to begin a new life journey, who recently divorced Jim and who will soon turn forty.
  3. Carol, nearly forty and divorced, she was nervous to begin a new life journey.
  4. Nearly forty, and recently divorced from Jim, Carol was nervous to begin a new life journey.

Correct answer: 4. The sentence should be rewritten, “Nearly forty, and recently divorced from Jim, Carol was nervous to begin a new life journey.” This correctly places the modifier by the subject, Carol.

Question 10

Which is the best version of this sentence?

Expecting a sold out show, the theater was opened an hour early to reduce crowds.

  1. Leave it as it is now
  2. The show was expected to sell out; therefore, the theater was opened an hour early to reduce crowds.
  3. The show was expected to sell out; as a result of, the theater was opened an hour early to reduce crowds.
  4. The show was expected to sell out; moreover, the theater was opened an hour early to reduce crowds.

Correct answer: 2. The sentence should be rewritten, “The show was expected to sell out; therefore, the theater was opened an hour early to reduce crowds. This is the only answer choice that appropriately links the two clauses.

 

Math: Arithmetic Placement Test

The Math: Arithmetic Placement Test has about 20 questions.

There are five broad categories:

  • Whole Number Operations
  • Fraction Operations
  • Decimal Operations
  • Percent
  • Number Comparisons and Equivalents

So, let’s talk about Whole Number Operations first.

Whole Number Operations

This category tests your ability to add, subtract, multiply, and divide whole numbers. You will need to use the order of operations. You’ll also need to know how to estimate values and round answer choices when necessary.

Let’s look at a concept.

Order of Operations

“PEMDAS” is the acronym which is often used to remember the order of operations. This acronym stands for Parentheses, Exponents, Multiplication/Division, and Addition/Subtraction. The mnemonic phrase, “Please Excuse My Dear Aunt Sally” can help you remember “PEMDAS” while taking the test.

Here’s a simple example problem:

7  + 5  – (4 × 2)

Solve what’s in the parentheses first:

7  + 5  – 8

Now, we add 7 and 5 to get 12:

12  – 8

Finally, we subtract 8 from 12 to get our answer:

4

Fraction Operations

This category tests your ability to solve problems with fractions and mixed numbers. On this section, you will need to apply your knowledge of the order of operations, estimation, and rounding. You will be asked to add, subtract, multiply, and divide fractions and mixed numbers.

Take a look at this concept.

Dividing Fractions by Fractions

When presented with a question that asks you to divide one fraction by another fraction, keep in mind that dividing fractions is just multiplying with reciprocals (inverses). Once your multiplication problem is set up, you will multiply the numerators, or values on top, by one another. Then you will multiply the denominators, or the values on bottom, by one another.

What might this look like on the test? Here’s an example:

8/6  ÷  3/2

First, set up your multiplication problem by flipping the second fraction (finding the reciprocal/inverse):

8/6  ×  2/3

Multiply across to find your answer:

16/18  or  8/9

You have your answer!

Decimal Operations

This category tests your ability to use the order of operations to add, subtract, multiply, divide, and round numbers that contain decimals. You may also need to convert decimals into fractions.

Let’s look at a concept that will more than likely appear on the test.

Converting Decimals to Fractions

On the test, it is likely that you will need to convert decimals to fractions. You may need to do this to add, subtract, multiply, or divide decimals and fractions within the same problem.

Converting decimals to fractions may sound complicated if you haven’t done it before, but we can do it in just three easy steps. Let’s practice by converting .55 to a fraction.

Step 1: Write the decimal as a numerator over the number 1:

.55/1

Step 2: Multiply both the numerator and the denominator by 10 for every digit to the right of the decimal. Since .55 has two digits to the right of the decimal, we will multiply both the numerator and the denominator by 10 twice:

(.55 x 10 x 10 = 55) / (1 x 10 x 10 = 100)

or

55 /100

Step 3: Now that we have converted .55 to a fraction, we need to simplify it. Look for the largest number that we can divide both 55 and 100 by to get two whole numbers.

In this case, 5 is the largest number that we can divide both 55 and 100 by to get whole numbers. When we divide both the numerator and denominator by 5, we get our final and simplified answer:

11 / 20

Percent

This category tests your ability to work with percentages in real-life contexts, such as calculating prices. You’ll need to be able to determine certain percentages of given numbers. You will also be asked questions that require you to calculate the percentage by which numbers decrease or increase.

Here is a concept you should know.

Calculating Percent Increase/Decrease

Whether calculating percent increase or decrease, you will be using the same formula. After all, you are really just calculating the percentage of change, regardless of whether there is an increase or decrease in value.

Here’s what the formula looks like:

So, let’s put the formula to work with an example:

Last semester, 75 freshmen enrolled in a drawing course. This semester, 81 students have enrolled in the same drawing course. What is the percentage increase in drawing course enrollment?

So, we will subtract the old value from the new value. Note that you do not need to worry about whether the answer is positive or negative. Just consider the absolute value of the answer:

81 – 75 = 6

Next, we will divide 6 by the old value:

6 ÷ 75 = .08

The final step is to multiply that answer by 100.

.08 x 100 = 8

Therefore, there was an 8% increase in student enrollment in the drawing course.

Remember that you will use the same formula to calculate percent decrease and percent increase. Isn’t it such a relief that you don’t have to memorize another formula?

Number Comparisons and Equivalents

This section will test your ability to compare and order numbers. You’ll need to remember how to use equality and inequality symbols, as well as how to correctly place numbers on a number line.

Let’s look at some concepts.

Ordering Numbers

The easiest way to order numbers is to convert any numbers containing fractions to decimals; then plot the numbers on a number line. Let’s look at an example together:

Let’s say you are given the following numbers:

-½, .5, 1¼ , 2.75

Let’s convert them each to decimals:

-.5, .5, 1.25, 2.75

Numbers are plotted on a number line by using a dot or circle for each number. Take a look at the following number line. Notice that the place of each of our numbers (-.5, .5, 1.25, 2.75) is indicated as a point on the number line.

Comparing Numbers

Whether you are comparing whole numbers, fractions, decimals, or mixed numbers, you will use the same comparison symbols:

Less than: <

Examples:

2 < 5

.10 < .20

5 ½ < 10 ¾

½ < ¾

Greater than: >

Examples:

21 > 5

.10 > .01

51 ½ > 2¾

½ > 

Equal to: =

Examples:

2 = 2

.5 = ½

3 ¾ = 3.75

⅓  = ⅓

It’s easy to remember the greater than and less than symbols, because the larger side of each symbol always points to the larger number!

And that’s some basic info about the Math: Arithmetic Placement Test.

Now, let’s look at a few practice questions to see how these concepts might actually appear on the real test.

Directions

Choose the best answer. Use paper if necessary.

Mathematics Arithmetic Practice Questions

Question 1

The varsity basketball team has 3 freshmen, 5 sophomores, 3 juniors, and 4 seniors. Approximately what percentage of the basketball team is comprised of sophomores?

  1. 30%
  2. 33%
  3. 25%
  4. 20%

Correct answer: 2. A percentage is a comparison. In this case, the percentage of the basketball team that is comprised of sophomores is found by dividing the number of sophomores on the team by the total number of students on the team. There are 5 sophomores on the team. There are a total of 3 + 5 + 3 + 4 = 15 students on the basketball team. Therefore, the ratio of sophomores to the whole team can be represented by 5:15 or 1:3. When the ratio is converted to a fraction and divided, the decimal form of ⅓ = 0.333… The decimal is converted to a percentage by moving the decimal point two places to the right and putting a percent symbol at the back. Therefore, the best answer is the rounded version of 33.333…%, or 33%.

Question 2

Which of the following is equivalent to 6 + (5 – 2)² × 3?

  1. 45
  2. 36
  3. 24
  4. 33

Correct answer: 4. This is the correct answer. You must work this problem by following the order of operations (parentheses, exponents, multiplication/division, addition/subtraction). First, solve within the parentheses: 5 – 2 = 3. The problem now reads 6 + 3² × 3. Then, solve for the exponents: 3² = 9. The problem now reads 6 + 9 × 3. Next, solve the multiplication problem: 9 × 3 = 27. The problem now reads 6 + 27. Finally, solve the addition problem: 6 + 27 = 33.

 

Question 3

Which of the following fractions is equal to 0.08?

  1. 1/8
  2. 1/80
  3. 8/10
  4. 8/100

Correct answer: 4. This is the corresponding fraction. It can further reduce to 2/25.

 

Question 4

A horse trotted 2/3 miles in 3/4 of an hour. What was the horse’s average trotting speed, in miles per hour?

  1. 11  5/6
  2. 9  1/2
  3. 8  5/7
  4. 10  2/9

Correct answer: 4. To determine the speed in miles per hour, the number of miles that the horse trotted must be divided by the number of hours the horse was trotting. In this case, 7 2/3 must be divided by 3/4. To do this computation by hand, the mixed number 7 2/3 should be converted to its improper fraction format: (7 × 3+2)/3 = 23/3. Therefore, 23/3 must be divided by 3/4. To divide by a fraction is the same as multiplying by that fraction’s reciprocal, and so 23/3 ÷ 3/4 becomes 23/3 × 4/3. Fractions should be multiplied straight across and then reduced, if possible, or factors common to the numerator and denominator can be identified and canceled before the multiplication is fully performed. In this case, there are no common factors, so 23/3 × 4/3 = (23 × 4)/ (3/3), and with no factors common to the numerator and denominator of the fraction, reducing will not be possible. The final answer is the simplified version of (23 × 4)/(3 × 3)92/9, written in its mixed number format (92 ÷ 9 = 10 with a remainder of 2): 10 2/9 miles per hour.

Question 5

Cierra purchased a car for $22,000. She is now trying to sell the car for $14,000. By approximately what percent did Cierra reduce the price of her car?

  1. 57%
  2. 8%
  3. 80%
  4. 36%

Correct answer: 4. The amount Cierra purchased the car for was $22,000. She is trying to sell it for $14,000, which is a difference of $22,000 – $14,000 = $8,000. In order to calculate the percent of change, the amount of change ($8,000) must be divided into the original value ($22,000). That division results in the repeating decimal 0.363636… When the decimal is converted to a percentage, it becomes the correct final answer of approximately 36%. Alternatively, the Amount of Reduction = Original Purchase Price × Percent Reduced, so 8,000 = 22,000 × p, where p is the percent by which the price is reduced. To solve the equation 8,000 = 22,000p, divide by 22,000 on each side. The result is 0.363636… = p, or ~36%.

Question 6

4/7,  5/11,  3/5,  7/9

Which of the fractions above is the greatest?

  1. 4/7
  2. 7/9
  3. 3/5
  4. 5/11

Correct answer: 2. First, estimation and benchmark fractions can be used to eliminate the answer option of 5/11. Because half of 11 is 5.5, 5/11 is clearly smaller than 0.5 or 1/2. When the numerators of the other fractions are compared to one-half of their own denominators, the other three fractions are quickly seen to be larger than 1/2, and therefore, 5/11 cannot be the largest fraction. The fraction 3/5 can also be eliminated as being the greatest value because it is clearly smaller than 7/9. (3/5 < 7/9 because each numerator is 2 units less than its denominator and because fifths are larger amounts than ninths, the fraction 3/5 is “missing more” of its whole than the fraction 7/9.) From there, 7/9 should seem to be easily more than 4/7 because the latter is so close to 1/2 and the former is closer to one.

Question 7

Which of the following points on a number line is the greatest distance from 0.5?

  1. 1.5
  2. 0
  3. -1.5
  4. 1

Correct answer: 3. To answer this question, the difference between each number in the answer choices and the number 0.5 can be considered. The order of the subtraction does not matter, as only the distance (in either direction) is of concern. 0.5 – (-1.5) = 0.5 + 1.5 = 2 0.5 – 0 = 0.5 1 – 0.5 = 0.5 1.5 – 0.5 = 1 Therefore, the option that is the greatest distance from 0.5 on a number line is the value -1.5 because it is 2 units away from 0.5, while the other numbers are just 0.5 or 1 unit away.

Question 8

What is 2,596 + 853?

  1. 3,449
  2. 3,349
  3. 3,309
  4. 4,449

Correct answer: 1. This is the correct sum.

Question 9

What is the value of 4.83 + 0.006 + 0.135?

  1. 6.24
  2. 5.24
  3. 4.971
  4. 4.961

Correct answer: 3. This is the correct sum.

Question 10

What is 0.5436 rounded to the nearest hundredth?

  1. 0.55
  2. 0.544
  3. 0.54
  4. 0.543

Correct answer: 3. This is the correct answer.

Math: Quantitative Reasoning, Algebra, and Statistics Placement Test

The Math: Arithmetic Quantitative Reasoning, Algebra, and Statistics Placement Test has about 20 questions. 

There are ten broad categories:

  • Rational Numbers
  • Ratio and Proportional Relationships
  • Exponents
  • Algebraic Expressions
  • Linear Equations
  • Linear Applications and Graphs
  • Probability and Sets
  • Descriptive Statistics
  • Geometry Concepts for Pre-Algebra
  • Geometry Concepts for Algebra 1

So, let’s talk about a few of the more challenging categories.

Algebraic Expressions

This category tests your ability to create and solve algebraic expressions and identify equivalent expressions. On this portion of the test, you will also need to combine like terms in order to simplify equations and determine solutions.

Take a look at this concept.

Identifying Equivalent Expressions

Equivalent expressions have the same solution, but they may be written differently. To solve for equivalent expressions, you’ll need to combine like terms and use the distributive property.

Here’s an example of two equivalent expressions:

2x + 2x and 2(x⁴ + x). 

The second expression has the same value as the first. If you distribute the 2 in the second expression, it will look like the first one.

In order to find out if two expressions are equivalent, just remember to simplify each expression as much as possible by combining like terms.

Linear Equations

This category tests your ability to solve and simplify linear equations and inequalities. You may be asked to solve systems of linear equations.

Let’s look at a concept.

Solving Linear Equations

A linear equation has simple variables, such as x and y. We will explore graphing linear equations later on. For now, let’s look at an example and solve it:

3x + 15 = x + 25

Now remember that the equals sign tells you that the values on each side are equivalent. If we are going to solve for x, we need to combine like terms. Let’s start by moving 15 to the other side of the equation.

When we move 15 to the other side, we need to use its inverse (-15).

3x + 15 = x + 25

-15           -15

3x = x + 10

Next, we need to get all of our x variables on one side of the equation. Remember that “x” is actually “1x.” Therefore, we need to subtract 1x from the left side of the equation:

3x = 1x + 10

-1x          -1x

2x = 10

Our last step is to get a single x variable alone. We can do this by dividing each side of the equation by 2:

x = 5

Linear Applications and Graphs

This category tests your ability to graph and describe linear equations. You will work with systems of equations, as well as linear inequalities.

Let’s look at a concept that will more than likely pop up on the test.

Graphing Linear Equations

To graph a linear equation, we must determine the value of each variable. Let’s graph x + 2y = 7 together. In order to graph this equation, we will set each of the variables to 0.

First, we will set x as 0. Our equation becomes:

0 + 2y = 7

In order to solve for y, we just need to divide each side by 2:

y = 3.5

Now that we have determined the value of y, we will determine the value of x. We can do this by setting y to 0 in the original equation:

x + 2(0) = 7

or
x + 0 = 7

therefore,

x = 7

We have now determined the value of both x and y: x = 7 and y = 3.5.

Now, let’s think about graphing. Remember that a coordinate on a graph is always shown as (x, y).

We know that two of our points will be (0,3.5) and (7,0). Using a graph, we can mark each of these two points and draw a line between them:

Probability and Sets

This category tests your ability to calculate probability, both when events are related and when they are independent from one another. You’ll show that you can answer questions about simple, compound, and conditional probability.

Here is a concept you should know.

Conditional Probability

Unlike compound probability, conditional probability refers to the likelihood that two related events will occur. When working with conditional probability, events are dependent, as opposed to independent. Let’s look at an example:

Several children go to a local fair and play a game. While playing the game, each child has two chances to throw a ball at a target.

Out of all the children, 35% hit the target on the first try, and 15% hit the target on both tries (these are the given conditions – that’s where the “conditional” part comes in).

Given these conditions, what is the probability that a child who hits the target on the first try also hits the target on the second try?

In order to answer this question correctly, we need to use an equation which can be used to determine conditional probability:

P(A|B) = P(AandB)/P(B)

This equation tells us that the probability of B occurring if A has occurred is the probability of both events occurring divided by the probability of B occurring.

Each time a child attempts to hit the target is an event. 15% of the children hit the target twice (event A and event B). 35% of the children hit the target on the first try (event A). Therefore, our formula will look like this:

P(A|B) = 15/35

When we divide 15 by 35, we get .4285.

Now, we need to show that decimal as a percentage: 42.85%. Our answer shows that any child who hits the target on the first try has a 42.85% chance of hitting the target on the second try.

Descriptive Statistics

This category tests your ability to work with statistics that describe various data sets. You will use measures of center (mean, median, and mode) to describe the data that is presented to you. You should also be able to interpret data presented in a variety of graphical displays.

Here is a concept you should know.

Measures of Center

On the test, you’ll answer questions about measures of center, such as mean, median, and mode. Take a look at the following set of numbers:

2, 2.5, 3, 3, 3.5, 7

The mean is the average. Add all of the numbers in the set together, and then divide by the amount of numbers in the set (in this case 6). The mean of this list is 3.5.

The mode is the number that occurs most in the set. In this case, the only number that appears more than once is 3. Therefore, 3 is the mode.

The median is the number that appears in the middle of the list when the numbers are ordered from least to greatest. What do you do if you have an even amount of numbers and no “middle” number? You add the two numbers in the center and divide them by two. Since 3 + 3 = 6 and 6/2 = 3, 3 is the median of this particular set.

Geometry Concepts for Pre-Algebra

This category tests your ability to understand and determine the radius, diameter, perimeter, and circumference of circles. You will also use formulas to determine the volume of prisms.

Let’s look at a concept that may appear on the test.

Circumference of a Circle

To calculate the circumference of a circle, you can use the formula C = 2πr. The “r” represents the radius, a segment that connects the center of the circle to the edge of the circle. Alternately, you can use the formula C = πd. The “d” in that formula represents diameter, which is twice the radius.

How might this appear on the test?

You have a circular flower bed with a radius of 4 feet.
What is the circumference of the flower bed?

A. 25.13 ft.

B. 50.27 ft.

C. 30.17 ft.

D. 21.22 ft.

The answer to this question is A. In order to find this answer, we plug the value of the radius into the formula C = 2πr. Our equation is C = 2Π4.

Geometry Concepts for Algebra 1

This category tests your ability to create expressions for area, perimeter, and volume problems. You’ll also use the distance formula and the Pythagorean Theorem to solve problems.

Take a look at this concept.

Pythagorean Theorem

You will use the Pythagorean equation (a² + b² = c²) to solve problems regarding right triangles.

According to this theorem, side c is the side of the triangle opposite the right angle; its name is the hypotenuse. Remember that a right angle is exactly 90°. Side a and side b are the other two sides (the legs) of the right triangle.

Here’s an example of what a related question might look like on the test:

Jane cuts a square piece of paper into 2 right triangles. She discards one triangle of paper and is left with the other.

Side a of the triangle is 4 inches, and side b is 5 inches. If side c is the hypotenuse, what is the length of side c?

To solve this problem, just plug the numbers into the Pythagorean equation:

(4² + 5² = c²)

Apply the exponents:

(16 + 25 = c²)

Add the values on the left:

(41 = c²)

And find the square root of each side:

(6.4 = c)

So, the hypotenuse of Jane’s triangle must be 6.4 inches!

And that’s some basic info about the Math: Quantitative Reasoning, Algebra, and Statistics Placement Test Test.

Now, let’s look at a few practice questions to see how these concepts might actually appear on the real test.

Directions

Choose the best answer. Use paper if necessary.

Mathematics Quantitative Practice Questions

Question 1

Which of the following is the equation of a line that passes through (2, -5) and (-3, 10)?

  1. y = -3x + 1
  2. y – 3x = 1
  3. 3x + 2y = 1
  4. y + 1 = -3x

Correct answer: A. m = (-5 – 10)/(2 – (-3)) = (-15)/5 = -3 So, y = -3x + b. Substitute one of the given points into the slope-intercept form of the equation: 10 = -3(-3) + b and b = 1. The equation in slope intercept form is y = -3x + 1.

Question 2

If 1/6x – 4 = 1, then x =

  1. 2
  2. 10
  3. 30
  4. -18

Correct answer: 3. The first step to solve this equation is to add 4 to each side, as +4 is the additive inverse of the -4 that is preventing the variable expression from being alone on its own side of the problem. The resulting equation is ⅙x = 5. At this point, 6 should be multiplied on each side of the equation in order to eliminate the factor ⅙ from in front of the x, as 6/1 is the multiplicative inverse of the fraction ⅙ that is preventing the variable from being alone. When 6 has been multiplied to each side, the result is x = 30.

Question 3

Which of the following is equivalent to the expression (4ab)(-5ab)?

  1. -20a²b²
  2. -ab
  3. a²b²
  4. -20ab

Correct answer: 1. This is the correct answer. (4ab)(-5ab) is the same as (4)(a)(b)(-5)(a)(b). We can rearrange the factors, because multiplication is commutative (order doesn’t matter): (4)(-5)(a)(a)(b)(b). Now, simplify: -20a²b².

Question 4

Which of the following correctly combines like terms for the following expression?

3x + 2(3x – 2) + 5 – 2(2x + 1)

  1. 5x – 1
  2. 15x – 1
  3. 4x + 6
  4. 13x + 11

Correct answer: 1. The response 5x – 1 is the only answer option that combines the terms correctly. First, each value from the front of each set of parentheses must be distributed into the parentheses and multiplied appropriately, with attention to positive and negative signs: 3x + 6x – 4 + 5 – 4x – 2. Next, like terms must be collected to be combined: (3x + 6x – 4x) + (-4 + 5 – 2). Finally, like terms can be combined: 5x – 1.

Question 5

A quilter is preparing to make a quilt for a baby’s crib that is 30 inches by 50 inches. The design of the quilt calls for a diagonal stripe of ribbon from one corner of the quilt to another corner as shown in the image above. What is the approximate length of that diagonal stripe of ribbon in the finished quilt?

  1. 80 inches
  2. 58 inches
  3. 40 inches
  4. 45 inches

Correct answer: 2. Since the quilt shown is rectangular, the diagonal ribbon cuts the rectangle into two congruent right triangles. The legs of the right triangles are known to be 30 inches and 50 inches. The diagonal can be considered the hypotenuse of the right triangles. Therefore, the Pythagorean Theorem can be applied in order to use the known leg lengths to find the unknown diagonal length. The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle will equal the square of the hypotenuse of that triangle. The formula for the Pythagorean Theorem is written a² + b² = c², where a and b are the leg lengths and c is the length of the hypotenuse. In this case, then, the equation to solve is 30² + 50² = c². Following the order of operations correctly makes the equation 900 + 2,500 = c², and then 3,400 = c². The equation is solved for c by taking the square root of each side and using only the positive version of the square root of 3,400, which is approximately 58.31. Therefore, the best final answer is that the diagonal stripe of ribbon will be ~58 inches.

Question 6

Which of the following expressions (in square units) is the most reasonable estimate for the area of the circle in the graph shown above?

  1. 41
  2. 128
  3. 23
  4. 27

Correct answer: 1. The area, A, of a circle with radius r is found using the formula A = πr². The radius of the circle in the diagram must be estimated as slightly larger than halfway between 3 and 4 units. Radius equals 3.6 is a reasonable approximation. Using 3.6 for r and using 3.14 as a decimal approximation for the irrational number π in the area formula leads to the calculation of A = 3.14(3.6)². Following the order of operations leads to a simplified equation of A = 3.14(12.96) and then A = 40.6944. The value 40.6944 becomes 41 when rounded to the nearest whole number because the tenths place digit is 6 (which is at least 5, and so requires rounding up).

 

Question 7

According to the table above, approximately how many stones are in one US ton?

  1. 224
  2. 9
  3. 8
  4. 143

Correct answer: 4. Because 1 stone = 224 ounces, and there are 16 ounces in 1 pound, the conversion should begin: 16 ounces/1 pound × 1 stone/224 ounces. The unit “ounces” cancels out and the resulting fraction shows 16 stones/224 pounds, which reduces to 1 stone/14 pounds. The fraction 1 stone/14 pounds can be multiplied by 2,000 pounds/1 US ton to eliminate the unit of pounds and to introduce the unit of US tons. The result is 2,000 stones/14 US tons. The decimal equivalent of that fraction is approximately 142.9 stones/1 US ton, and so it can be concluded that there are approximately 143 stones in one US ton. Or, the calculation can be performed: 1 ton × 2,000 pounds/1 US ton × 16 ounces/1 pound × 1 stone/224 ounces = (2,000 × 16) stones/224 ≈ 142.9 stones. 

Question 8

Marco completed a 130-page book in 100 minutes. The graph above shows the total number of pages he had read in 20-minute increments. Use the graph to determine how many pages Marco read during his last 40 minutes of reading.

  1. 42 pages
  2. 62 pages
  3. 13 pages
  4. 68 pages

Correct answer: 1. 42 pages is correct. To determine the number of pages that Marco read during his last 40 minutes of reading is to determine the number of pages read between t = 60 minutes and t = 100 minutes. At the 60 minute mark, Marco had read 88 pages. At 100 minutes, Marco had read 130 pages. To find the number of pages that Marco read in his last 40 minutes, then subtraction of 130 – 88 must be performed. Because 130 – 88 = 42, the correct final answer is 42 pages.

Question 9

A change purse contains 4 pennies, 3 nickels, 2 dimes, and the rest of the coins are quarters. If a person has a 1/3 probability of selecting a penny when randomly selecting a coin from the change purse, how many quarters are there?

  1. 4
  2. 3
  3. 12
  4. 2

Correct answer: 2. Because drawing a penny when randomly selecting a coin from the change purse has a probability of 1/3, the 4 pennies known to be in the change purse must represent one-third of the total number of coins in the change purse. Because 1/3 = 4/x is solved by cross-multiplication to get 1x = 12, there must be 12 coins in the change purse. The change purse contains 4 + 3 + 2 = 9 coins that are not quarters. The difference between the total number of coins and the number of coins that are not quarters, 12 – 9, must be equal to the number of quarters in the change purse. Therefore, there are 3 quarters in the change purse.

Question 10

The dot plot above shows the shoe size of the 25 students in Ms. Redmond’s kindergarten class. What is the median of these shoe sizes?

  1. 9
  2. 11
  3. 10.5
  4. 13

Correct answer: 3. The median of a set of data is the middle value (in a data set that contains an odd number of pieces of data) when the data is ordered. The dot plot given in the prompt shows the data of the students’ shoe sizes organized from least (size 9) to greatest (size 13). Each dot represents a single student with that size shoe. With 25 pieces of data, the 13th piece of data is the median (with 12 values that are smaller than or equal to it, and 12 values that are greater than or equal to it). In this case, the 13th piece of data is in the column for size 10.5, and so the median of the set is 10.5.

 

Math: Advanced Algebra and Functions Placement Test

The Math: Advanced Algebra Functions Placement Test has about 20 questions.

There are eleven broad categories:

  • Linear Equations
  • Linear Applications and Graphs
  • Factoring
  • Quadratics
  • Functions
  • Radical and Rational Equations
  • Polynomial Equations
  • Exponential and Logarithmic Equations
  • Geometry Concepts for Algebra 1
  • Geometry Concepts for Algebra 2
  • Trigonometry

So, let’s talk about some of the more challenging categories.

Factoring

This category tests your ability to recognize quadratic equations and cubic polynomials, convert those equations to standard form, and factor them out.

Let’s take a look at a concept that is likely to appear on the test.

Factoring Quadratic Equations

The standard form of a quadratic equation is a² + bx + c = 0. In quadratic equations, a, b, and c are known variables. Keep in mind that a is never equal to 0 in a quadratic equation.

Let’s look at an example of a quadratic equation and practice factoring it:

x² + 6 = 0 -5x

This is a quadratic equation in disguise. Once we move the x value to the other side, we can see the equation in standard form:

x² + 5x + 6 = 0

In this equation, a = 1, b = 5, and c = 6. The first step is to find out which factors (numbers multiplied by each other) will give us the c value (6 in this case). We can multiply 6 and 1 or 2 and 3 to get 6.

Next, we have to decide which set of factors will give us the b value (5) when added together:

6 + 1 = 7
3 + 2 = 5

Now we know that 3 and 2 must be the factors that we want to work with, because they add up to 5. In order to factor the quadratic equation, we need to plug the factors (3 and 2) into this equation:

(x +   )(x + )

Therefore, when the equation is factored, we get:

(x + 3)(x + 2)

or

(x + 2)(x + 3)

Notice that the factors may be plugged into the equation in any order.

Quadratics

This category tests your ability to create, simplify, and solve quadratic equations and systems of quadratic equations. Remember that a system of equations refers to two or more equations which work together.

Take a look at this concept.

Simplifying Quadratic Equations

As we’ve learned, the standard form of a quadratic equation is a² + bx + c = 0, but sometimes these equations appear in disguise. Let’s look at an example of a quadratic equation that requires multiple steps to simplify into standard form:

4/32 [(184t² – 168t²) + (-18t × 4)] + 2 = -13t +7

First, let’s deal with 4/32. We can easily simplify the fraction to 1/8:

1/[(184t² – 168t²) + (-18t x 4)] + 2 = -13t +7

Next, let’s complete the simple arithmetic in the brackets:

1/(16t² – 72t) + 2 = -13t +7

Now, let’s multiply the variables in parentheses by 1/8:

2t² – 9t + 2 = -13t + 7

This is starting to look much more like a quadratic equation now, but we still need to combine like terms. Let’s move 13t to the other side of the equation via addition:

2t² + 4t + 2 = 7

Next, we need to subtract 7 from each side in order for our equation to equal 0:

2t² + 4t – 5 = 0

Finally, we have our simplified quadratic equation. We know that a = 2, b = 4, and c = −5.

Functions

This category tests your ability to evaluate and graph linear and quadratic functions. You should be able to determine the function of a variable and use that information to create and identify points on a graph which correspond to that function.

Here is a concept you should know.

Graphing Functions

Let’s practice graphing a simple quadratic function together:

f(x) = x²

First, let’s list some points for the x and y values. The function of x will be our y value. We know that the y value of any point will just be the x value squared.

The table that we have created gives us the points for our graph. We can see that one point on the graph is (-2, 4), another point is (-1, 1), and so on. Let’s graph the points:

You just graphed a function! Wasn’t that easy?

Radical and Rational Equations

This category tests your ability to work with equations that contain radicals (such as squares, cubes, fourth roots, etc.). You will also work with equations that contain fractions with variables in their denominators. You should be able to create, simplify, and graph these equations.

Here is a concept you need to know.

Simplifying Radical Equations

Let’s practice working with radical equations by simplifying:

3√405t⁶y⁴  

First, we need to break down 405 into its prime factors:

405 = 3 • 3 • 3 • 3 • 5

Now, let’s factor out t⁶ and y⁴:

t⁶ = t • t • t • t • t • t

y⁴ = y • y • y • y

Next, we need to look at the index of the radical. In this case, our index is 3, because we are working with a cube root. This means that we need to identify any factors that appear in a group of three. These factors are underlined for you in the example below:

3√405 = 3 • 3 • 3 • 3 • 5

3√t⁶ = t• t• t• t• t• t

3√y⁴  = y • y • y

Now, all of the underlined factors need to be moved outside of the radical. We will move y and 3 out once each (because they each have one group of 3). Because there are two groups of t, we will move that factor out twice:

y (3• • y)

Notice that any variables that are not underlined remain inside the radical. The last step is to simplify the expression outside of the radical, as well as the expression inside of the radical. When we simplify by multiplication, our final expression is:

3t²y³√15y

Polynomial Equations

This category tests your ability to solve equations with multiple variables, as well as polynomial equations. You will also be expected to graph polynomial functions.

Let’s look at an important concept.

Solving Polynomial Equations

Here’s an example of a polynomial equation. We’ll solve it step by step:

5y² + 3(5y) = 0

First, we need to complete the arithmetic inside the parentheses:

5y² + 15y = 0

Next, we will factor out 5y, because both variables on the left side of the equation contain 5y:

5y (y + 3)= 0

Now, we will use the zero principle. Set each factor equal to 0:

5y = 0

y + 3 = 0

We can solve 5y = 0 by dividing each side by 5 to get y = 0. The equation y + 3 = 0 can be solved by subtracting 3 from each side: y = -3.

Therefore, y = 0 or y = -3.

Exponential and Logarithmic Equations

This category tests your ability to solve and graph equations that contain exponents or logs. Keep in mind that logs are the inverse of exponents. You will also demonstrate that you know how to graph equations that contain logs.

Let’s look at a concept.

²Solving Logarithmic Equations

Take a look at the following equation:

logb(x² – 30) = logb(x) 

Since the logs both have the same base, we can set both sides equal to each other:

x² – 30 = x

Let’s set the equation to 0 by subtracting the x from both sides. Then, we can factor:

x² – x – 30 = 30

(x – 6)(x + 5) = 0

x = 6 or -5

Now, we have to decide if x is equal to 6 or -5. Since a log cannot contain a negative number, x is equal to 6.

Geometry Concepts for Algebra 1

This category tests your ability to interpret and measure geometrical shapes and their dilations, rotations, translations, and reflections.

Here is an important concept.

Dilations

To dilate something means to make it larger. A dilation of a geometric shape is the same geometric shape, only its size has increased. Consider figure F on the graph below:

Figure L is a dilation of figure F. In order to get figure L, we multiply the distance between the points in figure F by 2.

Side a = side w

Side b = side x

Side c = side y

Side d = side z

Figures H and K are rotations of F. Figure P is larger than F, but it is not a dilation, because its proportions have been distorted.

Geometry Concepts for Algebra 2

This category tests your ability to work with equations and theorems in order to evaluate lines and angles. You’ll also determine the measures of shapes and look for instances of congruency.

Let’s take a look at a concept that may appear on the test.

Intersecting Line Theorems

Let’s explore theorems related to parallel lines and transversals. Transversals are lines that intersect parallel lines. Take a look at the example of a transversal and two parallel lines below:

In order to understand some theorems, let’s identify the angles in the example:

Alternate interior angle pairs: (3 and 6) (4 and 5). These angle pairs are on opposite sides of the transversal, and they are between the parallel lines.

Alternate exterior angles: (1 and 8) (2 and 7). These angle pairs are on opposite sides of the transversal, and they are outside of the parallel lines.

Corresponding angles: (1 and 5) (2 and 6) (3 and 7) (4 and 8). These angle pairs are corresponding, because each angle is in the same position in its group of four angles.

Same-side interior angles: (3 and 5) (4 and 6). These angle pairs are on the same side of the transversal, and they are between the parallel lines.

Same-side exterior angles: (1 and 7) (2 and 8). These angle pairs are on the same side of the transversal, and they are outside of the parallel lines.

Now that you can identify the angles, you can apply them to the following criteria in order to determine whether lines are parallel:

If parallel lines are intersected by a transversal,

1. corresponding angles are congruent
2. alternate interior angles are congruent
3. alternate exterior angles are congruent
4. same-side interior angles are supplementary
5. same-side exterior angles are supplementary

Let’s use Rule 4 as an example. If angles 3 and 5 add up to 180°, then we know that we are working with a set of parallel lines. If they add up to a different number, the lines aren’t parallel.

Trigonometry

This category tests your ability to evaluate and graph trigonometric equations, functions, and relationships. In order to do well on this portion of the test, you will need to be able to determine the length of arcs and use the law of sines and the law of cosines.

Take a look at this concept.

Determining Arc Length

Let’s start out with a visual that is similar to one you might see on the test:

An arc is usually represented by the symbol s. It is a portion of the circumference opposite a central angle. We can find out the degree of an arc by using the following equation:

Arc length = (central angle° ÷ 360°) circumference

Let’s try out the following question and solve it together:

Refer to the diagram of circle Q, which has a radius of 8 inches. What is the length of arc AB?

Now, we can plug the numbers into our formula:

s = (110° ÷ 360°) 2π(8)

s = .306 50.265

s = 15.38

The length of arc AB is equal to 15.38 inches.

And that’s some basic info about the Math: Advanced Algebra and Functions Placement Test.

Now, let’s look at a few practice questions to see how these concepts might actually appear on the real test.

Directions

Choose the best answer. Use paper if necessary.

Mathematics Advanced Practice Questions

Question 1

Which graph is not a function?

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 3. Graph C does not pass the vertical line test and is, therefore, not a function.

Question 2

What is the domain and range of f(x) = 3 – |x|?

  1. Domain: all real numbers; Range: f(x) ≥ 3
  2. Domain: x ≥ 3; Range: f(x) ≥ 0
  3. Domain: all real numbers; Range: f(x) ≤ 3
  4. Domain: x ≥ 3; Range: f(x) ≤ 3

Correct answer: 3. When graphed, this equation forms an inverted “V” with a peak at (0, 3). The slope of the lines is 1 and -1, respectively. There are no restrictions on x, but that f(x) has values of 3 down: x = any real number; f(x) ≤ 3.

Question 3

Which of the following is an equation for a line perpendicular to 3y – 5x = 10?

  1. 5y + 3x = 4
  2. y = (3/5)x – 6
  3. 3y + 0.2x = 8
  4. y = (5/3)x – 5

Correct answer: 1. First, change the equation to slope-intercept form by solving for y: 1. 3y = 5x + 10 2. y = (5/3)x + (10/3)y The slope for this line is 5/3. Perpendicular lines have slopes that are negative reciprocals of each other, so the slope of the perpendicular line, in this case, would be the negative reciprocal of 5/3 or (-3/5). The equation of the perpendicular line would be y = (-3/5)x + b. The value for b is arbitrary. The equation for our perpendicular line is currently in slope-intercept form and does not match any of the choices. Two of the choices are in standard form, so convert the equation: y = (-3/5)x = b to standard form. Multiplying by 5, we have 5y = -3x + 5b or 3x + 5b or 3x + 5y = 5b Since b is arbitrary, meaning it could be any real number, then this choice, 5y + 3x = 4, would be a correct solution. Remember that there are infinite perpendicular lines through any given line. This choice is the only choice with the correct slope for a line perpendicular to 3y 5x = 10.

Question 4

Which of the following equations is equivalent to y = -1/4(x–3)² + 5?

  1. 4y + 20 = x² + 9
  2. 4y = -(x² – 6x) + 11
  3. 4y = (x-3)² – 5
  4. 4y + 5 = x² + 9

Correct answer: 2. 4y = (x-3)² – 5 is incorrect, because the constant term is incorrect. The coefficient of y, 4, should be divided into all terms on both sides of the equation, but the 5 was not divided by 4. 4y + 5 = x² + 9 is incorrect because when put into vertex form, the equation becomes y = (1/4)x² + 1. 4y + 20 = x² + 9 is incorrect because when put into vertex form, this equation becomes y = (1/4)x² – (11/4).

nts.

Question 5

Simplify: (4x²y³z⁻³)² / (8x³y⁵z⁻⁴)

  1. z / 2xy²
  2. 8xy / z²
  3. 2xy / z²
  4. 2z² / xy

Correct answer: 3. Begin by raising the parenthetical expression in the numerator to the power 2 as indicated: 16x⁴y⁶z⁻⁶ / 8x³y⁵z⁻⁴. Then, simplify using the quotient of powers property: 2x⁽⁴ ⁻ ³⁾y⁽⁶ ⁻ ⁵⁾ z⁻⁶⁻⁽⁻⁴⁾= 2x¹y¹z⁻² = (2xy) / z².

Question 6

Which of the following is equivalent to log₈ 25?

  1. log₅/log₈
  2. 2log₅/log₈
  3. 2(log₈/log₅)
  4. 2 log₅ 8

Correct answer: 2. 2 log₅ 8 is incorrect, because 2log₅8 = log₅8² = log₅64 log₅/log₈ is incorrect, because log₅/log₈ = log₈(5) 2(log₈/log₅) is incorrect, because 2(log₈/log₅) = log₅8² = log₅64.

Question 7

What is the measure of arc AFD?

  1. 190°
  2. It cannot be determined
  3. 170°
  4. 135°

Correct answer: 1. m∠AXD = mBC + mAFD2. Substitute known values: 5° = 80 + mAFD2; mAFD = 190°.

Question 8

Solve the equation: 2x² + x – 28 = 0.

  1. x = -4, 3.5
  2. x = -3.5, 4
  3. x = 3.5, 4
  4. x = -4, -3.5

Correct answer: 1. 2x² + x – 28 = (2x – 7)(x + 4) = 0. So, x = 7/2 = 3.5 or x = -4.

 

Question 9

Solve the equation: x² – 11x + 24 = 0.

  1. x = -3, 8
  2. x = 3, 8
  3. x = -8, 3
  4. x = -8, -3

Correct answer: 2. This equation can be factored into (x-8)(x-3) = 0. Therefore, the solutions will be x = 3, 8.

 

Question 10

Triangle ABC undergoes a 90° counterclockwise rotation centered at B.  What are the coordinates of the image of A under this transformation?

  1. (1, 7)
  2. (-1, 1)
  3. (7, 1)
  4. (1, -1)

Correct answer: 3. The angle CBC’ is a right angle. This represents the counterclockwise rotation of point C about point B. It is also true that angle ABA’ is a 90° angle. Point B, of course, remains fixed. The coordinates for A’ are (7,1).

 

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