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Compass Practice Test Reading
Acid has been present in rain for millennia, naturally occurring from volcanoes and plankton. However, scientific research shows that the acid content of rain has increased dramatically over the past two hundred years, in spite of humanity’s recent attempts to control the problem.
Rain consists of two elements, nitrogen and sulphur. When sulphur is burned, it transforms into sulphur dioxide. Nitrogen also oxidizes when burned. Subsequently, both sulphur dioxide and nitrogen oxide react with the water molecules in rain to form sulphuric acid and nitric acid, respectively.
Factories and other enterprises have built high chimneys in an attempt to carry these gases away from urban areas. Yet, the effect of the structures has been to spread the gases more thinly and widely in the atmosphere, thereby exacerbating the problem.
The acid in rain also emanates from automobile exhaust, domestic residences, and power stations. The latter have been the culprit of the bulk of the acid in rainwater in recent years. Since the pollutants are carried by the wind, countries can experience acid rain from pollution that was generated in countries thousands of miles away.
1. Which one of the following phrases is closest in meaning to the latter have been the culprit of the bulk as it is used in the above text?
A Automobile exhaust has caused the majority of acid rain.
B. Automobile exhaust, domestic residences, and power stations have equally contributed to the creation of acid rain.
C. Power stations are more widespread geographically than other causes of acid rain.
D. Power stations generate a great deal of pollution that is carried by the wind.
E. Power stations have been the largest contributor to the problem.
2. Between paragraph 2 and 3, the writer’s approach shifts from
A. scientific explanation to current problems
B. chemical analysis to scientific inquiry
C. historical background to current problems
D. scientific inquiry to possible solutions
E. cause to effect
3. Which detail from the passage best supports the writer’s main idea?
A. When sulphur is burned, it transforms into sulphur dioxide.
B. Subsequently, both sulphur dioxide and nitrogen oxide react with the water molecules in rain to form sulphuric acid and nitric acid, respectively.
C. Yet, the effect of the structures has been to spread the gases more thinly and widely in the atmosphere, thereby exacerbating the problem.
D. The acid in rain also emanates from automobile exhaust, domestic residences, and power stations.
E. Since the pollutants are carried by the wind, countries can experience acid rain from pollution that was generated in countries thousands of miles away.
Compass Practice Test – Reading Answers
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Compass Practice Test – Writing
(1) In the fall of 1859, a discouraged man was sitting (2) in his run-down law office in Springfield Illinois. (3) He was fifty years old, and he had been a lawyer for twenty years, was earning on average 3,000 dollars a year. (4) His tangible possessions and property consisted in 160 acres of farm land in Iowa (5) and the house which he lived in Illinois. (6) When his monetary resources were limited and he was in debt, (7) this man can later go on to do great things for his country. (8) His name was Abraham Lincoln.
A. In the fall of 1859, a discouraged man was sitting
B. In the fall of 1859, a discouraged man is sitting
C. In the fall of 1859, a discouraged man sitting
D. In the fall of 1859, a discouraged man is seated
E. In the fall of 1859, a discouraged man seated
A. in his run-down law office in Springfield Illinois.
B. in his run-down law office in Springfield, Illinois.
C. in his run-down law office, in Springfield Illinois.
D. in his run-down law office, in Springfield, Illinois.
E. in his run-down, law office, in Springfield, Illinois.
A. He was fifty years old, and he had been a lawyer for twenty years, was earning on average 3,000 dollars a year.
B. He was fifty years old, and he had been a lawyer for twenty years, who earning on average 3,000 dollars a year.
C. He was fifty years old, and he had been a lawyer for twenty years, who is earning on average 3,000 dollars a year.
D. He was fifty years old, and he had been a lawyer for twenty years, earning on average 3,000 dollars a year.
E. He was fifty years old, and he had been a lawyer for twenty years, earned on average 3,000 dollars a year.
A. His tangible possessions and property consisted in 160 acres of farm land in Iowa
B. His tangible possessions and property consisted with 160 acres of farm land in Iowa
C. His tangible possessions and property consisted for 160 acres of farm land in Iowa
D. His tangible possessions and property consisted of 160 acres of farm land in Iowa
E. His tangible possessions and property consisted on 160 acres of farm land in Iowa
A. and the house which he lived in Illinois.
B. and the house in which he lived in Illinois.
C. and the house in that he lived in Illinois.
D. and the house in where he lived in Illinois.
E. and the house whose he lived in Illinois.
A. When his monetary resources were limited and he was in debt,
B. Although his monetary resources were limited and he was in debt,
C. Even his monetary resources were limited and he was in debt,
D. Thus his monetary resources were limited and he was in debt,
E. Hence his monetary resources were limited and he was in debt,
A. this man can later go on to do great things for his country.
B. this man could later go on to do great things for his country.
C. this man should later go on to do great things for his country.
D. this man would later go on to do great things for his country.
E. this man ought to later go on to do great things for his country.
A. His name was Abraham Lincoln.
B. His name is Abraham Lincoln.
C. His name Abraham Lincoln.
D. His name; Abraham Lincoln.
E. His name, Abraham Lincoln.
Compass Practice Test – Writing Answers
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Compass Practice Test – Pre-algebra
1) Find the value of x that solves the following proportion: 9/6 = x/10
A) 1.5
B) 15
C) .67
D) 67
E) 150
2) Carmen wanted to find the average of the five tests she has taken this semester. However, she erroneously divided the total points from the five tests by 4, which gave her a result of 90. What is the correct average of her five tests?
A) 64
B) 72
C) 80
D) 90
E) 110
3) 2/3 x 4/5 = ?
A) 6/15
B) 5/6
C) 12/10
D) 15/8
E) 8/15
4) 1/8 ÷ 4/3 = ?
A) 1/6
B) 32/3
C) 3/24
D) 4/24
E) 3/32
5) Professor Smith uses a system of extra–credit points for his class. Extra–credit points can be offset against the points lost on an exam due to incorrect responses. David answered 18 questions incorrectly on the exam and lost 36 points. He then earned 25 extra credit points. By how much was his exam score ultimately lowered?
A) –11
B) 11
C) 18
D) 25
Compass Practice Test – Pre-algebra Answers
1) The correct answer is B. Simplify the first fraction: 9 /6 = 3 /2
Then divide the denominator of the second fraction by the denominator of the simplified first fraction: 10 ÷ 2 = 5
Now, multiply this number by the numerator of the first fraction to get your result:
2) The correct answer is B. First find the total points by taking Carmen’s erroneous average times 4: 4 x 90 = 360
Now divide the total points by the correct number of tests in order to get the correct average: 360 ÷ 5 = 72
3) The correct answer is E. When you are asked to multiply fractions, you need to multiply the numerators together and the denominators together to get the new fraction.
2 x 4 = 8
3 x 5 = 15
So, the new fraction is 8 /15
4) The correct answer is E. When you are asked to divide fractions, first you need to invert the second fraction. This means that you change the numerator with the denominator. Then you multiply this inverted fraction by the first fraction given in the problem. 4 /3 inverted is 3 /4
Then multiply the numerators and the denominators together to get the new fraction.
1 x 3 = 3
8 x 4 = 32
So, the new fraction is 3 /32
5) The correct answer is B. Take the number of questions missed and add the extra credit points: –36 + 25 = –11
Since the question is asking how much the score was lowered, you need to give the amount as a positive number.
Compass Practice Test – Algebra
3) Find the midpoint between the following coordinates: (2, 2) and (4, -6)
4) If 4 + 3(2 Ö x – 3) = 25, then x = ?
A) -5
B) -25
C) 5
D) 25
E) 36
Compass Practice Test – Algebra Answers
The correct answer is: B
Remember: F-O-I-L method: First – Outside – Inside – Last
Then we add all of the above parts together to get:
2) Factor the following: 2xy – 8x 2 y + 6y 2 x 2
The correct answer is: C
In order to factor an equation, you must figure out what variables are common to each term of the equation. Let’s look at this equation:
We can see that each term contains x. We can also see that each term contains y. So, now let’s factor out xy:
Then, think about integers. We can see that all of the terms inside the parentheses are divisible by 2. Now let’s factor out the 2. In order to do this, we divide each term inside the parentheses by 2 to get our final answer:
3) Find the midpoint between the following coordinates:
The correct answer is: D
For coordinate geometry questions like this one, remember that you need to use the following formula:
Now calculate for x and y:
(2 + 4) ÷ 2 = midpoint x, (2 – 6) ÷ 2 = midpoint y
6 ÷ 2 = midpoint x, -4 ÷ 2 = midpoint y
3 = midpoint x, -2 = midpoint y
4) If 4 + 3(2 Ö x – 3) = 25, then x =
The correct answer is: D
In equations that have both integers and square roots, deal with the integers that are outside the parentheses first:
4 + 3(2 Ö x – 3) = 25
4 – 4 + 3(2 Ö x – 3) = 25 – 4
3(2 Ö x – 3) = 21
Then carry out the operations for the parenthetical terms:
3(2 Ö x -3) = 21
6 Ö x – 9 = 21
6 Ö x – 9 + 9 = 21 + 9
6 Ö x = 30
6 Ö x ÷ 6 = 30 ÷ 6
Ö x = 5
The correct answer is: C
First perform the division on the integers:
Then do the division on the other variables:
a 2 ÷ a = a
b 3 ÷ b 2 = b
c ÷ c 2 = 1 /c
Then multiply these together to get:
Compass Practice Test – Geometry, Trigonometry, and Advanced Math
1) If the first term of an arithmetic sequence is 5, and we can find subsequent terms by adding 8, what equation can be used to find the n th term of the sequence?
A) (n + 8) x 5
B) (n + 5) x 8
C) 5 + (n x 8)
D) 5 + [(n + 1) x 8]
E) 5 + [(n – 1) x 8]
2) Express the equation 2 5 = 32 as a logarithmic function.
A) 2 = log532
B) 5 = log232
C) 2 = log325
D) 5 = log322
E) 32 = log25
4) If circle A has a radius of 0.4 and circle B has a radius of 0.2, what is the difference in area between the two circles?
A) .04 p
B) .12 p
C) .16 p
D) .40 p
E) .60 p
Compass Practice Test – Geometry, Trigonometry, and Advanced Math Answers
1) If the first term of an arithmetic sequence is 5, and we can find subsequent terms by adding 8, what equation can be used to find the n th term of the sequence?
The correct answer is: E
For an arithmetic sequence, the n th term is calculated by taking the first term of the sequence (in this case 5) plus (n – 1) times the increment (in this case 8).
This results in the equation:
2) Express the equation 2 5 = 32 as a logarithmic function.
The correct answer is: B
Logarithmic functions are just another way of expressing exponents. Remember that y x is always the same as: x = logy
So 2 5 = 32 is the same as 5 = log232
3) For the functions f2(x) listed below, x and y are integers greater than 1. If f1(x) = x 2 , which of the functions below has the greatest value for f1(f2(x))?
The correct answer is: C
Two whole numbers that are greater than 1 will always result in a greater number when they are multiplied by each other, rather than when those numbers are divided by each other or subtracted from each other.
4) If circle A has a radius of 0.4 and circle B has a radius of 0.2, what is the difference in area between the two circles?
The correct answer is: B
The area of a circle is always:
p times the radius squared.
Therefore, the area of circle A is:
The area of circle B is:
To calculate the difference in area between the two circles, we then subtract:
0.16 p – 0.04 p =
0.12 p
5) If sin a = 11 /15 and cos a = 8 /15, then tan a = ?
The correct answer is: D
Remember the following important trigonometric formulas for calculating the sine, cosine and tangent of any given angle A:
sin A = x/z
cos A = y/z
tan A = x/y
Therefore, we use the measurements for sin a = 11 /15 and cos a = 8 /15 to calculate tan:
sin a = 11 /15; x = 11; z = 15
cos a = 8 /15; y = 8; z = 15
tan A = x/y, so tan A = 11 /8