ACT -- The Math Test Review (Draft)
The Mathematics Test (which is the second section of the ACT) is the only section of the ACT that lists
questions in order of difficulty. Generally, the lower the number of the question, the easier it is and the
higher the number, the harder it is. Question #1 is very easy, and question #60 is very hard. Notice in the
break out below that pre-algebra and basic algebra account for 24 of the 60 questions or 40% of the Math
test. Calculators may be used on the math section.
The Mathematics Test breaks out this way:
Content
Number of Questions
Pre-Algebra
14
Elementary Algebra
10
Intermediate Algebra
9
Coordinate Geometry
9
Plane Geometry
14
Trigonometry
4
You will most likely not have much trouble with the 14 pre-algebra questions. The plug-in method
shown below will help with the 19 elementary and intermediate algebra questions. You’ll probably need
to review coordinate and plane geometry. You may not have even had much trigonometry, but don’t
worry. There’s only four trig questions on the entire test.
Specific Testing Techniques for the Math Section: Remember, the 60 questions are arranged from
easy to hard. Don’t worry if you don’t get to all the questions. Your score will probably be higher if you
spend more time on the easy and medium questions so that you answer them correctly instead of rushing
through them to get to the hard ones that you have a high probability of missing.
Review your math: SAT strategies won’t help if you don’t have a basic foundation in math. You
should especially review exponents and basic geometry principles.
Don’t solve the problems the way you would in math class. In math class your teacher wants to
know that you know the proper procedures for finding the answer (Show your work!). The ACT grading
machine only sees the answer you bubbled in. It doesn’t care how you got the answer. There’s no partial
credit for showing your work. Your answer is either 100% right or 100% wrong. Find the right answers
the quickest and easiest way you can. On the ACT you’re looking for the correct (best) answer, not
demonstrating your math prowess.
Draw and write in your test booklet or on the scratch paper. For many questions especially
geometry questions that don’t have diagrams drawing a picture to understand the question is very
important.
Guesstimate. Almost every diagram, or graph on the ACT is drawn to scale, so you can use your
eyes to estimate some answers or more importantly to eliminate some of the answer choices. In the
diagram below, I can’t know the actual degrees of the angles, but I can estimate that angel a < 90 and
angle b > 90. If the question asks what is the degree measurement of angle b, I can eliminate all the
answer choices that are 90o or less.
a
b
1
THINGS TO REVIEW:
Coordinate Geometry: 9 Questions
(1) The signs of the quadrants.
(2) The slope-intercept formula.
XY+
X+
Y+
XY-
X+
Y-
y = mx + b
You’ll find two values, using this formula – the slope (m) and the y intercept (b). Here’s a sample
question.
What is the slope of any line parallel to the line 7x + 9y = 6?1 ACT #21p. 308
Any line parallel to this one would have the same slope as this line, so if you can find the slope of this
line, you’ll have the answer. You‘ll need to change the equation into the format y = mx + b. Use basic
algebra skills to change 7x + 9y = 6 to 9y = -7x + 6. Now change the 9y to y (by dividing by 9) resulting
in y = -7/9(x) + 6/9. The slope is -7/9 and the y intercept is 6/9. Note: To get the correct slope or y
intercept, you must change the equation you’re given into the y = mx + b format.
(3) The slope formula when you have only two points (coordinates) of a line. Use the formula
y1 - y2/x1- x2
Here’s a sample question.
What is the slope of a straight line that passes through the points (-3, 5) and (5, 4)?
Using the formula, you would have (5 - 4)/(-3 - 5) or 1/-8 or -1/8. Given that the slope in rise over run,
we can see in the graph below that the rise (change in y) is 1 and the run (change in x is 8). Rise over run
is 1 over 8 (1/8) and since the slope is negative the slope is -1/8.
(-3, 5)
(5, 4)
1
8
1
-6 -5 -4 -3 -2 -1
1 2 3 4 5 6
-3
-4
2
(4) The mid-point formula. (x1 + x2)/2 to find the x coordinate and (y1 + y2)/2 for the y coordinate,
where (x1, y1) and (x2, y2) are the endpoints of the line.
Sample question.
The coordinates of the endpoints of line GH, in the standard (x, y) coordinate plane, are
(-8, -3) and (2, 3). What is the x-coordinate of the midpoint of line GH?1 ACT #38 p.596
Using the formula, the x coordinate would be (-8 + 2)/2 or -6/3 = -3. What the formula does (and another
way to look at this) is simply take the average of the two x coordinates and the average of the two y
coordinates.
(5) The distance formula. If you plan to put formulas into your calculator you may want to use the
formula shown below, but it’s hard to remember and a little complicated to work with.
The formula is d = √(x2 – x1)2 + (y2 – y1)2
However, you can use the Pythagorean Theorem, which is easier for most people. Here’s a sample
question.
In the standard (x, y) coordinate plane, what is the distance, in coordinate units, between (-3, -2)
and (5, 5)?1 ACT #31 p.739
A. √13
B. √15
C. √113
D. 5
E. 15
Plot the points on the graph and then draw a right triangle.
7 units vertical
8 units horizontal
Next determine the length of each of the two legs that form the 90o angle. Use these lengths in the
Pythagorean theory to determine the hypotenuse, which is the distance you’re trying to determine.
a 2 + b2 = c 2
82 + 72 = C2
64 + 49 = 113, therefore, the length of the hypotenuse is √113.
(6) Circles. The formula for a circle is (x - h)2 + (y – k)2 = r2 where (h, k) are the coordinates of the
center of the circle and r is the radius. The best way to solve circle problems is to sketch out the figure in
your test booklet or on scratch paper. A sample problem is shown below.
A circle in the standard (x, y) coordinate plane is tangent to the x-axis at 5 and the y-axis at 5.
Which of the following is an equation for the circle?1 ACT #47 p.175
A. x2 + y2 = 5
B. x2 + y2 = 25
C. (x – 5)2 + (y - 5)2 = 5
D. (x – 5)2 + (y - 5)2 = 25
E. (x + 5)2 + (y + 5)2 = 5
3
For the circle to be tangent at x = +5 and y = +5, your sketch would look like the diagram shown below.
10
5
5
10
15
The graph shows that the center of the circle is (5, 5) and that the radius of the circle is 5. Plug these
numbers into the formulas for a circle to find the correct answer choice.
(x - h)2 + (y – k)2 = r2
(x - 5)2 + (y – 5)2 = 52
(x - 5)2 + (y – 5)2 = 25 The correct answer is D.
You should memorize the equation for a circle. Here’s another circle question.
In the standard (x, y) coordinate plane an equation of a circle is x2 + y2 = 81. At what point does
the circle intersect the y-axis?1 ACT #15 p.735
A. (0, 1) and (0, -1)
B. (0, 9) and (0, -9)
C. (0, 18) and (0, -18)
D. (0, 27) and (0, -27)
E. (0, 81) and (0, -81)
Since the equation for a circle is (x - h)2 + (y – k)2 = r2, the only way the left side of this equation could be
x2 + y2 is if the value for h and k (the coordinates of the center of the circle) was zero -- the center of the
circle is (0, 0). Therefore, (x - 0)2 + (y – 0)2 = x2 + y2. Using the equation in the question, you also know
that the radius of the circle in 9 (√81). Now you can draw a graph in your test booklet to see where the
circle intersects (crosses) the y axis.
9
6
3
3
6
9
-9
If the center of the circle is (0, 0) and the radius is 9, the circle will intercept the y axis at 9 and -9, so the
correct answer choice is B. (It will intercept the x axis at 9 and -9 also.) You had to know the formula
(equation) of a circle to answer this question. Have it memorized or programmed in your computer.
Continued on the next page.
4
(6) Parabolas. y = a(x – h)2 + k, where h & k are the vertex (h, k). Let’s look at a problem.
The graph of y = x2 is shown in the standard (x, y) coordinate plane below. For which of the
following equations is the graph of the parabola shifted 3 units to the right and 2 units down?1 ACT #48 p.744
F. y = (x + 3)2 + 2
G. (y = (x + 3)2 - 2
H. y = (x - 2)2 + 3
J. y = (x - 3)2 + 2
K. y = (x - 3)2 - 2
Using the formula for a parabola, the only way
the equation could be y = x2 is if (h, k) = (0, 0) and a = 1;
y = 1(x-0)2 + 0. When the vertex is shifted right 3 and down 2 the vertex becomes (3, -2), so the equation
become y = 1(x-3)2 + (-2) = (x – 3)2 – 2 (answer K).
Plane Geometry: 14 Questions. You’ve taken geometry, so I won’t review in detail what you need to
know. We’ll go through some problems later, but first. I’ll list the items you need to take with you into
the test.
(1) Triangles. The ACT asks lots of ∆ questions, so you need to review (or learn) the following.
3 angels = 180o
area of ∆: base x height/2
equilateral ∆: 3 equal sides and three equal angles (60o each)
isosceles ∆: 2 equal sides and two equal angles
right ∆; one 90o angle
Pythagorean theorem:
a2 + b2 = c2 (c is the hypotenuse)
ACT favorite sides for right ∆
3 - 4 - 5 (if the legs of a right ∆ are 3 & 4, the hypotenuse will be 5) or multiples 6 - 8 - 10,
15 - 20 - 25, etc.
5 - 12 – 13 and multiples. Use this for shortcuts instead of using Pythagorean theory.
ACT favorite right ∆: You can’t answer some question without knowing these.
30o - 60o – 90o
30o 2X
X√3
60o
X
45 - 45 – 90
o
o
o
X√2
X
X
similar ∆s: two ∆s that have the same angles; therefore, their sides are proportional
10
6
5
3
4
8
Now let’s look at a few triangle questions.
5
A chord 24 inches long is 5 inches from the center of a circle, as shown below. What is the
radius of the circle, to the nearest tenth inch?1 ACT #25 p.169
A. 29.0
B. 24.5
C. 16.9
D. 13.0
E. 10.9
r
5
24
12
We haven’t looked at circles yet, but this is really a ∆ problem. The question asks for the radius. We have
information about a ∆ within the circle that could give us the third side of the ∆, which is the radius. The
∆ has one leg of 5 and one leg of 12 (half the chord). You could use the Pythagorean theorem to find the
hypotenuse.
a2 + b2 = c2; 52 + 122 = c2 (which is the radius, r); 25 + 144 = 169.
√169 = 13; so the radius (and the answer) is 13. Or you could have remembered that a
5 - 12 - 13 ∆ is a favorite on the ACT and skipped the Pythagorean theorem business
altogether.
The ratio of the side lengths for a ∆ is exactly 12:14:15. In a second ∆ similar to the first, the
shortest side is 8 inches long. To the nearest tenth of an inch, what is the length of the longest
side of the second ∆?1 ACT #39 p.173
A. 11.0
B. 10.0
C. 9.3
D. 6.4
E. Cannot be determined from the given information
14
15
?
12
8
In your test booklet draw the ∆s. The ratio of the shortest sides is 12/8 (which is 3/2) so this same ratio
must hold for the longest side which is 15/? = 3/2 or 15/10. The answer is 10.0. You didn’t need to find
the third side of the smaller ∆ to answer the question.
A 6-inch-by-8-inch rectangle is inscribed in a circle as shown below. What is the area of the
circle, in square inches?1 ACT #52 p.316
F. 5π
G, 16π
H. 25π
J. 48π
K. 96π
6
8
6
We haven’t looked at circles or rectangles yet, but we can answer this question using a ∆. You’ll
remember the formula for the area of a circle as A = πr2. We need to find the radius. If a diagonal of the
rectangle is drawn (see the diagram below), it is the diameter of the circle. To find the diameter of the
circle, which is the hypotenuse of the right ∆, you could use the Pythagorean theorem, or better yet,
remember that the 3-4-5 right ∆ is a favorite on the ACT and take the ratio 3-4-5 and double it to make a
6-8-10 right ∆. If the diameter is 10, the radius is r = d/2 or 5, so the area of the circle is
πr2 = π52, or π25 or 25π (answer choice H)
10
6
6
8
(2) Circles. The ACT asks lots of circle questions, so you need to review (or learn) the following.
A circle has 360o
A (area) = πr2 (r = radius)
C (circumference) = 2πr or 2d (d = diameter)
Let’s try some circle questions.
The figure below shows 2 tangent circles such that the 10-centimeter diameter of the smaller
circle is equal to the radius of the larger circle. What is the area, in square centimeters, of the
shaded region?1 ACT # 30 p.310
F. 10
G. 75
H. 5π
J. 10π
K. 75π
10
The question asks for the area of the shaded part of the diagram. If we can find the area of the larger
circle and the area of the smaller circle, we can subtract the smaller from the larger and find the square
centimeters of the shaded area. The question tells us that the radius of the larger circle in 10cm; therefore,
the area of the larger circle is A = πr2 or A = π102 or 100π. The question also tells us the diameter of the
smaller circle is equal to the radius of the larger circle or 10, so its radius is 5. Therefore, the area of the
smaller circle is A = πr2 or A = π52 or 25π. The answer is 100π - 25π = 75π.
Points M and N are the endpoints of the diameter of a circle, with center O as shown below.
Point P is on the circle and angle MOP measures 60o. The shortest distance along the circle from
M to P is what percent of the distance along the circle form M to N?1 ACT #43 p.460
P
A. 75%
B. 60%
C. 50%
D. 33&1/3%
E. 16&2/3%
m
60O
n
O
7
The question is looking for the percent relationship between the two distances which are measured in
degrees of the angles. The angle MOP is given as 60o. You need to find the degrees made by the “angle”
MN. Since MN is half a circle (or a straight line) it has 180o. The degrees of the two angles are 60 and
180. The percentage of 60o compared to 180o is 60/180 or 1/3 which is 33&1/3% (1/3 = 33&1/3%).
(3) Squares and other four-sided figures: The ACT asks lots of questions that have four-sided
figures in them, so you need to review (or learn) the following.
Squares & Rectangles:
All four angles are 90o
A= l x w
W
W
L
L
Parallelogram:
Each set of lines is parallel
Opposite angels are equal
Adjacent angels = 180o
A=bxh
h
b
Trapezoid
Two sides are parallel
A= ½(base1 + base2)(h)
b2
h
b1
The trapezoid shown below is divided into 2 triangles and 1 rectangle. Lengths are given in
inches. What is the combined area in square inches, of the 2 triaingles?1 ACT # 47 p.315
A. 4
B. 6
C. 9
D. 12
E. 18
4
3
3
8
To find the combined area of the two triangles using the information given, we could find the area of the
trapezoid and then subtract the area of the rectangle. The area of the trapezoid (the entire figure) is
A= ½(base1 + base2)(h); A = ½(4 + 8)(3); A= ½(12)(3), A= 6x3; A = 18
The area of the rectangle is A = l x w; A= 4x3; A= 12. Therefore 18-12 = 6 (answer B).
(4) Volume:
Cube: V = L x W x H
Cylinder = πr2(h)
What is the volume of a cylinder that has a radius of 5 and a height of 6?1 ACT #43 p.597
A. 30π
B. 31π
C. 150π
D. 180π
E. 900π
5
6
Just plug the numbers into the formula. π(5)2(6) = π(25)(6) = 150π
8
(5) Lines and angles:
180o
60o
120o
o
A straight line is 180 .
Two intersecting lines form four angles
of which the opposite angles are equal.
a
a=c&b=d
d
b
a + b = 180o
c
A straight line intersecting two parallel
lines forms 8 angles of which there
are only two measures. In the diagram
a= c = e = g and b = d = f = h
d
a b
c
e f
h g
In the figure below, lines m and n are parallel. Transversals r and s intersect to form and angle of
measure xo, and 2 other angels measures are as marked. What is the value of x?1 ACT #17 p.307
A. 15
B. 25
C. 35
D. 65
E. 80
r
s
m
65o
100o
n
xo
We know that when a line intersects two
parallel lines that the angles formed result
in just two values. Using the information
you learned above, angle gs =100 and
angle gr = 65. Since angle gs = 100,
angle hs = 80 (since a straight line = 180).
Now we have two angles of a ∆ (80 & 65)
Since there are 180o in a ∆, the 3rd angle n
must be 35 (180 -145). Since angle x is
opposite the 35o angle it must be 35o also
(answer C)
r
s
m
65o
gr
100o
gs
xo hs
Continued on the next page.
9
Statistics:
(1) Averages: You have to know two out of three pieces of information to work average problems –
the total, the number of events and the average. You will be given two and have to determine the third. If
a student took 10 tests (number of events) and the scores on all 10 tests added up to 888 (the total), then
the average would be 88.8 (888/10). Total/# of events = average. Let’s look at a problem.
In a town of 500 people, the 300 males have an average age of 45 and the 200 females have an
average age of 35. To the nearest year, what is the average age of the town’s entire population?1
ACT #59 p.465
A. 40
B. 41
C. 42
D. 43
E. 44
To find the entire population average, we will need to find the total of all ages (male & female), which we
don’t know and the total number of events (the total number of people), which we do know (500). Is
there a way to find the total of all ages? Yes, solve for the missing information for male and female and
then add them to get the total of all ages.
Males: total/# of events = avg; total/300 = 45; ?/300 = 45; 13,500/300 = 45; note: 13,500 = 300 x 45
Females: total/200 = 35, ?/200 = 35; 7,000/200 = 35; note: 7,000 = 200 x 35
The total of ages for both males & females is 13,500 + 7,000 = 20,500, so
20,500/500 = ?; 20,500/500 = 41 the answer is B.
Warning: Don’t take the average of the averages (45 + 35)/2 = 40. Since the totals for the males and
females are different (300 v. 200) averaging the averages will give you the wrong answer. The way we
solved the problem took into account weighted averages (i.e., more males than females).
(2) Probability: To determine the probability of something happening, you need to know the total
number of outcomes and the number of outcomes that you’re specifically looking for. Simple example:
100 marbles - 25 blue, 25 green and 50 red. What is the probability that if a marble is drawn from a jar
containing all 100 marbles, it will be green?
How many marble outcomes can there be? 100, because there are100 marbles
How many green marble outcomes can there be? 25, because there are only 25 green marbles
Now take the ratio of the number of green marble outcomes and the number of total marble outcomes.
25/100 = ¼ or 25%
If a marble is randomly chosen from a bag that contains exactly 8 red, 6 blue and 6 white
marbles, what is the probability that the marble will NOT be white?1 ACT #12 p.166
F. 3/4
G. 3/5
H. 4/5
J. 3/10
K. 7/10
How any ways can any marble be chosen? 8 + 6 + 6 = 20
How many ways can a non-white marble (red or blue) be chosen? 14
14/20 is the probability, which can be reduced to 7/10
10
(3) Combinations: Questions about combinations ask you to determine how many different ways
you can combine different items. Look at this problem.
Kareem has 4 sweaters, 6 shirts and 3 pairs of slacks. How many distinct outfits (each consisting
of a sweater and a shirt and a pair of slacks) can Kareem select?1 ACT #18 p.168
F. 13
G. 36
H. 42
J. 72
K. 216
To get a handle on how to proceed with this question, let’s see how many combinations of only sweaters
and shirts Kareem can make. How many ways can Kareem wear a sweater? The answer is 4 (because he
has 4 sweaters). Once he selects a particular sweater how many ways can he wear a shirt with that one
sweater? The answer is 6 (because he has 6 shirts). Therefore, with one sweater he can create 6 sweatershirt combinations, and since he has 4 sweaters, he can create 6 sweater-shirt combinations with each
sweater for a total of 24 different sweater-shirt combinations. To find the answer, we multiplied 4 x 6.
This holds true for any amount of combinations, so for our problem the procedure is 4 x 6 x 3 = 72.
Percents: You’ll see some percent questions on the ACT. Don’t be afraid to plug-in for percent
problems. We discuss percents by solving a problem.
Jennifer’s best long jump distance increased by 10% from 1990 to 1991 and by 20% from 1991 to
1992. By what percent did her best long jump distance increase from 1990 to 1992?1 ACT #60 p.319
F. 32%
G. 30%
H. 20%
J. 15%
K. 2%
Let’s use a concrete number to solve this problem. Let’s make her best distance for the first year (1990)
100 feet ( the number doesn’t have to be realistic). From 1990 to 1991 she increased this distance by
10%. 10% of 100 is 10 (10% = .1 & .1 x 100 = 10), so her new record distance for 1991 was 110. Now
the next year (1992) she increases the new record distance (110) by 20% or 22 feet (20% = .2 & .2 x 110
= 22) making the new distance record 132 (110 + 22). The question asks what was the percentage
increase from 1990 to 1992 or what is the percentage increase from 100 feet to 132 feet. To find a
percentage change, subtract the beginning number from the ending number and divide that number by the
beginning number.
Ending number – Beginning number/Beginning number
132(the distance in 1992) – 100(the distance in 1990) = 32
32(the change in distance from 1990 to 1992)/100(the distance in 1990) = .32 or 32% (.32 x 100)
Continued the next page.
11
Algebra: In this section you will learn how to solve algebra (and other) problems using an alternative
method that saves time and cuts down on mistakes. It’s okay to use algebra the way you learned it in
school, but sometimes you won’t remember how to solve a problem and the plug-in method may allow
you to get the correct answer.
Use easy-to-work-with numbers for plugging in such as 2, 3, 4 or 10, 20.
However don’t use 0 or 1.
If a = b + 2, then (b - a)4 = ?1 ACT #34 p.172
F. -16
G. -8
H. 1
H. 8
K. 16
If you can manipulate the equations and find the answer, go for it, but if you’re having a problem getting
started, try the plug in method. Since a variable stands for a number (or numbers), let’s give the variable
a number. In this case if we make b = 2, we can find the value of a. a = b + 2; a = 2 + 2; a= 4, Now we
have values for a and b that we can use in the other equation. (b - a)4 = (2 – 4)4 = (-2)4 = 16 (answer E).
Let’s try some other plug-in number to see if the system works. How about b = 5.
a = b + 2; a= 5 + 2, a = 7; then (b - a)4 = (5 – 7)4 = (-2)4 = 16
Still not convinced? Try b = 17. a = b + 2; a= 17 + 2, a = 19; then (b - a)4 = (17 – 19)4 = (-2)4 = 16
If f(x) = x2 – 2, then f(x + h) = ?1 ACT #56 p.177
F. x2 + h2
G. x2 - 2 + h
H. x2 + h2 - 2
J. x2 + 2xh +h2
K. x2 +2xh + h2 - 2
Again if you see the math-class way to solve this problem, use it by all means. But if you’re having
trouble, try plugging in. We are working with two variables that are independent of each other so let’s
give each one a number, say x = 2 and h = 4.
f(x) = x2 – 2; f(2) = 22 – 2 = 4 – 2 = 2; f(x) = 2 when x = 2. Now let’s find what f(x + 4) or f(2 + 4) equals
f(6) = (x + h)2 – 2; f(6) = (2 + 4)2 – 2; f(6) = 62 – 2 = 36 – 2 = 34
Which of the equations in the answer choices results in 34 when x= 2 and b = 4? Plug in 2 and 4 for the
respective variables and solve the equations. Try them until you find the right one.
The answer is K. x2 +2xh + h2 -2 or 22 + 2(2)(4) + 42 – 2 = 4 + 16 + 16 – 2 = 36 – 2 = 34
Which of the following equations expresses c in terms of a for all real numbers a, b, and c such
that a3 = b and b2 = c.= ?1 ACT #36 p.312
F. c = a6
G. c = a5
H. c = 2a3
J. c = 1/2a
K. c = a
Continued on the next page.
12
Let’s plug in. Make a = 2; therefore if a = 2, b must be 23 since a3 = b or b = 8. Now if b = 8 then c = 64
because b2 = c or 82 = 64. Since all the answers are expressions of c, which of the answers = 64? Try
each answer choice until you find the right one. This time we’re lucky and save time because the correct
answer is F - the very first one.
c = a6; 64 = 26; since 26 is 64 the answer is F.
Let’s try another problem where we can plug-in the answers. When the answers to a question are
numbers, you can plug then in to see which one works.
Which real number satisfies (2x)(4) = 83?1 ACT #18 p.453
F. 2
G. 3
H. 4
J. 4.5
K. 7
This one is easy. You don’t even have to determine what number to plug in. We know that the right
answer has to be one of the answer choices, so let’s plug them in to see which one makes the equation
work.
83 = 512, so (2x)(4) must = 512. Pug in answer F. (22)(4) = (4)(4) = 16; 2 is not the right answer.
You should have noticed that x = 2 gave a very small number (16) when we need the equation to = 512.
Let’s jump up to answer K the largest number and see what we get. (27)(4) = (128)(4) = 512 the correct
answer.
Okay, one last question.
When Angela was cleaning her refrigerator, she found 2 bottles of catsup. Looking at the labels,
she noticed that the capacity of the larger bottle was twice the capacity of the smaller bottle. She
estimated that the smaller bottle was about 1/3 full of catsup and the larger bottle was about 2/3
full of catsup. She poured all the catsup from the smaller bottle into the larger bottle. Then,
about how full was the larger bottle?1 ACT #23 p.455
A. 2/9 full
B. 1/2 full
C. 5/6 full
D. Completely full
E. Overflowing
If you are comfortable multiplying fractions, etc., answer the math class way; however, if you’re not able
to use the math class method you can still get this one right. Just plug in a number for the capacity of the
larger bottle. At first, I thought about using the real-life amount and making the larger bottle 32 (ounces),
but I quickly noticed that 1/3 of 32 is an awkward number to work with, so, to keep it simple, I made the
larger bottle 30 and the smaller 15. Now the process is easy. Since the smaller bottle is 15, 1/3 of 15 is 5
ounces of catsup in the smaller container. Since the larger bottle is 30, 2/3 of 30 = 20, so there are 20
ounces in the larger container. How much catsup is there? 25 ounces. Now, what is the fraction of 25/30
equal to? The larger bottle had 20 ounces in it, and then Angela added the 5 ounces from the smaller
bottle for a total of 25 ounces in the bottle that can hold 30 ounces. 25/30 reduces to 5/6 (answer C).
Continued on the next page.
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In addition to the information presented make sure you review the following:
Prime numbers: 0 & 1 are not prime numbers; 2 is the smallest prime number and the only even one.
When multiplying fractions the answer is smaller: ½ x ½ = ¼
Exponents: Don’t be afraid to use your calculator.
Multiply two numbers with same base, add exponents: 23 x 24 = 27
Divide two numbers with the same base, subtract exponents: 24 / 22 = 22
Raise a power to a power, multiply exponents: (34)3 = 312
Raise a positive fraction smaller than one, it becomes smaller: (1/2)3 = 1/8
Negative number raised to even power becomes positive: (-2)2 = 4
Negative number raised to odd power becomes negative: (-2) 3 = -8
√x√y = √xy
√x/y = √x/√y
√50 = √25 x √2 = 5√2 look for these type questions
Probability: Remember this: # of outcomes you’re looking for /total number of possible outcomes
Example: Bag has 10 red and 20 blue marbles. If one marble is drawn what is probability of
getting a red one?
# of outcomes you’re looking for = 10
total number of possible outcomes = 30
answer 10/30 = 1/3
Plane & Coordinate Geometry:
Know the 3-4-5 and 5-12-13 right triangles. The ACT uses these or multiples of these (e. g.,
6-8-10) often.
Memorize the formulas presented the review earlier.
Review basic math concepts:
Integer, Digit, Factor, Multiple(s), Real, Imaginary
Simultaneous equations
Quadratic equations
Trigonometry:
sin x = opp/ hyp
cos x = adj/hyp
tan x = opp/adj
opp
hyp
x
adj
Don’t forget to fill in any blank answer ovals (using your designated column) when the five minute
warning is given. Then go back and work on as many questions as you can until time runs out. There is
no guessing penalty on the ACT. NEVER, NEVER, NEVER leave a blank oval on your answer sheet.
______________________________
1. These examples questions are from The Real ACT Prep Guide, 3rd Edition.
This presentation is for the students and staff of Granville County Schools for educational purposes only
and no guarantee of ACT results is offered. Factual information is derived from sources that
are considered reliable but have not been verified. Increased ACT scores by students who have
participated in this preparation and review program in the past are no guarantee of an increased score
by any individual student who may participate in this program in the future.
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