﻿ ﻿Discovering Some Problem-Solving Strategies - Getting into the Act: An Overview of ACT Math Basics - ACT Math For Dummies

ACT Math For Dummies (2011)

Chapter 3. Discovering Some Problem-Solving Strategies

In This Chapter

Putting your basic math skills to the test on the ACT

Considering which formulas to remember and understanding how to use them

Jotting and sketching your way to success with word problems

Every math test — especially the math portion of the ACT — requires some basic problem-solving skills. So in this chapter, I outline these skills. First, I discuss the basic math skills you should remember from before high school: the four basic operations (addition, subtraction, multiplication, and division), negative numbers, the order of operations (which is affectionately known by the mnemonic PEMDAS), and fractions. Next, I list the math formulas, along with some tips on how to use them, that appear most on the ACT. Finally, I discuss a few basic techniques for working with every student’s favorite type of math question: word problems.

You can use the skills I review in this chapter throughout the entire math portion of the ACT. No matter what the topic is, basic math problem solving can always help you out.

Identifying Basic Math Skills You Need to Know

The ACT tests your understanding of the math that most students take in high school: pre-algebra, algebra, geometry, and a bit of trigonometry. In this book, I focus specifically on these topics. However, to do well on the test, you also need a good grounding in the math basics that are taught in the earlier grades.

So in this section, I give you a look at these basic skills that you need to remember. Make sure you feel confident with this material. If you’re concerned that some of these skills may hang you up on the test, I recommend that you pick up Basic Math and Pre-Algebra For Dummies (Wiley) for thorough and detailed explanations of these skills. And for additional practice, you also can grab Basic Math and Pre-Algebra Workbook For Dummies (Wiley). Both of these are written by me, so I’m hoping you’ll like them and maybe even buy them.

Knowing how the four basic operations operate

The four basic operations — addition, subtraction, multiplication, and division — are the underpinning of everything in math. The ACT doesn’t directly test these skills, but you need to use them on virtually every question you face.

Fortunately, you can use your calculator, so you won’t need to do complicated calculations — such as long division — with paper and pencil. On the other hand, you don’t want to waste time doing every tiny calculation on your calculator.

At a minimum, make sure you know the following:

The addition table up to 9 + 9 = 18

All corresponding subtraction inverses up to 18 – 9 = 9

The multiplication table up to 9 × 9 = 81

All corresponding division inverses up to 81 ÷ 9 = 9

If you’re a little sketchy (and, be honest, who isn’t a little sketchy about 6 × 9 and 7 × 8?), a couple of days working with flashcards could be time well spent.

Negative numbers aren’t all that difficult, but they can be confusing — especially when you’re trying to move quickly on a test. Here are a few basic points to remember:

When you negate a number, you change its sign (either from + to –, or from – to +). For example:

The negation of 5 is –5. The negation of –8 is 8.

When you subtract a larger number from a smaller one, the result is negative. For example:

3 – 7 = –4 because 7 – 3 = 4

Adding a negative number is the same as subtracting a positive number. For example:

6 + (–1) = 6 – 1 = 5

When you subtract a negative number, the minus signs cancel out to become addition. For example:

9 – (–2) = 9 + 2 = 11

When you multiply or divide a positive number by a negative number (in either direction), the result is negative. For example:

3 × –6 = –18 8 ÷ –2 = –4

–5 × 10 = –50 –20 ÷ 4 = –5

When you multiply or divide a negative number by a negative number, the minus signs cancel out and the result is positive. For example:

–9 × –3 = 27 –45 ÷ –5 = 9

Evaluating with the order of operations

When you use more than one operation at a time, ambiguities arise that may cause you to arrive at the wrong answer. For example, suppose you have the following problem:

3 + 6 × 2 = ?

You might potentially approach this problem in two different ways. You could either do the addition first and the multiplication second or vice versa. But each of these approaches would give you a different answer:

(3 + 6) × 2 = 9 × 2 = 18 3 + (6 × 2) = 3 + 12 = 15

To avoid this problem, mathematicians have standardized the order of operations (also called order of precedence), which is a set of rules telling you the order in which to break down large problems. Here is the order of operations that you need for all math on the ACT (you can remember the order with the mnemonic PEMDAS):

1. Parentheses

2. Exponents

3. Multiplication and Division

Remember: For each of these four steps, work from left to right. Check out the following example problem to see how these steps work:

(3 + 7)2 – (2 × 4) × 23

To begin, evaluate what’s inside the parentheses, from left to right:

= 102 – 8 × 23

Next, evaluate the exponents, from left to right:

= 100 – 8 × 8

Now evaluate the multiplication:

= 100 – 64

Finally, evaluate the subtraction:

= 36

You probably won’t need to evaluate anything quite this complicated on the ACT, but here’s the point: If you remember the order of operations, you won’t get into trouble.

Making peace with fractions

Many calculators allow you to work with fractions. If your calculator has this functionality, spend some time getting comfortable with the following basic fraction manipulations:

Adding, subtracting, multiplying, and dividing simple fractions

Reducing fractions

Increasing a pair of fractions to a common denominator

Performing fraction-decimal conversions for common fractions

If you don’t know how to perform these functions, check out your calculator’s manual for guidance. On the other hand, if your calculator doesn’t have this capability (or you’re not prepared to figure out how it works), make sure you can do these basic manipulations quickly on paper.

Getting Comfortable with Formulas

A mathematical formula is an equation that allows you to find a value when you know one or more related values. For example, if you know the base and height of a triangle, you can find the area of that triangle using this formula:

You can use this formula to find any value (the area, base, or height) as long as you have the other two values. For example, you can find the base of a triangle by plugging the values of the area and height into the equation and solving.

In this section, I include a list of all the formulas I cover in this book. This list provides one-stop shopping to make sure you know these formulas before you take the ACT. But simply memorizing formulas isn’t enough, so I also discuss how to use the formulas in your arsenal.

This section is a mere preview of formulas. Throughout the book, I discuss each formula in much greater detail, show you how to use it, and then provide practice problems that reinforce this information.

Reviewing the formulas you need to know

In this section, I compile all the formulas that appear throughout the rest of the book and tell you what chapter discusses each in depth. This is pretty much a need-to-know list for the ACT, so be sure you memorize the ones you’re not familiar with. But having said that, I can assure you that this book provides plenty of practice understanding and using these formulas. And when you work with them for a while, you may find the memorizing part pretty much takes care of itself.

Pre-algebra

I review pre-algebra in Chapter 4. In that chapter, I introduce the following two important formulas:

Elementary algebra

In Chapter 5, where I talk about elementary algebra, I don’t introduce any formulas. However, elementary algebra often is necessary for finding a value in a formula, particularly when the value you’re trying to find isn’t isolated on one side of the formula. I discuss this idea in greater detail in the later section “Working with your arsenal of formulas.”

Intermediate algebra

Here are three useful formulas that make their way into the intermediate algebra discussion in Chapter 7:

Direct proportionality: Two variables, x and y, are directly proportional when the following equation is true for some constant k:

Inverse proportionality: Two variables, x and y, are inversely proportional when the following equation is true for some constant k:

The quadratic formula: For any quadratic equation — that is, an equation of the form ax2 + bx + c = 0 — the value of x is as follows:

Coordinate geometry

In Chapter 8, the topic is coordinate geometry. Lots of useful formulas emerge there. Here are the ones I focus on:

Midpoint formula: The formula for coordinates of the midpoint of a line segment between any two points (x1, y1) and (x2, y2) is:

Distance formula: The formula for the length of a line segment between any two points (x1, y1) and (x2, y2) is:

Rise-run slope formula: The formula for the slope of a line in terms of the rise and the run is:

Two-point slope formula: The formula for the slope of a line that includes the points (x1, y1) and (x2, y2) is:

Axis of symmetry of a parabola: The axis of symmetry is the vertical line that divides a parabola down the middle. For a quadratic function y = ax2 + bx + c, the formula for the equation of the axis of symmetry is:

Vertex of a parabola: The vertex of a parabola is the point where the parabola changes directions. It’s always either the lowest or the highest point on the parabola. Here’s the formula for the coordinates of the vertex for the quadratic function y = ax2 + bx + c:

Formula for graphing a circle: The formula for the equation of a circle of radius r centered at the point (h, k) is:

(xh)2 + (yk)2 = r2

Plane geometry

Plane geometry, which I discuss in Chapter 10, is chock-full of useful formulas. If you can’t commit all these formulas to memory and need to prioritize them, memorize everything up to the circumference of a circle. These are the must-know formulas for geometry. Then, if you have time, work your way forward from there.

Interior angles of a polygon (the total of all angles inside the polygon):

Total interior angles = 180(Number of angles – 2)

Area of a triangle:

Pythagorean theorem:

a2 + b2 = c2

Area and perimeter of a square:

A = s2

P = 4s

Area and perimeter of a rectangle:

A = lw

P = 2l + 2w

Area of a parallelogram:

A = bh

Area of a trapezoid:

Diameter, area, and circumference of a circle:

D = 2r

A = πr2

C = 2πr

Arc length of a circle:

Volume and surface area of a cube:

V = s3

A = 6s2

Volume and surface area of a box:

V = lwh

A = 2lw + 2lh + 2wh

Volume and surface area of a sphere:

Volume of a prism:

V = Abh

Volume of a pyramid:

Volume of a cylinder:

V = πr2h

Volume of a cone:

Trigonometry and other topics

In Chapter 11, I discuss trigonometry and a few other advanced math concepts that may show up on the ACT. Here’s a summary of the formulas I introduce in that chapter:

Six trig ratios:

Five trig identities:

Determinant: The formula for the determinant of a 2-by-2 matrix M = :

Expressing a logarithmic equation as an exponential equation: The logarithmic equation logbn = a is equivalent to the following equation:

ba = n

Formula for the imaginary number i:

i2 = –1

Working with your arsenal of formulas

Just as important as knowing the necessary formulas is knowing how to use them. The good news here is that in a certain sense, when you’ve seen one formula, you’ve seen them all. Every formula is simply an equation containing two or more variables. When you fill in all but one value, you can use algebra to solve for the missing value. In this section, I go over the most common skills you need for the ACT when working with formulas.

Solving for a missing value

Generally speaking, when you know all the values in a formula except for one, you can find the missing value. You do this by plugging in the values you know and then using algebra to find the remaining value. The following example illustrates this concept.

If the width of a rectangle is 4 and its perimeter is 28, what is its length?

(A) 7

(B) 8

(C) 9

(D) 10

(E) 11

Use the formula for the perimeter of a rectangle:

P = 2l + 2w

Plug in 4 for w and 28 for P:

28 = 2l + 2(4)

Solve for l:

28 = 2l + 8

20 = 2l

10 = l

Therefore, the correct answer is Choice (D).

Using two different formulas to solve one problem

To answer some questions, you may need to use two different formulas. Always start with the formula for which you can fill in all but one value, and then use this value to plug into the next formula. The following example illustrates this idea.

If the circumference of a circle is 9π, what is its area?

(A) 6π

(B) 18π

(C) 20.25π

(D) 40.5π

(E) 81π

The question gives you the circumference of the circle and asks for the area. So use the formula for the circumference first:

C = 2πr

Fill in 9π for C, and then solve for r:

9π = 2πr

4.5 = r

Now you can use the formula for the area of a circle:

A = πr2

Plug in 4.5 for r, and then solve for A:

A = πr2 = π(4.5)2 = 20.25π

So the correct answer is Choice (C).

Working with unfamiliar formulas

When you know how to use a formula, you don’t even need to know what the formula means. For example, here’s a question that uses a formula for a trig identity that I discuss in Chapter 11:

If cos x = 0.8 and tan x = –0.75, then what is the value of sin x, given the formula ?

(A) 0.6

(B) –0.6

(C) 0.75

(D) 0.8

(E) –0.8

If you haven’t studied trigonometry, you may not have a clue what this question is asking. Even so, you probably know you can plug in 0.8 for cos x and –0.75 for tan x:

Now use algebra to solve for sin x. Simply multiply both sides of the equation by 0.8:

–0.6 = sin x

Thus, the correct answer is Choice (B).

Sorting Through Word Problems

A word problem (also called a story problem or a problem in a setting) gives you information in words rather than in just equations and numbers. To answer an ACT word problem, you have to translate the provided information into one or more equations and then solve.

Throughout this book, I provide practice answering word problems that focus on specific math skills, such as algebra or geometry. In this section, I give you a few general tools for solving word problems.

Jotting down the numbers

You can solve some word problems fairly easily. Jotting down the numbers in the problem can be useful to help get you focused and moving in the right direction. The example word problems in this section show you how.

Seminar X brought in \$700 in revenue and had 20 participants, each of whom paid the same amount. Seminar Y brought in \$750 and had 15 participants, each of whom paid the same amount. How much more did each person pay for Seminar Y than Seminar X?

(A) \$5

(B) \$10

(C) \$15

(D) \$20

(E) The two seminars cost the same amount.

If you’re not immediately sure how to proceed, jot down the numbers in an orderly fashion:

X \$700 20

Y \$750 15

This step only takes a moment and gets your brain moving. When you organize the information in this way, you may see that the next step involves division:

X \$700 ÷ 20 = \$35

Y \$750 ÷ 15 = \$50

Now you can easily see that Seminar Y cost \$15 more than Seminar X, so the correct answer is Choice (C).

Jessica is in charge of stocking shelves at a supermarket. Today, she has already stocked 8 boxes that each contained 40 cans of soup, 12 boxes that each contained 24 cans of corned beef hash, and 4 boxes that each contained 60 cans of tuna. How many cans has Jessica stocked today?

(F) 148

(G) 310

(H) 624

(J) 848

(K) 1,020

Record the numbers in this question as follows:

8 40

12 24

4 60

Now multiply these numbers across (with or without your calculator, as needed) to get the number of cans in each set of boxes:

8 × 40 = 320

12 × 24 = 288

4 × 60 = 240

Finish by adding the results: 320 + 288 + 240 = 848. Therefore, the correct answer is Choice (J).

Sketching out problem information

Some word problems are much easier to solve when you draw a sketch to organize your thoughts. This technique is especially helpful if you’re a visual learner. So if you like to draw, paint, or play video games, lead with your strength and try to find a visual way to express math problems whenever possible. The following example shows how to use a sketch to your advantage.

The 12:00 p.m. eastbound train left the station at a constant speed of 40 miles per hour. At 12:45 p.m., the next eastbound train left the station at a constant speed of 60 miles per hour. Assuming neither train stops along the way, how far apart will the two trains be at 2:00 p.m.?

(A) 5 miles

(B) 10 miles

(C) 12 miles

(D) 15 miles

(E) 18 miles

This problem is difficult to visualize, so sketching out the information can help you arrive at the correct answer:

This figure helps illustrate a way into the problem: By 2:00 p.m., the 12:00 p.m. train has traveled for 2 hours at 40 miles per hour, so it’s 80 miles from the station. And the 12:45 p.m. train has traveled for 1 hour and 15 minutes at 60 miles per hour. The 1 hour accounts for 60 miles. The 15 minutes is a quarter of an hour, so this accounts for 15 miles. Thus, the 12:45 p.m. train is 75 miles from the station. As a result, the trains are 5 miles apart, making the correct answer Choice (A).

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