Mathematical Reasoning (SAT & ACT) - Level 1

SAT/ACT Test Prep: Introduction Math Skills and Strategies

This chapter will give you an overview of basic strategies and expectations for the SAT/ACT mathematical reasoning section as it stands in 2023. Going through this chapter, we will discuss some general and rudimentary strategies to tackle quick problem solving and trouble shooting for various kinds of basic, general math problems one might find on the SAT or ACT. Keep in mind that the SAT and ACT are highly similar, despite what the SAT and ACT themselves tell you. We will point out important differences when they arise, but generally, the strategies and skills you will learn herein are equally applicable on both tests.

Test Format, Contents, and Expectations

Below is a basic comparative chart of the SAT and ACT mathematical reasoning sections.

SAT	ACT
includes two math sections (part 3 & 4)	Includes one large math section
Topics ranging from pre-algebra through Algebra 2 and basic Trigonometry (contains NO logarithms or matrices)	Topics ranging from pre-algebra through Algebra 2 and basic Trigonometry (DOES contain logarithms or matrices)
Both multiple choice with FOUR possible selections and free response grid-in questions	All questions are multiple choice with FIVE possible selections.
1^st SECTION \| 2^nd SECTION	ONE SECTION ONLY
NO Calculator is allowed \| A Calculator is ALLOWED	Calculator ALLOWED
25 minutes \| 55 minutes	60 minutes
20 questions \| 38 questions	60 questions
c. 1’15” per question \| c. 1’30” per question	c. 1’ per question (some more some less)

There is NO PENALTY FOR INCORRECT ANSWERS, so ANSWER EVERY QUESTION

CONTENT BREAKDOWN:

Below is a fairly comprehensive chart outlining all of the concepts one should know to do well on either the SAT or ACT. There are a few additional concepts in the ACT, which have been noted in the chart; however, nearly 95% of the math content between the SAT and ACT is the same.

On both tests, you can expect that all the “Prerequisite” and “Pre-algebra” skills will be needed in all the problems to some extent and will be applied in multi-concept fashion. For example, you will unlikely find a question on the test that merely asks you to add two fractions; however, the pre-algebra knowledge of fraction addition will be necessary in problems wherein there are expressions with unknown variables given in relatively complex fractions (or division operations), which can only be solved with a fundamental understanding of how to add fractions (both concretely and abstractly).

The SAT uses silly classifications for these math concepts; we have noted the SAT terminology), but have categorized the topics into more widely understood categories (such as seen on ACT guides). Ultimately, the categorization of the content is not important, but rather the content itself and whether you know it. Look over the list of topics and note the things that seem unfamiliar or which you have forgotten. These will likely be where we need to spend the most time in math review.

CONTENT AREA	TOPICS COVERED	SAT	ACT
Prerequisites & Pre-Algebra (SAT: “Problem Solving & Data Analysis)	Prerequisite knowledge of numbers, decimals, fractions, and integers and operations thereof Factors of numbers Number systems Interpreting data (data collection, representation, and interpretation) Rudimentary statics (mean, median, mode, range, etc.) Elementary counting techniques Simple probability (independent and dependent)	c. 17 questions	20-25% 12-15 questions
Elementary Algebra (SAT: “Heart of Algebra”)	Solving equations of a single variable, Solving inequalities and absolute value equations, Solving equations with ratios and percentages Properties of exponents and roots Functions and their evaluation (domain and range)	c. 19 questions	15-20% 9-12 questions
Intermediate Algebra (SAT: “Passport to Advanced Math”)	Rewriting expressions Systems of equations/inequalities, Solving quadratic equations and inequalities Factoring polynomials (mostly 2^nd degree, but some higher) Quadratic formula Simplification and solving of radical expressions Solving and evaluating exponential equations Imaginary and complex numbers Matrices (ACT only) Logarithms (ACT only)	c. 16 questions	15-20% 9-12 questions
Coordinate Geometry (SAT: “Additional Topics in Math”)	Graphical representations of linear equations Graphical systems of linear equations Graphical representations of exponential functions, Graphical quadratics and higher-degree polynomials	c. 4-6 questions	15-20% 9-12 questions
Plane Geometry (SAT: “Additional Topics in Math”)	Calculating area and volume Solving problems involving lines and angles Properties and relations of plane figures Angles and relations among perpendicular and parallel lines Properties of basic polygons Properties of circles (circumference, area, arc length, sector area, central angle theorem)	c. 4-6 questions	20-25% 12-15 questions
Trigonometry (SAT: “Additional Topics in Math”)	Using basic trigonometry to solve problems with right triangles Values and basic properties of trigonometric functions Basic trigonometric identities Solving basic trigonometric equations.	c. 1-2 questions	5-10% 3-6 questions

Keep in mind that SAT/ACT questions often require the simultaneous implementations of two or more concepts in algebra to answer a question, even at a basic level. Thus, most questions on the SAT are at least two-parted.

Also keep in mind that the test makers know the kinds of mistakes that students often make; if you become familiar with these common mistakes and pitfalls, you can be advantaged by potentially foreseeing in many problems the expected mistakes. Since the test makers consider commonly made mistakes, many of the multiple choice answers will reflect these mistakes, making the selection that much more difficult.

Tip!

The questions in the SAT and ACT Math sections generally progress by level of difficulty, with the easier questions coming toward the beginning of the section and the harder questions coming toward the end of the section.

Also, when the section switches from multiple-choice to student-produced response (grid-in) questions on the SAT, the difficulty level starts back with easier questions first despite all the free-response questions being at the end of the section. If you can reasonably predict which math questions are likely more challenging than others, you can better strategize you time and efforts in completing the SAT and/or ACT math sections.

GRID-IN FREE RESPONSE QUESTIONS ON THE SAT:

In the “grid-in” questions on the SAT, there are no given solution options. Thus, you are on your own devices completely to solve these problems.

Rule #	HOW TO FILL THE GRID (SAT)
Rule 1	Start in the left column; consistency will help avoid mistakes.
Rule 2	You must fill in the bubbles!
Rule 3	If the number is a decimal, do not put the “0” before it (e.g. “0.45” should be gridded as “.45”.
Rule 4	Decimal answers that exceed the grid spacing, you can round or truncate. However, your answer must fill the entire grid.
Rule 5	You can grid fractions, but not mixed numbers.
Rule 6	There may be more than one answer to a question; just choose the one that is easiest to grid.
Rule 7	There are no answers with negative values.

GUESSING ON THE GRID-IN QUESTIONS: The “no guessing penalty” policy still holds true, so if you cannot find an answer, still give some guess. While you might think “how could I guess?! There are countless possibilities for the answer,” the answers to these free response questions still often tend to be simple answers like 1, 2, 100, 1/2 (or 0.5), etc.. Granted, there are also answers like 1547 and 3.495 and 11/7, so the likelihood of a correct guess is very low, but still guess! Give a guess that seems likely given what the problem is asking and try to stick to simple numbers and values for guessed answers.

A Note on Calculators

You should definitely bring a calculator when you take the SAT and/or ACT. Test administrators will not provide you with a calculator, even if yours malfunctions; thus bring extra batteries and an extra calculator if you have one.

Be sure to check the SAT and ACT websites to make sure your calculator is one that is approved for use on test day.

When allowed, use your calculator for any arithmetic calculation or function evaluation that you cannot compute faster or more accurately in your head. However, most problems can be completed more easily without excessive button pushing on one’s calculator. There are two situations when you should be especially careful using your calculator:

If you think that you must input a lot of numbers (excessive button pushing) into the calculator to solve a problem, you should double-check; there is likely an easier way to solve the problem.
Use your calculator one step of the problem at a time; do not input too much into your calculator at once, since this could lead to weird solutions and miscalculations.

The Most General Strategy (Easy-Medium-Hard)

Knowing the general progression of difficulty on the math sections of both the SAT and ACT,

answer all the easy questions that you definitely know how to solve first
along the way of answering all the easiest and most confidently answered questions, mark any slightly more difficult questions you know how to solve but suspect will take some time to work though. Return to these questions after you have gone through the entire section and answered all the easy, confident questions. Keep in mind that what is “medium”-level questions vs “easy”-level questions might be different for you than someone else, thus you might encounter a question you have no idea how to answer at the start of the test and have an easy questions at the end, merely because of what you know, do not know, and are relatively comfortable doing in algebra.
If you encounter any questions you have no idea how to solve at all, mark this question differently. Return to these clueless-questions only at the end after you have answered all the easy-confident questions and time-consuming-confident questions. You might find that you are nearly out of time at this point, but that is okay! Since you do not know how to answer these questions, you will likely be guessing anyways. With whatever time you have, attempt these questions and try to come to a reasonable guess.
When time runs out, be sure to put some answer for ALL questions, since there is no penalty for incorrect guesses.

Basic Math Strategies with Precautions:

There are three basic strategies for the math section of the SAT and ACT: drawing diagrams, working efficiently, substitution, back-solving; the first two we would highly recommend, though last other two, while potentially useful, come with some caution. Let us briefly examine each.

General Strategy 1: Drawing Diagrams

Many of the word problems on the SAT/ACT suggest diagrams, especially those dealing with principles of geometry. Sometimes diagrams will be given to you, but sometimes you will have to make the diagram yourself. Often a quickly scrawled rectangle or triangle can go a long way to making the problem more concrete and contractible.

Example: Drawing Diagrams

Farmer Joe wants to build a rectangular vegetable garden against his barn in a field that is 4 acres and use the remaining area outside the barn and garden to grow corn. The long side of the barn will serve as one side to the garden plot. Farmer Joe has 125 feet of fencing to cordon off the garden. If the short side of the barn is 40 feet long and the and the area of the barn is 3000 ft², what is the remaining area of the farmer’s field, in which he can grow corn, once the garden is appended to the barn? (Note: 1 acre = 43,560 ft²)

A) 1875 ft²

B) 4875 ft²

C) 43,560 ft²

D) 169,365 ft²

E) 174,240 ft²

To the right is the diagram one could construct from the information in the problem.

There are very many steps to this problem, which we can ascertain by thinking backwards through the problem:

To answer this question, we ultimately need the combine area of the barn and garden so that we can compare that to the area of 4 acres.
To find this combined area, we need to know the area of the garden.
To know the area of the garden, we need to know the length and width of the garden.
To know the width of the garden, we need to know the length of the garden.
To know the length of the garden, we need to know the length of the barn.

So, what is the length of the barn? Well, we know the width (short side) of the barn and the area of the rectangular barn; thus, we can use knowledge of rectangular area to find the length.

$$A=\text{length}\times\text{width}=l\times{w}\quad\quad\quad{A=3000;}\quad{w=40}$$

$$3000=(40)(\text{length})\quad\quad\frac{3000}{40}=\text{length}\quad\quad\text{length}=75$$

Thus, the long side of the garden must be 75 feet. Furthermore, we know that the garden is rectangular and that one side of the garden will be walled-in by the long side of the barn. Thus, of our 125 feet of fencing, we use 75 feet for the other long side of the garden, leaving us with w50 feet of fencing to divide among the two short sides of the garden.

Since the garden is rectangular, we know that both short sides must be the same length; thus 50/2 will successfully divide the remainder of the fencing for the two short sides.

Thus, the garden is cordoned off by 25 feet of fencing on the short sides and 75 feet of fencing on one of the long sides. Consequently, the area of the garden will be (25)(75) = 1875 ft².

Thus, the combined area of the barn and the garden will be 3000 ft² + 1875 ft² = 4875 ft².

Now, do not forget to complete the problem. The answer is NOT 4875 ft². We need to know the area of the remaining field, the total area of which is (4 x 43,560 ft²= 174,240 ft²). So we should subtract the combined area of the barn and garden from the area of the field.

$$174,240\times4,875=169,365$$

So, the answer is 169,365 ft².

Drawing diagrams when appropriate is highly recommended. It will not be necessary at any juncture; however, it can be extremely helpful to keep track of all the information in the problem and how it all fits together. Also, the path to using this information to find a solution might be more obvious when starting at a diagram rather than when starting at a jumble of words.

Diagrams will take some extra time, but might make the problem easier to solve and your work more accurate. Do not spend lots of time making your diagrams beautiful or highly legible, since you are not graded on showing your work or the aesthetic merits of your diagrams. Do just enough to be clear to yourself.

Practice Problems: Diagraming

1. Free Response (calculator allowed):

A new ramp is being built that will span a distance along the ground of 20 feet to allow people to reach the entrance to a store that is 2 feet above the parking lot where the ramp begins. What is the length of the surface of the ramp in meters?

2. Multiple Choice (calculator not allowed)

The Grand Inquisitor of Parking Spaces has decreed that the length of a rectangular parking space must be two meters less than twice its width. A section of parking lot that will be used exclusively for parking spaces has been measured and determined to be 1008 square feet. The wonder of the lot wants to fit 42 equally-sized parking spaces in this 1008 square foot space. Given the parking space ordinance and a typical car width of a car (approximately 5.8 feet) is it reasonable to make 42 parking spaces?

A) No because the width of the space Is not enough and so an average car will simply not fit.

B) No because the width of the space is exactly enough and so the driver of the average car will not have enough room to get out of the car in a crowded parking lot.

C) Yes because the width of the space required by law and given by the parameters of the problem is exactly enough to fit the average car width.

D) Yes because the width of the space required by law and given by the parameters of the problem is more than enough to park a car and have room to get out of the car in a crowded parking lot.

General Strategy 2: Be Efficient

Work as much of the problem as is absolutely necessary to determine the answer given the choices available. Sometimes, given the limitations of the answer choices, one only needs to work a small portion of the problem to eliminate all but one answer. Those with enough experience can readily recognize such situations. However, for those with less experience, simply keep in mind the answers you have as you work the problem; you might notice at some point you are clearly being forced into a specific answer before you have completed the problem. At this point, you might as well select that answer and move on!

Example: Be Efficient!

Which of the following is equivalent to $(x^2+2x-2)(8x^3+1)$

A) $8x^6+16x^4-8x^3-2x^2-2x-2$

B) $8x^5+16x^4-16x^3+x^2-+2×2$

C) $8x^6+16x^4+16x^3+x^2+2x+2$

D) $8x^5+16x^4-8x^3-2x^2+2x+2$

First, we could consider solving this problem using substitution, but this would clearly take a lot of arithmetic with a lot of exponentiation. Perhaps this should be avoided.

The only other option available to use is to do the direct distribution of these two polynomials. However, we can do this cleverly! The easiest part of any polynomial distribution are the first and last values (FOIL = First, Outside, Inside, Last). It turns out that this is all we need to narrow our possible answers to only one! Let’s do the distribution for the first and last values:

$$(x^2+2x-2)(8x^3+1)=x^2(8x^3)+…-2=8x^5+…-2$$

The only answer with these first and last values is B. Thus, we are done! No need to figure out the rest of the problem! Move on!

Practice Problem: Be Efficient!

$$\frac{x^2y^{\frac{2}{5}}z^5}{y^{-2}z^3}$$

Which of the following is a simplified form of the above expression:

A) $\frac{(y^2z^2)\sqrt[5]{y^2}}{x^2}$

B) $-x^2y^{\frac{-8}{5}}z^{\frac{5}{3}}$

C) $\frac{1}{2}xy^{\frac{1}{5}}z\sqrt[3]{z^2}$

D) $\frac{\sqrt[5]{y^2}}{x^2y^2z^2}$

General Strategy 3: Back-Solving

Sometimes answer choices can easily be substituted “back” into the problem in such a way that all the wrong answers will generate nonsense (i.e. 1=2, 7=4, 0=1, etc.) while the singular correct answer will generate a truth statement (i.e. 2=2, 1=1, 0=0, etc.).

The cautions with this strategy are:

This method does not work for all problems and does not necessary work well for all problems wherein it works.
Unless the problem is conveniently optimal for back-solving, this method will require a lot of tedious arithmetic, in which one has many opportunities to make silly mistakes, which will require reworking. This could take a lot of time, especially if one makes any mistakes.

Generally, most of the problems that can be solved using back-solving are more quickly done directly. Only a few problems on the SAT and ACT will be conveniently designed such that back-solving will be your optimal strategy to answer the problem; unfortunately, there is not general formula to determine if a problem is well-suited for back solving. Knowing whether this strategy is optimal is a matter of experience.

Thus, back-solving is most useful to you if you really do not know how to solve the problem directly and back-solving is an available alternative path to an answer. As a last resort, back-solving can be a quick save.

If you choose to use this strategy (either optimally or as a last resort), to the right are some basic tips and methods for use:

THE BACK-SOLVING METHOD:

If you have no clue as to which answer is likely the best, start with a medial value. Conveniently, if the answer choices are numbers, they will be listed from least to greatest; use this to your advantage! If you start with a middle value and find that it is too much or not enough to make the problem work, you know which direction in which to keep searching (either increasing or decreasing in value)
Plug in the chosen value to the problem given.
Assuming your middle choice does not work, increase the value or decrease the value as needed (move up or down the multiple choice list) and check the next value.
Continue this process until the answer choice generates a truth statement.

Example: Back-Solving

If $8^{\frac{1}{3}x+2}=2^7$ , then x is equal to which of the following?

A) 1

B) 2

C) 3

D) 5

E ) 7

Assuming you have no idea how to directly solve this problem, let us start by choosing one of the middle choices and evaluating.

Given $8^{\frac{1}{3}x+2}=2^7$, if $x=3$, then

$$8^{\frac{1}{3}(3)+2}=2^7$$

$$8^{\frac{3}{3}+2}=2^7$$

$$8^3=2^7$$

$$512=128$$

Since back-solving with $x=3$ results in a nonsense statement like $512=128$, we know this is not the answer. Furthermore, since 512 is way larger than 128, we know the value of $x$ should also be smaller. Now, we must choose between the options of A and B. Since, 512 is significantly larger than 128, we might try A first, since it will result in a much smaller quality than B. If A turns out to be wrong, then B must be the answer. there is only one choice that is smaller, we can simply choose that and be done! Let’s try $x=1$

Given $8^{\frac{1}{3}x+2}=2^7$, if $x=1$, then

$$8^{\frac{1}{3}(1)+2}=2^7$$

$$8^{\frac{1}{3}+2}=2^7$$

$$8^{\frac{7}{3}}=2^7$$

$$\sqrt[3]{8^7}=2^7$$

$$128=128$$

Great! However, the problem would be much easier to simply solve directly if we know how, as given below:

$$8^{\frac{1}{3}x+2}=2^7$$

$$(2^3)^{\frac{1}{3}x+2}=2^7$$

$$2^{3(\frac{1}{3}x+2)}=2^7$$

$$2^{x+6}=2^7$$

$$x+6=7$$

$$x=1$$

If we did not have a calculator on hand, values like $2^7$ and $sqrt[3]{8^7}$ might be difficult to evaluate. The method of back-solving is generally nonoptimal and fraught.

For example, a similar problem like $4^{x+2}=16^5$ , could present overwhelming problems if a calculator is not allowed, since $16^5$ is not easily arrived at with mental math. However, this problem is easily done with the same direct, algebraic method:

$$4^{x+2}=16^5$$

$$4^{x+2}=(4^2)^5$$

$$4^{x+2}=4^{10}$$

$$x+2=10$$

$$x=8$$

Practice Probles: Back-Solving

$$x-y=1$$

$$4x+3y=11$$

1) Which of the following ordered pairs (x, y) satisfies the system of equations above?

A) (1, 2)

B) (2, 1)

C) (3, 4)

D) (4, 3)

2) If $\frac{1}{x+3}=\frac{2}{3x+1}$, the value of $x$ is:

A) 2
B) 3
C) 5
D) 6
E ) 7

$$\sqrt{12}\times\sqrt[3]{81}$$

3) Which of the following is equivalent to the expression above?

A) $3(6)^{\frac{1}{6}}$

B) $6(3)^{\frac{1}{6}}$

C) $3(6)^{\frac{5}{6}}$

D) $6(3)^{\frac{5}{6}}$

4) Which value of x is the solution to the equation $\sqrt{x+1}=4$ ?

A) 12
B) 13
C) 14
D) 15
E ) 16

General Strategy 4: Substitution

If your problem does not have specific numbers for answers, back-solving is not as useful. In such cases, we could potentially use our own number as a substitution into the problem and answers to make some sense of the problem.

This method also has some flaws, particularly in the amount of work it might create for you and the very many places errors could happen in such work. Thus, generally it is still better to solve a problem directly with tools of algebra; however, if you have no idea how to solve a problem, substitution might be a good last resort. There are some problems, like with back-solving that are arguably best solved by using substitution; knowing which problems for which this is true is a matter of experience.

TTHE SUBSTITUTION METHOD

Step 1	Use substitution (generally as a last resort) in problems where there are variable in the question and answers and/or when the question involves percentages but does not give specific values.
Step 2	Thoughtfully choose your own number to substitute for the variables in the problem; make sure they make sense and will not cause you more work than is necessary. Some suggestions to consider: Do NOT choose 0 (zero) as a value for substitution ever. Zero often results in everything equaling everything, making all answers correct. Since this is not going to be the case, we should not waste our time with the trivial cases of zero. Do NOT choose 1 (one) as value for substitution. While one (1) might work okay, there are frequently cases, like with zero, where you get trivial answers. Thus, do not take the risk of wasting your time by choosing 1. Two (2) is an excellent number to use! Not only will it rarely, if never, results in trivial solutions, it is a very easy number to work with. One hundred (100) is an excellent number to use in the case of problems dealing with percentages, since perCENTages deal intrinsically with parts per “centum” (Latin for one hundred or “parts per one hundred”).
Step 3	Solve the problem using the number you chose (like 2 or 100, depending on the context). Note the result.
Step 4	Solve every answer using the same substitution number. Compare the results to each reduced answer to the reduced form of the question itself.
Step 5	The reduced answer choice that matches the reduced question is the correct answer. If any of the reduced answers have the same value, you need to choose a different substitution number and start again.

This method obviously will require a fair amount of arithmetic calculation, in which there are likely many places to make minor errors. This could potentially add significant time to completion of the problem. Thus, this method is not recommended expect as a last resort or if you can ascertain that the problem is optimally solved using substitution (not necessarily easy to determine).

Examples: Substitution

Example: Situation 1 – variable in the question and in the answer choices

Which of the following is equivalent to $(x^2+2x-2)(5x-1)$

A) $5x^3-x^2-2x+2$

B) $5x^3-x^2-10x+2$

C) $5x^3+9x^2-12x+2$

D) $5x^3+11x^2+12x+2$

This can, of course, be solved directly by distributing the polynomials. You might even notice that (should be try to work efficiently), we should check the middle values ($x^2$ and $x$) values for the polynomial, since those are the unique choices in our answers.

However, let us try solving this using substitution. First, to demonstrate why 0 and 1 are bad choices generally, consider if we chose them as our substitution value:

$f(x)=(x^2+2x-2)(5x-1)$

$$f(0)=(0^2+0-2)(0-1)=(-2)(-1)=2$$

Choice A if $x=0$ is $5x^3-x^2-2x+2=0-0-0+2=2$

Choice B if $x=0$ is $5x^3-x^2-10x+2=0-0-0+2=2$

Choice C if $x=0$ is $5x^3+9x^2-12x+2=0+0-0+2=2$

Choice D if $x=0$ is $5x^3+11x^2+12x+2=0+0+0+2=2$

$$f(1)=(1^2+2-2)(5-1)=(1)(4)=4$$

Choice A if $x=1$ is $5x^3-x^2-2x+2=5-1-2+2=4$

Choice B if $x=1$ is $5x^3-x^2-10x+2=5-1-10+2=-4$

Choice C if $x=1$ is $5x^3+9x^2-12x+2=5+9-12+2=4$

Choice D if $x=1$ is $5x^3+11x^2+12x+2=5+11+12+2=30$

While $x = 1$ gets us more unique solutions, we are still given two non-unique solutions, and so we can only be certainly that either A or C is the right answer; we would have to do the process again to know for sure which answer is truly correct in all cases.

Thus, let’s use $x = 2$ instead:

$f(x)=(x^2+2x-2)(5x-1)$

$$f(2)=(2^2+4-2)(10-1)=(6)(9)=54$$

Choice A if $x=2$ is $5x^3-x^2-2x+2=40-4-4+2=34$

Choice B if $x=2$ is $5x^3-x^2-10x+2=40-4-20+2=18$

Choice C if $x=2$ is $5x^3+9x^2-12x+2=40+36-24+2=54$

Choice D if $x=2$ is $5x^3+11x^2+12x+2=40+44+24+2=110$

Since the evaluation of the original problem is the same as the evaluation of answer choice C, the correct answer is C. Wow, that took a lot of tedious arithmetic!

Example: Situation 2 – question with percentages without specific values

60% of 75% of x is equal to which of the following?

A) 0.45x

B) 1.15x

C) 0.045x

D) 0.15x

E) 1.35x

Since this is a problem with percentages, substituting a value of $x = 100$ makes the most sense and will make the evaluations in this problem much easier. So, let’s solve the problem assuming $x$ is 100.

$$60\%\;\text{of}\;(75\%\;\text{of}\;100)=.6(.75\times100)=.6(75)=(\frac{3}{5})(75)=3(15)=45$$

Now, we should substitute 100 into each of the answer choices and see what matches with our evaluation of the question.

Choice A: $0.45(100) = 45$

Choice B: $1.15(100) = 115$

Choice C: $0.045(100) = 4.5$

Choice D: $(.15)(100) = 15$

Choice E: $1.35(100) = 135$

Since choice A matches, the answer is A.

Such problems as this are sometime optimal to solve using substitution. Generally, one could assume percentage problems are just as fast to solve with substitution as they are using direct methods.

To efficiently solve this problem directly without a calculator, we should change each percentage into a fraction and simplify the fraction multiplication as done below:

$$60\%=.6=\frac{3}{5}\quad75\%=.75=\frac{3}{4}$$

So the question could be alternatively understood as “3/5’s of 3/4’s of x is…”. So we could simply evaluate like:

$$\left(\frac{3}{5}\right)\left[\left(\frac{3}{4}\right)(x)\right]=\left(\frac{3}{5}\right)\left(\frac{3}{4}\right)(x)=$$

$$\left(\frac{9}{20}\right)(x)=\left(\frac{45}{100}\right)(x)\;\text{or 40% of x, which is 0.45x}$$

Practice Problems: Back-Solving

1) If $p$ is a positive number, which of the following would be equivalent to increasing $p$ by $10\%$, and then decreasing the result by $20\%$?

A) Increasing $p$ by $10\%$

B) Increasing $p$ by $8\%$

C) Decreasing $p$ by $10\%$

D) Decreasing $p$ by $12\%$

2) If $b>a$, which of the following must be true?

A) $b^2>a$

B) $-a>-b$

C) $b>2a$

D) $a>0.5b$

3) If $x$ is an odd integer, and $y$ is an even integer, which of the following must also be an odd integer?

A) $2(x+y)$

B) $2(x-y)$

C) $(xy)^3$

D) $x+y$

E ) $x+y+1$

4) If $k>1$, what is the value of:

$$\frac{2k+6}{k^2+2k-3}$$

A) $k$

B) $k+3$

C) $\frac{2}{k-1}$

D) $\frac{2k}{k-3}$

E ) $\frac{k-1}{2}$