Introduction to the Mathematics Section
The Digital SAT (2024) will have two sections in the Mathematics category. Each section will be 22 questions long, with 35 minutes of allotted time.
The module will have 16-17 multiple-choice questions of the 22, and the rest will be Student Produced Responses where you must write your own answer. If your answer to a Student Produced Response is a fraction, write it in as an improper fraction (7/3) or as a decimal rounded to the fourth digit (2.333). Do NOT write it as a mixed number (2 1/3) or a decimal truncated too quickly (2.3). Write only the number – don’t include symbols like commas, %, or $.
Questions in this section are ordered roughly according to the CollegeBoard’s view of their difficulty. You will have three sheets of scratch paper and can use your own pen and pencil. I highly recommend writing down each step for each question, even if it’s easy, as it helps with proofreading and avoiding silly mistakes. Calculators are permitted on every Math question unless stated otherwise, and the Digital SAT app includes a built-in calculator. The calculator should be similar to the one here (https://www.desmos.com/testing/cb-digital-sat/graphing). Since there are trap answers, make sure you understand the logic of the question/answer before plugging numbers away.
There are four different content areas this section will cover. They breakdown (per module) as follows:
· Algebra– 7-8 questions
· Advanced Math – 7-8 questions
· Problem-Solving and Data Analysis – 3-4 questions
· Geometry and Trigonometry – 3-4 questions
Over the following sections, we will address each Topic Area in the Mathematics section, and break down the Question Subtypes that you are likely to see.
Many of these questions will involve translating words or word problems into math terms as part of the process. The following table should help with translation:
Any unknown variable: Variable like x, y, z
Any unknown value: Constant like a, b, c
More, sum: +
x less y, difference between x and y, y subtracted from x: x-y
Of, times, product: * (multiply)
Ratio, quotient, out of, per: / (divide)
a% less than X: (1-a/100)*X
a% more than X: (1+a/100)*X
There is one general strategic rule to think about as you go through the next few sections:
Look for shortcuts in the tougher, more complex questions, but take care (actually write out the problem in equation form) for the easier questions.
This may seem counterintuitive, but remember that all questions are worth the same. The tougher questions are generally trying to test your deeper math knowledge and are less likely to be laden with reading comprehension tricks. In many cases, the math tested in the tougher questions can actually be avoided and brute-forced. The easier questions are more likely to be constructed in a way that leads to trap answers. Since the question is easier, brute-forcing doesn't save that much time, and is more likely to lead to a trap answer. A common mistake many test-takers make is focusing too much time/energy on the tougher questions, which can lead to a time crunch on the other, easier questions. This may mean you don’t finish the whole module, or simply that you don’t have enough time to double-check the easier questions so your accuracy drops. In either case, the time tradeoff usually isn’t worth it.
For sake of illustration, let’s take a set of four questions, where 3 are easy and 1 hard. Let’s say that using shortcuts / brute forcing reduces your accuracy from 90% to 60%. If you would have spent 10 seconds on an easy question with a shortcut but it takes 20seconds to do more diligently, doing the easy problem the diligent way means you would spend 10 seconds to gain 0.3 points on the easy question (30%accuracy boost times 1 point). On the other hand, if you spend 30 seconds on a tougher question with a shortcut, rather than 60 seconds writing it out, you may lose 0.3 points on average on the tough question, but you just gained 30seconds of time, which is enough to do 3 different easy questions the long way. If you take the advice I have given and take care with the easier questions, and try to find shortcuts with the harder one, you would expect to get 0.9+0.9+0.9+0.6= 3.3 points on average over 90 seconds. If, instead, you spend the full minute on the hard question and have to rush through the easier ones at 10 seconds each, you would only get 0.6+0.6+0.6+0.9 = 2.7 points on average over those same 90 seconds. On the SAT math section, this is the difference between a 680and an 800.
ALGEBRA SECTION
The algebra section will test your knowledge of high school algebra, and consists of the following question subtypes:
• Linear Equations in 1 Variable
• Linear Equations in 2 Variables
• Linear Functions
• Systems of 2 Linear Equations in 2 Variables
• Linear Inequalities in 1 or 2 Variables
Linear Equations in 1 Variable
Linear Equations in 1 Variable will test your fluency in a) constructing 1 variable linear equations from word problems and b) solving these equations.
A typical linear equation is of the form y = mx+b. X and Y are variables here, m is the coefficient, and b is the constant (intercept). SAT questions will often ask you to either construct such an equation (or sometimes just the mx+b side of the equation) or solve for either x or y (usually x) given such an equation (in which case you should be able to figure out the variable not asked for – usually y – from the question prompt).
Since these are some of the simpler equation types tested, many of the problems may be word problems, to add some difficulty during the translation process.
When you read a word problem, be on the lookout for when the prompt is signaling to you that something is a variable, a constant, or a coefficient.
For example:
When she was 11 years old, Alia was 53 inches tall. Since then, she has grown at a rate of 3 inches per year. If Alia is 65 inches tall now, how old is she?
A. 4
B. 12
C. 15
D. 21
“When she was 11 years old, Alia was 53 inches tall” – this part of the problem gives us the constant. 53 inches is the set point Alia starts at within the context of this problem. Then, the problem tells us that Alia has grown at a rate of 3 inches per year. Whenever a problem makes a value dependent on how much of something else has happened (“per” is a common clue here), it is likely signaling a variable, since we don’t know how many “years” to multiple the rate by. The rate (3) is the coefficient. Whenever a question gives you a rate, speed, variable cost, or other such number that needs to be multiplied by a different quantity, you should suspect that this is a coefficient. “If Alia is 65 inches tall now” gives us the “y” portion of a typical linear equation.
So we end up with the equation 65 = 3x + 53. We simply subtract 53 from 65 to get 12, and then divide by 3 to get x = 4.
So is the answer A)? No. Always double-check what the question is actually asking for. The prompt is asking for Alia’s current age. X gives us the number of years that have passed since she started growing from her initial height of 53 inches. However, she was already 11 at that age. The question is not asking for x, but rather for the value of x + 11. 4 + 11 = 15. C) is the answer to this question.
Linear Equations in 2 Variables
This subsection of questions will ask generally ask you to either 1) construct a linear equation of the form y = mx+b, 2) solve a system of linear equations of the form y = mx+b, or 3) consider the graphical characteristics of a linear equation.
The simplest form of this subtype will ask you to construct a linear equation based on a word problem or other information. See below:
When she was 11 years old, Alia was 53 inches tall. Since birth, she has grown at a rate of 3 inches per year. What equation best describes Alia’s height Y given age x?
A. Y = 53x + 3
B. Y = 11x + 53
C. Y = 3x + 11
D. Y = 3x + 20
I have intentionally used a similar word problem to the one in the prior section; the strategies for these types of problems are similar, except in this case we don’t have a number “65” to plug in for Y. Again, we know that 3 is the rate of growth and so should be the coefficient of x. This eliminates A) and B) as answers. We then need to see which of these two equations gives us the right value of Y (53) when Alia’s age (x) was 11. D is the only answer that satisfies this, and so is the correct choice.
For these kinds of problems, you may find it easier/faster to plug in values of x/y rather than build the equation from the ground up like above. For example, we could simply plug in 11 for x in each of the answer choices and use the provided calculator to find the corresponding value of Y. This turns an algebra problem into an arithmetic problem, and would also have gotten us to the answer D). Which strategy you use will depend on whether you are faster/more comfortable with algebra or arithmetic, and your general comfort level with the provided calculator.
There may also be systems of equations (see the last section in Algebra for this: "Systems of 2 Linear Equations in 2 Variables").
The final question type involving linear equations might ask you about graphical properties. For example:
Given the following system of equation, what is the x intercept?
3x + y = 30
A. 5
B. 10
C. 15
D. 25
Recall that the x intercept is the point at which the line crosses the y-axis, and therefore the point at which y = 0. The y intercept is where x = 0. For these questions, simply plug in 0 for the appropriate variable and do the arithmetic. Here the equation simplifies to 3x + 0 = 30, and then x = 10, giving us B) as the answer.
These questions may also ask you to find answers given two points on the line. Example:
Two points (-2, 3) and (3, -2) lie on a line. What is the slope of this line?
A. -1
B. 0
C. 1
D. 5
If two points lie on a line, the equation of the line is given by y = mx + b, as before. However, in this case we have the values for x and y at various points, and are instead solving for m or b. The two points on the line give us a system of equations 3 = m*(-2) + b and -2 = m*(3)+b. We can solve this with traditional substitution to get b = 2m+3, and plug that into equation 2 to get -2 = 3m + (2m + 3), which simplifies to -5 = 5m, and so m = -1 (A). If this works for you, great.
However, a faster way to do this is to simply think of the slope as the difference in y values over the difference in x values. In this case, between point 1 and point 2, the y value decreases by 5 while the x value increases by 5. m = (y1 – y2)/(x1 – x2) = 5/-5 = -1.
Why is this true? Because the intercept b is the same in both equations. So we can simply subtract y1 = mx1+b and y2 = mx2 + b to get (y1 – y2) = m(x1-x2), so m = (y1 – y2)/(x1 – x2).
If you are more of a visual thinker, try graphing the two points on a quick scratch pad graph sheet. The slope of the line will be immediately evident – it is the number of y that the line moves for every increment in x. The intercept will also be clear here, as you can see where the lines intersect with the axes.
The fact that systems of linear equations describe 2 lines brings us to the final question subtype here. The SAT may ask you how many points two lines intersect at, or to find their point of intersection. If you think about any two lines, there are three possibilities. The lines could intersect at exactly one point (criss-crossing). The lines could be parallel, in which case they would not intersect at any point. Or the lines could be the same, intersecting at infinite points.
Given the lines y = 2x + 3 and 2y – 4x = 7, how many points do these lines intersect at?
A. 0
B. 1
C. 2
D. Infinite
The first thing we do is simplify both equations to the typical y = mx + b format. Equation 1 is already in this format, and equation 2 simplifies to y = 2x + 7/2. If lines have the same slope but different intercepts, they are parallel, and do not intersect. In this case, the slope is 2 for both lines, but the intercepts are different (3 vs 3.5). Essentially, Line 2 is Line 1 shifted up by 0.5 units. These lines are parallel and will not intersect, so the answer is A).
What if Equation 2 was 2y – 4x = 6? In this case, it would simplify to y = 2x + 3, which is the same as equation 1. The Lines would be equivalent in this case, and so there would be infinite solutions to the system of equations.
In any other situation, there should be 1 and exactly 1 solution to a system of distinct non-parallel linear equations. If the question asks you “where do these lines intersect” it is essentially asking you to solve that system of equations.
Linear Functions
Linear function questions will be very similar to the linear equations and systems of equations questions we’ve already discussed, except instead of describing the second variable as y or any other letter, the equation will be expressed in terms of a function of x. For example:
A local artist sells handmade pottery and offers a discount on bulk orders. The price for ordering 10 pieces includes a 5% bulk order discount, totaling $475, while the price for ordering 20 pieces includes a 10% discount, totaling $900. If p represents the number of pottery pieces ordered, which of the following functions best represents the total price P, in dollars, before any bulk order discount is applied?
A. P(p) = 50p
B. P(p) = 45p + 50
C. P(p) = 47.5p
D. P(p) = 50p + 25
We solve this the same way we would approach a linear equation question, except instead of y we have a function P of p, written P(p). In this case we know that the price for ordering 10 pieces is P(10) minus a 5% discount. So P(10)*(1-0.05) = 475. This simplifies to P(10) = 500. Let’s plug in 10 into the answer choices to see what fits. Both A) and B) gives us a $500 answer after we plug in 10 for p. So we look at our second constraint. We know that P(20)*(1-0.10) = $900. This simplifies to P(20) = 1000. Let’s plug in 20 for p in the answers A) and B). A) gives us P(20) = 1000 while B) gives us P(20) = 950. This implies that A) is the answer.
There is no grand strategy change here – these are the linear equations you’ve known and loved all along, just with a variable y, a, z, etc. replaced with a function F(x), P(p), G(a), etc.
Linear Inequalities
This question subtype will be very similar to the previous systems of equations, but you will need to remember 1) how to translate word problems into inequalities and 2) rules for simplifying inequalities.
When it comes to translating word problems, note the following table:
- No more than, at most as much as, less than or equal to, etc.
- <=
- No less than, at least as much as, greater than or equal to, etc.
- >=
- Less than, fewer, etc.
- <
- Greater than, more than, etc.
- >
When simplifying inequalities, the process works similarly to simplifying/solving linear equations except for two caveats:
- When you multiply or divide by a negative number, the direction of the inequality flips.
- E.g. simplifying -3x < 6 gives us x > 2
- When you flip sides, the direction of the inequality flips.
- So if we arrive at 2 > x, this means that x < 2
While this seems self-evident, note that you can flip equations like 2 = x to x = 2 without much forethought, so this is important to remember when manipulating inequalities.
Two types of tape are sold at a general store. Masking tape is sold for $3 per foot, while duct tape is sold for $4 per foot. Peter needs at least 10 feet of tape, but does not want to spend more than $33. If m and d are nonnegative integers representing the number of feet of masking tape and duct tape respectively, which system of inequalities describes this scenario?
A. 4m + 3d < 10, m + d > 33
B. 3m + 4d < = 33, m + d <= 10
C. 33 >= 3m + 4d, m + d >= 10
D. 3m + 4d > = 33, m + d >= 10
This is similar to translating a word problem into a system of equations, except now we have to be careful about the inequality signs. Just like before, we identify the constraints. The price of the tape in total is given by the price of masking tape times the length of masking tape plus the price of duct tape times the length of duct tape (3m + 4d). We know that this value needs to be “not… more than $33”, which tells us that the inequality should be 3m + 4d <= 33. B) fits this description, but we need to be careful. 3m + 4d <= 33 is equivalent to 33 >= 3m + 4d (remember we flip the signs if we flip the inequality). So C) also fits this description. Our second constraint is that Peter needs at least 10 feet of tape in total. The total feet of tape is given by the length of masking tape + the length of duct tape, or m + d. Since this needs to be “at least” 10 feet, that tells us the inequality is >=, and so m + d >= 10. C) fits this constraint while the sign in B) goes the wrong way. So we pick C) as the correct answer.
What if the question asks us to select which value of m and d will satisfy the constraints? E.g. A) m = 10, d = 1, B) m = 9, d = 0, C) m = 8, d = 2, D) m = 0, d = 10. The simplest way to solve this kind of problem is to just substitute the answers into the inequalities and see what works. Start with the simplest inequality first to reduce your arithmetic load. We know that m + d must be >= 10, so we can immediately eliminate B. Now we know that d is the more expensive tape, so as long as the length requirement is satisfied (which we just checked), the way to minimize price (so that it is <= $33) is to maximize the mix of m vs d. We already know from this that D) is unlikely to be correct, and if we plug it in we get $40 price which is above Peter’s price constraint. Now A) has higher mix of m than C) does, but it also has more feet of tape altogether, so we should plug in to both answers and see if that clarifies things. The cost of the combination in A) is $30 + $4 which is greater than $33 constraint, so we eliminate A. This gives us C) as the answer. To double check, we plug the numbers in and get $24 + $8 = $32, which is less than $33.
What if the question asks us to graph the system of inequalities and pick the right solution? While a system of linear equations is given by two lines, and the solution is their point of intersection, and inequality is given by one side of the line shaded, and the solution is the area where these shadings overlap. In order to select the right answer here, transform each inequality into y = mx + b form. So (assuming m is graphed on the y axis), we would transform the inequalities into m <= (-4/3) d + 11 and m >= -d + 10. Graph the lines the same way you would if these were linear equations. Then, if the inequality is less than like in constraint 1, shade the half of the graph that is below the line (including or not including the line depending on whether “=” is present). If the inequality is greater than (like in constraint 2), shade the half that is above the line. I have graphed these below – the green shaded region represents the first inequality while the blue shaded region represents the second. The darkest region, the sliver in the middle where both are shaded, is the solution to this system of inequalities.
System of 2 Linear Equations in 2 Variables
In this section, the problem will give you two equations (either explicitly, or in the context of a word problem), which you can use to solve for the required answer. Example:
Given the following system of equations, what is the value of x:
3x + y = 30
6x + 3y = 15
A. 15
B. 25
C. 35
D. 75
The “proper” way to solve a system of equations like this is to solve one of them to find either x or y in terms of the other. In this case, we can subtract “3x” from both sides of Equation 1 to get y = 30 – 3x. We then substitute that expression for y in Equation 2. So 6x + 3y = 15 becomes 6x + 3*(30-3x) = 15. We simplify this to 6x -9x + 90 = 15 by multiplying out the parenthetical term, and then to -3x = -75 by subtracting 90 from both sides, and then to x = 25 by dividing both sides by -3. This gives us B) as the answer.
Note, however, that there are often shortcuts to solving these equations, particularly when the question asks for an expression rather than a singular variable. Example:
Given the following system of equations, what is the value of y-x:
3x + y = 30
6x - 2y = 15
A. 1
B. 5
C. 15
D. 25
Now we could go through the whole process above and get the correct answer. But you might also notice that you can simply subtract these equations outright for a shortcut. If we subtract Equation 2 from Equation 1, we get -3x + 3y = 15, or 3(y-x) = 15. We then simply divide by 3 on both sides to get y-x = 5. This got us to answer B) with simply three subtractions and a division, rather than the 6-7 operations required if we went through the standard substitution process.
Identifying this shortcut will become easier with pattern recognition. I recommend doing a number of linear equation systems problems and starting out by looking at the equations to see if you can add or subtract them to get to some multiple of the expression asked for in the answer. If this becomes second nature, it could save you a lot of time on the exam. This will not work for everyone, however, and if you find after significant practice that you are still more accurate/quick solving these with traditional substitution, stick with that process.
There will also be word problems that essentially ask you to build a system of equations like above and then potentially solve it. Example:
A coffee shop sells a blend of two types of coffee beans: Arabica and Robusta. Arabica beans cost $15 per pound, and Robusta beans cost $9 per pound. A customer wants to purchase a total of 10 pounds of beans, consisting of both Arabica and Robusta, with a total cost of $120. How many pounds of each type of bean does the customer want to buy? Let a represent the pounds of Arabica beans and r represent the pounds of Robusta beans.
Which of the following systems of equations could be used to solve for a and r ?
A. 15a + 9r = 120 and a + r = 10
B. 15(a + r) = 120 and 9(a + r) = 10
C. a + r = 120 and 15a + 9r = 10
D. a + 15r = 120 and r + 9a = 10
The key here is to immediately identify the constraints that the problem is giving you; these constraints are the constants in the equations. Just like in prior problems, any value that scales with a variable (like a rate, price, etc.) will be a coefficient. Our first constraint is that Arabica beans cost $15 per pound (telling us that 15 is the coefficient for a) and Robusta beans cost $9 per pound (telling us that 9 is the coefficient for r). The total price of the purchase must equal $120, so 15a + 9r = 120 is our first constraint. This alone is enough to select A) in the question; you should do Process of Elimination after your first equation construction to narrow the choices and see if you have the answer already. Do not do more work than necessary, as time is of the essence. If D) had instead been 9r + 15a = 120 and r + 9a = 10, then we would still need to differentiate between A) and D). Our second constraint in this problem is that the pounds of beans need to total 10. This tells us that a + r = 10, and narrows the answer choices down to A).
What if the question had instead asked us how many dollars the customer should spend on Arabica? We would then solve the system of equations in answer A) to find the value of variable a (the pounds of Arabica beans) and then multiply by the price of the Arabica beans. In this case, we solve Equation 2 for r = 10 – a, and substitute this for r in Equation 1 to get 15a + 9*(10-a) = 120, simplified to 15a – 9a + 90 = 120, and then 6a = 30, so a = 5. We multiply a by 15 to get $75 of arabica beans. The trick here (and in many of the SAT math questions) is to always double check what the answer is asking for. A student who is perfectly comfortable with the material of solving systems of equations could easily have solved for a = 5, and then selected the answer “5” without double-checking and gotten the question wrong. While you might wait until the end of the module for a full proof-reading session, you should do these basic checks of “did I answer what the question was actually asking” after each individual question; it should take <10 seconds and save you from silly mistakes that will be more difficult to catch later on.
ADVANCED MATH
The advanced math section will test your knowledge of nonlinear equations and expressions, including absolute value, quadratic, exponential, polynomial, rational, radical, and others.
The questions here will include the same general subtypes as the Algebra section (equations in 1 variable, systems of equations in 2 variables, and functions), so this chapter will not be broken down by question subtype, but rather by the individual equation types and the content.
Absolute Value
When faced with an absolute value equation or inequality, first simplify the equation so that the absolute value expression is on one side and the rest of the equation is on the other. Then, split the equation into two separate equations with an OR between them. For example:
Which of the following is a solution to the equation |3x + 2| = 5?
A. -1
B. 0
C. 7/3
D. -7/3
We can interpret this equation as 3x + 2 = 5 OR 3x + 2 = -5. We solve each equation separately, to get the answers 1 and -7/3. Only -7/3 is in the answer list, so D) is the correct answer. The trick is to be careful about what part of the expression is within absolute value. One common mistake here could be to solve for 3x + 2 = 5, get 1 as the answer, and then simply flip that to -1 and pick A). This would have been the correct answer if only x was within the absolute value sign ( 3|x| + 2 = 5), but that is not the equation we are faced with.
When dealing with inequalities and absolute values, flip the sign when considering the negative scenario. So |3x + 2| < 5 becomes 3x + 2 < 5 OR 3x + 2 > -5. This is because we can think of the absolute value signal as a conditional multiplication by -1; it multiplies the expression by -1 if the expression is negative, to make it positive again. But if we multiply an inequality by a negative number, we have to flip the sign.
Rational Equations
Rational equations are any equations of the form y = f(x)/g(x). Simple linear/polynomial/exponential equations are rational as well (g(x) is simply 1 in those cases), but the more interesting setups involve equations like y = (2x+1)/(3x+2).
If the rational equation is in two variables, most of the time you will simply want to brute force plug in the answers to see what gets you the right value of the 2nd variable.
There are more elegant ways, most of which involve multiplying expressions and going through multi-step simplification. I will walk through this with the example below. However, it is evident that the elegant way to answer these questions isn’t always the fastest, and you should consider brute forcing with arithmetic and getting to the answer quickly if simplification isn’t going well.
Example:
Which of the following is a value of x that satisfies the equation (12/(x-1)) + (24/(x+1)) = 120?
A. 3
B. 1-√(3)
C. (1-√(369))/2
D. (3 ± √(369))/20
To do this “properly,” the first step is to multiply both fractions by the other’s denominator, just like you would when trying to add different fractions together. This turns the equation into(12/(x-1))*((x+1)/(x+1)) + (24/(x+1))*((x-1)/(x-1)) = 120. This simplifies to(12*(x+1) + 24*(x-1))/((x+1)*(x-1)) = 120. We multiply by the denominator and distribute the numerator to get 12x + 12 + 24x – 24 = 120*(x+1)*(x-1). This simplifies further to 36x-12 = 120*(x2-1), and finally 120x2 -36x - 108 = 0, which simplifies to 12*(10x2-3x-9)=0. We can use the quadratic formula on this monster expression to get x = (3 ± √(369))/20, which gives us answer D). But that is an exhausting way to answer this problem, and will suck a lot of time from the rest of the module.
Instead, you could get rough estimates for the answer choices with your calculator and plug them into the expression to see what fits. A) is simply 3. B) is roughly -0.73. C) is roughly -9.1. D)is roughly -0.81. Now we plug these into the expression and see which one gives us a value that is closest to 120. A) gives us 12, a whole order of magnitude off. B) gives us ~82. C) gives us ~-4. D) gives us a number very close to 120. Now this took some time, but not nearly as much time as simplifying that rational expression and then solving the quadratic would have taken. There is a lot of beauty in math and in working with these expressions in the proper way, but sometimes that beauty is best saved for outside the SAT environment. There is no need to be a hero here; just do what works well and works fast.
Radical Expressions
There are three main aspects of radicals that the SAT will test you on.
1. Simplifying radicals
2. Remember both signs
3. Solving radical expressions
The exam may ask you to simplify a radical expression that looks dauntingly large. The key is to factor the number inside the radical step by step until you can simplify it.
Example 1:
Which expression is equivalent to √(832)?
A. 3√90
B. 8√13
C. 29
D. 30
If this wasn’t a multiple choice question, we could factor 832 to simplify it and pull out the squared terms from there. First we try dividing by 2, this gives us 416. This looks further divisible by 2, or even 4. If we keep dividing by 2 until we can no longer, we get 832 = 2*2*2*2*2*2*13. This is 8*8*13, as we can group the 2’s. 13 is a prime and not a square so we can’t really divide any further. This gives us √((82)*13),or 8√(13), so answer B) is correct.
However, this is multiple choice, and so there is an easier way to finish this without trying to factor a massive number. Simply “un-simplify” the answers by squaring the expression outside of the radical, to see what product equates to 832. So answer A) 3√90 becomes √(32* 90) = √9*90. 9*90 is 810, not 832, so this is wrong. With answer B, 64*13very easily gives us 832, so we know that’s the correct answer. Doing a handful of multiplication operations with a calculator is typically faster than factoring a large number, so while you should know how to do the former, and practice your skills as such, feel free to use brute force in the real exam.
Example 2:
Which statement is true about the values A and B where x2-49 = 0 and:
A = x
B = 7
A. A > B
B. A < B
C. A = B
D. The answer cannot be determined
This is a trap question. If we simply brought 49 to the right side and took the square root, we might be tempted to say x = 7, and so A = B. This would be incorrect. When taking the square root to find a solution, both the positive AND negative roots are correct solutions. In this case, x could be 7 or -7 (which is more obvious if we factored the quadratic instead, into (x+7)*(x-7) = 0). Since x could be -7 as well, we do not know if A = x = 7 = B or A = x = -7 < B. The answer cannot be determined, and D) is the correct choice.
Example 3:
Solve the following expression for x:
3 + √(x-1) = 7
A. 4
B. 7
C. 15
D. 17
The general strategy for most types of equations is to do the opposite of whatever has been done to the variable x to "free” it, and mirror this on both sides. In this case, we first subtract 3 to get √(x-1) = 4. Then we square both sides to get (x-1) = 16. This leaves us with x = 17.
Example 4:
Solve the following expression for x:
x - 14√(x) + 56 = 7
A. √7
B. 2√7
C. 49
D. 56
First, we simplify to x - 14√(x) + 49= 0. This looks a lot like a quadratic, doesn’t it? With pattern recognition fromthe quadratic drills, we can see that this expression looks exactly like z2 -14z + 49, except with √x as the variable, instead of z. We solve the quadratic equation with z, and then substitute x back in. z2 - 14z + 49 simplifies to (z-7)2= 0, so z = 7. Now z = √x = 7, so x = 72 = 49. The answer is C).
Exponential
The SAT will sometimes test exponential equations (of the general form y = axr + b). Recall the basic operations we can do with exponents from the Content Review:
Exponent Operations
When simplifying exponents with a common substrate, you can basically do to the exponent term the one-level-down version of the operation done to the substrate. For example, if two exponential terms with the same substrate are being multiplied, you can add the exponents. If they are being divided, subtract. If an exponent is raised to the power of another exponent, you can multiply the terms.
Recall the following:
o x0 = 1
o x1 = x
o xb =x*x*x….*x, b times
o x-b =1/(xb)
o x1/b =bth root of x = b√x
o (xa)*(xb)= xa+b
o (xa)/(xb)= xa-b; Note that x-b = 1/(xb) is the same formula as above but with a=0
o (xa)b= xa*b
The exam may test your familiarity with these operations in a straightforward way, e.g.:
Let a = 3 and b = 2. If (xa)*(xb)= xc, what is the value of c?
A. 2
B. 3
C. 5
D. 6
This is simply testing your knowledge that when multiplying exponential expressions, you can add the exponents. So c= a+b = 3+2 = 5, and answer C) is correct.
The most common form of complex exponential question you will see on the exam will be related to rate of change, or interest. The key here is translating the word problems into an exponential equation. The actual arithmetic will typically be too complex for you to do effectively on your own, but can easily be done using a calculator. Example:
Warren deposits $1000 into a bank account that earns an annual compound interest rate of 5%. IF Warren does not make any additional deposits or withdrawals, how many years (rounded up) will it take him to double his money?
A. 5
B. 14
C. 15
D. 20
Growth rate type questions (whether compound interest, or population growth, etc.) are typically of the form F(t) =A*(1+r)t where A is the original amount, r is the growth rate, and tis the number of periods (if the growth rate is annual, this should be measured in years). In this problem, the initial amount is $1000. The rate of growth is5%, or 0.05. We are asked to solve for t, but we know that F(t) should be double the initial amount, or $2000. So we set up the equation 2000 =1000*(1.05)t. Divide both sides by 1000 to get 2 = 1.05t.Now the “proper” way to solve this is to set up a logarithm, since t = log1.05(2).But that is unnecessarily complicated for the SAT. The fastest way to go from here is to simply plug in the answers to this equation. Raise 1.05 to the power of each answer using your calculator, and see which answer gets you 2 or more.5 is too low. 14 is close but not enough. 15 is just enough for Warren to have doubled his money. 20 years is too long. C) is the answer.
How do we identify when a problem is hinting at an exponential equation rather than the linear growth word problems we did in past chapters (e.g. the ones with price per pound)? The key is that exponential growth rates grow based on the total amount at the end of each period, and not just based on the variable t or x itself. If an account is growing with compound interest, in the first year it will grow by a simple 1.04 times the original amount, which seems linear. But the next year, the growth rate is not applied to the original amount, but to the total amount at the end of the prior year – which is A*1.04. The growth rate in these problems scales with the total value of the function in the prior period and not just with a single linear variable. You should think of exponential equations whenever compound interest, population growth, or other % growth rates or multipliers are mentioned. In this case, answer D) is a trap answer for students who did not realize the question was implying an exponential equation, and instead used a simple interest formula. With simple interest, Warren would be making 5% of just the original amount each year, and so he would earn $50 per year, and take 20 years to double his money. With the power of compound interest, he gets there 5 years faster.
Polynomial
You may also see higher order polynomial functions on the SAT (functions where a variable is raised to a power greater than 2). There are two ways to approach these.
Sometimes there is an easy way to simplify the equation to a product of lower-order equations.
Example 1: y = x3-2x2-3x.We can factor out x, since it is included in every term of the expression, to get y = x*(x2-2x-3). Now we have a simple quadratic that we can factor. The equation thus becomes y = x*(x-3)*(x+1). Setting y = 0, we get the root -1, 0, and 3.
Example 2: y = x4 - 14x2+ 49. This kind of looks like a quadratic, except with x2 as the term instead of x. We can factor this like a normal quadratic, but replacing x2for x. As 49 is a perfect square, we get y = (x2-7)2.Setting y = 0, we get x2 = 7, so x = +/-√(7).
If you can factor or simplify the polynomial like in the above scenarios, that’s great. However, it is not worth trying to be “proper” about these solutions if they don’t come easy. Especially with higher order and polynomial equations, sometimes the most effective strategy is to brute-force it and try all the answers. With the help of a calculator, simply doing the arithmetic for each answer choice may be a fine strategy, and one that I would recommend if obvious simplifications don’t come to mind.
Quadratics
Solving quadratics is the first portion of the math section where you will likely need to do some memorization, and practice these questions until the relevant formulas are second-nature.
Recall from the Content Review:
Remember the following formulae for a quadratic equation of the form y = ax2+bx+c
- Quadratic formula for finding the roots (solutions where y = 0):
- x = [-b ± √(b2 – 4ac)]/2a
- Sum of the roots of a quadratic equation:
- -b/a
- Product of the roots of a quadratic equation:
- c/a
- X-coordinate of the vertex of the parabola given by a quadratic equation:
- -b/2a
It is important to know the above formulae like the back of your hand since they can substantially speed up the answering process in questions that ask for roots of quadratic equations. However, the Quadratic formula in particular is a bit large/slow so it may be better to see if you can factor a quadratic first.
Factoring Methods:
- sum-product:
- If a quadratic equation doesn’t have an a term (so is of the form x2 + bx + c), you can factor it simply by trying to find numbers that add up to b and multiply together to form c
- For example, if we have x2 + 7x + 12, we can see that 3*4=12 and 3+4=7, so we factor this into (x+3)*(x+4)
- Grouping:
- In a more complex quadratic expression where there is an a term, we want to first find the factors of the product of ac that add up to b. Then we separate b into these factors, and group.
- For example, if we have 2x2 + 7x + 6, we want to find factors of 2*6 = 12 that add up to 7. 3 and 4 seem to do the trick, as 3*4 = 12 and 3+4 = 7. We then separate the middle term into these factors so that we have 2x2 + 4x + 3x + 6. If we group both halves, we should get a common expression. This turns into 2x*(x + 2) + 3*(x+2). We can then reverse the distributive property to sort this into (2x+3)*(x+2).
- Personally, I think this starts to get unwieldy, and doesn’t offer much of a time benefit vs. the quadratic formula. If this method makes more sense to you and/or you can’t reliably remember the quadratic formula, then by all means stick to it. But in more complex quadratic equations like this, I usually just use the quadratic formula.
- Remember the following formulae for squaring expressions
- (a + b)2 = a2 + 2ab + b2
- It follows from this that (a - b)2 = a2 - 2ab + b2 since we can treat (-b) as the second part of the expression in the original formula, and the minus sign gets squared away in (-b)2
- In practice, if you have an expression ax2 + bx + c and c looks like a perfect square, you should see if b is actually twice the square root of c so that this formula fits. For example, if we were given x2 - 14x + 49, we can recognize that 49 is 72 and 14 is 7*2, and so can immediately fit this into the pattern above and factor to (x-7)2
- It follows from this that (a - b)2 = a2 - 2ab + b2 since we can treat (-b) as the second part of the expression in the original formula, and the minus sign gets squared away in (-b)2
- (a+b)*(a-b) = a2 – b2
- You should look out for this pattern whenever you see a quadratic expression where there are two squares. For example, if we saw x2 – 49, we can immediately factor this to (x+7)*(x-7) using the above formula without going through a lengthy quadratic formula process.
- (a + b)2 = a2 + 2ab + b2
Let’s try a few problems just to run through them. My best advice for this section is to read the formulae above until they are second nature, and practice the relevant problems with multiple drills until the process is intuitive. You can get by on just using quadratic formula everywhere, but it is slow, and time is your enemy on this test. So the best way to solve these problems is to have enough pattern recognition to quickly see factors and squared expressions when they are in front of you.
Example 1:
Solve the quadratic equation y = x2 – 9
A. -3 and 3
B. 3
C. -1 and 1
D. 9
We see immediately that 9 is a square (3*3) and so this equation looks to be of the form a2-b2 we saw earlier, where a = x and b = 3. We can immediately factor this into (a+b)*(a-b), which in this case is (x+3)*(x-3). To find the roots, we set this expression = 0 (since the solution is where y = 0). This gives us x = 3 or x = -3. A) is the correct answer. Now we could have done this with the quadratic formula, and plugged in 1 for a, 0 for b, and -9 for c, but the pattern recognition here is much faster than trying to plug in numbers into that large expression.
Example 2:
Solve the quadratic equation y = x2 + 2x - 3
A. -3 and 1
B. 3 and -1
C. 2
D. 3
Again, we can use the quadratic formula as a failsafe, but it is much faster to try and factor this, especially since there are only a few factors of -3. It could only be 3 and -1, or -3 and 1. 3 and -1 in this case add up to 2, so we have our factored expression: (x+3)*(x-1). We set this equal to zero to find the roots, and get x = -3 or +1, which corresponds to A).
Example 3:
What is the sum of the roots of the quadratic equation y = x2 + 4x + 3?
A. -4
B. -3
C. 3
D. 4
Now, again we could use the quadratic formula (or factoring into (x+3)*(x+1)) to find the two roots -3 and -1, which add up to -4 and give us A) as the answer. However, the shortcut formula for sum of roots is -b/a which very quickly gives us -4 within seconds, without having to go through either of those more laborious processes. This is particularly useful for more complex quadratic equations, where factoring or using the formula may take too long.
Example 4:
The trajectory of a ball, in meters, is given by the equation y = -x2 + 2x + 3. What is the highest point the ball reaches?
A. -1
B. 1
C. 2
D. 4
Remember the shortcut for finding the vertex of a parabola: x = -b/2a. The equation for the ball’s trajectory is a quadratic with negative a, and so the vertex in this case represents the maximum value of the quadratic equation. Using the x = -b/2a formula, we get x = -2/-2 = 1. This is the trap answer. The x-axis simply tells us how far the ball went when it reached its maximum point. The actual maximum height is given by the value of y when x = 1. Plugging in 1 for x, we get -1 + 2 + 3 = 4. The answer is D).
Watch out for the general pattern of quadratics. Sometimes an equation will look like it mirrors a quadratic, but isn’t one, but the same techniques can be used. For example, z - 4√z + 3 = 0 and v4 – 4v2 + 3 = 0 both seem to resemble quadratic equations, even though there isn’t a square term. These are, in fact, quadratic equations, except with variable “x” terms that are not in the first order. The first equation is a simple quadratic of the form x2-4x+3, except that x = √z. The second equation is also of the form x2-4x+3 except in this case x = v2. We can solve these by replacing z with x, solving like a normal quadratic, and then putting z back in its place. x2+4x+3 factors to (x-3)*(x-1) = 0, so x = 1 or 3. We substitute back z, so that the first equation becomes x = √z = 1 or 3, and square both sides to get z = 1 or 9. The second equation substitutes to x = v2 = 1 or 3, so v = ±1 or ±√3
PROBLEM SOLVING
The problem solving section will test your knowledge of a variety of general math concepts, including ratios, percentages, probability, and basic statistics/graphs. We will go through these problem types in order below.
Evaluating Statistical Claims: Observational Studies and Experiments
This section is similar to the prior in that it tests your ability to draw precise inferences from experimental setups and data. There isn’t much mathematical knowledge required here – it’s mostly a test of logic and careful thought.
There are two types of studies that will be presented in this section:
1. Observational studies – these studies draw inferences from a sample to a population where the independent variable is not under the researcher’s control. These scenarios will be similar to the ones seen in the prior section, as that covered sampling and surveys. E.g. a researcher sampling 300 specimens of local wildlife to check for rates of disease to draw conclusions about the broader population.
2. Experiments – these are controlled studies where the researcher assigns two or more groups and tests them against each other based on a controlled independent variable. E.g. a drug trial offering one group of patients the experimental drug, and the other group of patients a placebo sugar pill, to be able to compare the results of the drug arm vs. a control set.
In general, a controlled experiment is much better for drawing conclusions about correlation or generating hypotheses about causation based on a specific variable, while an observational study is better for understanding the statistics of a population. Sometimes observational studies may try and separate the population into two distinct groups base on an independent variable (e.g. surveying smokers vs. non-smokers), but findings from such studies are typically only suggestive, since there are a number of extraneous variables involved, and the there is higher risk of adverse selection.
For example, an observational study may divide individuals into those who wear Rolex watches and those who don’t. Maybe the study finds that Rolex-wearers have higher expected life expectancy than non-Rolex wearers. Does this mean that your doctor should prescribe you a watch for your health? No. What is most likely happening here is that there is a common third variable (wealth) that both correlates with the independent variable (Rolex ownership) and also causes the change in the dependent variable (life expectancy). This is one of the most common pitfalls of an observational study, and the first thing to look out for when drawing conclusions from one. Any time an observational study concludes there is a correlation between two variables, you should be particularly skeptical about extrapolating this to causation.
Experiments help reduce this risk, because the independent variable is assigned/split between the two groups randomly. When done in a randomized, controlled, blinded fashion, the only difference between the experimental groups should be the tested independent variable. This makes it much easier to be firm in conclusions about correlation, and even get some confidence in causation. In the example above, if researchers had randomly assigned participants to join Group A or Group B and then given every participant of Group A their own Rolex, the study’s power to claim causation would be much higher. Of course, this experiment would likely show no difference between the two groups, because giving someone a Rolex is unlikely to have a meaningful impact on their life expectancy. See the example below:
An undergraduate psychology study claims that students who exercise for at least 30 minutes per week get higher grades than those who do not. The study involved 500 teenagers, half of whom were categorized as active (>30 minutes of cardio per week), and half of whom were categorized as sedentary. The study found the active cohort performed 25% better on average GPA compared to the sedentary cohort.
Given this information, which of the following statements accurately assesses the study’s claim?
A) The study conclusively proves that exercise improves academic performance in teenagers.
B) The study’s findings suggest a correlation between moderate exercise and improved academic performance, but they do not establish causation.
C) The difference in grades between active and sedentary teenagers is likely due to chance, and the study’s claim is statistically insignificant.
D) The study's design does not allow for any meaningful conclusions because it did not distinguish between cardio and strength training.
First we identify this as an observational study. The study did not take a random cohort of teenagers and get half of them to start exercising while the other stayed sedentary. Instead, the study took a random sample and split the sample up based on exercise frequency. This comes with the same pitfalls that so often affect observational studies; causation is difficult to establish because there can be numerous extraneous variables driving the correlation. In this case, it is possible that exercise has a direct positive impact on GPA. But it is also possible that the high-exercise cohort is simply more energetic, more Type A, more conscientious, or otherwise has personality characteristics that drive both the active lifestyle as well as the academic success.
From this discussion you can probably already tell that Answer B) is the best choice, but let’s go through the others one by one. Answer A) is too strong of a statement. Due to the lack of randomized control, it is difficult to leap to a causation conclusion from this study, let alone to claim that the study “conclusively proves” such causation. Answer C) is not a claim that has any evidence in the text. It is theoretically possible, but the text only tells us the difference in means between the two groups and not their standard deviation or p-value, so we can’t make any claims about statistical significance here. Additionally, a 25% delta between the two groups appears, at least intuitively, to be enough of an effect size to be significant, if standard deviations were reasonably tight. The dichotomy in answer D) might add nuance to the study and could be an interesting follow up, but it isn’t necessary to draw the correlation between physical activity and academic achievement.
Another pitfall to watch out for with both observational studies and experiments is the study design and recruitment. Is the study sampling the right population in the right way? See the example below:
A local Philadelphia cookie shoppe wants to see how interested its customers would be in a potential new addition to the shoppe’s menu: a large cookie cake. The shop owners post a poll in a Facebook group of cookie lovers from across the nation and get 1000 responses. 85% of respondents view the menu addition favorably. Which of the following weaknesses most clearly prevents the shoppe owners from extrapolating the survey results to a business decision?
A. The sample size is too small
B. The study does not test for the right variable
C. The sample is not representative of the population
D. The standard deviation is too high
Answer A) does not seem correct since 1000 responses is actually a fairly large sample size, especially compared to the likely size of the customer base of this local cookie shoppe. Answer D) also does not have any proof within the text of the question, as we are not given the standard deviation of the 85% sample mean. Answer B) seems compelling. Indeed the shoppe may have found it more useful to ask about willingness to pay rather than a general “favorable” or “unfavorable rating,” when it comes to making a business decision. Still, this doesn’t seem like an egregious error, since overwhelming demand and positivity for the new menu item may still be enough to give the shoppe confidence to add it, and adjust pricing later. Answer C) gets to the crux of the issue with this study and is the correct choice. The cookie shoppe’s customers are local Philadelphians. The study’s sample, however, is a group of cookie enthusiasts all across the country. More importantly, the group is specifically composed of people who love cookies enough to join a whole Facebook group about it, so they are likely much more positively inclined towards cookies than the average Philadelphian. This means that the sample is very unlikely to be representative of the population that the shoppe cares about.
Let’s look at some of the limitations of experimental studies next:
Researchers conducted a study on the impacts of caloric intake prior to sleep on sleep quality. They randomly selected 500 participants and separated them (in a randomized and blinded fashion) into two groups. Group A was given 100 calories of warm milk before bed while Group B was given nothing. The study found that Group A had better quality of sleep, measured by 15% greater hours of REM sleep per night, than Group B. Which of the following conclusions can be drawn from the study?
A. People should drink a glass of warm milk before bedtime.
B. Drinking a small portion of warm milk is correlated with improved sleep quality.
C. 100 calories of any substance is likely to improve sleep quality in a measurable way.
D. No conclusions can be drawn as the effect size is too small to be statistically significant.
Again, we can discard answer D) because the text does not give us any information to determine whether this effect size is significant or not. Answer C) is too broad. The study only tested warm milk, and it is entirely possible that 100 calories of any other beverage or snack may not have worked as well (perhaps milk contains specific pro-sleep compounds). Answer A) is also too broad, and uses the wrong comparator. This is a placebo-controlled study, so the experimental group with warm milk is compared to a group that does not consume anything before bedtime. In the real world, people may be consuming any number of substances before bedtime, such as dessert or supplements. There is no evidence in the study that warm milk is a better pro-sleep substance than whatever else a person might consume before bed. We cannot use comparison to placebo to justify a blanket recommendation. (This is particularly relevant in drug trials – a new experimental drug may work better than placebo, but still be inferior or inconclusive in comparison to the plethora of existing drugs for a condition).
Inference from Sample Statistics and Margin of Error
This section will test your ability to interpret data from sample statistics. There are a few basic types of questions to understand here: 1) margin of error, 2) drawing conclusions, and 3) extrapolation from sample statistics.
Margin of Error/Standard Error:
Margin of error is a statistic expressing the amount of random sampling error in the results of a survey. It is specified by percentile usually (e.g. 90% margin of error) but if unspecified it is conventional to assume 95%. This represents the bounds around the sample mean that we would expect the population value to fall within, with 95% confidence.
For a confidence level C, MOEC = zC * √(S2/N) where zC is the quantile, S2 is the population variance, and N is the sample size. The √(S2/N) term is also known as the standard error. You do not need to remember this formula exactly – the point of showing this is to demonstrate that margin of error should scale with the variance of the overall population, and inversely with the sample size of the survey.
SAT questions may ask you why the margin of error is different between two different surveys. Example:
Two samples of voters were polled, resulting in the following % votes for the incumbent: Sample 1: 54%, Sample 2: 49%. The margins of error on these proportions were calculated in the same way, and were as follows: Sample 1: 5%, Sample 2: 2%. Which of the following is the most appropriate reason that the margin of error for Sample 1 may be greater than the margin of error for sample 2?
A) Sample 1 had a larger number of uncountable votes
B) Sample 1 had a higher percentage of votes in favor
C) Sample 1 had a smaller sample size
D) Sample 1 had a larger sample size
Remember that MOE scales inversely with sample size (this is intuitive, as having a larger sample for the survey should give you more confidence in the results), so Answer C) looks promising from the start. But let’s eliminate the others. Answer B) is false because the mean (% votes in favor) of the survey should not be relevant for the margin of error, or confidence around that mean. Answer A) looks intuitively promising, since one would think a survey with many unusable votes is worse, except that uncountable votes are excluded from the sample size to begin with (and a large enough initial survey could have both higher sample size and higher number of uncountable votes compared to a survey with smaller reach).
Drawing conclusions:
Another question subtype here will set up the sample statistics of an experiment and ask you to draw conclusions. In general, think about 1) separating correlation from causation and 2) making sure the standard deviation is small enough to justify the conclusion.
A mobile app development company conducted an A/B test to compare the effectiveness of two app designs in retaining users. Design A was tested on 200 users, resulting in an average retention rate of 40% with a standard deviation of 15%. Design B was tested on another 200 users, showing an average retention rate of 41% with a standard deviation of 20%. Assuming normal distribution for retention rates, which conclusion can the company most accurately draw from these test results?
A) Design B is more effective than Design A at retaining users, as indicated by the higher average retention rate.
B) The difference in retention rates is not significant enough to determine which design is more effective.
C) The smaller standard deviation for Design B indicates that it is not only more effective but also more consistent in retaining users.
D) Without knowing the variance within each user group, no conclusion can be drawn about the effectiveness of the designs.
This survey looks generally well constructed, and there is no obvious extraneous variable that would explain the difference in means other than the Design change. However, the standard deviations of the results are massive compared to the differences in the means. If the standard deviations of the sample means are 15-20%, then the 1% difference between 41% and 40% starts to look like random chance. Thus we conclude that the difference in retention rates isn’t significant enough to determine which design is more effective (answer B).
What if the results were instead 40% and 61%, with standard deviations of 5-10%? There the difference in means looks much more significant. In a normal distribution, 95% of values should fall within 2 standard deviations, and the higher mean in this case is well over 2 standard deviations greater than the 40% number, so we have >95% confidence that the difference is meaningful. In that case, we would have picked Option A).
Extrapolation
The final question subtype here involves extrapolation from the sample to the population. This is mainly a logic question, where you have to be careful about the specific parameters of the study and make sure the conclusions you draw are in-scope.
A study was conducted on the animal life in Glacier National Park. A random and representative sample of animals in the park were tested for reproductive health. The sample contained 625 deer, of which 15% were infertile. Which of the following conclusions is best supported by the data?
A) Approximately 15% of all deer in America are infertile.
B) Approximately 85% of deer in Glacier National Park should be healthy.
C) Approximately 15% of deer in Glacier National Park should be infertile.
D) Approximately 15% of animals in Glacier National Park should be infertile.
We can eliminate the answers one by one based on how they exceed the scope of the study. Answer A) tries to draw a conclusion about all deer in America. While it may be true that the deer in Glacier are representative of all deer, we do not know this to be the case. The study did not randomly sample from America at large; it only sampled from Glacier. There may be extraneous variables that cause the deer at Glacier to have higher or lower rates of infertility than the rest of America (e.g. higher tourist activity, or less plastic in the dirt). So A) is a stretch and we can eliminate it. Answer B) tries to extrapolate the conclusion to health, which is too broad. 85% of the deer in the park may be free of infertility, but that does not mean that they don’t have other diseases. Answer C) is a more accurate and restrictive claim, and is the correct one in this case. Answer D) tries to extrapolate to all animals in the Park. This is incorrect because the 15% number is only a percentage of the deer in the Park that were infertile, and the sample means for other species may be wildly different.
One-Variable Data – Distributions and Measures of Center/Spread
Some of the questions in this section will be fairly straightforward calculations of measures of central tendency, and so the most important part of preparation here is reviewing the content guide.
If you know these formulae, many of the problems resolve to simple series of arithmetic, which the provided calculator can meaningfully help with.
Still, there a few tricks to consider before we go over example problems:
• When finding the median, put the set of data in increasing order, and then strike off numbers on each side until you get to the middle 1 or 2: {1, 2, 3, 4, 5, 6}
• Sanity check the answers first – you can often use Process of Elimination to narrow down the answer choices substantially. If the answer choice for the mean looks way off from your ballpark estimate of the center of the data set, it can probably be eliminated. If the standard deviation looks too large or too small, it can be eliminated. Remember that in a normally distributed dataset, 68% of the values will fall within one standard deviation. While a random non-normal dataset will not follow this rule, it shouldn’t be dramatically off as long as the data isn’t full of extremes and outliers. E.g. if the question gives us {1, 3, 5, 7, 9, 11, 13}, and the answer choices for standard deviation include 1 and 20, those can probably be eliminated.
• Use the visual conceptualization of these concepts discussed in the content review (repeated below) to get intuitive answers for how changing a dataset might affect its parameters.
Recall:
If you are a visual thinker, it can be helpful to think about these concepts as descriptions of data in space. Imagine your data as a cloud of dots on a graph.
The median is the middle dot. The mean is more like the center of mass of the entire cloud of dots. If we increased x = 4 to x = 7, the median would stay the same, since x=1 is still the middle dot, but the mean would increase, since the weight of the overall cloud has shifted to the right. The range is a measure of how spread out the dots are, but only at the edges. So the range of 6 here tells us that the rightmost dot is 6 away from the leftmost dot. However, it does not tell us how spread apart the whole cloud is. The other dots could all be huddled up at x = 1, or they could be spread far apart within that range. This difference is noted in the variance and standard deviation, which tell us about the spread of the cloud of values as a whole.
Certain SAT questions may ask you about which of these values change, and how, if the data is shifted in certain ways. This is where this visualization can be helpful. If the question asked, for example, what would happen if you added 2 to every value, you can think about what happens to this cloud of points if they are moved to the right two units. The mean and median would shift, of course, but the measures of spread (both the range and the variance/stdev) would not change since the dots are in the same position relative to each other. On the other hand, if we moved x=2 up 1 and x=0 down 1, the variance/stdev would change since the spread of the dots from each other has widened (the cloud has dispersed, in a sense), but the median and range would remain the same since the center and edges of the group haven’t changed.
With these concepts in mind, let’s try out a few questions.
Example 1:
A pediatrician records the heights of her patients as {34, 42, 55, 56, 60, 62} in inches. If all of the students grow by 3 inches next year, how will this impact the mean, median, and standard deviation of the dataset?
A) The mean, median, and standard deviation increase
B) The mean, median, and standard deviation stay the same
C) The mean increases, but the median and standard deviation stay the same
D) The mean and median increase, but the standard deviation stays the same
Now we can use the formulae for mean, median, and standard deviation to get proper answers here. The Mean is the sum of all datapoints divided by the number of datapoints, and if we add 3 to every single data point, the mean would definitionally increase. The median datapoint here would also increase, since every point is increasing. If we look at the standard deviation formula, we realize that while the xi term increases by 3, the mean increases as well, so that the difference between each data point and the mean stays the same. Since N doesn’t change, there is nothing in the standard deviation formula that would result in an increase here. Using the formulae, then, gives us answer (D).
In the context of a time crunch (and depending on how your brain works best) it can be helpful to conceptualize these terms in alternate ways rather than going through the full labor of writing out the formula.
You can think of these measures of central tendency in the visual way described above. If the entire cloud of datapoints moves by 3 units, then the value of the median data point would have increased by 3, and the center of mass of the entire dataset would also have increased by 3. However, the dispersion/variance of the data wouldn’t change as the points are just as far apart from each other as they were before.
What if only one patient, patient #5, grew by 3 inches? In this case the mean would still change, as the sum of the values has increased by 3 and the N hasn’t changed. The standard deviation would increase, since a datapoint is moving further away from the mean, while the others stay put. The range would increase, since the highest value would now be 63” rather than 62”. The median would actually stay the same.
We have to be careful here; the median stays the same because the ordering of the data in both the before {34, 42, 55, 56, 60, 62} and the after {34, 42, 55, 56, 62, 63} scenarios leaves us with 55” and 56” as the median datapoints. Changing the value of the datapoints on either side does not matter unless the change makes a datapoint flip to the other side of the prior median. E.g. if we’d added 20 inches to the 42” patient instead, then the new dataset would have been {34, 55, 56, 60, 62, 62} median datapoints would have flipped to 56” and 60”. But since adding a few inches to the 60” patient doesn’t change the ordering in a way that would matter, the median stays the same.
Percentages
Percentages questions are another form where the trick lies in reading the question properly and formulating the steps in the right way, rather than knowing obscure mathematical formulae. Read these questions carefully, and make sure you are applying the percentages to the right value. See the examples below for ways the test can trick you.
Example 1:
A record store increases the price of a vinyl by 10% and later realizes the price is too high. To make the sale attractive, the store decides to offer a 10% discount on the new price. If the original price of the vinyl was $50, what is the final price of the vinyl after the discount?
A) $49.00
B) $49.50
C) $50.00
D) $50.50
The trap here is for students taking the lazy way out and simply assuming that the price going up 10% and then down 10% would leave the record at the original price of $50. This is wrong, because the 10% price hike is applied to the original price of $50, while the 10% discount is on the new price. So we need to take this step by step. The original price is $50. The store hikes by 10%, so the new price is $55. The store then offers a discount of 10% on the new price, so to find the final price we need to multiple $55 by (1-10%) = 55*0.9 = $49.50, or Answer B.
Another question format will have you work backwards from a percentage conversion.
Example 2:
A habitat’s population of rabbits increased 25% from 2012 to 2013. If the population was 150,000 rabbits in 2013, what was the population in 2012?
A) 120,000
B) 150,000
C) 187,500
D) 200,000
The best way to avoid making mistakes in word problems generally is to set up the equation with variables first, and then substitute in. In this case, let’s say p(t) is the population of rabbits at time t. The first sentence tells us that p(2013) = p(2012)*(1+25%) = p(2012)*1.25. However, we are given p(2013) and need to work backwards to find p(2012). So we substitute in as follows: p(2013) = 150,000 = p(2012)*1.25. Then we divide both sides by 1.25 to get p(2012) = 150,000/1.25 = 120,000 (answer A).
Note that this would have helped us keep the values in order for example 1 as well. We could have set up the equations as p(final) = p(hike)*(1 – 10%) = p(hike)*0.9 and p(hike) = p(original)*1.1. We could then substitute to get p(final) = 0.9*p(original)*1.1 = 0.9*$50*1.1 = $49.50. While this may not have been fully necessary in 1 or 2 step problems like the above, it becomes imperative in more complex problems as it is more difficult to keep all of the steps in order in your head. Setting up the equations and double checking them adds a few seconds to your time taken, but substantially lowers your chances of making a silly mistake. We can pay for these seconds by using tricks like brute forcing or factoring in other more complex questions such as the ones we’ve discussed in the Algebra section.
Example 3:
A college volleyball team won 70% of its first 40 games of the season. To finish the season with a win rate of at least 75%, how many of the remaining 10 games must the team win?
A) 75%
B) 80%
C) 95%
D) 100%
Like many of the previous questions in this section, this one benefits from writing out the problem instead of short-cutting it. Work backwards. There are clearly 40 + 10 = 50 games in the season. To get 75% win rate, how many games does the team need to win in total? 75% * 50 = 37.5, so we take the ceiling (“at least”) to get 38. How many games has the team already won? They won 70% of 40 games so far, which gives us 28 wins. This means that the team needs 38 – 28 or 10 more wins to get to their target. How many games are left? The question itself tells us 10 games are left. This means they need 10/10 or 100% win rate in the final set of games to reach their 75% win rate target, and the answer is D).
Probability and Conditional Probability
Probability questions can seem straightforward on the surface, but answers here can be counterintuitive at times. We will go over the step-by-step process with a few examples, and then cover a common brain teaser / trick question. Review the content section for probability and make sure you understand AND, OR, and conditional rules.
Example 1:
In a deck of 52 playing cards, there are 4 suits (hearts, diamonds, clubs, and spades) and each suit has 13 cards (from 2 to Ace). If a card is drawn at random from the deck, what is the probability of drawing a spade or an ace?
A) 4/13
B) 1/13
C) 4/52
D) 17/52
It is important to break probability questions down into their components. Before doing any calculations, we should think about how these probabilities relate to one another. Are they disjoint? NO, because you could draw an Ace of Spades – they are not mutually exclusive. Are they independent? YES, because the probability of drawing an Ace is the same whether you are only looking at spades, or any other suit, or the entire deck – and vice versa. The fact that we have drawn a particular suit or a particular number does not change the probability of the other variable. Since the variables are not disjoint, we need to use the general form of the union formula (we can’t just add the probabilities and call it a day).
The central question here is P(Ace OR Spade), which we know from the general union formula is equal to P(Ace)+P(Spade)-P(Ace AND Spade). P(Ace) is 1/13 since each suit has 13 cards. P(Spade) is 1/4 since there are four suits. However we can’t stop there. We need to subtract the probability of both, because Ace and Spades are not mutually exclusive – we could draw a card that is both, and we don’t want to double count that possibility. So we must add 1/13 and 1/4 and then subtract the probability of Ace AND Spades. Now the full formula for AND is P(Ace AND Spades) = P(Ace)*P(Spades|Ace). However, these variables are independent, so we can simplify this to just the product P(Ace)*P(Spades), or 1/13 * 1/4. Thus we get (1/13)+(1/4)-(1/52) = (4/52)+(13/52)-(1/52) = 16/52 which simplifies to 4/13, or Answer A).
Example 2:
A jar contains 8 lime-flavored candies, 10 orange candies, and 6 grape candies. If John draws three candies from a jar without replacement, what is the probability that the first two are lime and the final candy is grape?
A) 1/8
B) 1/36
C) 7/253
D) 11/12
The key phrase to notice in this question is “without replacement.” The trap answer here is to simply AND the probabilities of getting three different candies based on the original subset; this would give us answer B and would be wrong. Since the candies are chosen without replacement, this means that every time John picks a candy, he changes the composition of the jar and thus changes the probability distribution for future picks. This is a three-step AND formula, but the choices are NOT independent, since the result of each pick influences the probabilities of future picks. Thus we cannot simplify the formula to P(Lime)*P(lime)*P(grape). We instead must use conditional probabilities and figure out how the conditions change, for the general AND formula of P(Lime)*P(Lime|Lime)*P(Grape|Lime AND Lime). Let’s break this down into steps.
The probability of getting a lime candy on his first draw is simple – 8 lime candies divided by 24 total candies in the jar implies 8/24 or 1/3rd chance. However, the probability of drawing a lime candy on the 2nd draw is NOT 1/3, because the distribution of the jar has changed (so P(Lime|Lime) is NOT equal to P(Lime)). What does the jar look like on the second draw? Now there are 7 lime candies in the jar rather than 8, and the total is 23 rather than 24. This means that P(Lime|Lime) is 7/23 rather than 1/3. We do this again for the third draw, P(Grape|Lime AND Lime). Now there are only 6 Lime candies in the jar and 22 total candies. The probability of drawing a grape candy out of this jar is now 6/22 rather than 6/24. This simplifies to 3/11.
Once we have the components, we multiply them together to get the answer: 1/3 * 7/23 * 3/11 = 7/253, or Answer C).
Example 3:
If Angela rolls a fair six-sided dice twice, what is the probability that the product of the dice rolls is odd?
A) 1/8
B) 1/4
C) 1/2
D) 3/4
We first need to figure out what situations lead to an odd product. If an even number is multiplied by an odd number, the product is even (see 2*3 = 6). If an even number is multiplied by an even number, the product is even (see 2*2 = 4). This is because even numbers are defined by having 2 among their factors, and if you multiply them by anything, 2 will necessarily be a factor of the product. This means that the only way to get an odd product is to multiple an odd number by an odd number. So what the question is actually asking is what is the probability of getting an odd number on both dice rolls? Well there are 3 even and 3 odd numbers on the die, so the chance P(Odd) of getting an odd number on any single roll is ½. Now we need an AND formula, but in this case the events are independent (rolling an odd number on the first roll doesn’t impact the probability of rolling an odd number on the second roll), so we can simplify the AND formula to a simple product P(Odd)*P(Odd) = (1/2)*(1/2) = ¼. This gives us Answer B.
Example 3:
A game show contestant Paul is shown three doors. Every door except one has a trash can behind it, except for one, which has a cut diamond. Whichever door Paul picks, he can keep the item behind it. Paul chooses Door 1 at first. The game show host opens one of the remaining doors at random, and reveals that it leads to a trash can. He then gives Paul the choice to either reaffirm his Door 1 choice, or switch and pick the remaining unopened door. What is the probability that Paul wins a diamond if he switches doors?
A) 1/3
B) 1/2
C) 2/3
D) 1
This question is a formulation of the famous brain teaser Monty Hall problem. It is fairly difficult and somewhat unlikely you will see it on the SAT, but is worth going through in case it shows up, since it is difficult to get right without some familiarity. It is also fairly likely to show up in job interviews in certain industries, like consulting.
There are a few trap answers here. For notation, let’s call the host-opened door Door 3, and Door 2 the unopened and unpicked one. The first mistake is thinking that the act of opening up Door 3 shouldn’t have changed anything about the probabilities of each door, and so Door 2 should still be 1/3 chance as in the beginning. This is clearly false, as obviously revealing more information does change the conditional probabilities. If the host had opened two doors and revealed two trash cans, for example, the probability of a diamond sitting behind the final door would suddenly be 100%.
The second mistake is thinking that since Door 3 was opened, the possibilities for Doors 1 and 2 are 1 trash can and 1 diamond, and so the likelihood of getting a diamond behind Door 2 is ½ or 50%. This is, quite unintuitively, false. The reason this is false is because the host’s choice of door to open is not random and is not independent vs the contestant’s choice, since the host will only open a door the contestant has not chosen. We can think of this problem as two questions, one within the other. The first is Is the Diamond behind Door 1 or not?. The baseline probability of the diamond being behind Door 1 is 1/3rd, and the probability of it not being behind Door 1 is 2/3rds. The second question is, if the diamond is NOT behind Door 1, what is the probability of it being behind Door 2? This is ½ at baseline, since the diamond could equally be behind Door 2 or Door 3. We can nest these (multiply probability of diamond NOT behind door 1 by probability that it is behind door 2 rather than 3) to get 2/3 * ½ = 1/3rd baseline probability of the diamond behind Door 2. Now what happens when Door 3 is revealed? The baseline probability of the diamond NOT being behind door 1 is unchanged – it is still 2/3rds. However, the conditional probability of the diamond being behind Door 2 is now 100%, given it is not behind Door 1, since we know that Door 3 covers a trash can. Essentially, the entire 2/3rds probability of the diamond not being behind Door 1 has now gotten concentrated behind Door 2. Thus the probability of the diamond being behind door 2 is 2/3, or answer C.
Ratios, rates, proportional relationships, and units
Basic problem types in this section will simply test your ability to construct ratio questions out of word problems. The easier questions here will not be particularly mathematically intensive, but may be tricky if you read through the problem too quickly, so take the time to make sure you have the right setup here.
Example 1:
A recipe for soup requires 1 cup of chicken stock per serving. If stock is sold at $5 per half-gallon container, how much money does the chef need to spend on chicken stock to make soup for a party of 10 (assuming one serving per person)? (Note: 1 gallon is approximately equal to 16 cups.)
A) $2
B) $5
C) $6
D) $10
The core math operations here are simple arithmetic, so the main challenge is setting up the problem properly and not getting confused with unit conversions. First, we figure out how many cups of chicken stock the chef needs. If the recipe calls for 1 cup of stock per person, and the party has 10 people, then the chef needs 10 cups of stock. Then, we have to remember to convert this into half-gallons. Note that the stock is sold in half-gallon containers and the conversion rate given is for full gallons. So while 1 gallon = 16 cups, a half-gallon is only 8 cups. To find the number of half-gallons, we have to divide the 10 cups we found earlier by 8, not 16. This gives us 1.25 half-gallon containers. Now if the stock was sold in bulk we would simply need to multiply 1.25 by the $5 rate to get $6.25. However, the stock is sold in discrete half-gallon containers; the chef can’t simply buy a quarter of a container. This means that to get enough stock for the party, he or she needs to buy 2 containers, not 1.25 (round up). Multiplying 2 by 5 gets us $10 as the answer D). Typically the SAT will not be so cruel as to include $6.25 as an answer here, since that would come down to a very specific reading of the question. But $6 is included as an answer; before selecting this based on an approximation, we need to look back and think about whether 1 container would actually be enough for the party, and it would not.
Example 2:
A recipe for making trail mix calls for blending nuts and granola in a 6:4 ratio. If a hiker wants to prepare 8 pounds of this mix, how many ounces of granola are needed (16 ounces make up 1 pound)?
A) 3.2
B) 5.3
C) 51.2
D) 85.3
Explicit ratio questions will also be common on the SAT. Any ratio of the form a:b implies that for (a+b) quantity of the mixture, you need a amount of substance 1 and b amount of substance 2. In this case, for every 10 pounds of trail mix, the hiker needs 6 pounds of nuts and 4 pounds of granola. Since the hiker needs 8 pounds of trail mix, we multiply 8 by (4/10) to get 3.2 pounds of granola. This is the trap – remember that the question asks for the ounces of granola needed. So we can’t stop at 3.2 pounds – we need to convert this to ounces by multiplying 3.2 by 16 oz/lb. This gives us 51.2 ounces, which is the answer C).
When dealing with dimensional changes (from side lengths in meters to volumes in cubic meters, for example), remember to convert to the right dimension before making unit conversions. See below.
Example 3:
A rectangular swimming pool is 35 meters long and 10 meters wide. If the depth of the pool is 5 meters and water weighs approximately 1,000 kilograms per cubic meter, what is the total weight of water, in kilograms, needed to fill the swimming pool completely?
A) 250,000 kilograms
B) 500,000 kilograms
C) 750,000 kilograms
D) 1,750,000 kilograms
In this case, the conversion is in kilograms per cubic meter, so we need to find the volume of the pool in cubic meters before converting. It would be wrong, for example, to multiple each of the values 35, 10, and 5 by 1000 kg/m3. So we start by finding the volume of the pool as 5 * 35 * 10 to get 1,750 cubic meters. Then we convert to kg by multiplying by 1000 to get 1,750,000 kg (answer D).
Let’s try a similar question but with the opposite setup below.
Example 4:
A rectangular swimming pool is 35 meters long and 10 meters wide. If the depth of the pool is 5 meters and there are approximately 3.28 feet in a meter, what is the total volume of the pool in cubic feet, rounded to the nearest whole number?
A) 5,740 cubic feet
B) 61,753 cubic feet
C) 100,000 cubic feet
D) 129,544 cubic feet
Here, the conversion is from feet to meters, so we need to convert the units before multiplying and making them cubic. 35 * 3.28 = 114.8 feet for the length of the pool. 10 * 3.28 = 32.8 feet for the width of the pool. 5*3.28 = 16.4 meters for the depth of the pool. We then multiply these values to get 61,753 cubic feet for the volume (answer B).
The other way to approach these questions is to change the conversion ratio to match the dimension needed. In other words, if there are 3.28 feet per meter, then there are 3.283 cubic feet per cubic meter, or 35.29 cubic feet per cubic meter. Then we can simply find the volume of the pool and convert using the new ratio, so that 1,750 cubic meters * 35.29 cubic feet per cubic meter gives us 61,753. The key point is to make sure the conversion ratio you are using matches the units you are working with, including the dimensions of those units.
Some of the trickier questions in this section involve rates and will look like the example below.
Example 5:
Jack is traveling from City A to City B, a distance of 360 miles. For the first 180 miles of the trip, Jack takes a bus and travels at an average speed of 40 miles per hour. For the second half of the trip, Jack rents a car and travels at an average speed of 80 miles per hour. What is Jack’s average speed for the entire trip, rounded to the nearest whole number, in miles per hour?
A) 45
B) 53
C) 60
D) 80
Whenever you see rate questions like this, try to break down the travel times and distances separately, then add them up, and recalculate the rate. If we simply looked at the fact that Jack is going 40 mph for half of the distance, and 80 mph for the other half of the distance, we might have averaged the two to get 60 mph. This is incorrect. This is because while the distances for the two speeds are equal, the fact that Jack is going slower at first means he spends more time at the slower speed (specifically twice as much time), and so the distance spent on the bus has twice the weight. The easiest way to deal with these questions without getting confused is just breaking up the two dimensions (distance / time) and calculating separately, and then combining them at the very end. So if Jack goes 40 mph for 180 miles, that means he spends 180/40 = 4.5 hours in a bus. Similarly, he spends 180/80 = 2.25 hours in his car. This means that in total, he has traveled 360 miles over the course of 4.5+2.25 = 6.75 hours. We combine these at the end to get an average speed of 360/6.75 = 53 miles per hour, rounded to the nearest whole number.
Two-Variable Data: Models and Scatterplots
This question sub-type will test your ability to read models and scatterplots and draw appropriate inferences. Review the content section on common graphs; most of the questions should be fairly straightforward inferences from there. Seethe example below:
Example:
The scatterplot above shows hypothetical data for relative likelihood of cancer risk in basis points on the y-axis plotted against pack-years of cigarettes smoked on the x-axis. Based on data and the line of best fit, how many pack-years must an individual have smoked for their cancer risk to double from baseline?
A) 5
B) 7
C) 10
D) 15
First, we have to figure out what the baseline risk is. This is the y-intercept, as it shows the risk for an individual with 0 pack-years smoked, and seems to be ~7.5basis points in the hypothetical data. To double this, the risk would need to be 15 bp. We follow the line of best fit upwards to see that it crosses y = 15 roughly where x = ~7. This gives us the answer B). In charts like this, the actual number may not be immediately obvious from the gridlines, but we can narrow it down by eliminating the numbers on either side. Both 5 and 10 years are clearly delineated, and drive ~12.5 and 20 bp of risk, so the answer should bein between, and that leaves only B).
GEOMETRY AND TRIGONOMETRY
This section will require a lot more math knowledge / formula memorization than the prior two. I would suggest reviewing the Content section again if you feel the need to refresh. Strategically, a lot of the difficulty here lies in setting up the problem and figuring out which equations to use to tackle it in the most time-efficient manner. We will go through specific problem types in each subsection so you have some basic idea of strategies that are available.
Area and Volume
Review the area and volume formulae in the Content section. While some of the shapes/solids described are a bit less intuitive / more complicated than others (e.g. cones, pyramids), these are also less likely to come up on the actual exam. More commonly, you will see compound problems that require you to calculate the area of multiple, simpler shapes, and then subtract or add these values. If you have the ability to memorize formulas for the more complex shapes I would still do so, but if not I would focus on the tooltips explaining how those formulas are derived, so you can construct them on the test if necessary even if memory fails.
Example 1: Compound Area Question
A rectangular swimming pool is 18 meters long and 12 meters wide. To prepare for a party, the owner decides to inflate a large circular raft that will cover an area equal to one-fourth of the pool's surface area. What is the radius of the circular raft, in meters?
A) 3π/2 meters
B) 3 meters
C) 3√(6π) meters
D) 3√(6/π) meters
This is a very typical area/volume question; the underlying formulas are for simple shapes, but the question requires a number of steps beyond just plugging them in. To begin with, we calculate the area of the rectangular swimming pool. For a rectangle, A = lw =18*12 = 216. Now the area of the raft needs to be 1/4th the area of the swimming pool, so we divide 216 by 4 to get 54. Then we need to use the Area of Circle formula A = πr2, set A = 54, and solve backwards for r. We divide by π to get r2 = 54/π, and then take the square root to get r = √(54/π). Now we try to see if 54 can be factored to simplify the expression. We divide by 2 to get 54 = 27*2. 27 is a common cube of 3, so we can factor further to get 54 = 9*3*2. Since 9 is a perfect square, √(54/π)simplifies to 3√(6/π), which is Answer D.
Example 2: Borders
A circular pool has a radius of 2 meters. The designers decide to surround the tub with a section of rough porous stone, creating an annular (ring-shaped) region that extends 1 meter outward from the pool’s edge. What is the total area of the porous strip surrounding the pool, in meters?
A) π
B) 4π
C) 5π
D) 9π
The question asks us for the area of the strip, and the strip is described as a ring. Any kind of “border” question is at its core a problem of subtracting areas; you need to subtract the area of the inner object from the outer shape. In this case, the “ring” is the difference between two circles. One circle has radius of 3 = 2+1 = the radius of the pool plus the additional meter taken up by the porous rock. The inner circle is simply the pool itself, with radius 2. So the area of the ring is π*(3)2 + π*(2)2 = (9-4)π = 5π. This gives us Answer C).
Example 3: Volume
A cylindrical water tank has a height of 20meters and a diameter of 3 meters. It is filled with water to a height of 12meters. A spherical ball with a diameter of 2 meters is fully submerged in the tank. What is the new height of the water level in the tank after the ball is submerged, to the nearest tenth of a meter?
A) 7.5 meters
B) 9.5 meters
C) 10.3 meters
D) 12.6 meters
This is a particularly interesting problem. After the ball is submerged, the water will be displaced so that the volume of the cylinder formed by the water + the ball is equal to the initial volume of water + the volume of the ball. First, we need to figure out the initial volume of water and the initial volume of the ball. The water is initially in a cylindrical shape with diameter 3 meters (so radius 3/2 meters)and height of 12 meters. The volume of a cylinder is πr2h; if you forget this formula, you can back-calculate it by simply multiplying the area of the base (in this case a circle) by the height. Thus the volume of the water, initially, is π*(3/2)2*12 = π*(9/4)*12 = 27 π. We then need to figure out the volume of the sphere. This is given by (4/3) πr3. Since the diameter of the sphere is 2meters, its radius is 1m, and the volume is (4/3)π. Now we know that the new cylinder former by the water + the sphere should have a volume of 27 π + (4/3)π, or (85/3)π combined. We plug this into the cylinder volume formula, with the same radius as before, to get πr2h = π(3/2)2h = (85/3)π.Simplify to get 9h/4 = 85/3, and simplify further to get h = 340/27 or 12.6meters.
As is evident from the examples above, knowing the formulae for area/volume of shapes/solids is just the basic requirement for answering these questions; the real crux of many of the questions lies in problem solving and figuring out how to model the word problem and arrive at the right answer. This is what makes it particularly important that you can remember (or derive) the relevant formulas quickly; it will give you more time to think through the actual logic of the problem.
Circles
For this section, review the formulas for the area/volume of a circle as well as the chord formulas and concepts. This section may also include a bit of trigonometry from the prior subsection.
Example 1:
Father James’ Pizza serves a specialty slice that is 3 feet in diameter. The pizza is cut into equal slices such that the angle at the tip of each slice is 20 degrees. How many square inches of pizza are in each slice?
A. 3π
B. 9π
C. 18π
D. 27π
This question is fundamentally asking us to find the area of a sector of a circle. Remember that the area of a sector is just the area of the circle times the fraction of the circle the sector covers, given by the interior angle over 360. A = (θ/360)*πr2. In this case, we know that the diameter is 3 feet, or 36 inches. That means the radius is 18 inches. We also know that the interior angle is 20 degrees. So the area of the sector becomes A = (20/360)*(182)*π = 18 π square inches.
Example 2:
A line intersects a circle at points A and B. If the diameter of the circle is 6 units and the line is a chord of the circle that is 2 units away from the center of the circle, what is the length of the chord AB?
A. 2π
B. 4√2
C. 4
D. 2√5
This is a basic test of the chord distance formula. We are given the diameter as 6 units, which implies the radius is 3. The distance from the chord to the center of the circle is 2 units. This is everything we need to fill in the formula for chord length 2√(r2-d2) = 2√(32-22) = 2√(9-4) = 2√5, or Answer D).
Example 3:
Two Chords AB and CD of Circle O intersect at point P and are perpendicular to each other. If AP is 6 units, PB is 4 units, and, CP is 8 units, what is the distance between point B and point D?
A. 2√3
B. 4
C. 5
D. 5√2
This is a compound question that will use some of our triangle knowledge from before. The first portion we need to solve for is the missing chord piece PD. The intersecting chords theorem tells us that the product of each section of one chord must equal the product of each section of the other. This means that 4*6 = 8*PD, so PD = 24/8 = 3. Now the question asks for the distance between BD, but note that BD forms a triangle with PB and PD. Additionally, the question tells us that the chords are perpendicular to each other, which means that the angle BPD must be 90 degrees. We are left, then, with a right triangle, where one side is PB = 4 units, and the other side is PD = 3 units. BD forms the hypotenuse of this triangle, so Pythagorean theorem gives us the answer. BD = √(32+42) = √(9+16) = √25 = 5, or Answer C).
Lines, Angles, and Triangles
You should review the content for this section, and also the basic right-triangle forms like 30-60-90 and 45-45-90. This section is less formula intensive than the prior, but question setup can be extensive, and I suggest drawing out each problem in detail to keep track of relevant information.
Example 1: Triangle Rules
A triangle ABC has three angles. If Angle A is 4 times Angle B, and Angle B is 1/3rd Angle C, what is the value of Angle A in degrees?
A. 23
B. 45
C. 68
D. 90
Remember that the sum of the angles of a triangle must add up to 180 degrees. This means that A + B + C = 180. The problem then becomes simple substitution within a set of linear equations. We could try substituting into A, but this introduces fractions earlier than necessary, since B is just 1/4th of A. Let’s substitute into B first as this appears to be the smallest angle. A = 4B, and C = 3B, from the text, so 4B + B + 3B = 180. This means that 8B = 180 and B = 45/2. Since A = 4B, angle A = (45/2)*(4) or 90 degrees. This gives us Answer D).
Example 2: Angle Rules
Lines AB and CD cross at point E. If the Angle AEC formed at the intersection is 60 degrees, what is the value of the adjacent angle AED in degrees?
A. 60
B. 120
C. 300
D. 360
In an intersection of lines, the adjacent angles should add up to 180 degrees, since they would add to span one entire side of the line. Opposite angles should be congruent, and the entire set of four (or more) angles should add to 360 degrees. In this case, the question asks us for the adjacent angle to AEC, which means that AED + AEC should add to 180. 180 – 60 = 120, which leaves us with Answer B.
Example 3: Polygon angles
A rhombus ABCD has angle A of 60 degrees. What is the value of adjacent angle B?
A. 60
B. 120
C. 300
D. 360
Since a rhombus is a special form of parallelogram, or four sided polygon, the sum of the interior angles should add to (4-2)*180 = 360 degrees. However, we also know that the opposite angles (and opposite sides) of a parallelogram are congruent. So in reality, we can think of this as two pairs of angles that each add up to 180 when combined with their adjacent angle. 180 – 60 = 120, which leaves us with Answer B.
Example 4: Similar Triangles
The sun beams over a 1ft sundial standing next to a 6ft tall statue. The sundial casts a shadow of 6 inches on the ground. How long is the shadow of the statue, in feet?
A. 1
B. 3
C. 6
D. 12
The rays of the sun hitting both the sundial and the statue are almost parallel (because the source is so far away), and the ground underneath both objects is on the same plane. This means that the angle between the sun’s rays and each object should be equivalent, as should the angle between the sun’s rays and the ground. This means that the triangles formed by the object, shadow, and ray in each situation should have two congruent angles. Angle-Angle congruence tells us that the two triangles, then, are similar. Similar triangles should have proportional side lengths. This means that if the statue is 6x the height of the sundial, the shadow cast by the statue should be 6x the length of the shadow cast by the sundial. The sundial shadow is 6 inches, or 0.5 feet, so the shadow of the statue should be 6*(0.5) or 3 ft. long.
Right Triangles and Trigonometry
You should review this section in Content Review before proceeding if not 100% confident in the material, as this section of the exam is content-dense. I recommend focusing on a few basic points, which are recapped below:
Right triangle basic rules:
- Pythagorean Theorem: a2+b2= c2 where c is the length of the hypotenuse
- 45-45-90 triangle (side lengths s, s, and s√2)
- 30-60-90 triangle (side lengths s, s√3,and 2s)
Trigonometry basics:
- SOHCAHTOA to remember trigonometric functions in the context of right triangle
- Sin = opposite/hypotenuse
- Cos = adjacent/hypotenuse
- Tan = opposite/adjacent
- Law of Sines: a/sinα = b/sinβ =c/sinγ
- This will be useful if you are given two sides and an angle and need to figure out the remaining opposing angle, or given two angles and a side and need to figure out the remaining opposing side.
- Law of Cosines: c = √(a2+b2-2abcos(γ))
- Use this to find the final side of a triangle given two sides their included angle, or triangulate the angles of a triangle given only the sides.
- Finally, whenever you are using your calculator for trigonometric calculations, remember to check whether you are calculating in radians or degrees.
Example 1:
Circle C is intersected tangentially by line segment AB at its midpoint P. If the radius of the circle is 2 units and the distance from the center of the Circle C and point A is 4 units, what is the length of AB?
A) 2 meters
B) 2√3 meters
C) 4 meters
D) 4√3 meters
I strongly recommend drawing out this diagram in your notes. Even in SAT questions where the diagram is included, it’s important to have a paper copy to be able to notate and adjust as needed, and it helps keep track of all of the information. A diagram for this question would look something like this:
Not all of this information was explicit in the question, of course. We could have drawn the radius in any direction. But it makes sense to draw it between C and P since those are established points in the diagram, and drawing the radius this way creates a triangle. Creating a triangle allows us to call upon our arsenal of triangle and trigonometry rules to better understand the situation. The question also does not specify that this triangle would be Right, but we know that tangent lines to a circle are perpendicular to the radius at the point of intersection. So not only have we converted this mess of points and lines into a triangle we can work with, but it’s a right triangle, which gives us even more existing math to utilize to solve the problem.
Now you could use the Pythagorean theorem from here to calculate AP and then double this value to get AB since P is the midpoint. However, this is where memorizing the 45-45-90 and 30-60-90 triangles becomes useful. In a 30-60-90 triangle the sides are s and s√3 and the hypotenuse is 2s. We notice in triangle ACP that the hypotenuse of 4 is twice the value of one of the sides. This is our “s” and “2s” – it should remind you immediately of a 30-60-90 triangle. This tells us that AP is 2√3 (substituting 2 in for “s” in the 30-60-90 triangle formula) without having to go through the entire Pythagorean theorem. Pattern recognition with these two triangle types will save you substantial time on the SAT.
Example 2:
A ladder is leaning against a wall, forming a 70 degree angle with the ground. If the top of the ladder touches the roof of a building that is 20 feet high off the ground, how long is the ladder?
A) 19.88 feet
B) 20.00 feet
C) 21.00 feet
D) 21.28 feet
The ladder, ground, and wall of the building form a right triangle. Let’s call this triangle ABC, where C is the right angle formed between the wall and the ground. We know that side a (the side connecting points BC) is 20 feet long, and that angle A is 70 degrees. Now we have an angle and the length of its opposite side, and we are trying to find the length of the ladder (which forms the hypotenuse of the triangle). We know from SOHCAHTOA that sin(A) = opposite/hypotenuse, and so the length of the hypotenuse should be the length of the opposite side divided by sin A. This means that the length of the ladder should be 20/sin(70) feet. Using our calculator, this gives 21.28 ft.
What if the question told us that the wall was actually slanted, and the angle between the wall and the ground was actually 100 degrees? SOHCAHTOA only applies to right triangles, so it is not easily usable here. This is where the law of sines and law of cosines can be useful. Now we have the length of side a and Angle A, as well as the value of Angle C (100 degrees). Using the law of sines, we know that a/sinA = c/sinC, so 20/sin70 = c/sin100. We simplify to c = 20*(sin100/sin70) and calculate to get 20.96 feet.
Example 3:
Centreville is 100 miles away from Albany and 120 miles away from Buffalo. If the roads connecting Centreville to Albany and Centreville to Buffalo form a 50 degree angle, how far is Albany from Buffalo, to the nearest mile?
A) 90 miles
B) 95 miles
C) 100 miles
D) 110 miles
First, we try to diagram the problem to get a better sense of what we’re working with. We have three points (A, B, C for Albany, Buffalo, and Centreville). We have the length of line segment AC and line segment BC and the value of the included angle ACB.
This should immediately clue us in that the Law of Cosines should be useful. Remember that the law of cosines says that c = √(a2+b2-2abcos(γ)). In this case, we are looking for the value of c, and have a = 100, b = 120, and γ = 50 degrees. Plugging this into our calculator gives a value for c = 94.7 miles, which rounds to 95 miles, or answer B.