# SAT / ACT Prep Online Guides and Tips

Want to test yourself against the most difficult SAT math questions? Want to know what makes these questions so difficult and how best to solve them? If you’re ready to really sink your teeth into the SAT math section and have your sights set on that perfect score, then this is the guide for you.

We’ve put together what we believe to be the 13 most difficult questions for the new 2016 SAT, with strategies and answer explanations for each. These are all hard SAT math questions from College Board SAT practice tests, which means understanding them is one of the best ways to study for those of you aiming for perfection.

Image: Sonia Sevilla/Wikimedia

## Brief Overview of the SAT Math Section

The SAT math section is broken into two subsections (always the third and fourth sections of the test). The first math subsection (labeled as “3”) does NOT allow you to use a calculator; the second math subsection (labeled as “4”) does allow the use of a calculator. Don’t fear: if it doesn’t allow for a calculator, it means you don’t need a calculator to answer the question.

Each math subsection is arranged in order of ascending difficulty, so on each subsection, question 1 will be “easy” and question 15 will be considered “difficult.” However, the ascending difficulty resets from easy to hard on the grid-ins. Hence, the multiple choice will be arranged in increasing difficulty (questions 1 and 2 being the easiest, questions 14 and 15 being the hardest) and then the difficulty level resets for the grid-in section (meaning questions 16 and 17 will again be “easy” and questions 19 and 20 will be very difficult).

The way the SAT classifies “difficulty” is by how long it takes an average student to solve a problem, as well as the percentage of students who answer the question correctly. As you might expect, the longer it takes to solve a problem and the fewer people who answer it correctly, the more difficult the problem.

So, with very few exceptions, the most difficult SAT math problems will be clustered in the far end of the multiple choice segments or the later half of the grid-in questions. Other than their placement on the test, these questions share a few other commonalities. Let us look at what these types of questions have in common, then look at example questions and how to solve them.

## But First, Should You Be Focusing on the Hardest Math Questions Right Now?

If you’re just getting started in your study prep (or if you’ve simply skipped this first, crucial step), definitely stop and take a full practice test to gauge your current scoring level. Check out our guide to all the free SAT practice tests available online and then sit down to take a test all at once. The absolute best way to assess your current level is to simply take the SAT practice test as if it were real, keeping strict timing and working straight through with only the allowed breaks (we know--probably not your favorite way to spend a Saturday).

Once you’ve got a good idea of your current level and percentile ranking, you can set milestones and goals for your ultimate score. If you’re currently scoring in the 200-400 range on the Math section or the 400-600 range, your best bet is first to check out our guides to improving your math score to where you want it to be before you start in trying to tackle the most difficult math problems on the test.

If, however, you're already scoring above a 600 on the Math section and want to test your mettle for the real SAT, then definitely proceed to the rest of this guide. If you’re aiming for perfect (or close to), then you’ll need to know what the most difficult SAT math questions look like and how to solve them. And luckily, that’s exactly what we’ll do.

WARNING: since there are a limited number of official SAT practice tests, you may want to wait to read this article until you've attempted all or most of the first four official practice tests (since the questions below were taken from those tests). If you're worried about spoiling those tests, stop reading this guide now; come back and read it when you've completed them.

Now let's get to our list of questions (whoo)!
Image: Niytx/DeviantArt

## The 13 Hardest SAT Math Questions

Now that you’re sure you should be attempting these questions, let’s dive right in!

### No Calculator Questions

#### Question 1

ANSWER EXPLANATION: If you think of the equation as an equation for a line

\$\$y=mx+b\$\$

where

\$\$C= {5}/{9} (F−32)\$\$

or

\$\$C={5}/{9}F −{5}/{9}(32)\$\$

you can see the slope of the graph is \${5}/{9}\$, which means that for an increase of 1 degree Fahrenheit, the increase is \${5}/{9}\$ of 1 degree Celsius. Therefore, statement I is true. This is the equivalent to saying that an increase of 1 degree Celsius is equal to an increase of \${9}/{5}\$ degrees Fahrenheit. Since \${9}/{5}\$ = 1.8, statement II is true.

On the other hand, statement III is not true, since a temperature increase of \${9}/{5}\$ degrees Fahrenheit, not \${5}/{9}\$ degree Fahrenheit, is equal to a temperature increase of 1 degree Celsius.

#### Question 2

ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by \$ax-2\$ (so you can get rid of the fraction). When you multiply each side by \$ax-2\$, you should have:

\$\$24x^2 + 25x - 47 = (-8x-3)(ax-2) - 53\$\$

You should then multiply \$(-8x-3)\$ and \$(ax-2)\$ using FOIL.

\$\$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x + 6 - 53\$\$

Then, reduce on the right side of the equation

\$\$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x - 47\$\$

Since the coefficients of the \$x^2\$-term have to be equal on both sides of the equation, \$−8a = 24\$, or \$a = −3\$.

The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and I do not recommend it for the actual SAT as it will waste too much time.

#### Question 3

ANSWER EXPLANATION: One approach is to express

\$\${8^x}/{2^y}\$\$

so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting \$2^3\$ for 8 in the numerator of \${8^x}/{2^y}\$ gives

\$\${(2^3)^x}/{2^y}\$\$

which can be rewritten

\$\${2^3x}/{2^y}\$\$

Since the numerator and denominator of have a common base, this expression can be rewritten as \$2^(3x−y)\$. In the question, it states that \$3x − y = 12\$, so one can substitute 12 for the exponent, \$3x − y\$, giving that the

\$\${8^x}/{2^y}= 2^12\$\$

#### Question 4

ANSWER EXPLANATION: To rewrite \${8-i}/{3-2i}\$ in the standard form \$a + bi\$, you need to multiply the numerator and denominator of \${8-i}/{3-2i}\$ by the conjugate, \$3 + 2i\$. This equals

\$\$({8-i}/{3-2i})({3+2i}/{3+2i})={24+16i-3+(-i)(2i)}/{(3^2)+(2i)^2}\$\$

Since \$i^2=-1\$, this last fraction can be reduced simplified to

\$\$ {24+16i-3i+2}/{9-(-4)}={26+13i}/{13}\$\$

which simplifies further to \$2 + i\$. Therefore, when\${8-i}/{3-2i}\$ is rewritten in the standard form a + bi, the value of a is 2.

#### Question 5

ANSWER EXPLANATION: Triangle ABC is a right triangle with its right angle at B. Therefore, \$\ov {AC}\$ is the hypotenuse of right triangle ABC, and \$\ov {AB}\$ and \$\ov {BC}\$ are the legs of right triangle ABC. According to the Pythagorean theorem,

\$\$AB =√{20^2-16^2}=√{400-256}=√{144}=12\$\$

Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of \$\angle ∠ {F}\$ equals the measure of \$\angle ∠ {C}\$. Therefore, \$sin F = sin C\$. From the side lengths of triangle ABC,

\$\$sinF ={opposite side}/{hypotenuse}={AB}/{AC}={12}/{20}={3}/{5}\$\$

Therefore, \$sinF ={3}/{5}\$.

The final answer is \${3}/{5}\$ or .6.

### Calculator Questions

#### Question 6

ANSWER EXPLANATION: In order to solve this problem, you should create two equations using two variables (\$x\$ and \$y\$) and the information you’re given. Let \$x\$ be the number of left-handed female students and let \$y\$ be the number of left-handed male students. Using the information given in the problem, the number of right-handed female students will be \$5x\$ and the number of right-handed male students will be \$9y\$. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be true:

\$\$x + y = 18\$\$

\$\$5x + 9y = 122\$\$

When you solve this system of equations, you get \$x = 10\$ and \$y = 8\$. Thus, 50 of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is \${50}/{122}\$, which to the nearest thousandth is 0.410.

#### Questions 7 & 8

Use the following information for both question 7 and question 8.

#### Question 7

ANSWER EXPLANATION: Since the question states that Little’s law can be applied to any single part of the store (for example, just the checkout line, then the average number of shoppers, \$N\$, in the checkout line at any time is \$N = rt\$, where \$r\$ is the number of shoppers entering the checkout line per minute and \$T\$ is the average number of minutes each shopper spends in the checkout line. Since 84 shoppers per hour make a purchase, 84 shoppers per hour enter the checkout line. However, this needs to be converted to the number of shoppers per minute (in order to be used with \$t = 5\$). Since there are 60 minutes in one hour, the rate is \${84  shoppers per hour}/{60  minutes} = 1.4 shoppers per minute\$. Using the given formula with \$r = 1.4\$ and \$t = 5\$ yields

\$\$N = rt = (1.4)(5) = 7\$\$

Therefore, the average number of shoppers, \$N\$, in the checkout line at any time during business hours is 7.

#### Question 8

ANSWER EXPLANATION: According to the original information given, the estimated average number of shoppers in the original store at any time is 45. In the question, it states that, in the new store, the manager estimates that an average of 90 shoppers per hour enter the store, which is equivalent to 1.5 shoppers per minute. The manager also estimates that each shopper stays in the store for an average of 12 minutes. Thus, by Little’s law, there are, on average, \$N = rt = (1.5)(12) = 18\$ shoppers in the new store at any time. This is

\$\${45-18}/{45} x 100 = 60 \$\$

percent less than the average number of shoppers in the original store at any time.

#### Question 9

ANSWER EXPLANATION: The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed (a cylinder and two cones). The silo is made up of a cylinder (with height 10 feet and base radius 5 feet) and two cones (each with height 5 ft and base radius 5 ft). The formulas given at the beginning of the SAT Math section:

Volume of a Cone

\$\$V={1}/{3}πr^2h\$\$

Volume of a Cylinder

\$\$V=πr^2h\$\$

can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by

\$\$V_{silo}=π(5^2)(10)+(2)({1}/{3})π(5^2)(5)=({4}/{3})(250)π\$\$

which is approximately equal to 1,047.2 cubic feet.

#### Question 10

ANSWER EXPLANATION: Since the average (arithmetic mean) of 2 numbers is equal to the sum of the 2 numbers divided by 2, the equations \$x={m+9}/{2}\$, \$y={2m+15}/{2}\$, \$z={3m+18}/{2}\$ are true. The average of \$x\$, \$y\$, and \$z\$ is given by \${x + y + z}/{3}\$. Substituting the expressions in m for each variable (\$x\$, \$y\$, \$z\$) gives

\$\$[{m+9}/{2}+{2m+15}/{2}+{3m+18}/{2}]/3\$\$

This fraction can be simplified to \$m + 7\$.

#### Question 11

ANSWER EXPLANATION: The equation \$f(x) = k\$ gives the solutions to the system of equations

\$\$y = f(x) = x^3-x^2-x-{11}/{4}\$\$

and

\$\$y = k\$\$

A real solution of a system of two equations corresponds to a point of intersection of the graphs of the two equations in the \$xy\$-plane. The graph of \$y = k\$ is a horizontal line that contains the point \$(0, k)\$, and the line with equation \$y = −3\$ is a horizontal line that intersects the graph of the cubic equation three times. Therefore, the equation \$f(x) = −3\$ has three real solutions.

#### Question 12

ANSWER EXPLANATION:  To solve this problem, you need to set up to equations with variables. Let \$q_1\$ be the dynamic pressure of the slower fluid moving with velocity \$v_1\$, and let \$q_2\$ be the dynamic pressure of the faster fluid moving with velocity \$v_2\$. Then

\$\$v_2 =1.5v_1\$\$

Given the equation \$q = {1}/{2}nv^2\$, substituting the dynamic pressure and velocity of the faster fluid gives \$q_2 = {1}/{2}n(v_2)^2\$. Since \$v_2 =1.5v_1\$, the expression \$1.5v_1\$ can be substituted for \$v_2\$ in this equation, giving \$q_2 = {1}/{2}n(1.5v_1)^2\$. By squaring \$1.5\$, you can rewrite the previous equation as

\$\$q_2 = (2.25)({1}/{2})n(v_1)^2 = (2.25)q_1\$\$

Therefore, the ratio of the dynamic pressure of the faster fluid is

\$\${q2}/{q1} = {2.25 q_1}/{q_1}= 2.25\$\$

The final answer is 2.25 or 9/4.

#### Question 13

ANSWER EXPLANATION: If the polynomial \$p(x)\$ is divided by \$x − 3\$, the result can be written as

\$\${p(x)}/{x-3}=q(x)+{r}/{x-3}\$\$

where \$q(x)\$ is a polynomial and \$r\$ is the remainder. Since \$x − 3\$ is a degree 1 polynomial, the remainder is a real number. Therefore, \$p(x)\$ can be rewritten as \$p(x) = (x − 3)q(x) + r\$, where \$r\$ is a real number. In the question it said that \$p(3) = −2\$ so it must be true that

\$\$−2 = p(3) = (3 − 3)q(3) + r = (0)q(3) + r = r\$\$

Therefore, the remainder when \$p(x)\$ is divided by \$x − 3\$ is \$−2\$.

You deserve all the naps after running through those questions.

## What Do the Hardest SAT Math Questions Have in Common?

It’s important to understand what makes these hard questions “hard.” By doing so, you’ll be able to both understand and solve similar questions when you see them on test day, as well as have a better strategy for identifying and correcting your previous SAT math errors.

In this section, we’ll look at what these questions have in common and give examples of each type. Some of the reasons why the hardest math questions are the hardest math questions is because they:

#### #1: Test Several Mathematical Concepts at Once

Here, we must deal with imaginary numbers and fractions all at once.

Secret to success: Think of what applicable math you could use to solve the problem, do one step at a time, try each technique until you find one that works!

#### #2: Involve a Lot of Steps

Remember: the more steps you need to take, the easier to mess up somewhere along the line!

We must solve this problem in steps (doing several averages) to unlock the rest of the answers in a domino effect. This can get confusing, especially if you're stressed or running out of time.

Secret to success: Take it slow, take it step by step, double check your work, so you don't make mistakes!

#### #3: Test Concepts That You Have Limited Familiarity With

For example, many students are less familiar with functions than they are with fractions and percentages, so most function questions are considered “high difficulty” problems.

If you don't know your way around functions, this would be a tricky problem.

Secret to success: Review math concepts that you don't have as much familiarity with such as functions. We have a lot of great SAT Math review guides.

#### #4: Are Worded in Unusual or Convoluted Ways

It can be difficult to figure out exactly what some questions are asking, much less figure out how to solve them. This is especially true when the question is located at the end of the section, and you are running out of time.

Because this question provides so much information without a diagram, it can be difficult to puzzle through in the limited time allowed.

Secret to success: Take your time, analyze what is being asked of you, draw a diagram if it's helpful to you.

#### #5: Use Many Different Variables

With so many different variables in play, it is quite easy to get confused.

Secret to success: Take your time, analyze what is being asked of you, consider if plugging in numbers is a good strategy to solve the problem (it wouldn't be for the question above, but would be for many other SAT variable questions).

## The Take-Aways

The SAT is a marathon and the better prepared you are for it, the better you'll feel on test day. Knowing how to handle the hardest questions the test can throw at you will make taking the real SAT seem a lot less daunting.

If you felt that these questions were easy, make sure not underestimate the effect of adrenaline and fatigue on your ability to solve problems. As you continue to study, always adhere to the proper timing guidelines and try to take full tests whenever possible. This is the best way to recreate the actual testing environment so that you can prepare for the real deal.

If you felt these questions were challenging, be sure to strengthen your math knowledge by checking out our individual math topic guides for the SAT. There, you'll see more detailed explanations of the topics in question as well as more detailed answer breakdowns.

## What’s Next?

Felt that these questions were harder than you were expecting? Take a look at all the topics covered in the SAT math section and then note which sections were particular difficulty for you. Next, take a gander at our individual math guides to help you shore up any of those weak areas.

Running out of time on the SAT math section? Our guide will help you beat the clock and maximize your score

Aiming for a perfect score? Check out our guide on how to get a perfect 800 on the SAT math section, written by a perfect-scorer.

Want to improve your SAT score by 160 points?

Check out our best-in-class online SAT prep program. We guarantee your money back if you don't improve your SAT score by 160 points or more.

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Courtney Montgomery

Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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