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500 New SAT Math Problems – Just 19.99 Today Only

June 30, 2024

$500 New SAT Math Problems$

500 New SAT Math Problems
Just 19.99 on Amazon

Hi everyone! The latest edition of 500 New SAT Math Problems is now available in paperback from Amazon. This edition just has been modified from the previous edition to account for the changes on the Digital SAT.

~~The paperback is now on sale on Amazon for only $19.99. Note that once the sale ends (by the end of today), the price of this book will go up to $42.99.~~

The promotion has ended. Thanks to everyone who participated. The book is now available at its regular price here: 500 New SAT Math Problems

If you have any questions, feel free to contact me at steve@SATPrepGet800.com

Thank you all for your continued support!

A Trick For Free Two Day Shipping

I would like to finish this post with a little trick you can use to get free 2 day shipping on any of the books you decide to purchase without making any additional purchases. If you have never used Amazon Prime you can sign up for a free month using the following link.

If you have already had a free trial of Amazon Prime you can simply open up a new Amazon account to get a new free trial. It just takes a few minutes! You will need to use a different email address than the one you usually use.

This next part is very important! After you finish your transaction, go to your Account, select “Manage my prime membership,” and turn off the recurring billing. This way in a month’s time Amazon will not start charging you for the service.

After shutting off the recurring billing you will still continue to receive the benefit of free 2 day shipping for one month. This means that as long as you use this new Amazon account for your purchases you can do all of your shopping on Amazon for the next month without having to worry about placing minimum orders to get free shipping.

Just be aware that certain products from outside sellers do not always qualify for free shipping, so please always check over your bill carefully before you check out.

Well I hope you decide to take advantage of this very special offer, or at the very least I hope you will benefit from my Amazon “free 2 day shipping trick.” Here is the link one more time:

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Thank you all for your continued support!

New SAT Verbal Prep Book for Reading and Writing Mastery

July 8, 2016

New SAT Verbal Prep Book

The New SAT Verbal Prep Book for Reading and Writing Mastery gives you the most effective tips, tricks and tactics from Get 800, a prep company of doctors dedicated to their students achieving their dream SAT scores.

Click the following link to get to the book’s Amazon page: New SAT Verbal Prep Book

If you would like to take a look at a free sample of this book, click the following link: New SAT Verbal Free Sample

Take a look at my product page to see all of my books: Get 800 Product Page

A Trick For Free Two Day Shipping

Just be aware that certain products from outside sellers do not always qualify for free shipping, so please always check over your bill carefully before you check out.

Well I hope you decide to take advantage of this very special offer, or at the very least I hope you will benefit from my Amazon “free 2 day shipping trick.” Here is the link one more time:

Please share this information with your Facebook friends:

The Features You Should Learn How to Use On Your Graphing Calculator

July 7, 2016

General Graphing Calculator Advice for the SAT/ACT

I always recommend that you use a TI-84 or comparable graphing calculator for the SAT or ACT.
It is important that you are comfortable with your calculator on test day, so make sure that you are consistently practicing with the calculator you plan to use.
Make sure that your calculator has fresh batteries the day of the test. Nobody will supply a calculator for you if yours dies.
You may have to switch between DEGREE and RADIAN modes during the test. If you are using a TI-84 (or equivalent) calculator press the MODE button and scroll down to the third line when necessary to switch between modes.

Graphing Calculator Features You Should Know for the SAT/ACT

Below are the most important things you should practice on your graphing calculator.

(1) Practice entering complicated computations in a single step, and know when to insert parentheses. In general, there are 4 instances when you should use parentheses in your calculator. Around numerators of fractions Around denominators of fractions Around exponents Whenever you actually see parentheses in the expression

Around numerators of fractions
Around denominators of fractions
Around exponents
Whenever you actually see parentheses in the expression

Examples: We will substitute a 5 in for x in each of the following examples.

TI-84 Graphing Calculator Examples

(2) Clear the screen before using it in a new problem. The big screen allows you to check over your computations easily.

(3) Press the ANS button (2^nd (-) ) to use your last answer in the next computation.

(4) Press 2^nd ENTER to bring up your last computation for editing. This is especially useful when you are plugging in answer choices, or guessing and checking.

(5) You can press 2^nd ENTER over and over again to cycle backwards through all the computations you have ever done.

(6) Know where the ,√ ,π and ^ buttons are so you can reach them quickly.

(7) Change a decimal to a fraction by pressing MATH ENTER ENTER.

(8) Press the MATH button – in the first menu that appears you can take cube roots and nth roots for any n.

(9) Know how to use the SIN, COS and TAN buttons as well as SIN^-1, COS^-1 and TAN^-1.

ACT only

The following features are necessary for the ACT only.

(10) Press the MATH button. Scroll right to NUM and you have lcm( and gcd(. Scroll right to PRB and you have nPr, nCr, and ! to compute permutations, combinations and factorials very quickly.

(11) Know where LOG button is so you can reach it quickly.

Advanced Features

The following items are less important but can be useful for both the SAT and ACT.

(12) Press the Y= button to enter a function, and then hit ZOOM 6 to graph it in a standard window.

(13) Practice using the WINDOW button to adjust the viewing window of your graph.

(14) Practice using the TRACE button to move along the graph and look at some of the points plotted.

(15) Pressing 2nd TRACE (which is really CALC) will bring up a menu of useful items. For example selecting ZERO will tell you where the graph hits the x-axis, or equivalently where the function is zero. Selecting MINIMUM or MAXIMUM can find the vertex of a parabola. Selecting INTERSECT will find the point of intersection of 2 graphs.

And now it’s time to start practicing lots of math problems. For this you may want to take a look at the Get 800 collection of test prep books. Click on the picture below for more information about these books.

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See you tomorrow!

Hard Geometry Problem with Solution for the SAT

July 3, 2016

Hard Geometry Problem with Solution

for the SAT

Yesterday I gave you a Level 4 Geometry problem for the revised SAT to try. Today I will provide a solution for this problem. If you have not yet attempted the problem go back and take a look at it first so you can try it on your own. Here is the link: Hard Geometry Problem for the Revised SAT

Level 4 – Geometry

Here is the problem once again followed by a solution:

The head of a copper “hexagon head screw bolt” (one cross section of which is shown above) has the shape of a cylinder with a hole shaped like a regular hexagon. The cylindrical head is 2 cm thick with a base diameter of 3 cm. The hexagonal hole is only half the thickness of the entire head, and each side of a hexagonal cross section has a length of 1 cm. Given that the density of copper is 8.96 grams per cubic cm, and density is mass divided by volume, find the mass of the head to the nearest gram.

Solution: We first compute the volume of the head. There are two parts to the volume.

The bottom half of the head is a cylinder with height 2/2 = 1 cm and base radius 3/2. It follows that the volume is

V = πr²h = π(3/2)²(1) = 9π/4 cm³.

The top half of the head consists of the same cylinder as the bottom half, but this time we have to subtract off the volume of a hexagonal prism. The regular hexagonal face can be divided into 6 equilateral triangles, each with area

A = s²√3/4 = 1²√3/4 = √3/4.

So the volume of the hexagonal prism is

V = Bh = (6√3/4)(1) = 3√3/2 cm³

and the volume of the top half of the head is

9π/4 – 3√3/2 cm³

It follows that the total volume of the head is

9π/4 + (9π/4 – 3√3/2) = 18π/4 – 3√3/2 = (9π – 3√3)/2 cm³.

Finally,

D = M/V → 8.96 = M / [(9π – 3√3)/2] → M = 8.96 ∙ (9π – 3√3)/2 ≈ 103.39 grams.

To the nearest gram, the answer is 103.

For an even more detailed solution to this problem check out New SAT Math Problems arranged by Topic and Difficulty Level.

Feel free to add your own solutions to the comments.

AP Calculus BC Geometric Series Problem

With Solution

Yesterday I gave an AP Calculus BC problems involving a geometric series. You can find that post here: AP Calculus BC Problem – Geometric Series.

Today I will provide a solution to that problem. First, here is the question one more time.

Level 2 Series

$g(x)=\Sigma_{n=1}^{\infty}(sin^2 x)^n$

If g is the function given above, then g(π/3) =

(A) 3
(B) 1
(C) 3/4
(D) 1/2

Solution:
$g(\frac{\pi}{3})$
$=\Sigma_{n=1}^{\infty}(sin^2(\frac{\pi}{3}))^n$
$=\Sigma_{n=1}^{\infty}((\frac{\sqrt{3}}{2})^2)^n$
$=\Sigma_{n=1}^{\infty}(\frac{3}{4})^n$
$=\frac{\frac{3}{4}}{1-\frac{3}{4}}$
$=\frac{3}{4}\div \frac{1}{4}$
$=\frac{3}{4}\cdot 4$
$=3$
Note:
$\Sigma_{n=1}^{\infty}(\frac{3}{4})^n$

is an infinite geometric series with first term a = 3/4 and common ratio r = 3/4. Watch the video above for more information on geometric series.

AP Calculus BC Problem- Geometric Series

AP Calculus BC problems involving series tend to be difficult for students. I thought I would start things off easy today with a level 2 problem involving geometric series. Those of you that have forgotten the basics of geometric series should find the video above helpful.

Let’s now try an AP Calculus BC series problem. AB students do not need to worry about this one. I will provide a full explanation for this problem tomorrow. In the meantime, feel free to leave your own solutions in the comments.

Level 2 Series

$g(x)=\Sigma_{n=1}^{\infty}(sin^2 x)^n$

If g is the function given above, then g(π/3) =

(A) 3
(B) 1
(C) 3/4
(D) 1/2

Remainder Problems – Part 3

Welcome to the third and final part of this thread on remainder problems. In the first part we discussed some calculator algorithms for finding remainders. You can review that article here: SAT Remainder Problems – Part 1

Yesterday we discussed the cyclical nature of remainders. You can review that article here: SAT Remainder Problems – Part 2

I decided to spend three posts on this subject because many of my students in a mid-level scoring range seem to have a difficult time with ACT and GRE remainder questions.

Solution To Last Week’s Remainder Problem

So where were we? I left you with an ACT math problem to solve using the information I provided regarding the cyclical nature of remainders. Let’s find out how you did.

The problem again:

What is the least positive integer greater than 4 that leaves a remainder of 4 when divided by both 6 and 8?

Solution: We first find the least positive integer greater than 4 that is divisible by both 6 and 8. This is the least common multiple of 6 and 8 which is 24.

We now simply add the remainder.

24 + 4 = 28.

Thus, the answer is 28.

If you don’t know how to find the least common multiple of two numbers, it’s okay. If you’re taking the ACT, you can just let your TI-84 calculator do it for you. In your calculator press MATH, scroll right to NUM, and press 8 for lcm. Then type 6,8) and you will get an output of 24. So lcm(6,8)=24.

If you are taking the GRE, or you just want to learn other ways to compute the lcm, see the following article: Greatest Common Divisor (GCD) And Least Common Multiple (LCM)

Remainders In Disguise

Test makers’ love to put their remainder problems in disguise. In fact, many remainder problems on standardized tests do not even have the word “remainder” in the question! So how can you be expected to know when to find a remainder? The answer is quite simple. If the problem mentions some kind of sequence that keeps repeating over and over, then the problem may be asking you to find a remainder.

Remember this and you will recognize a remainder problem every time. In fact, this is such a key point I am going to repeat this in bold for you!

If the problem mentions some kind of sequence that keeps repeating over and over, then the problem may be asking you to find a remainder.

Let’s take a look at a standard example of such a question:

Example

Cards numbered from 1 through 2013 are distributed, one at a time, into nine stacks. The card numbered 1 is placed on stack 1, card number 2 on stack 2, card number 3 on stack 3, and so on until each stack has one card. If this pattern is repeated, each time beginning with stack one, on which stack will the card numbered 2013 be placed?

A. 3^rd stack
B. 4^th stack
C. 5^th stack
D. 6^th stack
E. 7^th stack

Solution: We are actually being asked to find the remainder when 2013 is divided by 9. Let’s do this using the first calculator algorithm that I showed you.

Step 1: Perform the division in your calculator

2013/9 ~ 223.667

Step 2: Multiply the integer part of this answer by the divisor:

9*223 = 2007

Step 3: Subtract this result from the dividend to get the remainder:

2013 – 2007 = 6.

So the card numbered 2013 will be placed on the 6^th stack, choice D.

Notice that the word remainder is never mentioned in this problem. The giveaway that this is a remainder problem in disguise is at the beginning of the last sentence where it says “If this pattern is repeated…”

So this is the end of this three part thread on remainders.

Can We Solve Remainder Problems

Without Using Long Division?

For the next three days I would like to help you to understand math problems involving remainders. These types of problems appear on standardized tests such as the ACT, GRE, and SAT math subject tests.

ACT and GRE math problems with remainders seem to give students a difficult time. This is mostly because a fundamental step in solving the problem is often missed – performing long division.

You cannot solve a remainder problem by simply dividing in your calculator. There are, however, calculator algorithms that can give you the answer very quickly. We will talk about these a bit later in this post.

A common error that students make is to perform a division calculation on their calculator, and simply take the first number after the decimal point and use this digit as the answer to the problem. This will usually result in a wrong answer.

Example

Suppose we are asked “Find the remainder when 14 is divided by 4.”

In your calculator, 14/4 = 3.5.

Penciling the number 5 as your answer would be incorrect – seriously incorrect. This “5” is actually part of the answer to the question “What is 14 divided by 4?” But it has nothing to do with the remainder.

Students that are currently scoring in the mid-range in ACT/GRE math seem to have the biggest problem with remainder questions. Many cannot perform long division correctly, and some do not even realize that long division is needed. This keeps many students from getting their scores to the next level.

Let’s take a look at how we can solve the problem of finding the remainder when 14 is divided by 4 in three different ways. This problem may seem very basic for some of you, but let’s go over it anyway to make sure we have a strong foundation before solving more difficult remainder problems.

Method 1 – Long Division: So 4 goes into 14 three times with 2 left over. In other words, 14 = 4(3) + 2. So we see that the remainder when we divide 14 by 4 is 2.

Let’s break this down step by step:

$SAT math long division explanation$

Count the number of groups of 4 objects that can be formed from 14 objects. Place this number over the 4 that is inside the division symbol.

$SAT math long division explanation$

Multiply this resulting number 3 by the divisor 4. Place this number below the 4 that is inside the division symbol.

$SAT math long division explanation 2$

Subtract the resulting number 12 from the dividend 14.

$SAT math long division explanation 3$

The quotient 3 appears above the division symbol and the remainder 2 appears at the bottom.

Below is a visual representation of 14 divided by 4. In the figure below we are grouping 14 objects 4 at a time. Note that we wind up with 3 groups (the quotient) and 2 extra objects (the remainder).

$sat math long division remainder diagram$

Method 2 – First Calculator Algorithm:

Step 1: Perform the division in your calculator:

14/4 = 3.5

Step 2: Multiply the integer part of this answer by the divisor:

4*3 = 12

Step 3: Subtract the above result from the dividend to get the remainder:

14 – 12 = 2.

Method 3 – Second Calculator Algorithm:

Step 1: Perform the division in your calculator:

14/4 = 3.5

Step 2: Subtract off the integer part of this result:

ANS – 3 = .5

Step 3: Multiply this result by the divisor:

4*ANS = 2.

Note that these calculator algorithms work exactly the same no matter how large the numbers are that you are dividing.

Exercise

As an additional exercise you should try to find the remainder when 15,216 is divided by 73 by using one of these calculator algorithms. Figure out which of the two algorithms you prefer. You can find solutions here: SAT Remainder Problems – Part 2

AP Calculus AB Related Rates Problem with Solution

Yesterday I posted a Level 4 related rates problem for the AP Calculus AB exam (although this problem is a pretty standard Calculus problem that you may come across in any Calculus course). You can see that post here: AP Calculus AB Problem – Related Rates

Today I would like to provide a full explanation for this problem. Here is the problem one more time:

The radius of a spherical balloon is decreasing at a constant rate of 0.5 centimeters per second. At the instant when the volume V becomes 288π cubic centimeters, what is the rate of decrease, in square centimeters per second, of the surface area of the balloon?

(A) 24π
(B) 48π
(C) 64π
(D) 72π

Solution: Recall that the volume of a sphere with radius r is

V = 4/3 πr³,

and the surface area of a sphere with radius r is

S = 4πr².

We are given that dr/dt = –0.5, and we are being asked to find dS/dt when V = 288π.

First note that when V = 288π, we have

288π = 4/3 πr³
3/4 ⋅ 288π = πr³
216 = r³
6 = r

Now,

dS/dt = 8πr ⋅ dr/dt = 8πr(–0.5) = –4πr.
dS/dt (at r = 6) = –4π(6) = –24π.

So the surface area of the balloon is decreasing at a rate of 24π cm²/sec. Therefore, the answer is choice (A).

Notes: (1) Observe that one can get the formula for the surface area of a sphere by differentiating the formula for the volume of a sphere. This is just a little trick that can be used to reduce the number of formulas to memorize.

(2) Remember that the word rate generally indicates a derivative. “Increasing at a rate of” indicates a positive derivative, and “decreasing at a rate of” indicates a negative derivative.

(3) A radius is a length and is therefore measured in single units (in this case centimeters). Area (and in particular, surface area) is measured in square units. Volume is measured in cubic units.

The type of units being mentioned usually gives a big hint as to what measurement is being given or asked for.

(4) This is a related rates problem. In a related rates problem we differentiate the independent and dependent variables with respect to a new variable, usually named t, for time.

(5) A related rates problem can be pictured as a dynamic (moving) process that gets fixed at a specific moment in time.

For this problem we can picture a sphere shaped balloon deflating. We then freeze time at the moment when the volume is 288π cm³.

At this moment in time we want to know what the rate of decrease of the surface area is. The word “rate” indicates that we want the derivative of surface area with respect to time, dS/dt.

Don’t forget to apply the chain rule when differentiating the right hand side. In this case, the derivative of r is dr/dt.

AP Calculus AB – Related Rates

Students that are taking AP Calculus (or any Calculus course) usually find related rates word problems to be difficult. The concept however is really not so hard to understand. The idea is that you have two or more quantities that are changing with time, and those two quantities are related in some way. In a Precalculus problem, we would explore the relationship between those two quantities. But in a Calculus problem we explore the relationship between the rates at which those quantities are changing.

Let’s try a typical AP Calculus AB related rates problem (note that any Calculus AB problem can also appear on the BC exam). I will provide a full explanation for this problem tomorrow. In the meantime, feel free to leave your own solutions in the comments.

Level 4 Related Rates Problem

(A) 24π
(B) 48π
(C) 64π
(D) 72π

The Differences Between the ACT and the SAT

The ACT and SAT are both standardized tests that are used for college admissions. Students should be aware that they have a choice in which standardized test they choose to take. Years ago, the SAT was the more popular choice on the east and west coasts, and the ACT was more popular throughout the rest of the country. Last year, when the College Board debuted a new version of the SAT, these numbers changed significantly. In 2016, more than 2 million high schools students took the ACT and slightly less than a million-and-a-half students took the “new” SAT. But every student should realize that the final decision as to which test to focus on is entirely up to them. This article provides important information about the differences between the two tests so that you can make a more well-informed decision.

Test Length and Structure

The ACT consists of 5 sections. Only 4 of these sections must be completed. The fifth section – writing – is optional. For the required sections, there is a 45 minute english section, a 60 minute math section, a 35 minute reading section, and a 35 minute science section. You are given 40 minutes for the optional writing section. This gives a total required testing time of 45 + 60 + 35 + 35 = 175 minutes, or 2 hours and 55 minutes. With the optional writing section, the total testing time increases to 3 hours and 35 minutes.

The SAT is a longer test than the ACT. It also consists of 5 sections, one being optional (the essay). For the required sections, there is a 65 minute reading section, a 35 minute grammar section, a 25 minute math section where a calculator is not allowed, and a 55 minute math section where a calculator is allowed. You are given 50 minutes for the optional essay section. This gives a total required testing time of 65 + 35 + 25 + 55 = 180 minutes, or 3 hours. With the optional essay, the total testing time is 3 hours and 50 minutes.

On both the ACT and SAT, the essay is optional, and it is NOT factored into the final score. Many schools do however require the essay, so if you are not yet certain about all the schools you will be applying to you may want to take the essay as a precaution.

Scoring

The ACT is graded out of 36 points. The overall score is the average of the scores on each of the 4 required sections. Each of these is also graded out of 36 points.

The SAT is graded out of 1600 points. The overall score is the sum of the scores for verbal and math each consisting of a total of 800 points (note that the verbal score is made from both the reading and grammar sections, and the math score is made from the two math sections).

Conversion charts are easy to find for comparing ACT and SAT scores. For example, take a look here: ACT/SAT Conversion Chart (pages 7 and 15)

Pacing

The ACT requires a faster pace than the SAT to get through each section. Let’s look at math as an example, and assume that a student is pacing themselves to answer every single math question. The student has an average of 1 minute to complete each math question on the ACT, compared to an average of 1 minute and 15 seconds per question on the “no calculator” math section on the SAT, and an average of about 1 minute and 25 seconds per question on the “calculator allowed” math section of the SAT.

It should be noted that for many students it may be counterproductive to attempt every question on a standardized test, but this kind of advice lies outside the scope of this article.

Science Section

The ACT has a science section and the SAT does not. It should be noted that the science section is really more about reading comprehension and interpreting data found in charts, figures and tables. Very little knowledge of science is actually required (although it could be argued that students with a stronger science background are at an advantage here).

Free Response Questions

Both math sections on the SAT have free response questions, also known as “grid-ins.” In these questions the student must give a numerical answer without having any choices to choose from.

Calculator Policy

On the ACT you can use a calculator for every question. There are two math sections on the SAT, one where a calculator can be used, and one where a calculator cannot be used. Students that are very calculator dependent tend to have a lot of trouble with the “no calculator” SAT math section.

Difficulty versus Trickiness and Complexity in Math

ACT math problems tend to be more straightforward than SAT problems, whereas some SAT math problems are more complex and may take more time to answer. It is often much easier to understand ACT questions upon a first reading, whereas SAT questions tend to seem more confusing at first.

Many SAT math problems now go out of their way to model “real life” situations. In my opinion these types of questions tend to be more confusing and harder to read. The SAT also has more “multi-step” problems than the ACT (especially the more difficult problems that appear later in each section). You may have to complete 2, 3, or 4 distinct tasks to get to a solution.

Mathematics Covered

The ACT and SAT cover many of the same topics, but there are some differences. The following topics are covered on the ACT, but not on the SAT.

Arithmetic
Divisibility, Primes, GCD, LCM
Sequences and Series (especially arithmetic and geometric sequences)
Counting Problems
Laws of Sines and Cosines
Logarithms

Although there are no specific topics on the SAT that are not on the ACT, the SAT does place a heavier emphasis on Data Analysis, systems of equations and inequalities, and some more advanced precalculus concepts such as completing the square and divison of polynomials.

The ACT places a much heavier emphasis on Geometry than the SAT. Up to 50% of the ACT math questions can be Geometry and Trigonometry questions (much more Geometry than Trigonometry), whereas just a few math questions on the SAT will come from these topics.

Final Note

The best way to decide which test you should focus on is to take a practice test for both the ACT and the SAT. I strongly recommend using official practice tests from the testmakers themselves. Use a score comparison chart to compare your scores and you will see firsthand where your strengths lie.

Once you decide which test you will focus on make sure you take some time to prepare. I always recommend preparing for 10 to 20 minutes per day over a period of 3 to 4 months.

If you decide that you will be taking the SAT, I suggest you take a look at my “Get 800” collection of SAT math prep books, and if you decide that you will be taking the ACT, check out 320 ACT Math Problems arranged by Topic and Difficulty Level. Click the picture below for details.

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Newer posts

Older posts

500 New SAT Math Problems Just 19.99 on Amazon

A Trick For Free Two Day Shipping

New SAT Verbal Prep Book

General Graphing Calculator Advice for the SAT/ACT

Graphing Calculator Features You Should Know for the SAT/ACT

ACT only

Advanced Features

Hard Geometry Problem with Solution

for the SAT

Level 4 – Geometry

More Hard SAT Math Practice Problems

AP Calculus BC Geometric Series Problem

With Solution

Level 2 Series

More AP Calculus Problems with Explanations

AP Calculus BC Problem- Geometric Series

Level 2 Series

More AP Calculus Problems with Explanations

Remainder Problems – Part 3

Solution To Last Week’s Remainder Problem

Remainders In Disguise

Example

More Remainder Problems with Explanations

Can We Solve Remainder Problems

Without Using Long Division?

Example

Exercise

More Remainder Problems with Explanations

AP Calculus AB Related Rates Problem with Solution

More AP Calculus Problems with Explanations

AP Calculus AB – Related Rates

Level 4 Related Rates Problem

More AP Calculus Problems with Explanations

The Differences Between the ACT and the SAT

Test Length and Structure

Scoring

Pacing

Science Section

Free Response Questions

Calculator Policy

Difficulty versus Trickiness and Complexity in Math

Mathematics Covered

Final Note

500 New SAT Math Problems
Just 19.99 on Amazon