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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:

If you think your friends might be interested in this special offer, please share it with them on Facebook:

Thank you all for your continued support!

Is Your SAT Math Score Above 600 But You Want To Get 800?

December 13, 2016

$28 New SAT Math Lessons - Advanced$

Is Your SAT Math Score Above 600 But You Want To Get 800?

“28 SAT Math Lessons to Improve Your Score in One Month – Advanced Course – For Students Currently Scoring Above 600 in SAT Math and Want to Score 800.”

Click the image above to view the book’s page on Amazon.

AP Physics Spiderman Problem with Solutions

December 6, 2016

AP Physics Spiderman Problem
with Solutions

Today I would like to give solutions to the AP level free response physics question I gave you yesterday: AP Physics Spiderman Problem

Spiderman Free Response Question with Solutions

As you know, Spiderman saves a runaway subway by standing at the front of the train and attaching his web to nearby buildings. After five city blocks the train comes to a halt. Assume that the mass of the train is 3.2 × 10⁵ kg and its initial velocity is 90 km/h. Assume also that the web is unstretched when initially attached to the building. If the distance to the buildings is 10 m on each side of the track, and a city block is 50 m long, estimate the spring constant of Spiderman’s web.

Quick solution: The situation looks something like the below figure, where the dark rectangle on the left is the initial position of the train, and the light rectangle on the right is the final position. If a city block is 50 m, then the train has moved 250 m before halting. This is much greater than the 10 meters to the buildings, so for a quick approximation, one can merely say that the web has stretched 250 m. The initial energy of the train is its kinetic energy E₀= 1/2 mv². By energy conservation, the initial kinetic energy must be the energy stored in the web when the train has finally halted. Assuming that the web acts like a spring, we have 1/2 mv²= kx², where x = 250 m and k is the spring constant. (We have not put a 1/2 before the expression on the right because there are two strands of web.) Converting km/h to m/s and plugging in the numbers gives k = 1600 N/m, which is high, but not inconceivable.

Rigorous solution: If s is the hypotenuse of the right triangle, the energy in the web is actually k(s – s₀)², where s – s₀ is the amount it has stretched from its initial position, s₀= 10 m. With the Pythagorean theorem, we get s = 250.2 m, almost exactly the previous answer, and (s – s₀) = 240.2 m. Setting mv²/2(s – s₀)²= k yields k = 1733 N/m, which is only 7.6% from the Quick solution, all of which shows the value of making approximations.

More Practice for the AP Physics Exams

If you are preparing for the AP Physics 1 or 2 exams, you may want to check out Physics Mastery. Click the image below to see the book’s Amazon page.

If you think your friends might be interested in this article, please share it with them on Facebook:

Thank you all for your continued support!

AP Physics Spiderman Problem

December 5, 2016

AP Physics Spiderman Problem

Today I would like to give an AP level free response physics question for you to try. I will post a full solution tomorrow.

Spiderman Free Response Question

More Practice for the AP Physics Exams

If you are preparing for the AP Physics 1 or 2 exams, you may want to check out Physics Mastery. Click the image below to see the book’s Amazon page.

If you think your friends might be interested in this article, please share it with them on Facebook:

Thank you all for your continued support!

320 SAT Chemistry Subject Test Problems – Coming Soon

December 1, 2016

320 SAT Chemistry Subject Test Problems

Coming Soon

Today I would like to announce a new prep book for the SAT chemistry subject test: 320 SAT Chemistry Subject Test Problems arranged by Topic and Difficulty Level

This book will be released in a few weeks at a special promotional price.

To be added to the notification list, simply send an email to steve@SATPrepGet800.com with “Notify me” written in the subject line.

In the meantime, you can take a look at my product page to see all of the other books from the Get 800 collection: Get 800 Product Page

If you think your friends might be interested in being notified of this special offer, please share it with them on Facebook:

Thank you all for your continued support!

SAT Exponential Growth Problem with Solution

November 25, 2016

SAT Exponential Growth Problem with Solution

Today I would like to give a solution to the SAT exponential growth problem presented in this post: Exponential Growth and Decay

Here is the problem once again, followed by a solution.

Level 4 SAT Exponential Growth

On January 1, 2015, a family living on an island releases their two pet rabbits into the wild. Due to the short gestation period of rabbits, and the fact that the rabbits have no natural predators on this island, the rabbit population doubles each month. If P represents the rabbit population years after January 1, 2015, then which of the following equations best models the rabbit population on this island over time?

A) P = 2^{t/12 + 1}
B) P = 2^{t + 1}
C) P = 2^12t
D) P = 2^{12t + 1}

Solution using the exponential growth model formula: As seen in example (4) from this post, a quantity that continually doubles over a fixed time period can be modeled by the exponential function P = a(2)^t/d where a is the quantity at time t = 0, and d is the doubling time in years. In this case, there are initially 2 rabbits, so that a = 2, and the doubling time is every month, or every 1/12 year.

It follows that P = 2(2)^{t ÷ 1/12}= 2(2)^12t= 2¹2^12t= 2^{1 + 12t}= 2^{12t + 1}, choice (D).

Notes: (1) For a review of the laws of exponents used here, see the following post: Laws of Exponents

(2) For additional methods for solving this problem, you may want to take a look at 28 New SAT Math Lessons – Advanced Course

Exponential Growth and Decay

The current version of the SAT gives problems on exponential growth and decay. Here are the basics that you should know if you want to get a perfect SAT score:

A general exponential function has the form f(t) = a(1 + r)^ct, where a = f(0) is the initial amount and r is the growth rate. If r > 0, then we have exponential growth and if r < 0 we have exponential decay.

Examples: (1) The exponential function f(t) = 300(2)^t can be used to model a population with a growth rate of 1 = 100% each year that begins with 300 specimens. The growth rate of 100% tells us that the population doubles each year.

(2) The exponential function f(t) = 50(3)^2t can be used to model a population with a growth rate of 2 = 200% every 6 months that begins with 50 specimens. The growth rate of 200% tells us that the population triples. Since c = 2, the tripling occurs every 1/2 year or 6 months.

(3) The exponential function f(t) = 120(0.75)^t/3 can be used to model a substance which is decaying at a rate of 1 – 0.75 = 0.25 = 25% every 3 years. The initial amount of the substance might be 120 grams. Since c = 1/3, the 25% decay occurs every 3 years.

(4) A quantity that continually doubles over a fixed time period can be modeled by the exponential function f(t) = a(2)^t/d where a is the quantity at time t = 0, and d is the doubling time in years.

Now try the following SAT math problem. I will post a solution after Thanksgiving, but feel free to post your own solutions in the comments meanwhile.

Level 4 SAT Exponential Growth

A) P = 2^{t/12 + 1}
B) P = 2^{t + 1}
C) P = 2^12t
D) P = 2^{12t + 1}

Hard SAT and ACT Math Problem with Solution

Quadratic Equations

Today I would like to provide a solution to yesterday’s hard math problem involving quadratic equations. Here is the problem once again, followed by a solution.

Level 5 Quadratic Equations

Which of the following is a quadratic equation that has –5/7 as its only solution?

A. 49x²– 70x + 25 = 0
B. 49x²+ 70x + 25 = 0
C. 49x²+ 35x + 25 = 0
D. 49x²+ 25 = 0
E. 49x²– 25 = 0

* If -5/7 is the only solution, then x + 5/7 is the only factor (by the factor theorem). So we have

(x + 5/7)(x + 5/7) = 0

x² + 5/7 x + 5/7 x + (5/7)² = 0

x² + 10/7 x + 25/49 = 0

49x² + 70x + 25 = 0

This is choice B.

Notes: (1) The factor theorem says that r is a root of the polynomial p(x) if and only if x – r is a factor of the polynomial.

(2) In factored form, a quadratic equation always has two factors. Since x + 5/7 is the only factor in this problem, it must appear twice.

(3) In going from the first equation to the second equation, we multiplied the two polynomials (x + 5/7 times itself). To do this you can use FOIL or any other method you like for multiplying polynomials.

(4) In going from the second equation to the third equation, we simply added 5/7 x + 5/7 x = 10/7 x.

(5) In going from the third equation to the last equation, we multiplied both sides of the equation by the least common denominator, which is 49. More precisely, on the left hand side we have

49(x² +10/7 x + 25/49)

= 49x² + 49(10/7 x) + 49(25/49)

= 49x² +7(10x) + 25

= 49x² +70x + 25.

Hard SAT and ACT Math Problem

Quadratic Equations

Today I would like to give a hard math problem involving quadratic equations. This question type can show up on both the SAT and ACT. I will provide a full explanation for this problem soon. Feel free to leave your own solutions in the comments.

Level 5 Quadratic Equations

Which of the following is a quadratic equation that has –5/7 as its only solution?

A. 49x²– 70x + 25 = 0
B. 49x²+ 70x + 25 = 0
C. 49x²+ 35x + 25 = 0
D. 49x²+ 25 = 0
E. 49x²– 25 = 0

Physics Mastery – Promotion Ended

Today I would like to announce the release of Physics Mastery for Advanced High School Students. This book is ideal if you are preparing for the SAT physics subject test or the AP physics 1 or 2 exam.

~~The paperback is now on sale on Amazon for only $7.77.~~

~~This sale will run for just a few hours, and once the sale ends the price will go up to $29.99. If you want to take advantage of the sale price I strongly recommend you purchase the book right away.~~

The promotion has ended. Thanks to everyone that participated. The book is now available at its regular price. Click the following link to get to the book’s Amazon page: Physics Mastery

I will still honor the promotion below however until the end of the week:

As an additional incentive to purchase this book today, I will also give you an additional book for FREE as a downloadable PDF file. You can choose ANY of my other books. So if you would like a different book, go ahead and purchase this one (as soon as it is available), forward me your Amazon confirmation email (to steve@SATPrepGet800.com) and let me know which of my books you would like for free. You will be provided with a link to download your additional free book. This offer is available until the end of ~~today~~ the week (November 6, 2016).

You can choose any book on my product page.