Master the Toughest SAT Math Questions from December 2023

Table of Contents

1. Introduction
2. Quadratics Question from the May Test
3. Tricky Linear Equation Questions
4. Percent Increase and Decrease
5. Exponent Questions
6. Graphing Questions
7. Trigonometry Concepts
8. Area and Ratio Questions
9. Unit Conversion Questions
10. Solving Absolute Value Equations
11. Additional Resources and Courses

Introduction

Hey, Mike from Prep Pros here! For those of You who don't already know me, I have been a full-time SAT tutor for the past eight years. I've scored perfectly on the SAT myself and have published what I believe is the best SAT Math book out there. I've helped students achieve perfect scores and obtain scores in the 99th percentile. In this article, I will go through a range of difficult SAT Math questions from recent tests and provide you with helpful tips, tricks, and strategies. By the end of this article, you will also learn about additional resources that can further assist you in achieving that perfect SAT score.

Quadratics Question from the May Test

Let's start with a challenging quadratic question from the May test. The question provides an equation where B and C are constants:

negative B plus root B squared minus four C equals 18

To solve this equation, let's relate it back to the quadratic formula, which is a must-know concept for the SAT. The quadratic formula is:

x = negative B plus or minus the square root of B squared minus 4AC all over 2A

In this equation, we Notice that A equals 1 since we only have x squared. Therefore, we can rewrite the given equation as:

x = negative B plus or minus the square root of B squared minus 4C over 2

By simplification, we can see that 2x is equal to 10 and 2x is equal to 18. Solving for x, we find that x equals 5 or x equals 9.

Tricky Linear Equation Questions

Another common Type of difficult question on the SAT involves tricky linear equations. Let's take a look at an example:

A fitness membership costs $45 per month. All new members receive a discount of 20% off the cost of the first month of membership. The function C gives the total cost C of T in dollars that a new member pays after T months of membership.

This question asks us to find the value of the function C. To approach this, let's plug in some values to make it simpler. For example, if we consider the first month of membership, the cost would be $25 (after applying the 20% discount). For the Second month, the cost would be $70 ($45 for the regular cost plus $25 from the first month). Similarly, for the third month, the cost would be $115. By plugging in these values, we can eliminate incorrect answer choices until we find the correct one.

Percent Increase and Decrease

Percent increase and decrease questions are frequently seen on the SAT. Let's tackle one of these questions:

The population of a city in 2000 was 2.6 times its population in 1999. The population of this city increased by P percent from 1999 to 2000. What is the value of P?

To solve this question, we need to think about percent increases or decreases represented in decimal form. In this case, since the population in 2000 is 2.6 times the population in 1999, the increase can be calculated as 2.6 - 1 = 1.6, which is equal to 160%. Therefore, we have a 160% increase.

Exponent Questions

Exponent questions are another challenging area tested on the SAT. Let's look at an example involving fractional exponents and reducing bases:

Simplify the expression: 32^(2/5) - 64^(6/5)

To solve this question, we start by putting the bases in the same form. 32 can be written as 2^5, and 64 can be written as 2^6. With fractional exponents, we can simplify the expression to (2^(5/2)) - (2^(6/5)). Combining the bases, we get (2^(37/10)). Next, we simplify further since 37/10 is equivalent to 3^(1/20). Therefore, the final answer is 2^3 * (3^(1/20)).

Graphing Questions

Graphing questions are a regularly tested concept. Let's look at an example:

In the figure shown, point B in the center of each circle lies on AC. The ratio of AB to BC is 4:1. The area of the small circle is 72. What is the area of the shaded region?

To find the area of the shaded region, we need to subtract the area of the small circle from the area of the larger circle. Since the ratio of AB to BC is 4:1 and AC represents the diameter of the larger circle, AC will be 5 times the radius of the small circle. By using this Scale factor, we can calculate the radius of the small circle and find its area. Subtracting this area from 72 will give us the area of the shaded region.

Trigonometry Concepts

Trigonometry questions often appear in difficult SAT Math sections. Let's explore one:

In Triangle D, Point G lies on DE. The measure of DFG is X degrees, and the measure of GFE is Y degrees. What is the value of Cosine(X) - sine(Y)?

This question tests knowledge of the trigonometric identity that the sine and cosine of complementary angles are equal. By knowing that X and Y are complementary angles, we can conclude that cos(X) = sin(Y). Therefore, cos(X) - sin(Y) will always equal zero.

Area and Ratio Questions

The SAT frequently tests area and ratio questions. Let's consider an example:

In the figure shown, point B in the center of each circle lies on AC. The ratio of AB to BC is 4:1. The area of the small circle is 72. What is the area of the shaded region?

To solve this question, we must find the area of the larger circle and subtract the area of the smaller circle. First, we determine that AC represents a scale factor of 5:1 for the radius or diameter of the circles. By calculating the area of the small circle using this scale factor, we can subtract it from the given area, 72, to find the area of the shaded region.

Unit Conversion Questions

Unit conversion questions can be tricky if not approached correctly. Let's look at an example:

The area of a rectangular region is increasing at a rate of 20 square feet per hour. Which expression represents the rate in square meters per minute?

To convert from square feet per hour to square meters per minute, we need to follow specific conversion steps. Many students make the mistake of canceling out units, which is incorrect. To convert correctly, we must multiply by the appropriate conversion factors and consider the units squared. By using the correct conversion factor and canceling out units, we can determine the rate in square meters per minute.

Solving Absolute Value Equations

Solving absolute value equations is a concept often tested on the SAT. Let's explore an example:

What are all the possible solutions to the equation |x - 5| = 2x?

To solve for x, we first isolate the absolute value term by getting |x - 5| on one side of the equation. Then, we split the equation into two parts: x - 5 = 2x and x - 5 = -2x. Through solving each part, we obtain two potential values for x. However, we need to check for extraneous solutions to ensure that they satisfy the original equation. In this case, only one of the potential values, Five-Thirds, satisfies the equation when plugged back in.

Additional Resources and Courses

To further enhance your SAT Math skills, I recommend utilizing additional resources and courses. Here are a few options:

• My Math Book: A comprehensive guide covering a range of SAT Math topics. Each chapter in the book corresponds to a particular concept or question type, allowing you to focus on filling knowledge gaps.
• Advanced SAT Math Course: This course targets the most challenging SAT Math questions, providing strategies, techniques, and content necessary to tackle these questions. You can access a free trial with 12 practice questions on my Website.
• Ultimate SAT Course: This comprehensive course offers a structured program to improve your SAT Math skills across all difficulty levels. With this course, you'll gain access to a range of content and resources designed to help you achieve a perfect SAT score.

I hope you've gained valuable insights and strategies from this article. Feel free to drop any questions or comments in the section below. Thank you for joining me today, and best of luck in your SAT Math preparation!