For many students, the SAT is one of the most difficult and stressful times of high school. The test is comprehensive and challenging. Because of this, many students want to prepare the best way they can: by practicing the very questions and problems they will face on exam day. By seeing these questions ahead of time, students can get a better understanding of what is going to be on the actual test, and gain an important sense of confidence that’s essential to success.

The reassuring confidence that students grasp from seeing questions ahead of time is invaluable. There are a variety of questions to cover, so let’s jump right in.

y ≤ -15x + 3000

y ≤ 5x

If a point with coordinates (a,b) lies within the above solution in the xy-plane, what is the maximum possible value of b?

Answer: 750

As we know the coordinates (a,b) lay within the solution of these equations, we must solve for maximum value. This means that we can solve it just like we would solve a system of equations via the substitution method.

- 5x ≤ -15x + 3000
- Move variables to the same side to get: 20x ≤ 3000
- Divide each side by twenty to get x: x ≤ 150
- Plug x into either equation to solve for value of b: 5 x 150 = 750

g(x) = (x – 3)^{2}

If the equation g(a) = a^{2}, is true, what is the value of a?

Answer: 3/2

It’s key to first recognize g(x) as the g function’s output, while also recognizing x as its input. At the same time, a represents a specific singular input for the equation. Similar to the last problem, we solve this with a system of equations, specifically using ‘a’ as a plugin for x.

- g(a) = a
^{2}, and g(a) = (a – 3)^{2} - Set them equal to one another: (a – 3)
^{2}= (a)^{2} - Solve for a: a
^{2}– 6a +9 = a^{2} - Shift everything to one side: a
^{2}– 6a + 9 – a^{2}= 0 - a
^{2}cancels, leaving you with: -6a + 9 = 0 - Add 6a to both sides: 9 = 6a
- Solve for final answer: a = 9/6 or 3/2

(2x + 6)(ax – 5) – x^{2} + 30

In the expression seen above, a is a constant. If the expression is equal to bx, and b represents a constant, what is the value of b?

A) -7

B) -4

C) 0

D) 5

Answer: -7

As we are told that one expression equals another, we can begin by making them equal and proceed to solve using FOIL.

- (2x +6)(ax – 5) – x
^{2}+30 = bx - Remove parenthesis: 2ax
^{2}– 10x + 6ax – 30 – x^{2}+ 30 = bx - Combine like terms: 2ax
^{2}– x^{2} - Set equal to zero: 2ax
^{2}– x^{2}= 0x^{2} - Simplify: 2a – 1 = 0
- Simplify more by adding 1: 2a = 1
- Divide by 2: a = ½
- Combine x terms and set equal to b: -10x + 6ax = bx
- Divide by x = -10 + 6a = b
- Plug in ½ for a: -10 + 6(1/2) = -10 + 3 = -7

Greta opened a bank account that earns 2 percent interest compounded annually. Initially, she deposited $100. She uses the expression 100(x)t to find the value of the account after a total of t years. Greta’s friend Tom opened an account that earns 2.5 percent interest compounded annually. He also deposits $100 into the account initially.

This is a two-part question: First, what is the value of x in the expression 100(x)t ? Next, after 10 years, how much more money will Tom have in his account than Greta has in hers? (Round your answer to the nearest whole cent.)

Answer: 1.02, and 6.11

The answer to the first problem can be found pretty easily. Since we know Greta’s account earns 2% interest, we can simply convert it to a decimal or .02. Then you add 1 to account for the initial deposit, giving us an answer of x = 1.02.

Before you can find out the difference in account earnings, you must create an equation for Tom.

- Since Greta’s equation is 100(1.02)t, we can plug the 2.5 % into an equation for Tom, just the same we did for Greta: 100(1.025)t
- Plug 10 in for t in Tom’s equation: 100(1.025)10 = $128.008
- Plug 10 in for t in Greta’s equation: 100(1.02)10 = $121.899
- Find difference and round to nearest cent: $128.008 – $121.899 = $6.11

x + r = 4x – 9

y + s = 4y – 9

If r is s + ½, and r and s are constants in the equations above, which of the following is true?

A.) x is y – 1/8

B.) x is y – 1

C.) x is y + 1/6

D.) x is y + 1/4

Answer: x is y + 1/6

The first important thing to recognize is that only x and y remain in the answer choices above. This means our first step is to remove r and s. We can do this using the elimination method and substitute r with s + ½ to eliminate r.

- x + s + ½ = 4x – 9, and y + s = 4y – 9
- Multiply the second equation by -1 so s goes away: -y – s = -4y + 9
- Then add equations: x – y + ½ = 4x – 4y
- Isolate x variables: -y + 4y + ½ = 4x – x
- Simplify: 3y + ½ = 3x
- Divide both sides by 3 to get solution: x = y + 1/6

x^{2} + y^{2} + 4x – 2y = -1

The above equation represents a circle in the xy-plane. What is the radius of the circle?

A.) 2

B.) 3

C.) 4

D.) 9

Answer: A

We’ve been given the information that the above equation is a circle, however, it is not written in the standard form of an equation which is (x – h)^{2} + (y – k)^{2} = r^{2}. Out first step is to rewrite the equation into standard form. This will help us easily find the radius.

- Rewritten, the equation looks like this: (x + 2)
^{2}– 4 + (y – 1)^{2}– 1 = -1 - Now we need to get our constants on one side: (x + 2)
^{2}+ (y – 1)^{2}= 4 - We can now see that r
^{2}is represented by 4 in our equation. - To find the radius, find the square root of 4, which is: 2

X = 2y + 5

Y = (2x – 3)(x + 9)

How many different ordered pairs, represented as (x, y), could satisfy the system of equations shown above?

A.) 0

B.) 1

C.) 2

D.) Infinitely many

Answer: C

The first step is to recognize that we can solve this equation with substitution.

- Substitute the second equation into the first: x = 2[(2x – 3)(x + 9)] + 5
- Expand quadratic equation: x = 2(2x
^{2}+ 15x – 27) + 5 - Simplify further: x = 4x
^{2}+ 30x -49 - Set equation equal to zero by subtracting x: 0 = 4x
^{2}+ 29x – 49

Now that the equation is completely broken down, we must remember the question didn’t ask for a specific set of coordinates, but rather a number of possible coordinate solutions. To find this we can use the discriminant of the quadratic formula.

- Discriminant = b
^{2}– 4ac - Plug in correlating values: 29
^{2}– 4(4)(49) = 841 – 784 = 57

The law in mathematics that governs the discriminant of the quadratic formula says if the discriminant is positive, there are 2 solutions; if the discriminant is negative, there are no real solutions; if the discriminant is 0, there is 1 real solution (or repeating solution).

Since our answer (57) is positive, we can conclude that there are 2 solutions.

The SAT is designed to test students’ abilities through a variety of problems, with an underlining goal of hitting the most important areas of mathematics. While there are multiple factors that play into a student's success during the test, the most important thing to do is to study the expected hardest SAT math questions in advance.

Students can also benefit from the help of some free tools in preparation for the SAT. Using an application like MathMaster, students will be able to enter equations into a problem solver to get instant feedback on their work, and even work through some extended exam prep. The hardest SAT math problems can seem insurmountable at first, but with the right preparation, anyone can find the confidence and knowledge necessary for success.

they can: by practicing the very questions and problems they will face on exam day.