5 Hardest SAT Math Questions
Students achieve perfect scores on the math section of the SAT by practicing the most difficult problems. I have chosen the five questions that challenge my SAT students the most.
There is a standard set of question types and math topics that show up on the math section. This gives students the opportunity to study them and feel fully prepared when they take the test. The questions I have chosen are question types that appear on official SAT practice tests from College Board.
The questions appear in the concept modules I have created and taught to students on these different question types and math topics. For each question, I have identified which math topic they belong to and whether you can use a calculator. I have also included the answer and answer explanation so that you can check your work.
The vast majority of the students I tutor do not get these questions right the first time around. Can you figure them out?
Question 1: Grid-in response, calculator not permitted
Topic: Functions
Answer: 3/2
Answer Explanation:
First realize that g(x) is the output and x is the input for the function g. By contrast, a represents a specific input that was put into the equation, giving us the output g(a) = a^2. In order to get the value of a, we need an alternative value for g(a) so that we then have a system of equations.
We get that alternative value for g(a) by plugging a into the equation for function g, g(x). That gives us g(a) = (a – 3)^2.
We then can set (a – 3)^2 equal to a^2 because they both represent the same value, g(a). From there, we solve for a first by FOILing the left side:
(a – 3)^2 = a^2
a^2 – 6a + 9 = a^2
Because it is a quadratic equation, we need to move everything to one side to solve. Moving the a2 over to the left and simplifying gives us the following:
a^2 – 6a + 9 - a^2 = 0
-6a + 9 = 0
We are left with a two-step equation to solve. Add 6a to both sides and then divide by 6 on both sides to get a:
9 = 6a
a = 9/6 = 3/2
Question 2: Multiple choice, calculator not permitted
Topic: Quadratics
Answer: -7
Answer Explanation:
In this question type, we are told that an expression is equal to another expression. We can therefore start by setting them equal:
(2x + 6)(ax - 5) - x^2 + 30 = bx
We want to get rid of the parentheses so that we can see all of the coefficients. So our next step will be to FOIL:
2ax^2 – 10x + 6ax – 30 – x^2 + 30 = bx
Then we can take the terms with x^2 on the left and put them together:
2ax2 – x^2
We know that there is no x^2 on the right, so in order for the two expressions in our problem to be equal, the following must be true:
2ax^2 – x^2 = 0x^2
We can then solve for a by dividing out the x^2 on both sides:
2a – 1 = 0
Then add 1 and divide by 2 on both sides:
2a = 1
a = 1/2
To get b, we can take the terms with x on the left and put them together:
-10x + 6ax
We know that there is a bx on the right, so in order for the two expressions in our problem to be equal, the following must be true:
-10x + 6ax = bx
Dividing out the x on both sides gives us the following:
-10 + 6a = b
Then we can substitute 1/2 for a, which gives us -10 + 6(1/2) = -10 + 3 = -7.
Question 3: Multiple choice, calculator not permitted
Topic: Systems of Equations
Answer: x is y plus 1/6
Answer Explanation:
Notice in the answer choices that only x and y remain. So we can start by eliminating r and s using the elimination method. First we can substitute r with s + 1/2 so that we now only have to get eliminate s:
x + s + 1/2 = 4x – 9
y + s = 4y – 9
Next, we can multiply the bottom equation by -1 and then add the equations so that the s goes away.
x + s + 1/2 = 4x – 9
-y – s = -4y + 9
Adding the equations gives us the following:
x – y +1/2 = 4x – 4y
Lastly, we want to rewrite this remaining equation above. The answer choices indicate we want to isolate the x variable. We can start by moving the x’s to the right side and the y’s to the left side by subtracting x and adding 4y on both sides:
-y + 4y + 1/2 = 4x – x
Simplifying gives us the following:
3y + 1/2 = 3x
Then to get x by itself, divide by 3 on both sides:
x = y + 1/6
Question 4: Multiple choice, calculator permitted
Topic: Polygons
Answer: 14.5
Answer Explanation:
This problem is tricky because it does not include a picture and it ties in two advanced topics: polygons and trigonometry. Let’s start by drawing a circle of radius 10 cm inscribed in a regular pentagon:
We can make a right triangle that includes the radius, half the pentagon side, and a hypotenuse extending from the pentagon corner to the center:
Then we can use the formula for the sum of the interior angles of a polygon to get the measure of each angle in the pentagon. The sum of interior angles of a polygon = (n- 2) * 180, where n = number of sides. So because a pentagon has 5 sides, the sum of the interior angles of a pentagon = (5 – 2) * 180 = 540. Each angle is then 540 / 5 = 108 degrees.
Because the hypotenuse of our right triangle divides that 108-degree angle, one of the angles in the triangle is 108 / 2 = 54 degrees.
That gives us the following information on our right triangle:
With right triangles, we have a few options we can consider to solve for x, which represents half the pentagon side. There is Pythagorean Theorem, but we need two known sides to use that. Then there is special right triangles, but we would need a 30-60-90 or 45-45-90 triangle to use that. The other option is SohCahToa.
Because the sides we are looking at are opposite and adjacent to the 54-degree angle, we would use the tan function and write tan(54) = 10/x.
Solving for x gives us x = 10 / tan(54). We multiply that by 2 to give us the length of the full pentagon side: 2 * 10 / tan(54) = 14.5.
Question 5: Grid-in response, calculator permitted
Topic: Unit Conversion
Answer: 8
Answer Explanation:
Start with what you want to convert. Because we want to see the images per day Peter can upload, we are converting his upload speed. The upload speed given to us is 7 kilobits per second. So we will start with that and do dimensional analysis, crossing out units as we go along.
We want to go from seconds to days. There are 60 seconds in a minute, 60 minutes in an hour, and Peter spends 5 hours in a day uploading. In our dimensional analysis, it will look like this:
We also want to go from kilobits to images. We know that there are 1024 kilobits per megabit and 15 megabits per image. We can use this information to continue our dimensional analysis:
Everything is cancelled out except image and day, which is what we want because we want to find how many images per day Peter can upload.
Now put the numbers in the calculator:
(7 * 60 * 60 * 5) / (1024 * 15) = 8 (rounded to the nearest whole number)