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# Conquering the Hardest SAT Math Questions

By Akshay R on April 30, 2024

The SAT Math section can be a formidable challenge, especially when you encounter those mind-boggling questions towards the end of each section. Don't let them intimidate you! This blog will dissect the types of questions that often give students the most trouble and provide strategies to help you tackle them.
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## Why are Some SAT Math Questions So Hard?

There are several reasons why some SAT math questions are designed to be challenging:
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• Unfamiliar Concepts: The hardest questions may test less common mathematical concepts, like advanced function analysis, complex trigonometry, or matrices.
• Multiple Steps and Word Problems: Often these questions require you to combine several math skills or translate word problems into equations, adding another layer of complexity.
• Tricky Wording: Some questions are intentionally worded to confuse you. They may try to throw you off by including unnecessary information or using misleading language.
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## Common Question Types That Cause Trouble

Many mathematical concepts can be tricky, but certain questions tend to consistently trip up students. Here's a breakdown of some common question types that cause trouble, along with strategies to tackle them:
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• Advanced Functions: Questions on graphs of complex functions, function transformations, and interpreting unusual function notation.
• Trigonometry and Radians: Problems involving less common trigonometric identities, angles in radians, or graphs of advanced trigonometric functions.
• Word Problems with Hidden Information: Systems of equations, probability questions, and problems involving rates or ratios can disguise crucial information in wordy setups.
• Data Interpretation in Unusual Formats: Questions asking you to analyze less common graphs and charts (like histograms, scatter plots, or box plots) or extract information presented in unusual ways.
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## Hard SAT Math Questions with Answers

Ready to put your skills to the test? Here's a collection of example questions covering some of the trickiest concepts in mathematics.
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## Problem 1: Systems of Equations

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$$48x - 64y = 48y + 24$$$$$ry = \dfrac{1}{8} - 12x$$$
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In the given system of equations, $$r$$ is a constant. If the system has no solution, what is the value of $$r$$?
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Solution: This problem looks intimidating because of all the variables! Remember this: if you graph a two-variable system of solutions, and the two lines are distinct and parallel, then the system has no solutions. So let’s try to make these two equations look the same, so that we can figure out which value of $$r$$ makes them parallel.
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For the first equation, let’s rearrange the terms to get $$48x - 112y = 24$$, and divide both sides by 4 to get $$12x - 28y = 6$$. This way, we have a $$12x$$ on both equations and we can compare them more easily.
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Now, let’s rearrange the second pair of equations to get $$12x + ry = 1/8$$. For the two lines to be parallel, the x and y terms need to match up. It looks like $$r = -28$$ will be the right answer!
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$$f(x) = (1.84)^{\frac{x}{4}}$$$﻿ The function $$f$$ is defined by the given equation. The equation can be rewritten as $$f(x) = (1 + p/100)^x$$, where $$p$$ is a constant. Which of the following is closest to the value of $$p$$? ﻿ A) 16 B) 21 C) 46 D) 96 ﻿ Answer: A) 16 ﻿ Solution: Using the rules of exponents, we can rewrite this equation as $$(1.84)^{\frac{x}{4}} = \left((1.84)^{1/4}\right)^x = \sqrt[4]{1.84}^x = 1.164^x = (1+0.164)^x = (1+16.4/100)^x$$. It follows that the answer is A) 16. ﻿ ﻿ ## Problem 3: Circle Geometry In the xy-plane, a circle has center C with coordinates $$(h, k)$$. Points A and B lie on the circle. Point A has coordinates $$(h + 1, k + \sqrt{102})$$, and $$\angle ACB$$ is a right angle. What is the length of $$\overline{AB}$$ ? ﻿ A) $$\sqrt{206}$$ B) $$2\sqrt{102}$$ C) $$103\sqrt{2}$$ D) $$103\sqrt{3}$$ ﻿ Answer: A) $$\sqrt{206}$$ ﻿ Solution: Since points A and B lie on the circle, both AC and BC are radii of the circle, and AB is the hypotenuse of a right triangle with AC and BC as legs. ﻿ So first, let’s find the radius of the circle. We can use the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ ﻿ The coordinates of point A are $$(h + 1, k + \sqrt{102})$$ and the coordinates of the center C are $$(h, k)$$. Substituting these into the distance formula gives us the radius $$r$$: ﻿ $$r = \sqrt{(h + 1 - h)^2 + (k + \sqrt{102} - k)^2}$$ $$r = \sqrt{1^2 + (\sqrt{102})^2}$$ $$r = \sqrt{1 + 102}$$ $$r = \sqrt{103}$$ ﻿ So now, we know that $$AC = BC = \sqrt{103}$$. To find the length of $$\overline{AB}$$, we can use the Pythagorean theorem. ﻿ $$AB^2 = AC^2 + BC^2$$ $$AB^2 = 103 + 103$$ $$AB = \sqrt{206}$$ ﻿ Therefore, the answer is A) $$\sqrt{206}$$. ﻿ ﻿Here is a relevant Khan Academy unit on circles, but this problem also requires knowing the distance formula and Pythagorean theorem! ﻿ ## Problem 4: Quadratic Equations ﻿ Which quadratic equation has no real solutions? ﻿ A) $$x^2 + 14x - 49 = 0$$ B) $$x^2 - 14x + 49 = 0$$ C) $$5x^2 - 14x - 49 = 0$$ D) $$5x^2 - 14x + 49 = 0$$ ﻿ Answer: D) $$5x^2 - 14x + 49 = 0$$ ﻿ Solution: For a quadratic equation of the form $$ax^2 + bx + c = 0$$ to have no real solutions, the discriminant $$b^2 - 4ac$$ must be negative. To understand why, remember how the quadratic equation has $$\sqrt{b^2 - 4ac}$$ in it? Well, you can’t take the square root of a negative number—so if the discriminant is negative, then the equation has no solutions. ﻿ Let's calculate the discriminant for each equation: ﻿ A) Discriminant: $$14^2 - 4(1)(-49) > 0$$ B) Discriminant: $$(-14)^2 - 4(1)(49) = 0$$ C) Discriminant: $$(-14)^2 - 4(5)(-49) > 0$$ D) Discriminant: $$(-14)^2 - 4(5)(49) < 0$$ ﻿ Only option D yields a negative discriminant, which means it has no real solutions. ﻿ ﻿ ## Problem 5: Exponential Population Model ﻿ The function $$P(t) = 260(1.04)^{\left(\frac{6}{4}\right)t}$$ models the population, in thousands, of a certain city $$t$$ years after 2003. According to the model, the population is predicted to increase by 4% every $$n$$ months. What is the value of $$n$$? ﻿ A) 8 B) 12 C) 18 D) 72 ﻿ Answer: A) 8 ﻿ Solution: This is a funny question—you might think you need to do a lot of complicated math here, but the answer is hidden in the question! First, notice that 1.04 already means increasing by 4%. So the question is basically asking which value of $$t$$ makes the value in the exponent equal $$1$$. This way, we’re left with $$1.04^1$$, or a $$4%$$ increase. ﻿ Therefore, we just need to solve $$(6/4) t = 1$$, which has a solution of $$t = 2/3$$ years, which is the same as A) 8 months. ﻿ ﻿ ## Problem 6: Electric Flux ﻿ For an electric field passing through a flat surface perpendicular to it, the electric flux of the electric field through the surface is the product of the electric field’s strength and the area of the surface. A certain flat surface consists of two adjacent squares, where the side length, in meters, of the larger square is 3 times the side length, in meters, of the smaller square. An electric field with strength 29.00 volts per meter passes uniformly through this surface, which is perpendicular to the electric field. If the total electric flux of the electric field through this surface is 4,640 volts · meters, what is the electric flux, in volts · meters, of the electric field through the larger square? ﻿ Answer: 4176 ﻿ Solution: This problem is trying to intimidate you with all the science terminology, but let’s cut through the fluff and figure out what this question is really asking. It looks like we have two squares, one 3 times larger than the other, and we also know that flux is directly proportional to area. ﻿ You could write some complicated equations to solve this problem, but there’s a simpler way to do it! Because the larger square is three times as big as the smaller square, its area is 9 times as big—in other words, if the area of the smaller square is A, then the larger square has area 9A. That means that the larger square is getting 9/10 of the total flux, so the answer is $$9/10 \cdot 4640 = 4176$$. ﻿ ﻿ ## Problem 7: Quadratic Equations ﻿ Given the system of equations: $$y = 2x^2 - 21x + 64$$$
$$y = 3x + a$$\$
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where $$a$$ is a constant. The graphs of the equations in the given system intersect at exactly one point $$(x, y)$$, in the xy-plane. What is the value of $$x$$?
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A) $$-8$$
B) $$-6$$
C) $$6$$
D) $$8$$
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Answer: A) $$-8$$
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Solution: Remember this: If a quadratic equation has one solution, then it can be expressed in the form $$a(x-b)^2$$. In other words, it’s a square.
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So let’s set the two equations equal to each other, and figure out which value of $$a$$ makes it a square.
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$$2x^2 - 21x + 64 = 3x + a$$
$$2x^2 - 24x + 64 - a= 0$$
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If we factor out the 2, we see a $$2(x^2 - 12x)$$ term, which means that our secret square term must be $$(x-6)^2 = x^2 - 12x + 36$$.
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This means that our equation must be $$2(x^2-12x+36) = 2x^2 - 24x + 72$$. This is a quadratic equation with only one solution ($$x = 6$$).
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Last but not least, we need to find $$a$$. We can do this by setting $$64 - a = 72$$, which results in $$A = -8$$.
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## Problem 8: Factoring Quadratic Expressions

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The expression $$4x^2 + bx - 45$$, where $$b$$ is a constant, can be rewritten as $$(hx + k)(x + j)$$, where $$h$$, $$k$$, and $$j$$ are integer constants. Which of the following must be an integer?
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A) $$\frac{b}{h}$$
B) $$\frac{b}{k}$$
C) $$\frac{45}{h}$$
D) $$\frac{45}{k}$$
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Answer: D) $$\frac{45}{k}$$
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Solution: Let's expand the factored form to compare it with the given quadratic expression.
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Expanding $$(hx + k)(x + j)$$ we get:
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$$hx \cdot x + hx \cdot j + k \cdot x + k \cdot j = h \cdot x^2 + (hj + k) \cdot x + kj$$
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Matching this with $$4x^2 + bx - 45$$, we get:
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$$h = 4, \quad hj + k = b, \quad kj = -45$$
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From the first equation, since $$h = 4$$ is an integer, the last equation $$kj = -45$$ tells us that $$k$$ and $$j$$ must be factors of $$-45$$. Because a factor of $$45$$ is also a factor of $$45$$, it follows that D) $$\frac{45}{k}$$ must be an integer as well.
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## Problem 8: Graph Translation

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The graph of $$9x - 10y = 19$$ is translated down 4 units in the xy-plane. What is the x-coordinate of the x-intercept of the resulting graph?
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Answer: $$59/9$$
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Solution: Translating a graph down by 4 units in the xy-plane means we will subtract 4 from the y-coordinate of every point on the graph. To find the new x-intercept, we set the new y value to 0 (since the x-intercept occurs where y = 0).
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The equation of the line after translating down 4 units becomes:
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$$9x - 10(y + 4) = 19$$
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(Why does this work? All the $$y$$ values have moved down. For example, plugging in $$y = 2$$ into this new equation is the same as plugging $$y = 6$$ into the original equation.)
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Expanding this gives us:
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$$9x - 10y - 40 = 19$$
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Simplifying further:
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$$9x - 10y = 19 + 40$$
$$9x - 10y = 59$$
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To find the x-intercept, we set $$y = 0$$ and solve for $$x$$:
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$$9x - 10(0) = 59$$
$$9x = 59$$
$$x = 59/9$$
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## Problem 9: Simplifying Rational Expressions

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Which expression is equivalent to $$\dfrac{y + 12}{x - 8} + \dfrac{y(x - 8)}{x^2y - 8xy}$$?
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A) $$\dfrac{xy + y + 4}{x^3y - 16x^2y + 64xy}$$
B) $$\dfrac{xy + 9y + 12}{x^2y - 8xy + x - 8}$$
C) $$\dfrac{xy^2 + 13xy - 8y}{x^2y - 8xy}$$
D) $$\dfrac{xy^2 + 13xy - 8y}{x^3y - 16x^2y + 64xy}$$
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Answer: C) $$\dfrac{xy^2 + 13xy - 8y}{x^2y - 8xy}$$
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Solution: First, we find the common denominator for the two fractions. The denominators are $$(x - 8)$$ and $$(x^2y - 8xy) = xy(x-8)$$, which suggests a common denominator of $$xy(x-8) = x^2y - 8xy$$.
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Now we can rewrite the fractions with the common denominator:
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$$\dfrac{y + 12}{x - 8} \cdot \dfrac{xy}{xy} + \dfrac{y(x - 8)}{x^2y - 8xy}$$
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Simplify and combine the numerators:
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$$\dfrac{xy(y + 12) + y(x - 8)}{x^2y - 8xy}$$
$$\dfrac{xy^2 + 12xy + xy - 8y}{x^2y - 8xy}$$
$$\dfrac{xy^2 + 13xy - 8y}{x^2y - 8xy}$$
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By comparing the simplified expression to the options given, it's evident that option C is the equivalent expression:
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C) $$\dfrac{xy^2 + 13xy - 8y}{x^2y - 8xy}$$
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In fact, once we found the common denominator, we could have eliminated all the other answers and chosen C!
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## Problem 10: Density

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A sample of oak has a density of 807 kilograms per cubic meter. The sample is in the shape of a cube, where each edge has a length of 0.90 meters. To the nearest whole number, what is the mass, in kilograms, of this sample?
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Solution: Remember that the equation for density is $$D = m / V$$, but you can also find this equation on your reference sheet.
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The mass of the sample is $$m = 0.90^3 = 0.729$$, so you can find $$m = D \cdot V = 807 \cdot 0.729 = 588.303$$, or $$588$$ to the nearest whole number.
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## Strategy is Just as Important as Knowledge

Remember, it's not just about the answers! After each question, analyze your approach. Did you understand what the problem was asking for? Did you take more or less time than you expected? Did you employ and test-taking strategies? Understanding how you think is the secret to long-term success.
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Tough math problems might seem scary now but don't despair. Consistent practice and strategic thinking will make them much easier to handle. Embrace the challenge, and don't fear mistakes–they're how we learn!
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If you need more help with SAT Math, we got you covered—sign up for an SAT Bootcamp on Schoolhouse!