Student Reported Difficult SAT problems

These are very difficult Math SAT problems based on student reports of problems on the exam. This material is not available in the official practice tests, question bank or anywhere else. This is extremely valuable material for anyone going to 650-800 on the Math SAT, and it is available nowhere else.

Hard SAT Math Problems: Answers and Solutions

1. The equation $x(x - 6)=c$ has a solution of $3+4\sqrt{2}$. What is the value of $c$ ?

Show correct answerHide correct answer

2. The equation $2x^2+5x - 9=0$ has solutions of the form $\dfrac{m\pm \sqrt{n}}{4}$. What is the value of $n$?

Show correct answerHide correct answer

3. A pyramid with a square base has a base side length of 4 and a height of 7. Which is closest to its total surface area?

Show correct answerHide correct answer

4. A sphere has a diameter of $10$ cm and a density of $8$ grams per cubic centimeter. Which is closest to its mass in kilograms?

Show correct answerHide correct answer

5. Which is not a factor of $81x^{11} - 16x^7$?

$3x - 2$
$3x+2$
$9x^2 - 2$
$9x^2 + 4$

Show correct answerHide correct answer

6. Two perpendicular lines, $j$ and $k$, intersect at $(9, 4)$. Line $j$ has an $x$-intercept at $(3,0)$ and line $k$ has an $x$-intercept at $(g, 0)$. What is the value of $g$?

Show correct answerHide correct answer

$\frac{35}{3}$

7. A right circular cone has a height of $10$ cm and a slant height of $12$ cm. Which is closest to its volume in cubic centimeters?

Show correct answerHide correct answer

8. The volume of a cube is $658,503$. What is its surface area?

Show correct answerHide correct answer

$45414$

9. The surface area of a cube is $1350$. What is its volume?

Show correct answerHide correct answer

$3375$

10. Given line $l$ with equation $y=3x+2$, line $m$ is perpendicular to $l$ and passes through $(0,k)$ and $(20, -k)$. What is the value of $k$?

Show correct answerHide correct answer

$\frac{10}{3}$

11. A sector of a circle has an arc length of $3\pi$ and a central angle of $40^{\circ}$. What is the radius of the circle?

Show correct answerHide correct answer

$\frac{27}{2}$

12. What is the area of a square inscribed in a circle of radius 12?

$144$
$288$
$432$
$576$

Show correct answerHide correct answer

13. A triangle has sides of length 17 and 23. Which of the following cannot be the length of the third side?

$7$
$10$
$38$
$40$

Show correct answerHide correct answer

14. A pyramid with a square base has a lateral surface area of 144 and a total surface area of 180. Which is closest to its height?

$5$
$6$
$9$
$12$

Show correct answerHide correct answer

15. $80$ feet per second is closest to how many miles per hour (5280 feet in a mile)?

$50$
$55$
$60$
$75$

Show correct answerHide correct answer

16. A cube has a side length of $14$ cm and a mass of $10$ kg. Which is closest to its density in grams per cubic centimeter?

$4$
$6$
$8$
$12$

Show correct answerHide correct answer

17. A pipe has an outer diameter of 60 inches, a thickness of 1 inch, and a length of 50 feet. Which is closest to the volume of the pipe in cubic inches?

$10,000$
$30,000$
$70,000$
$110,000$

Show correct answerHide correct answer

18. $f(t)=70,000\cdot 1.02^{4t}$, where $t$ is the number of years. Which statement is accurate?3

$f(t)$ increases $2\%$ every 3 months
$f(t)$ increases $2\%$ every 4 months
$f(t)$ increases $2\%$ every 4 years
$f(t)$ increases $20\%$ every 4 years

Show correct answerHide correct answer

19. A box measuring $3$ ft by $4$ ft by $7$ ft is made of a material that weighs 300 grams per square foot. What is the weight of the box in kilograms?

Show correct answerHide correct answer

$36.6$

20. The expression $6x^2+37x+56$ has a factor of the form $ax+b$, where $a$ and $b$ are positive integers. What is one possible value of $a+b$?

Show correct answerHide correct answer

$9$ or $11$

21. What is the $x$-coordinate of the $x$-intercept of the tangent to the circle $(x - 2)^2+(y+6)^2=25$ at the point $(-2, -3)$?

$4$
$-\dfrac{1}{4}$
$\dfrac{1}{4}$
$\dfrac{3}{4}$

Show correct answerHide correct answer

22. What is the surface area in square meters of a cube with no lid and a side length of $83$ cm?

Show correct answerHide correct answer

$3.4445$

23. One gallon of paint costs $\$27$ and covers 300 square feet. Which is closest to the paint cost for painting a $3\times 4$ yard deck with 2 coats of this paint?

$\$10$
$\$14$
$\$19$
$\$39$

Show correct answerHide correct answer

24. The ratio of the lengths of the diagonals of a rhombus is $5:12$. The perimeter of the rhombus is $754$. What is the length of the longer diagonal?

Show correct answerHide correct answer

$348$

25. The quadratic equation $x^2+bx+c=0$ with rational coefficients has a solution of $3+\sqrt{5}$. What is the value of $b+c$?

Show correct answerHide correct answer

$-2$

26. Lucia traveled 27 miles in 2 hours. Part of the time she ran at 9 mph and part of the time she bicycled at 20 mph. What is the number of minutes she spent running, rounded to the nearest minute?

Show correct answerHide correct answer

$71$

Answer Key : 1.D, 2. $97$, 3. C, 4. A, 5. C, 6. $\frac{35}{3}$, 7. C, 8. $45414$, 9. $3375$, 10. $\frac{10}{3}$, 11. $\frac{27}{2}$, 12. B, 13. D, 14. D, 15. B, 16. A, 17. D, 18. A, 19. $36.6$, 20. $9$ or $11$, 21. C, 22. $3.4445$, 23. C, 24. $348$, 25. $-2$, 26. $71$.

Solutions

1. (D) Rewrite the equation in standard form: $x^2 - 6x - c = 0$. Apply the quadratic formula and set the positive root equal to the given solution: \begin{align*} & \dfrac{6 + \sqrt{36 + 4c}}{2} = 3 + 4\sqrt{2} \\ \implies & 3 + \dfrac{\sqrt{36 + 4c}}{2} = 3 + 4\sqrt{2} \\ \implies & \sqrt{36 + 4c} = 8\sqrt{2} \\ \implies & 36 + 4c = 128 \\ \implies & 4c = 92 \\ \implies & c = 23. \end{align*}

2. $97$. Apply the quadratic formula to find the solutions: $$ x = \dfrac{-5\pm \sqrt{5^2 - 4(2)(-9)}}{2(2)} = \dfrac{-5\pm \sqrt{25+72}}{4} = \dfrac{-5\pm \sqrt{97}}{4}. $$ Compare this result to the given form to determine that $n = 97$.

3. (C) Calculate the area of the base, then find the slant height using the Pythagorean theorem: \begin{align*} \text{Base} = 4^2 = 16, \ \ \text{Slant height}^2 = 2^2+7^2 \implies \text{Slant height} = \sqrt{53}. \end{align*} Determine the area of the triangular faces, and add the base area to calculate the total surface area: \begin{align*} & \text{Each face} = 4 \cdot \dfrac{\sqrt{53}}{2} = 2\sqrt{53} \implies \text{All 4 faces} = 4 \cdot 2\sqrt{53} = 8\sqrt{53} \\ \implies & \text{Total area} = 16 + 8\sqrt{53} \approx 74.2. \end{align*} This value is closest to 75.

4. (A) Calculate the radius to find the volume of the sphere: $$ r = \dfrac{10}{2} = 5 \implies V = \dfrac{4\pi (5)^3}{3} = \dfrac{500\pi}{3} \approx 523.6. $$ Multiply the volume by the density to find the mass in grams, then divide by 1000 to convert to kilograms: $$ \text{Mass} = 8 \cdot 523.6 \approx 4189 \implies \dfrac{4189}{1000} \approx 4. $$

5. (C) Factor the expression completely: \begin{align*} 81x^{11} - 16x^7 = x^7\big(81x^4 - 16\big) = x^7\big(9x^2+4\big)\big(9x^2 - 4\big) = x^7\big(9x^2+4\big)\big(3x+2\big)\big(3x - 2\big). \end{align*} The expression $9x^2 - 2$ is not a factor.

6. $\frac{35}{3}$. Find the slope of line $j$: \begin{align*} m_j = \dfrac{4 - 0}{9 - 3} = \dfrac{4}{6} = \dfrac{2}{3}. \end{align*} Determine the slope of line $k$, which is the negative reciprocal: $m_k = -\frac{3}{2}.$ Write the equation for line $k$, substitute $y=0$ and $x=g$, and solve for $g$: \begin{align*} 0 - 4 = -\dfrac{3}{2}(g - 9) \implies -4 = -\dfrac{3g}{2} + \dfrac{27}{2} \implies -\dfrac{35}{2} = -\dfrac{3g}{2} \implies g = \dfrac{35}{3}. \end{align*}

7. (C) Use the Pythagorean theorem to find the radius of the cone: \begin{align*} \text{radius}^2+\text{height}^2=\text{slant height}^2 \implies r^2+10^2=12^2 \implies r^2+100=144 \implies r=\sqrt{44}=2\sqrt{11}. \end{align*} Calculate the volume of the cone: \begin{align*} \text{Volume} = \dfrac{\pi r^2 h}{3} = \dfrac{\pi \big(\sqrt{44}\big)^2(10)}{3} = \dfrac{440\pi}{3} \approx 460.8. \end{align*} This value is closest to 460.

8. $45,414$. Find the side length of the cube: \begin{align*} \text{Side length} = \sqrt[3]{658,503} = 87. \end{align*} Calculate the surface area: \begin{align*} \text{Surface area} = 6 \cdot 87^2 = 45,414. \end{align*}

9. $3,375$. For a cube with side length $s$, use the given surface area to solve for $s$: \begin{align*} 6s^2=1350\implies s^2=225\implies s=15. \end{align*} Calculate the volume of the cube: \begin{align*} \text{Volume} = s^3=15^3=3,375. \end{align*}

10. $\frac{10}{3}$. Identify the slope of line $l$ to determine the perpendicular slope for line $m$: \begin{align*} m_l = 3 \implies m_m = -\dfrac{1}{3}. \end{align*} Use the slope formula with the points $(0,k)$ and $(20, -k)$ to solve for $k$: \begin{align*} \dfrac{k - (-k)}{0 - 20}=-\dfrac{1}{3}\implies \dfrac{2k}{-20}=-\dfrac{1}{3}\implies -\dfrac{k}{10}=-\dfrac{1}{3}\implies k=\dfrac{10}{3}. \end{align*}

11. $\frac{27}{2}$. For a circle with radius $r$ and central angle $\theta$ in degrees, use the arc length formula: \begin{align*} \text{Arc length} = \dfrac{2\pi r \cdot \theta}{360}. \end{align*} Substitute the given arc length and angle to solve for $r$: \begin{align*} 3\pi = \dfrac{2\pi r (40)}{360} \implies 3\pi = \dfrac{2\pi r}{9} \implies r = 3\pi \left(\dfrac{9}{2\pi}\right) \implies r = \dfrac{27}{2}. \end{align*}

12. (B) Determine the diagonal of the square, which is equal to the diameter of the circle: $$ \text{Diagonal} = 2(12) = 24. $$ Use the diagonal to find the side length of the square, then calculate the area: $$ \text{Side length} = \dfrac{24}{\sqrt{2}} \implies \text{Area} = \left(\dfrac{24}{\sqrt{2}}\right)^2 = \dfrac{576}{2} = 288. $$

13.(D) Apply the triangle inequality theorem to the given side lengths to establish the valid boundary for the third side, $x$: \begin{align*} 23 - 17 \lt x \lt 23 + 17 \implies 6 \lt x \lt 40. \end{align*}

14. (D) Calculate the area of the square base to determine its side length: \begin{gather*} \text{Base area} = 180 - 144 = 36 \implies \text{Side} = \sqrt{36} = 6. \end{gather*} Find the slant height using the area of a single lateral face: \begin{align*} & \text{Each face} = \dfrac{144}{4} = 36 \\ \implies & \dfrac{\text{side} \cdot \text{slant height}}{2} = 36 \\ \implies & \dfrac{6 \cdot \text{slant height}}{2} = 36 \\ \implies & \text{slant height} = 12. \end{align*} Apply the Pythagorean theorem to calculate the height of the pyramid: \begin{align*} & \text{height}^2 + \left(\dfrac{\text{side}}{2}\right)^2 = \text{slant height}^2 \\ \implies & \text{height}^2 + 3^2 = 12^2 \\ \implies & \text{height}^2 + 9 = 144 \\ \implies & \text{height}^2 = 135 \\ \implies & \text{height} = \sqrt{135} = 3\sqrt{15} \approx 11.62. \end{align*} This value is closest to 12.

15. (B) Convert the speed from feet per second to miles per hour: \begin{align*} \dfrac{80\cdot 60\cdot 60}{5280}\approx 54.54. \end{align*} This value is closest to 55.

16. (A) Calculate the volume of the cube: \begin{align*} \text{Volume} = 14^3 = 2744. \end{align*} Convert the mass from kilograms to grams and divide by the volume to calculate the density: \begin{align*} \text{Density} = \dfrac{10 \cdot 1000}{2744} \approx 3.64. \end{align*} This value is closest to 4.

17. (D) Determine the outer and inner radii of the pipe in inches: \begin{gather*} R = \dfrac{60}{2} = 30, \quad r = 30 - 1 = 29. \end{gather*} Calculate the cross-sectional area of the pipe: \begin{gather*} A = \pi R^2 - \pi r^2 = \pi(30^2 - 29^2) = \pi(900 - 841) = 59\pi. \end{gather*} Convert the length of the pipe from feet to inches and calculate the volume: \begin{gather*} \text{Length} = 50 \cdot 12 = 600 \implies \text{Volume} = 59\pi \cdot 600 = 35,400\pi \approx 111,212. \end{gather*} This value is closest to $110,000$.

18. (A) Identify the growth rate from the base of the exponential function: \begin{gather*} \text{Rate} = 1.02 - 1 = 0.02 = 2\%. \end{gather*} Analyze the exponent $4t$ to determine the frequency of this increase. Since $t$ represents years, the $2\%$ growth occurs $4$ times per year. Calculate the number of months in each growth period: \begin{gather*} \text{Period} = \dfrac{12}{4} = 3. \end{gather*} Therefore, $f(t)$ increases $2\%$ every 3 months.

19. $36.6$. Calculate the surface area of the box: \begin{gather*} \text{Surface area} = 2(lw+wh+lh) = 2(3\cdot 4 + 4\cdot 7 + 3\cdot 7) = 2(12 + 28 + 21) = 2(61) = 122. \end{gather*} Multiply the surface area by the weight per square foot to find the total weight in grams, then divide by 1000 to convert to kilograms: \begin{gather*} \text{Weight} = 122 \cdot 300 = 36,600 \text{ grams} \implies \text{Weight} = \dfrac{36,600}{1000} = 36.6 \text{ kilograms}. \end{gather*}

20. $9$ or $11$. Factor the quadratic expression: \begin{gather*} 6x^2+37x+56 = (3x+8)(2x+7). \end{gather*} Calculate the sum $a+b$ for each factor of the form $ax+b$: \begin{gather*} 3x+8 \implies 3+8 = 11, \ 2x+7 \implies 2+7 = 9. \end{gather*} Thus, the answer is $9$ or $11$.Alternatively, find the roots of the equation $6x^2+37x+56=0$ by graphing or using the quadratic formula: \begin{gather*} x = -\dfrac{8}{3}, \quad x = -\dfrac{7}{2}. \end{gather*} Rearrange these roots to deduce the corresponding integer factors: \begin{align*} x = -\dfrac{8}{3} \implies 3x = -8 \implies 3x+8 = 0, \\ x = -\dfrac{7}{2} \implies 2x = -7 \implies 2x+7 = 0. \end{align*}

21. (C) Identify the center of the circle from its equation as $(2, -6)$ and calculate the slope of the radius connecting the center to the point of tangency, $(-2, -3)$: \begin{gather*} m_r = \dfrac{-3 - (-6)}{-2 - 2} = \dfrac{3}{-4} = -\dfrac{3}{4}. \end{gather*} Determine the slope of the tangent line, which is the negative reciprocal of the radius slope: $m_t = \frac{4}{3}.$ Write the equation of the tangent line, substitute $y=0$, and solve for $x$ to find the $x$-intercept: \begin{gather*} 0 + 3 = \dfrac{4}{3}(x + 2) \implies 3 = \dfrac{4}{3}(x + 2) \implies \dfrac{9}{4} = x + 2 \implies x = \dfrac{1}{4}. \end{gather*}

22. $3.4445$. Calculate the surface area of the cube with no lid, which consists of 5 faces, in square centimeters: \begin{gather*} \text{Surface area} = 5 \cdot 83^2 = 5 \cdot 6889 = 34,445. \end{gather*} Convert the area from square centimeters to square meters by dividing by $100^2$: \begin{gather*} \text{Surface area} = \dfrac{34,445}{10,000} = 3.4445. \end{gather*}

23. (C) Convert the dimensions of the deck from yards to feet and calculate the base area: \begin{gather*} \text{Area} = (3 \cdot 3) \cdot (4 \cdot 3) = 9 \cdot 12 = 108. \end{gather*} Multiply the area by 2 to account for the two coats of paint required: \begin{gather*} \text{Total coverage} = 108 \cdot 2 = 216. \end{gather*} Calculate the final cost by multiplying the price per gallon by the required fraction of a gallon's coverage: \begin{gather*} \text{Cost} = 27 \cdot \dfrac{216}{300} = 19.44. \end{gather*} This value is closest to $\$19$.

24. $348$. Relate the half-diagonals of the rhombus, $5x$ and $12x$, to its side length using the Pythagorean theorem: \begin{gather*} \text{Side length}^2 = (5x)^2+(12x)^2 = 169x^2 \implies \text{Side length} = 13x. \end{gather*} Set the perimeter equal to the given value and solve for $x$: \begin{gather*} \text{Perimeter} = 4(13x) = 52x \implies 52x = 754 \implies x = 14.5. \end{gather*} Calculate the full length of the longer diagonal, which corresponds to twice the longer half-diagonal: \begin{gather*} \text{Longer diagonal} = 2(12x) = 24(14.5) = 348. \end{gather*}

25. $-2$. Apply the conjugate root theorem to identify the second root, since the quadratic equation has rational coefficients: $x = 3 - \sqrt{5}$. Construct the quadratic polynomial by multiplying the factors associated with these roots, grouping terms to efficiently expand using the difference of squares: \begin{align*} & \left(x - (3+\sqrt{5})\right) \left(x - (3-\sqrt{5})\right) = \left((x - 3) - \sqrt{5}\right) \left((x - 3) + \sqrt{5}\right) \\ \implies & (x - 3)^2 - \left(\sqrt{5}\right)^2 = (x^2 - 6x + 9) - 5 = x^2 - 6x + 4. \end{align*} Identify the corresponding coefficients $b$ and $c$ from the expanded form to calculate their sum: \begin{gather*} b = -6, \quad c = 4 \implies b+c = -6 + 4 = -2. \end{gather*}

26. $71$. Define $t$ as the time in hours Lucia spent running, and set up an equation for the total distance traveled: \begin{gather*} 9t + 20(2 - t) = 27. \end{gather*} Solve the equation to find the time spent running in hours: \begin{gather*} 9t + 40 - 20t = 27 \implies 13 = 11t \implies t = \dfrac{13}{11}. \end{gather*} Convert the time from hours to minutes and round to the nearest integer: \begin{gather*} \text{Time in minutes} = \dfrac{13}{11} \cdot 60 = \dfrac{780}{11} \approx 70.9. \end{gather*} This value rounds to 71.

Student Reported Difficult SAT problems

Hard SAT Math Problems: Answers and Solutions

Solutions

The best and most challenging ACT math problems on the web in 2024.