SAT Math Full-Length Practice Test 2

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Module 1

Question 1

If \(3(x - 2) + 4 = 13\), what is the value of \(x\)?




Show solution

Distribute: \(3x - 6 + 4 = 13\), so \(3x - 2 = 13\), \(3x = 15\), \(x = 5\).

    1. 3: Forgot to distribute the 3, solving \(3x - 2 + 4 = 13\) instead of \(3(x-2)+4=13\).
    1. 7: Sign error distributing: treated \(3(x-2)\) as \(3x + 2\) instead of \(3x - 6\).
    1. 9: Forgot to subtract 4 at the end, reporting the value after only subtracting 6.

Answer: B

Question 2

If \(|x - 5| = 3\), what is the greatest possible value of \(x\)?




Show solution

\(|x-5|=3\) means \(x - 5 = 3\) or \(x - 5 = -3\), giving \(x = 8\) or \(x = 2\). The greatest value is \(8\).

    1. 2: Found only the smaller solution (\(x=2\)) instead of the greater one.
    1. 3: Reported the value inside the absolute value bars (3) instead of solving for \(x\).
    1. 5: Reported the constant being subtracted (5) rather than a solution to the equation.

Answer: D

Question 3

What is the value of \(\dfrac{2}{3} + \dfrac{5}{6}\)?




Show solution

Convert to a common denominator of 6: \(\dfrac{4}{6} + \dfrac{5}{6} = \dfrac{9}{6} = \dfrac{3}{2}\).

    1. \(\dfrac{7}{9}\): Added numerators and denominators separately (\(\frac{2+5}{3+6}\)) instead of finding a common denominator.
    1. \(\dfrac{7}{6}\): Used a common denominator of 6 but only converted one fraction, computing \(\frac{2}{6}+\frac{5}{6}\) instead of \(\frac{4}{6}+\frac{5}{6}\).
    1. \(\dfrac{5}{3}\): Arithmetic slip after finding the common denominator, computing \(\frac{4+5}{6}\) as \(\frac{10}{6}\) instead of \(\frac{9}{6}\).

Answer: C

Question 4

A gym charges a \(\$40\) enrollment fee plus \(\$25\) per month. If Marcus has paid a total of \(\$215\), how many months has he been a member?




Show solution

Let \(m\) = number of months. \(40 + 25m = 215\), so \(25m = 175\), \(m = 7\).

    1. 5: Forgot the enrollment fee entirely, dividing the full total by 25: \(215/25 = 8.6\), then likely rounded down incorrectly.
    1. 8: Subtracted the fee but divided by the wrong amount, e.g. treated the fee as \(\$15\).
    1. 9: Added the enrollment fee back on instead of subtracting it: \((215+40)/25 = 10.2\), then rounded.

Answer: B

Question 5

The first term of an arithmetic sequence is 4, and each term after that increases by 5. What is the 6th term of the sequence?

Enter your answer:

Show solution

The \(n\)th term of an arithmetic sequence is \(a_n = a_1 + (n-1)d\). Here \(a_1 = 4\), \(d = 5\), \(n = 6\): \(a_6 = 4 + (5)(5) = 4 + 25 = 29\).

Answer: 29

Question 6

What is the value of \(\sqrt{72}\) in simplest radical form?




Show solution

Factor out the largest perfect square: \(72 = 36 \times 2\), so \(\sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}\).

    1. 2: Divided 72 by the perfect-square factor (36) instead of taking the square root of that factor.
    1. \(2\sqrt{18}\): Correctly pulled out a perfect-square factor of 4 but didn’t fully simplify the remaining radical, which still contains a perfect-square factor of 9.
    1. \(8\sqrt{9}\): Misfactored 72 as \(8 \times 9\) and moved the wrong factor outside the radical instead of extracting \(\sqrt{9} = 3\).

Answer: C

Question 7

Points \(A\) and \(B\) lie on a number line at \(-8\) and \(6\). What is the distance between points \(A\) and \(B\)?

0 -8 A 6 B




Show solution

Distance between two points on a number line is \(|{-8} - 6| = |-14| = 14\).

    1. 2: Sign error: computed \(-8+6\) instead of the distance.
    1. 6: Reported one of the two coordinates rather than the distance between them.
    1. 8: Reported the absolute value of only one coordinate (\(|-8|\)).

Answer: D

Question 8

A rectangular garden has a perimeter of 44 feet. If the length is 4 feet more than the width, what is the width, in feet, of the garden?

w + 4 w




Show solution

Let \(w\) be the width; length is \(w+4\). Perimeter: \(2w + 2(w+4) = 44\), so \(4w + 8 = 44\), \(4w = 36\), \(w = 9\).

    1. 11: Divided the perimeter by 4 directly, ignoring the length-width relationship.
    1. 13: Used the length instead of the width as the final answer.
    1. 18: Forgot to divide by 2 in the perimeter formula, solving \(4w + 4 = 44\) instead of \(2w + 2(w+4) = 44\).

Answer: A

Question 9

If \(g(x) = 2x^2 - 3\), what is the value of \(g(-3)\)?

Enter your answer:

Show solution

Substitute \(x = -3\): \(g(-3) = 2(-3)^2 - 3 = 2(9) - 3 = 18 - 3 = 15\).

Answer: 15

Question 10

A recipe calls for \(2\dfrac{1}{2}\) cups of flour to make 20 cookies. How many cups of flour are needed to make 32 cookies?

Enter your answer:

Show solution

Set up a proportion: \(\dfrac{2.5}{20} = \dfrac{f}{32}\). Solving: \(f = \dfrac{2.5 \times 32}{20} = \dfrac{80}{20} = 4\).

Answer: 4

Question 11

A survey asked 250 people their favorite season. The results are shown in the table below.

Season Number of People
Spring 60
Summer 55
Fall 70
Winter 65




Show solution

From the table, 55 people chose Summer out of 250 total. \(\dfrac{55}{250} = \dfrac{11}{50}\).

    1. \(\dfrac{2}{25}\): Divided the number who chose Fall (20) by the total instead of Summer’s count.
    1. \(\dfrac{55}{250}\): Left the fraction unsimplified, which is numerically the same value but not in the requested simplest form.
    1. \(\dfrac{11}{25}\): Doubled the correct numerator by mistake, using 110 instead of 55.

Answer: B

Question 12

The function \(h\) is defined by \(h(x) = \sqrt{12 - 2x}\). What is the domain of \(h\)?




Show solution

The radicand must be non-negative: \(12 - 2x \geq 0\), so \(-2x \geq -12\). Dividing by \(-2\) (flipping the inequality): \(x \leq 6\).

    1. \(x \geq 6\): Solved the inequality but flipped the direction, forgetting that dividing by a negative number reverses it.
    1. \(x \leq -6\): Made a sign error, solving \(12 + 2x \geq 0\) instead of \(12 - 2x \geq 0\).
    1. \(x \geq -6\): Combined both errors: sign error on setup and a flipped inequality direction.

Answer: A

Question 13

If \(y = 3x - 2\) and \(x + y = 18\), what is the value of \(x\)?




Show solution

Substitute \(y = 3x-2\) into \(x+y=18\): \(x + (3x-2) = 18\), so \(4x - 2 = 18\), \(4x = 20\), \(x = 5\).

    1. 4: Sign error substituting, treating the equation as \(x + 3x + 2 = 18\) instead of \(x + 3x - 2 = 18\), giving \(4x = 16\).
    1. 13: Correctly solved the system but reported the value of \(y\) (\(3(5)-2=13\)) instead of \(x\).
    1. 20: Correctly reduced the equation to \(4x = 20\) but forgot to divide by 4, reporting the value of \(4x\) instead of \(x\).

Answer: B

Question 14

What are the solutions to \(x^2 + 3x - 18 = 0\)?




Show solution

Factor: \(x^2 + 3x - 18 = (x+6)(x-3) = 0\), so \(x = -6\) or \(x = 3\).

    1. \(x = 6\) and \(x = -3\): Flipped the signs of both roots.
    1. \(x = 6\) and \(x = 3\): Correctly found 6 but dropped the negative sign on the second root.
    1. \(x = -9\) and \(x = 2\): Chose a factor pair of \(-18\) (9 and \(-2\), signs swapped) without checking that the pair sums to \(3\).

Answer: A

Question 15

A store’s weekly revenue and its advertising spending over 6 weeks are shown below. Which of the following best describes the association between advertising spending and revenue?

0 2 4 6 8 10 0 10 20 30 Advertising (hundreds of dollars) Revenue (thousands of dollars)




Show solution

As advertising spending increases across the 6 weeks, revenue also tends to increase, with the points following this upward trend closely but not perfectly. This describes a strong positive association.

    1. A strong negative association: Misread the general upward trend as downward.
    1. No association: Failed to recognize the consistent upward pattern across the data points.
    1. A perfect positive association: Overstated the strength of the relationship — the points follow a clear upward trend but are not perfectly aligned on a single line.

Answer: B

Question 16

A cylindrical water tank has a radius of 4 feet and a height of 9 feet. What is the volume of the tank, in cubic feet, in terms of \(\pi\)?

[Figure: A right circular cylinder labeled with radius r = 4 ft and height h = 9 ft.]




Show solution

Volume of a cylinder \(= \pi r^2 h = \pi (4)^2 (9) = \pi(16)(9) = 144\pi\).

    1. \(36\pi\): Computed \(\pi r h\) instead of \(\pi r^2 h\), forgetting to square the radius.
    1. \(72\pi\): Used the diameter (8) in place of the radius but forgot to square it, computing \(\pi(8)(9)\).
    1. \(324\pi\): Swapped the radius and height values, computing \(\pi(9)^2(4)\) instead of \(\pi(4)^2(9)\).

Answer: C

Question 17

The equation of line \(p\) is \(y = -\dfrac{2}{5}x + 3\). Which of the following equations represents a line that is parallel to line \(p\) and passes through the point \((5, -1)\)?




Show solution

Parallel lines share the same slope: \(m = -\dfrac{2}{5}\). Using point \((5,-1)\): \(-1 = -\dfrac{2}{5}(5) + b \Rightarrow -1 = -2 + b \Rightarrow b = 1\). Equation: \(y = -\dfrac{2}{5}x + 1\).

    1. \(y = \dfrac{5}{2}x - 1\): Used the negative reciprocal of the slope instead of the same slope, treating the lines as perpendicular rather than parallel.
    1. \(y = -\dfrac{2}{5}x - 1\): Correctly used the same slope but made a sign error solving for the intercept.
    1. \(y = \dfrac{2}{5}x - 1\): Used the correct intercept but flipped the sign of the slope.

Answer: A

Question 18

The graph of \(f(x) = x^2\) is reflected over the \(x\)-axis and then shifted up 5 units to form \(g(x)\). Which equation defines \(g(x)\)?




Show solution

A reflection over the \(x\)-axis negates the function: \(-x^2\). Shifting up 5 adds 5 to the result: \(g(x) = -x^2 + 5\).

    1. \(g(x) = -x^2 - 5\): Correctly reflected over the \(x\)-axis but shifted down instead of up.
    1. \(g(x) = -(x+5)^2\): Confused a vertical shift with a horizontal shift, applying the \(+5\) inside the function instead of outside.
    1. \(g(x) = x^2 + 5\): Applied the vertical shift correctly but forgot to reflect over the \(x\)-axis.

Answer: A

Question 19

A city’s population was 45,000 in 2015 and grew at a constant rate to 52,200 in 2020. If the population continued to grow at this same constant rate, which of the following best describes the growth from 2015 to 2020?




Show solution

Total growth \(= 52{,}200 - 45{,}000 = 7{,}200\) people over 5 years. At a constant rate: \(7{,}200/5 = 1{,}440\) people per year — a linear (constant-rate) trend.

    1. The population grew by 7,200 people per year, which is a linear trend.: Used the total change (7,200) as the annual rate instead of dividing by the number of years.
    1. The population grew by 1,440 people total over the 5 years.: Reported the correct per-year rate but mislabeled it as the total change over 5 years.
    1. The population grew exponentially each year.: Assumed growth was exponential without checking whether the change is consistent with a constant additive rate.

Answer: C

Question 20

Line segment \(\overline{AB}\) has endpoints \(A(2, 3)\) and \(B(8, 11)\). What are the coordinates of the midpoint of \(\overline{AB}\)?

x y 2 6 3 6 A(2, 3) B(8, 11)




Show solution

Midpoint formula: \(\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right) = \left(\dfrac{2+8}{2}, \dfrac{3+11}{2}\right) = (5, 7)\).

    1. \((3, 4)\): Computed the slope-related difference (6, 8) and halved it incorrectly, not using the midpoint formula.
    1. \((6, 8)\): Computed the differences \(B - A = (6, 8)\) instead of averaging the coordinates.
    1. \((10, 14)\): Added the coordinates but forgot to divide by 2: \((2+8, 3+11) = (10, 14)\).

Answer: B

Question 21

The equation \(3x^2 + kx + 12 = 0\) has exactly one real solution. What is the positive value of \(k\)?

Enter your answer:

Show solution

For exactly one real solution, the discriminant is zero: \(k^2 - 4(3)(12) = 0\), so \(k^2 = 144\), \(k = \pm 12\). The positive value is \(k = 12\).

Answer: 12

Question 22

Triangle \(XYZ\) is similar to triangle \(RST\), where angle \(X\) corresponds to angle \(R\). If \(XY = 5\), \(YZ = 7\), \(XZ = 9\), and \(RT = 18\), what is the length of \(YZ\)’s corresponding side, \(ST\)?

X Z Y 5 7 9 R T S 18




Show solution

The scale factor is \(\dfrac{RT}{XZ} = \dfrac{18}{9} = 2\) (since \(XZ\) corresponds to \(RT\)). Since \(YZ\) corresponds to \(ST\): \(ST = YZ \times 2 = 7 \times 2 = 14\).

    1. 10: Inverted the scale factor, using \(\dfrac{XZ}{RT} = \dfrac{9}{18} = \dfrac{1}{2}\) applied backwards to \(YZ\).
    1. 21: Correctly computed the scale factor \(\dfrac{RT}{XZ} = 2\), but mistakenly applied it to \(XY\) instead of \(YZ\).
    1. 25.2: Multiplied \(YZ\) by the wrong ratio, using \(\dfrac{18}{5}\) (RT over XY) instead of \(\dfrac{RT}{XZ}\).

Answer: B

Module 2

Question 1

If \(4x + 9 = 7x - 6\), what is the value of \(x\)?




Show solution

Subtract \(4x\) from both sides: \(9 = 3x - 6\). Add 6: \(15 = 3x\). Divide by 3: \(x = 5\).

    1. 1: Sign error on the constant, solving \(4x+9=7x+6\) instead of \(4x+9=7x-6\), giving \(3=3x\).
    1. 3: Correctly reduced the equation to \(15=3x\) but divided by the wrong number (5 instead of 3).
    1. 15: Correctly reduced the equation to \(15=3x\) but forgot to divide by 3, reporting the value of \(3x\) instead of \(x\).

Answer: C

Question 2

What is the value of \(\dfrac{3^6}{3^2 \cdot 3^1}\)?




Show solution

Combine the denominator: \(3^2 \cdot 3^1 = 3^3\). Then \(\dfrac{3^6}{3^3} = 3^{6-3} = 3^3 = 27\).

    1. 3: Subtracted all exponents in sequence incorrectly, computing \(6-2-1-1\) instead of \(6-(2+1)\).
    1. 9: Computed \(6 - 2 - 1 = 3\) correctly as an exponent but then evaluated \(3^2\) instead of \(3^3\).
    1. 81: Added the denominator exponents into the numerator instead of subtracting: \(3^{6+2+1}\) reduced incorrectly.

Answer: C

Question 3

A car travels 270 miles using 9 gallons of gas. At this rate, how many gallons of gas are needed to travel 420 miles?




Show solution

Rate: \(270\) miles per \(9\) gallons \(= 30\) miles per gallon. Gallons needed: \(420 / 30 = 14\).

    1. 12.6: Inverted the ratio, using \(\dfrac{9}{270} \times 420\) divided incorrectly.
    1. 30: Computed the car’s miles-per-gallon rate (30) and reported it directly instead of using it to answer the question.
    1. 46.7: Divided 420 by 9 directly without accounting for the 270-mile baseline.

Answer: B

Question 4

A ladder leans against a wall, forming a right triangle with the ground. The base of the ladder is 5 feet from the wall, and the ladder is 13 feet long. How high up the wall does the ladder reach, in feet?

P Q R ? 5 13




Show solution

By the Pythagorean theorem: \(5^2 + h^2 = 13^2\), so \(h^2 = 169 - 25 = 144\), \(h = 12\).

    1. 8: Subtracted the two given values directly (\(13-5=8\)) instead of applying the Pythagorean theorem.
    1. 9: Arithmetic slip computing \(\sqrt{169-25}\).
    1. 18: Added the two given values (\(13+5=18\)) instead of applying the Pythagorean theorem.

Answer: C

Question 5

If \(3x - 2y = 7\) and \(x + 2y = 9\), what is the value of \(x\)?




Show solution

Add the two equations: \((3x-2y)+(x+2y)=7+9\), giving \(4x=16\), so \(x=4\).

    1. -4: Correctly found \(4x=16\) but divided by \(-4\) instead of \(4\) (sign error in the division step).
    1. 2: Correctly found \(4x=16\) but divided by the wrong number (8 instead of 4).
    1. 16: Correctly found \(4x=16\) but forgot to divide by 4, reporting the value of \(4x\) instead of \(x\).

Answer: C

Question 6

If \(p(x) = 2x^3 - 5x^2 + x - 3\), what is the remainder when \(p(x)\) is divided by \((x - 2)\)?

Enter your answer:

Show solution

By the Remainder Theorem, the remainder equals \(p(2)\): \(p(2) = 2(2)^3 - 5(2)^2 + (2) - 3 = 16 - 20 + 2 - 3 = -5\).

Answer: -5

Question 7

A survey of 15 employees found their average commute time was 28 minutes. When 5 more employees were surveyed, their average commute time was 40 minutes. What is the average commute time, in minutes, for all 20 employees?




Show solution

Total commute time \(= 15(28) + 5(40) = 420 + 200 = 620\). Average for all 20: \(620/20 = 31\).

    1. 30.8: Made an arithmetic slip in the weighted sum, undercounting one group’s contribution.
    1. 34: Computed the simple (unweighted) average of the two group averages, \(\dfrac{28+40}{2} = 34\).
    1. 35: Mistakenly reported the midpoint between 28 and 40 using an incorrect method.

Answer: B

Question 8

In a circle, a chord subtends a central angle of \(100°\). What is the measure of the inscribed angle that subtends the same arc as this chord, from a point on the major arc?

O A B P 100°




Show solution

The Inscribed Angle Theorem states the inscribed angle is half the central angle subtending the same arc: \(100°/2 = 50°\).

    1. 100°: Assumed the inscribed angle equals the central angle.
    1. 130°: Used a supplementary-style relationship (\(180° - 50°\)) rather than the inscribed-angle theorem.
    1. 200°: Doubled the central angle instead of halving it.

Answer: A

Question 9

A concert venue sells floor tickets for $75 and balcony tickets for $50. If 280 tickets were sold for a total of $18,500, how many floor tickets were sold?




Show solution

Let \(f\) = floor tickets, \(b\) = balcony tickets. \(f+b=280\) and \(75f+50b=18{,}500\). Substituting \(b=280-f\): \(75f+50(280-f)=18{,}500\), so \(25f+14{,}000=18{,}500\), \(25f=4{,}500\), \(f=180\).

    1. 100: Reported the number of balcony tickets instead of floor tickets.
    1. 140: Assumed the tickets split evenly (280/2), ignoring the price difference.
    1. 225: Arithmetic slip in solving the system, computing \(25f=5{,}625\) instead of \(25f=4{,}500\).

Answer: C

Question 10

Which of the following is equivalent to \(\dfrac{x^2 - 4}{x^2 + 5x + 6}\) for \(x \neq -2, -3\)?




Show solution

Factor: \(\dfrac{(x-2)(x+2)}{(x+2)(x+3)}\). Cancel the common factor \((x+2)\): \(\dfrac{x-2}{x+3}\).

    1. \(\dfrac{x+2}{x+3}\): Cancelled the \((x+2)\) factor from the denominator but left an \((x+2)\) in the numerator instead of \((x-2)\).
    1. \(\dfrac{x-2}{x-3}\): Made a sign error factoring the denominator, using \((x-3)\) instead of \((x+3)\).
    1. \(\dfrac{(x-2)(x+2)}{x+3}\): Factored both numerator and denominator correctly but forgot to cancel the common \((x+2)\) factor.

Answer: A

Question 11

A jar contains 5 red candies, 4 blue candies, and 3 green candies. If one candy is drawn at random and not replaced, then a second candy is drawn, what is the probability that both candies are blue? (Enter your answer as a decimal, rounded to the nearest hundredth.)

Enter your answer:

Show solution

Total candies \(= 12\). \(P(\text{both blue}) = \dfrac{4}{12} \times \dfrac{3}{11} = \dfrac{12}{132} = \dfrac{1}{11} \approx 0.09\). Recomputing carefully: \(\frac{1}{11} = 0.0909...\), which rounds to \(0.09\).

Answer: 0.1

Question 12

If \(\dfrac{3}{4}(x + 8) = 21\), what is the value of \(x\)?

Enter your answer:

Show solution

Multiply both sides by \(\dfrac{4}{3}\): \(x + 8 = 28\). Subtract 8: \(x = 20\).

Answer: 20

Question 13

A radioactive substance decays such that its mass is modeled by \(M(t) = 200 \cdot (0.85)^t\), where \(t\) is the number of years and \(M(t)\) is the mass in grams. What does the value \(0.85\) represent in this context?




Show solution

In a model \(M(t) = a \cdot b^t\), the base \(b\) is the decay factor. Since \(b = 0.85 = 1 - 0.15\), the substance retains \(85\%\) of its mass each year, meaning it loses \(15\%\) each year.

    1. The substance loses \(0.85\) grams each year.: Treated the decay factor as an additive amount lost per year rather than a multiplicative factor.
    1. The substance loses \(85\%\) of its mass each year.: Misread the decay factor \(0.85\) directly as the percent lost, rather than computing \(1 - 0.85 = 0.15\).
    1. The initial mass of the substance was \(0.85\) grams.: Confused the decay factor with the initial mass, which is actually 200 grams.

Answer: A

Question 14

In right triangle \(DEF\), the right angle is at \(E\). If \(DE = 7\) and \(EF = 24\), what is \(\cos(D)\)?

D E F 7 24




Show solution

The hypotenuse is \(\sqrt{7^2+24^2} = \sqrt{625} = 25\). Angle \(D\) has adjacent side \(DE = 7\), so \(\cos(D) = \dfrac{DE}{DF} = \dfrac{7}{25}\).

    1. \(\dfrac{24}{25}\): Used the opposite side (\(EF\)) instead of the adjacent side (\(DE\)) in the cosine ratio.
    1. \(\dfrac{24}{7}\): Computed the tangent ratio (opposite/adjacent) instead of cosine.
    1. \(\dfrac{7}{24}\): Computed the reciprocal of tangent using the wrong leg placement.

Answer: B

Question 15

For what value of \(k\) does the system of equations below have infinitely many solutions?

\[\begin{aligned} 6x + 9y &= 18 \\ 4x + 6y &= k \end{aligned}\]




Show solution

The coefficients of the second equation are \(\dfrac{2}{3}\) of the first equation’s coefficients (\(4 = \frac{2}{3}(6)\) and \(6 = \frac{2}{3}(9)\)). For the system to represent the same line, the constant terms must follow the same ratio: \(k = \dfrac{2}{3}(18) = 12\).

    1. 8: Divided 12 by 1.5 in the wrong direction instead of multiplying, effectively inverting the scale factor.
    1. 18: Assumed the constant term must match the first equation directly without accounting for the scale factor between the equations.
    1. 27: Multiplied 18 by 1.5 instead of dividing, applying the scale factor backwards.

Answer: B

Question 16

If \(c = 16\), how many distinct real solutions does the equation \(x^2 - 8x + c = 0\) have?




Show solution

The discriminant is \((-8)^2 - 4(1)(16) = 64 - 64 = 0\). A discriminant of zero means the equation has exactly one distinct real (repeated) solution.

    1. Zero: Assumed a discriminant of 0 means no real solutions, confusing it with a negative discriminant.
    1. Two: Assumed all quadratics have two solutions without checking the discriminant.
    1. Infinitely many: Confused a repeated root with an identity that holds for all \(x\).

Answer: A

Question 17

If \(f(x) = x^2 + 2\) and \(g(x) = 3x - 1\), what is the value of \(g(f(2))\)?




Show solution

First find \(f(2) = 2^2 + 2 = 6\). Then \(g(f(2)) = g(6) = 3(6) - 1 = 17\).

    1. 5: Correctly computed \(f(2) = 6\), but forgot the coefficient on \(g\), computing \(6 - 1\) instead of \(3(6) - 1\).
    1. 11: Added \(f(2) + g(2) = 6 + 5\) instead of composing the functions.
    1. 27: Reversed the order of composition, computing \(f(g(2))\) instead of \(g(f(2))\): \(g(2) = 5\), then \(f(5) = 5^2 + 2 = 27\).

Answer: C

Question 18

A gym’s monthly membership graph shows a steady increase in members from January to April, then the vertical axis increments change from 50 to 10 for the remaining months. If the visual slope from April to August looks the same as from January to April, which of the following is true?

100 150 200 210 220 230 Jan Feb Mar Apr May Jun Jul Aug Month Members




Show solution

Because the axis increments shrank from 50 to 10, a visually similar slope after April represents a much smaller actual change in membership per month than the same visual slope before April.

    1. The rate of membership growth from April to August is the same as from January to April.: Assumes visual slope directly reflects actual rate of change, ignoring that the axis scale changed.
    1. The gym lost members from April to August.: Misreads an increasing trend as a decrease, when the graph still shows growth, just at a different scale.
    1. The change in axis scale has no effect on how the data should be interpreted.: Ignores the fact that inconsistent axis scaling distorts the visual comparison between two segments of a graph.

Answer: A

Question 19

The solution to the system of equations below is \((x, y)\). What is the value of \(2x - y\)?

\[\begin{aligned} 3x + y &= 14 \\ x - y &= 2 \end{aligned}\]

Enter your answer:

Show solution

Add the two equations: \((3x+y)+(x-y)=14+2\), giving \(4x=16\), so \(x=4\). Substitute into \(x-y=2\): \(4-y=2\), so \(y=2\). Then \(2x-y = 2(4)-2 = 8-2 = 6\).

Answer: 6

Question 20

The polynomial function \(k\) is given by \(k(x) = 3x^4 - 2x^3 + x - 5\). As \(x\) approaches negative infinity, what happens to \(k(x)\)?




Show solution

End behavior is determined by the leading term, \(3x^4\). Since the degree is even and the leading coefficient is positive, both ends of the graph rise: as \(x \to -\infty\), \(k(x) \to +\infty\).

    1. \(k(x)\) approaches negative infinity.: Assumed all polynomials fall to negative infinity on the left without checking the degree’s parity.
    1. \(k(x)\) approaches zero.: Mistakenly assumed the polynomial’s value approaches zero as \(x\) decreases, confusing end behavior with a horizontal asymptote.
    1. \(k(x)\) approaches \(-5\).: Assumed the constant term determines long-run behavior, ignoring the dominant leading term.

Answer: A

Question 21

What is the value of \(x\) that satisfies \(\dfrac{4}{x+1} - \dfrac{1}{3} = 1\), given that \(x \neq -1\)? Enter your answer as a decimal.

Enter your answer:

Show solution

Add \(\dfrac{1}{3}\) to both sides: \(\dfrac{4}{x+1} = \dfrac{4}{3}\). Since the numerators match, the denominators must be equal: \(x+1 = 3\), so \(x = 2\).

Answer: 2

Question 22

A circle is defined by the equation \(x^2 + y^2 + 4x - 10y + 20 = 0\). What is the radius of the circle?

Enter your answer:

Show solution

Complete the square: \((x^2+4x+4) + (y^2-10y+25) = -20+4+25\), giving \((x+2)^2+(y-5)^2=9\). The radius is \(\sqrt{9}=3\).

Answer: 3