Full-Length Practice Test 2

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

Module 1

Question 1

If \(5x - 3 = 17\), what is the value of \(10x\)?




Show solution

Solve for \(x\): \(5x - 3 = 17 \Rightarrow 5x = 20 \Rightarrow x = 4\).

Then \(10x = 10(4) = 40\).

    1. \(4\) — this is just \(x\), not \(10x\)
    1. \(20\) — this is \(5x\), not \(10x\)
    1. \(80\) — doubles \(10x\) again; arithmetic error

Answer: C

Question 2

A store sells notebooks for \(\$3\) each and pens for \(\$1.50\) each. A customer buys a total of 10 items and spends \(\$21\). Which system of equations can be used to find the number of notebooks \(n\) and pens \(p\) the customer bought?




Show solution

The total number of items is 10, so \(n + p = 10\).

The total cost is \(\$21\). Notebooks cost \(\$3\) each and pens cost \(\$1.50\) each, so \(3n + 1.5p = 21\).

    1. Swaps which equation represents count vs. cost
    1. Swaps the prices of notebooks and pens
    1. Swaps both the count and the prices

Answer: B

Question 3

Which of the following is equivalent to \(3(2x - 4) - (x + 5)\)?




Show solution

Distribute and combine like terms:

\[3(2x - 4) - (x + 5) = 6x - 12 - x - 5 = 5x - 17\]

    1. \(5x - 7\): forgot to distribute the negative sign to the \(+5\), giving \(-12 + 5 = -7\) instead of \(-12 - 5 = -17\)
    1. \(7x - 17\): didn’t subtract \(x\); treated \(-(x + 5)\) as \(+(x - 5)\)
    1. \(7x - 7\): both errors combined

Answer: B

Question 4

Line \(\ell\) passes through the points \((0, 5)\) and \((4, 13)\). What is the slope of line \(\ell\)?




Show solution

\[m = \frac{13 - 5}{4 - 0} = \frac{8}{4} = 2\]

    1. \(\frac{1}{2}\): inverted the slope formula (ran \(\Delta x / \Delta y\))
    1. \(\frac{9}{4}\): computed \(13 - 4 = 9\) in the numerator instead of \(13 - 5 = 8\)
    1. \(\frac{18}{4}\): doubled the numerator; likely added \(13 + 5\) instead of subtracting

Answer: B

Question 5

If \(\dfrac{x}{4} + 7 = 10\), what is the value of \(x\)?

Enter your answer:

Show solution

\[\frac{x}{4} + 7 = 10\] \[\frac{x}{4} = 3\] \[x = 12\]

Answer: 12

Question 6

The graph of \(y = f(x)\) is a line with slope \(-3\) and \(y\)-intercept \(6\). Which of the following is the equation of \(f(x)\)?




Show solution

Slope-intercept form: \(y = mx + b\), where \(m\) is slope and \(b\) is the \(y\)-intercept.

Here \(m = -3\) and \(b = 6\), so \(f(x) = -3x + 6\).

    1. Swaps the slope and intercept
    1. Uses the correct slope but a negative \(y\)-intercept
    1. Uses a positive slope instead of negative

Answer: C

Question 7

In the \(xy\)-plane, the equation \(4x + ky = 20\) represents a line. If the line passes through the point \((2, 4)\), what is the value of \(k\)?




Show solution

Substitute \((2, 4)\) into \(4x + ky = 20\):

\[4(2) + k(4) = 20\] \[8 + 4k = 20\] \[4k = 12\] \[k = 3\]

    1. \(-3\): sign error — subtracted \(8\) from both sides incorrectly
    1. \(1\): divided \(4k = 20\) by 4 without subtracting 8 first; computed \(k = 20/4 = 5\)… actually this arises from \(4k = 4 \Rightarrow k = 1\) via arithmetic error
    1. \(5\): skipped the subtraction step; divided \(20\) by \(4\) directly

Answer: C

Question 8

The function \(f\) is defined by \(f(x) = 2x^2 - 5\). What is the value of \(f(3)\)?




Show solution

\[f(3) = 2(3)^2 - 5 = 2(9) - 5 = 18 - 5 = 13\]

    1. \(7\): computed \(2(3) - 5 = 1\) then added \(6\) (forgot to square \(x\), used \(2x\) instead of \(2x^2\))
    1. \(18\): computed \(2(9) = 18\) but forgot to subtract \(5\)
    1. \(31\): computed \((2 \cdot 3)^2 - 5 = 36 - 5 = 31\), squaring \(2x\) instead of just \(x\)

Answer: B

Question 9

Which of the following is equivalent to \(x^2 - 10x + 25\)?




Show solution

Recognize the perfect square trinomial: \(x^2 - 10x + 25 = (x - 5)^2\).

Check: \((x-5)^2 = x^2 - 10x + 25\). ✓

    1. \((x+5)^2 = x^2 + 10x + 25\): sign of middle term is wrong
    1. \((x-5)(x+5) = x^2 - 25\): difference of squares, not a perfect square trinomial
    1. \((x-25)(x+1) = x^2 - 24x - 25\): incorrect factoring

Answer: A

Question 10

A rectangle has a length that is 3 more than twice its width. If the perimeter of the rectangle is 54, what is the width of the rectangle?

Enter your answer:

Show solution

Let width \(= w\). Then length \(= 2w + 3\).

Perimeter \(= 2(\text{length} + \text{width})\):

\[2(2w + 3 + w) = 54\] \[2(3w + 3) = 54\] \[6w + 6 = 54\] \[6w = 48\] \[w = 8\]

Answer: 8

Question 11

A survey of 200 students asked whether they prefer reading or watching videos to study. The results are shown in the table below.

Prefers Reading Prefers Videos Total
Grade 11 54 66 120
Grade 12 28 52 80
Total 82 118 200

What fraction of the Grade 12 students surveyed prefer reading?




Show solution

Of the 80 Grade 12 students, 28 prefer reading:

\[\frac{28}{80} = \frac{7}{20}\]

    1. \(\frac{7}{50}\): used \(28/200\) — divided by all students surveyed, not just Grade 12
    1. \(\frac{41}{100}\): used 82 total readers out of 200 students
    1. \(\frac{14}{41}\): used \(28/82\) — divided by total readers, not Grade 12 total

Answer: B

Question 12

A triangle has a base of 10 and a height of 7. What is the area of the triangle?




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\[\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2}(10)(7) = 35\]

    1. \(17\): added base and height instead of multiplying
    1. \(70\): forgot the \(\frac{1}{2}\) factor; computed \(10 \times 7\)
    1. \(140\): computed \(10 \times 7 \times 2\) — doubled instead of halving

Answer: B

Question 13

In the \(xy\)-plane, the lines \(y = 2x + 1\) and \(y = -x + 7\) intersect at point \(P\). What are the coordinates of \(P\)?




Show solution

Set the equations equal:

\[2x + 1 = -x + 7\] \[3x = 6\] \[x = 2\]

Substitute: \(y = 2(2) + 1 = 5\).

Point \(P = (2, 5)\).

    1. \((3, 4)\): likely solved \(3x = 9\) from an arithmetic error (\(1 + 7 = 9\)? No — \(7 - 1 = 6\))… this arises from \(3x = 9 \Rightarrow x = 3\), then \(y = 7 - 3 = 4\)
    1. \((3, 7)\): found \(x = 3\) with same error, then substituted into \(y = -x + 7 = 4\), so this represents a further error or using the \(y\)-intercept directly
    1. \((6, 1)\): found \(3x = 6\) correctly but forgot to divide by 3; used \(x = 6\)

Answer: A

Question 14

A scientist models the number of bacteria in a culture using the equation \(N = 500 \cdot 2^t\), where \(N\) is the number of bacteria and \(t\) is time in hours. Which of the following best describes the meaning of the value \(500\) in this equation?




Show solution

At \(t = 0\): \(N = 500 \cdot 2^0 = 500 \cdot 1 = 500\).

So 500 is the initial population (when \(t = 0\)).

    1. Misidentifies 500 as the doubling period (that role belongs to the base \(2\))
    1. Describes linear growth, not the initial value; the model is exponential
    1. Confuses 500 as a value reached at \(t = 2\), rather than the starting value

Answer: C

Question 15

If \(x^2 - 6x + 8 = 0\), what is one possible value of \(x\)?

Enter your answer:

Show solution

Factor: \(x^2 - 6x + 8 = (x - 2)(x - 4) = 0\)

So \(x = 2\) or \(x = 4\).

Either value is acceptable.

Answer: 2 (or 4)

Question 16

In the figure below, two parallel lines are cut by a transversal. The measure of one angle is \((3x + 10)°\) and the measure of its co-interior angle is \((2x + 30)°\). What is the value of \(x\)?

(3x + 10)° (2x + 30)°

Note: Figure not drawn to scale.




Show solution

Co-interior angles (same-side interior angles) are supplementary when lines are parallel:

\[(3x + 10) + (2x + 30) = 180\] \[5x + 40 = 180\] \[5x = 140\] \[x = 28\]

    1. \(32\): set the angles equal instead of supplementary: \(3x + 10 = 2x + 30 \Rightarrow x = 20\)… actually \(32\) comes from \(5x = 160\) (used \(180 - 20 = 160\))
    1. \(36\): set the sum equal to \(220\) (adding incorrectly)
    1. \(140\): gave the value of \(5x\), not \(x\)

Answer: A

Question 17

The function \(g\) is defined by \(g(x) = \dfrac{x^2 + 3x - 10}{x - 2}\), where \(x \neq 2\). Which of the following is equivalent to \(g(x)\)?




Show solution

Factor the numerator: \(x^2 + 3x - 10 = (x + 5)(x - 2)\).

\[g(x) = \frac{(x+5)(x-2)}{x-2} = x + 5 \quad (x \neq 2)\]

    1. \(x - 5\): factored as \((x - 5)(x + 2)\) instead of \((x + 5)(x - 2)\)
    1. Did not factor; attempted polynomial division incorrectly
    1. Performed incorrect polynomial division with an erroneous remainder

Answer: B

Question 18

The scatterplot below shows the study time (in hours) and exam score (out of 100) for 10 students. A line of best fit is drawn.

40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 Study Time (hours) Score

Note: Figure not drawn to scale.

Based on the line of best fit, what is the best estimate of the exam score for a student who studies for 5 hours?




Show solution

Using the line of best fit equation \(y \approx 4.9x + 49.4\):

At \(x = 5\): \(y \approx 4.9(5) + 49.4 = 24.5 + 49.4 \approx 73.9 \approx 74\).

    1. \(72\): reads the actual data point at \(x = 5\) rather than the line of best fit
    1. \(76\): overestimates the line slightly; this is closer to the value at \(x = 5.5\)
    1. \(80\): significantly overestimates; closer to the line at \(x = 6\)

Answer: B

Question 19

If \(2x + y = 11\) and \(x - y = 1\), what is the value of \(x + y\)?




Show solution

Add the two equations:

\[(2x + y) + (x - y) = 11 + 1\] \[3x = 12 \Rightarrow x = 4\]

Substitute into \(x - y = 1\): \(4 - y = 1 \Rightarrow y = 3\).

Then \(x + y = 4 + 3 = 7\).

    1. \(5\): computed \(x - y = 1\) as the answer instead of solving for \(x + y\)
    1. \(9\): arithmetic error when substituting back to find \(y\)
    1. \(11\): gave the value of \(2x + y\) from the first equation, not \(x + y\)

Answer: B

Question 20

In a right triangle, one leg has length 9 and the hypotenuse has length 15. What is the length of the other leg?

Enter your answer:

Show solution

By the Pythagorean theorem: \(a^2 + b^2 = c^2\)

\[9^2 + b^2 = 15^2\] \[81 + b^2 = 225\] \[b^2 = 144\] \[b = 12\]

This is a 3-4-5 triple scaled by 3: \((9, 12, 15)\).

Answer: 12

Question 21

For what value of \(x\) does \(\dfrac{x^2 - 4}{x + 2} = 0\)?




Show solution

A fraction equals zero when its numerator equals zero (and the denominator is non-zero).

Factor the numerator: \(x^2 - 4 = (x-2)(x+2)\).

\[\frac{(x-2)(x+2)}{x+2} = x - 2 \quad (x \neq -2)\]

Set \(x - 2 = 0 \Rightarrow x = 2\).

Check: at \(x = 2\), the denominator is \(2 + 2 = 4 \neq 0\). ✓

    1. \(-2\): sets the denominator to zero — the expression is undefined at \(x = -2\), not zero
    1. \(0\): substitutes \(x = 0\); expression gives \(\frac{-4}{2} = -2 \neq 0\)
    1. \(4\): confuses the constant in \(x^2 - 4\) with the answer

Answer: B

Question 22

A circle in the \(xy\)-plane has center \((3, -2)\) and radius \(5\). Which of the following is the equation of the circle?




Show solution

Standard form: \((x - h)^2 + (y - k)^2 = r^2\) where \((h, k)\) is the center and \(r\) is the radius.

Center \((3, -2)\) and \(r = 5\):

\[(x - 3)^2 + (y - (-2))^2 = 5^2\] \[(x - 3)^2 + (y + 2)^2 = 25\]

    1. Signs on \(h\) and \(k\) are both flipped, and uses \(r\) instead of \(r^2\)
    1. Signs on \(h\) and \(k\) are both flipped (uses \(-h\) and \(-k\) in the formula)
    1. Correct signs but uses \(r = 5\) instead of \(r^2 = 25\)

Answer: C

Module 2

Question 23

If \(\dfrac{3x + 9}{3} = 8\), what is the value of \(x\)?




Show solution

Multiply both sides by 3: \(3x + 9 = 24\). Then subtract 9: \(3x = 15\), so \(x = 5\).

    1. \(\frac{5}{3}\): divided \(15\) by \(9\) instead of by \(3\)
    1. \(7\): subtracted \(3\) from \(24\) instead of \(9\) when isolating \(3x\): \(3x = 24 - 3 = 21\), giving \(x = 7\)
    1. \(9\): used the constant in the numerator as the answer directly

Answer: B

Question 24

In the \(xy\)-plane, line \(\ell\) has a slope of \(\dfrac{2}{3}\) and passes through the point \((6, 7)\). What is the \(y\)-intercept of line \(\ell\)?




Show solution

Use \(y = mx + b\) with \(m = \frac{2}{3}\) and point \((6, 7)\):

\[7 = \frac{2}{3}(6) + b = 4 + b \Rightarrow b = 3\]

    1. \(4\): gave the value of \(mx = \frac{2}{3}(6) = 4\) at the given point, not the \(y\)-intercept
    1. \(5\): substituted \(x = 1\) instead of \(x = 6\): \(b = 7 - \frac{2}{3}(1) \approx 6.3\), then rounded down; or used \(b = y - m = 7 - 2 = 5\)
    1. \(11\): added \(mx\) to \(y\) instead of subtracting: \(b = 7 + 4 = 11\)

Answer: A

Question 25

The number of subscribers to a podcast, in thousands, is modeled by \(S(t) = 12(1.08)^t\), where \(t\) is the number of months since the podcast launched. Which of the following best describes what happens to the number of subscribers each month?




Show solution

In an exponential model \(S(t) = a \cdot b^t\), the base \(b = 1.08\) represents a growth factor of \(1 + 0.08\), meaning an \(8\%\) increase per time period.

    1. \(1{,}080\): misreads \(1.08\) as an additive amount and multiplies by \(1{,}000\) (the unit)
    1. \(12{,}000\): identifies the initial value \(12\) (thousands) but interprets it as the monthly change
    1. \(12\%\): confuses the initial coefficient \(12\) with the growth rate

Answer: B

Question 26

If \(3y - 5 = y + 11\), what is the value of \(y\)?

Enter your answer:

Show solution

\[3y - 5 = y + 11\] \[2y = 16\] \[y = 8\]

Answer: 8

Question 27

The graph of \(y = f(x)\) is a parabola in the \(xy\)-plane. The function \(f\) is defined by \(f(x) = (x - 3)^2 - 4\). Which of the following is the vertex of the parabola?




Show solution

In vertex form \(f(x) = (x - h)^2 + k\), the vertex is \((h, k)\).

Here \(h = 3\) and \(k = -4\), so the vertex is \((3, -4)\).

    1. \((-3, -4)\): negated \(h\), ignoring that vertex form uses \((x - h)\)
    1. \((-3, 4)\): negated both \(h\) and \(k\)
    1. \((3, 4)\): correct \(h\) but negated \(k\)

Answer: C

Question 28

In the \(xy\)-plane, the system of equations below has no solution. What is the value of \(k\)?

\[2x - 5y = 8\] \[4x - ky = 3\]




Show solution

A system has no solution when the lines are parallel — same slope, different intercepts.

The first equation \(2x - 5y = 8\) has slope \(\frac{2}{5}\).

For the lines to be parallel, the second equation must have the same slope. Rewrite \(4x - ky = 3\): slope \(= \frac{4}{k}\).

Set slopes equal: \(\frac{4}{k} = \frac{2}{5} \Rightarrow 2k = 20 \Rightarrow k = 10\).

Verify the lines are not identical: \(\frac{8}{3} \neq \frac{2}{1}\)… check by comparing \(\frac{8}{3}\) to \(\frac{3}{?}\). With \(k = 10\): equations are \(2x - 5y = 8\) and \(4x - 10y = 3\). Multiply first by 2: \(4x - 10y = 16 \neq 3\). ✓ Parallel, not coincident.

    1. \(5\): set \(k = 5\) to match the coefficient in the first equation directly (ignored the factor of 2)
    1. \(8\): used the constant from the right-hand side of the first equation
    1. \(16\): doubled the constant \(8\) instead of the coefficient \(5\)

Answer: C

Question 29

The table below shows selected values of the function \(g\).

\(x\) \(g(x)\)
\(-2\) \(3\)
\(0\) \(5\)
\(2\) \(7\)
\(4\) \(9\)
\(6\) \(11\)

Which of the following defines \(g(x)\)?




Show solution

From the table, as \(x\) increases by 2, \(g(x)\) increases by 2. So the slope is \(\frac{2}{2} = 1\).

At \(x = 0\), \(g(0) = 5\), so the \(y\)-intercept is \(5\).

Therefore \(g(x) = x + 5\).

Check: \(g(-2) = -2 + 5 = 3\) ✓, \(g(4) = 4 + 5 = 9\) ✓.

    1. \(g(x) = 2x + 5\): used the raw difference of 2 in \(g(x)\) as the slope without dividing by the change in \(x\) (also 2)
    1. \(g(x) = x + 7\): read the \(y\)-intercept as \(g(2) = 7\) rather than \(g(0) = 5\)
    1. \(g(x) = 2x + 3\): incorrect slope and used \(g(-2) = 3\) as the \(y\)-intercept

Answer: A

Question 30

If \(p(x) = x^3 - 4x^2 + x + 6\) and \(p(3) = 0\), what is one other value of \(x\) for which \(p(x) = 0\)?

Enter your answer:

Show solution

Since \(p(3) = 0\), \((x - 3)\) is a factor. Divide \(p(x)\) by \((x - 3)\):

\[x^3 - 4x^2 + x + 6 \div (x - 3) = x^2 - x - 2\]

Factor: \(x^2 - x - 2 = (x - 2)(x + 1)\).

So \(p(x) = (x - 3)(x - 2)(x + 1)\), and the zeros are \(x = 3\), \(x = 2\), and \(x = -1\).

The other positive zero is \(x = 2\).

Answer: 2

Question 31

Which of the following is equivalent to \(\dfrac{1}{x - 3} + \dfrac{2}{x + 1}\), for \(x \neq 3\) and \(x \neq -1\)?




Show solution

Find a common denominator \((x-3)(x+1)\):

\[\frac{1}{x-3} + \frac{2}{x+1} = \frac{x+1}{(x-3)(x+1)} + \frac{2(x-3)}{(x-3)(x+1)}\]

\[= \frac{x + 1 + 2x - 6}{(x-3)(x+1)} = \frac{3x - 5}{(x-3)(x+1)}\]

    1. \(\frac{3}{(x-3)(x+1)}\): added numerators directly (\(1 + 2 = 3\)) without adjusting each fraction for the common denominator
    1. \(\frac{3x+1}{(x-3)(x+1)}\): sign error — expanded \(2(x-3)\) as \(2x + 6\) instead of \(2x - 6\), giving \(x + 1 + 2x + 6 = 3x + 7\); or dropped the \(-3\), giving \(x + 1 + 2x = 3x + 1\)
    1. \(\frac{3}{2x-2}\): added both numerators and both denominators directly, which is not valid fraction addition

Answer: B

Question 32

A geologist records the depth of rock samples, in meters, and the temperature of each sample, in degrees Celsius. The data can be modeled by a linear equation. According to the model, the temperature increases by \(3°C\) for every \(100\) meters of depth, and a sample at \(200\) meters deep has a temperature of \(24°C\). What is the predicted temperature, in degrees Celsius, of a sample at \(500\) meters deep?




Show solution

The rate is \(3°C\) per \(100\) meters, so going from \(200\) m to \(500\) m is an increase of \(300\) meters.

Additional temperature: \(\frac{300}{100} \times 3 = 9°C\).

Predicted temperature: \(24 + 9 = 33°C\).

    1. \(27\): only added \(3°C\) (one \(100\)-meter increment instead of three)
    1. \(30\): added \(6°C\) (two increments instead of three; used \(400 - 200 = 200\) m)
    1. \(39\): added \(15°C\) (used \(500\) m total depth instead of the \(300\) m difference)

Answer: C

Question 33

The expression \(4x^2 + 12x + 9\) can be written in the form \((ax + b)^2\). What is the value of \(a + b\)?

Enter your answer:

Show solution

Recognize the perfect square: \(4x^2 + 12x + 9 = (2x + 3)^2\).

Check: \((2x+3)^2 = 4x^2 + 12x + 9\) ✓

So \(a = 2\) and \(b = 3\), giving \(a + b = 5\).

Answer: 5

Question 34

In the figure below, a right triangle has legs of length \(5\) and \(12\). What is \(\sin \theta\), where \(\theta\) is the angle opposite the leg of length \(5\)?

θ 12 5 A B C

Note: Figure not drawn to scale.




Show solution

By the Pythagorean theorem, the hypotenuse \(= \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).

\(\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{5}{13}\).

The leg opposite \(\theta\) has length \(5\), and the hypotenuse is \(13\).

    1. \(\frac{12}{13}\): used the adjacent leg (12) instead of the opposite leg (5); this is \(\cos \theta\)
    1. \(\frac{5}{12}\): used leg over leg; this is \(\tan \theta\)
    1. \(\frac{12}{5}\): inverted \(\tan \theta\)

Answer: A

Question 35

The function \(f\) is defined by \(f(x) = x^2 - 6x + 5\). For what values of \(x\) is \(f(x) < 0\)?




Show solution

Factor: \(f(x) = (x - 1)(x - 5)\).

The parabola opens upward (positive leading coefficient), so \(f(x) < 0\) between the roots.

Roots: \(x = 1\) and \(x = 5\). The function is negative for \(1 < x < 5\).

    1. \(x < 1\) or \(x > 5\): describes where \(f(x) > 0\) (outside the roots for an upward parabola)
    1. \(x < -5\) or \(x > -1\): negated the roots; factored as \((x+1)(x+5)\) incorrectly
    1. \(-5 < x < -1\): negated the roots and selected the between-roots interval

Answer: B

Question 36

The equation \(x^2 + y^2 - 6x + 4y = 12\) represents a circle in the \(xy\)-plane. What is the radius of the circle?




Show solution

Complete the square for both \(x\) and \(y\):

\[x^2 - 6x + y^2 + 4y = 12\] \[(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4\] \[(x - 3)^2 + (y + 2)^2 = 25\]

The radius is \(\sqrt{25} = 5\).

    1. \(\sqrt{12}\): used only the right-hand side without completing the square
    1. \(\sqrt{25}\): technically correct in value (\(= 5\)) but left unsimplified; the clean answer is \(5\)
    1. \(\sqrt{37}\): added the completing-the-square terms incorrectly (\(12 + 9 + 16 = 37\), using \(4^2 = 16\) instead of \(2^2 = 4\))

Answer: B

Question 37

A line in the \(xy\)-plane passes through the points \((-1, 4)\) and \((3, -4)\). What is the \(x\)-intercept of the line?

Enter your answer:

Show solution

Find the slope: \(m = \dfrac{-4 - 4}{3 - (-1)} = \dfrac{-8}{4} = -2\).

Use point-slope with \((-1, 4)\): \(y - 4 = -2(x + 1) \Rightarrow y = -2x + 2\).

Set \(y = 0\): \(0 = -2x + 2 \Rightarrow x = 1\).

Answer: 1

Question 38

The function \(h\) is defined by \(h(x) = \sqrt{3x - 6}\). For what value of \(x\) does \(h(x) = 6\)?




Show solution

\[\sqrt{3x - 6} = 6\] \[3x - 6 = 36\] \[3x = 42\] \[x = 14\]

    1. \(6\): multiplied \(6\) by \(3\) instead of squaring: solved \(3x - 6 = 18\), giving \(3x = 24\), \(x = 8\); or solved \(3x - 6 = 6 \times 3 = 18\) directly
    1. \(10\): squared correctly to get \(3x - 6 = 36\), but solved \(3x = 36\) (forgot to add 6), giving \(x = 12\); or computed \(3x = 6^2 - 6 = 30\), giving \(x = 10\)
    1. \(42\): correctly found \(3x = 42\) but forgot to divide by \(3\)

Answer: C

Question 39

In the \(xy\)-plane, the parabola \(y = x^2 - 5x + 4\) intersects the line \(y = x - 1\) at two points. What is the sum of the \(x\)-coordinates of the two intersection points?




Show solution

Set equal to find intersection:

\[x^2 - 5x + 4 = x - 1\] \[x^2 - 6x + 5 = 0\] \[(x - 1)(x - 5) = 0\]

The \(x\)-coordinates are \(x = 1\) and \(x = 5\). Their sum is \(1 + 5 = 6\).

By Vieta’s formulas, for \(x^2 - 6x + 5 = 0\), the sum of roots \(= 6\).

    1. \(4\): used \(-b/a\) from the original parabola (\(y = x^2 - 5x + 4\)) before subtracting the line; sum of roots of original = 5, not the system
    1. \(5\): sum of roots of the original parabola \(x^2 - 5x + 4 = 0\)
    1. \(8\): added the constants incorrectly; used \(5 + 4 - 1 = 8\)

Answer: C

Question 40

A data set of 8 positive integers has a mean of 10 and a median of 9. A ninth value of 19 is added to the data set. Which of the following must be true?

I. The mean of the new data set is greater than 10.
II. The median of the new data set is greater than 9.




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Statement I — Mean increases:

Original sum \(= 8 \times 10 = 80\). New sum \(= 80 + 19 = 99\). New mean \(= 99/9 = 11 > 10\). True.

Statement II — Median increases:

Original data set has 8 values with median \(= 9\), meaning the average of the 4th and 5th values is 9. When 19 (a large value) is added, it becomes the 9th value when sorted. The new median is the 5th value of the sorted 9-value set. The 5th value could equal 9 if the data were arranged as, for example, \(\{7, 8, 9, 9, 9, 9, 10, 11, 19\}\) — where the new median is still 9. So the median does not necessarily increase. Not necessarily true.

    1. II only: incorrectly concludes the median must increase while missing that the mean definitely increases
    1. I and II: over-concludes that the median must change
    1. Neither: misses the certain increase in mean caused by adding a value above the mean

Answer: A

Question 41

The polynomial \(p(x) = x^3 + ax^2 - 7x - 3\) has a factor of \((x + 3)\). What is the value of \(a\)?




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By the Factor Theorem, if \((x + 3)\) is a factor, then \(p(-3) = 0\):

\[(-3)^3 + a(-3)^2 - 7(-3) - 3 = 0\] \[-27 + 9a + 21 - 3 = 0\] \[9a - 9 = 0\] \[a = 1\]

Verify no distractor collision — check other choices:

  • \(a = -1\): \(9(-1) - 9 = -18 \neq 0\) ✓

  • \(a = 3\): \(9(3) - 9 = 18 \neq 0\) ✓

  • \(a = 5\): \(9(5) - 9 = 36 \neq 0\) ✓

    1. \(-1\): sign error when combining constants — computed \(-27 + 21 - 3 = -9\) correctly but then flipped: \(9a = -9\), giving \(a = -1\)
    1. \(3\): substituted \(x = 3\) instead of \(x = -3\) (confused the factor sign): \(27 + 9a - 21 - 3 = 0 \Rightarrow 9a = -3\)… not clean; more likely confused sign on the \(-7x\) term
    1. \(5\): sign error on the \(-7x\) term — used \(-7(-3) = -21\) as \(+7(-3) = -21\)… actually \(a = 5\) arises from \(-27 - 21 + 3 = -45\), so \(9a = 45\)

Answer: B

Question 42

In a right triangle, one acute angle measures \(30°\). The side adjacent to the \(30°\) angle has length \(6\sqrt{3}\). What is the length of the hypotenuse?

Enter your answer:

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In a 30-60-90 triangle, the sides are in ratio \(1 : \sqrt{3} : 2\).

The side adjacent to the \(30°\) angle is the longer leg (opposite the \(60°\) angle), with length \(\sqrt{3}\) in the ratio.

If the longer leg \(= 6\sqrt{3}\), then the scale factor \(= 6\).

Hypotenuse \(= 2 \times 6 = 12\).

Answer: 12

Question 43

The function \(f\) is defined by \(f(x) = \dfrac{3x}{x - 2}\). Which of the following is an expression for \(f^{-1}(x)\), the inverse of \(f\)?




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To find \(f^{-1}(x)\), swap \(x\) and \(y\) in \(y = \dfrac{3x}{x-2}\) and solve for \(y\):

\[x = \frac{3y}{y - 2}\] \[x(y - 2) = 3y\] \[xy - 2x = 3y\] \[xy - 3y = 2x\] \[y(x - 3) = 2x\] \[y = \frac{2x}{x - 3}\]

So \(f^{-1}(x) = \dfrac{2x}{x - 3}\).

    1. \(\frac{x-2}{3x}\): simply flipped numerator and denominator of \(f(x)\) without solving for the inverse properly
    1. \(\frac{3x}{x+2}\): changed the sign of \(-2\) in the denominator without going through the inversion process
    1. \(\frac{x+2}{3}\): solved a linear approximation, ignoring the \(x\) in the denominator of \(f\)

Answer: B

Question 44

For the equation \(kx^2 + 6x + 3 = 0\), where \(k\) is a nonzero constant, the equation has exactly two real solutions. Which of the following describes all possible values of \(k\)?




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For exactly two real solutions, the discriminant must be positive: \(b^2 - 4ac > 0\).

Here \(a = k\), \(b = 6\), \(c = 3\):

\[36 - 4(k)(3) > 0\] \[36 - 12k > 0\] \[12k < 36\] \[k < 3\]

But \(k \neq 0\) (given). Also, if \(k = 0\) the equation becomes linear with one solution, not two.

So the condition is \(k < 3\) and \(k \neq 0\), which means \(k < 0\) or \(0 < k < 3\).

    1. \(k < 0\): only captures the negative portion; misses the valid range \(0 < k < 3\)
    1. \(k < 3\): correct inequality but includes \(k = 0\), which is excluded
    1. \(0 < k < 3\): misses the valid negative values of \(k\)

Answer: D