SAT Math Full-Length Practice Test 3

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

Module 1

Question 1

If $4x - 3 = 17$, what is the value of $8x$?

A) $20$
B) $34$
C) $40$
D) $80$

Show solution

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

Then $8x = 8(5) = 40$.

Alternatively, $8x = 2(4x) = 2(20) = 40$ directly.

1. $20$: this is the value of $4x$, not $8x$
1. $34$: doubled the right-hand side of the original equation ($2 \times 17$) without solving for $x$ first
1. $80$: solved $4x = 20$ to get $x = 20$ (divided wrong), then computed $8(20)$

Answer: C

Question 2

A taxi charges a flat fee of $\$3.50$ plus $\$2.25$ per mile. Which of the following expressions represents the total charge, in dollars, for a ride of $m$ miles?

A) $2.25m$
B) $3.50 + m$
C) $3.50m + 2.25$
D) $3.50 + 2.25m$

Show solution

The flat fee is a fixed cost of $\$3.50$ (constant term), and the per-mile charge of $\$2.25$ multiplies by the number of miles $m$.

Total charge $= 3.50 + 2.25m$.

1. $2.25m$: omits the flat fee entirely
1. $3.50 + m$: ignores the per-mile rate of $2.25$ (treats rate as $1$)
1. $3.50m + 2.25$: swaps the role of the flat fee and the per-mile rate

Answer: D

Question 3

What is the slope of the line that passes through the points $(2, 5)$ and $(6, 13)$?

A) $\dfrac{1}{2}$
B) $1$
C) $2$
D) $4$

Show solution

\[\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\]

1. $\dfrac{1}{2}$: inverted the formula, computing $\dfrac{\Delta x}{\Delta y} = \dfrac{4}{8}$
1. $1$: subtracted both numerator and denominator incorrectly ($13 - 6 = 7$ and $5 - 2 = 3$) or averaged the coordinates
1. $4$: used only $\Delta x = 6 - 2 = 4$ without dividing into $\Delta y$

Answer: C

Question 4

Which of the following is equivalent to $3(2x - 4) - 2(x + 1)$?

A) $4x - 14$
B) $4x - 10$
C) $8x - 14$
D) $8x - 10$

Show solution

Distribute each term:

\[3(2x - 4) - 2(x + 1) = 6x - 12 - 2x - 2 = 4x - 14\]

1. $4x - 10$: sign error on the second term: computed $-2(x+1) = -2x + 2$ instead of $-2x - 2$
1. $8x - 14$: added $6x + 2x = 8x$ instead of subtracting
1. $8x - 10$: combined both errors above

Answer: A

Question 5

A recipe calls for $2\dfrac{1}{2}$ cups of flour for every 12 cookies. At this rate, how many cups of flour are needed to make 48 cookies?

Enter your answer:

Show solution

Set up a proportion. The ratio of flour to cookies is $\dfrac{2.5}{12}$.

\[\frac{2.5}{12} = \frac{x}{48} \Rightarrow x = \frac{2.5 \times 48}{12} = \frac{120}{12} = 10\]

Answer: 10

Question 6

Which of the following is equivalent to $x^2 - 10x + 25$?

A) $(x - 5)(x + 5)$
B) $(x - 5)^2$
C) $(x + 5)^2$
D) $(x - 25)(x - 1)$

Show solution

Recognize the perfect square trinomial: $a^2 - 2ab + b^2 = (a-b)^2$.

Here $a = x$, $b = 5$: $x^2 - 2(5)x + 5^2 = (x-5)^2$.

Verify: $(x-5)^2 = x^2 - 10x + 25$ ✓

1. $(x-5)(x+5)$: difference of squares pattern; expands to $x^2 - 25$, not $x^2 - 10x + 25$
1. $(x+5)^2$: expands to $x^2 + 10x + 25$; wrong sign on the middle term
1. $(x-25)(x-1)$: expands to $x^2 - 26x + 25$; two numbers that multiply to 25 but don’t sum to $-10$

Answer: B

Question 7

The table below shows the number of students who chose each lunch option on a given day.

Lunch Option	Number of Students
Pizza	42
Salad	18
Sandwich	30
Soup	10

If a student is selected at random, what is the probability that the student chose pizza?

A) $\dfrac{7}{50}$
B) $\dfrac{3}{7}$
C) $\dfrac{7}{16}$
D) $\dfrac{21}{50}$

Show solution

Total students: $42 + 18 + 30 + 10 = 100$.

\[P(\text{pizza}) = \frac{42}{100} = \frac{21}{50}\]

1. $\dfrac{7}{50}$: used $14$ (roughly one-third of 42) in the numerator rather than 42
1. $\dfrac{3}{7}$: used only the pizza and salad counts ($42 + 18 = 60$) as the denominator
1. $\dfrac{7}{16}$: used the 4 category count as a denominator of 16 instead of total students

Answer: D

Question 8

In the $xy$-plane, the line $y = kx - 4$ passes through the point $(3, 8)$. What is the value of $k$?

A) $\dfrac{4}{3}$
B) $3$
C) $4$
D) $12$

Show solution

Substitute $(x, y) = (3, 8)$:

\[8 = k(3) - 4 \Rightarrow 12 = 3k \Rightarrow k = 4\]

1. $\dfrac{4}{3}$: substituted into $8 = 3k$ without first adding 4 to both sides
1. $3$: used the $x$-coordinate of the given point as the answer
1. $12$: reported the value of $3k$ ($= 12$) instead of $k$ itself

Answer: C

Question 9

The function $f$ is defined by $f(x) = x^2 - 3x + 1$. What is the value of $f(4)$?

Enter your answer:

Show solution

\[f(4) = (4)^2 - 3(4) + 1 = 16 - 12 + 1 = 5\]

Answer: 5

Question 10

A right triangle has legs of length $5$ and $12$. What is the length of the hypotenuse?

A) $\sqrt{17}$
B) $13$
C) $17$
D) $60$

Show solution

By the Pythagorean theorem: $c^2 = 5^2 + 12^2 = 25 + 144 = 169$, so $c = 13$.

This is the well-known 5–12–13 Pythagorean triple.

1. $\sqrt{17}$: computed $\sqrt{5 + 12}$ instead of $\sqrt{5^2 + 12^2}$
1. $17$: added the legs ($5 + 12 = 17$) without squaring
1. $60$: multiplied the legs ($5 \times 12 = 60$) instead of applying the Pythagorean theorem

Answer: B

Question 11

A researcher surveyed 200 residents of a city. Of those surveyed, 80 said they exercised at least 3 times per week. Based on this sample, which of the following is the best estimate of the number of residents who exercise at least 3 times per week in a city of 15,000 people?

A) $600$
B) $3{,}000$
C) $6{,}000$
D) $9{,}000$

Show solution

The sample proportion who exercise at least 3 times per week: $\dfrac{80}{200} = 0.40$.

Estimated number in the full city: $0.40 \times 15{,}000 = 6{,}000$.

1. $600$: scaled the sample count of 80 by only a factor of 7.5 instead of 75
1. $3{,}000$: used 20% instead of 40% (perhaps used $\frac{80}{200}$ as $\frac{1}{5}$ rather than $\frac{2}{5}$)
1. $9{,}000$: used 60% instead of 40%

Answer: C

Question 12

In the $xy$-plane, line $\ell$ has slope $3$ and passes through the point $(-1, 2)$. Which of the following is an equation of line $\ell$?

A) $y = 3x - 1$
B) $y = 3x + 1$
C) $y = 3x + 2$
D) $y = 3x + 5$

Show solution

Use point-slope form with slope $3$ and point $(-1, 2)$:

\[y - 2 = 3(x - (-1)) = 3(x + 1) = 3x + 3\]

\[y = 3x + 5\]

Verify: $y(-1) = 3(-1) + 5 = 2$ ✓

1. $y = 3x - 1$: used the $x$-coordinate of the given point ($-1$) as the $y$-intercept
1. $y = 3x + 1$: computed $b = 2 - 3(1)$ instead of $b = 2 - 3(-1)$; ignored the negative sign on $x$
1. $y = 3x + 2$: used the $y$-coordinate of the given point directly as the $y$-intercept without adjusting

Answer: D

Question 13

A line in the $xy$-plane has the equation $6x - 2y = 14$. What is the $y$-intercept of the line?

Enter your answer:

Show solution

Solve for $y$ to find slope-intercept form:

\[6x - 2y = 14 \Rightarrow -2y = -6x + 14 \Rightarrow y = 3x - 7\]

The $y$-intercept is $-7$.

Answer: -7

Question 14

The scatterplot below shows the relationship between the number of hours studied and the score on a math test for 12 students. A line of best fit is shown.

Note: Figure not drawn to scale.

According to the line of best fit shown, which of the following best approximates the predicted score for a student who studied for 7 hours?

A) $88$
B) $93$
C) $95$
D) $100$

Show solution

The line of best fit has the approximate equation $\text{score} \approx 43 + 7.5 \times \text{hours}$.

At 7 hours: $\text{score} \approx 43 + 7.5(7) = 43 + 52.5 = 95.5 \approx 95$.

1. $88$: reads from the line at 6 hours instead of 7
1. $93$: underestimates the slope of the line during extrapolation
1. $100$: assumes the score reaches the maximum at exactly 7 hours, but the extrapolated line gives approximately 95–96

Answer: C

Question 15

In the figure below, lines $m$ and $n$ are parallel and are cut by transversal $t$. The measure of angle 1 is $(4x + 10)°$ and the measure of angle 2 is $(6x - 20)°$, as shown.

Note: Figure not drawn to scale.

What is the measure of angle 2?

A) $55°$
B) $70°$
C) $110°$
D) $125°$

Show solution

Angles 1 and 2 are alternate interior angles formed by parallel lines cut by a transversal. Alternate interior angles are equal:

\[4x + 10 = 6x - 20\]

\[30 = 2x \Rightarrow x = 15\]

\[\angle 2 = 6(15) - 20 = 90 - 20 = 70°\]

Verify: $\angle 1 = 4(15) + 10 = 70°$ ✓

1. $55°$: solved $6x - 20 = 4x + 10$ but made an arithmetic error getting $2x = 10 \Rightarrow x = 5$ rather than $2x = 30 \Rightarrow x = 15$
1. $110°$: computed the supplement of the correct answer ($180° - 70°$), treating the angles as supplementary instead of equal
1. $125°$: used $x = 15$ but computed $\angle 2 = 6(15) + 5 = 95$ or otherwise made an arithmetic error

Answer: B

Question 16

If $f(x) = 2x^2 + 3$ and $g(x) = x - 4$, what is the value of $f(g(3))$?

A) $-1$
B) $5$
C) $21$
D) $25$

Show solution

First evaluate the inner function: $g(3) = 3 - 4 = -1$.

Then evaluate $f$ at that result:

\[f(-1) = 2(-1)^2 + 3 = 2(1) + 3 = 5\]

1. $-1$: stopped after computing $g(3)$ without evaluating $f$
1. $21$: evaluated $f(3)$ directly: $2(3)^2 + 3 = 21$; ignored the composition with $g$
1. $25$: evaluated $g(f(3))$ instead of $f(g(3))$: $f(3) = 21$, $g(21) = 17$… or computed $f(g(3))$ by squaring the whole expression $2(-1)^2 + 3 + \text{extra}$

Answer: B

Question 17

A rectangular garden has a length that is 4 feet more than twice its width. If the perimeter of the garden is 56 feet, what is the width, in feet, of the garden?

Enter your answer:

Show solution

Let the width be $w$. Then the length is $2w + 4$.

Perimeter formula: $2(\text{length} + \text{width}) = 56$

\[2((2w + 4) + w) = 56\] \[2(3w + 4) = 56\] \[6w + 8 = 56\] \[6w = 48 \Rightarrow w = 8\]

Answer: 8

Question 18

The system of equations below has solution $(x, y)$.

\[2x + 3y = 16\] \[4x - y = 4\]

What is the value of $y$?

A) $1$
B) $2$
C) $3$
D) $4$

Show solution

From the second equation, solve for $y$: $y = 4x - 4$.

Substitute into the first equation:

\[2x + 3(4x - 4) = 16\] \[2x + 12x - 12 = 16\] \[14x = 28 \Rightarrow x = 2\]

Then $y = 4(2) - 4 = 4$.

Verify in the first equation: $2(2) + 3(4) = 4 + 12 = 16$ ✓

1. $1$: solved for $x$ and reported $x$ instead of $y$, or made an arithmetic error in substitution
1. $2$: reported the value of $x$ (not $y$)
1. $3$: arithmetic error when substituting; solved $14x = 28$ as $x = 3$ instead of $x = 2$

Answer: D

Question 19

Which of the following is equivalent to $\dfrac{x^2 - 16}{x^2 + 2x - 8}$ for $x \neq 2$ and $x \neq -4$?

A) $\dfrac{x - 4}{x - 2}$
B) $\dfrac{x + 4}{x + 2}$
C) $\dfrac{x - 4}{x + 2}$
D) $\dfrac{x + 4}{x - 2}$

Show solution

Factor numerator and denominator:

\[x^2 - 16 = (x + 4)(x - 4)\]

\[x^2 + 2x - 8 = (x + 4)(x - 2)\]

Cancel the common factor $(x + 4)$ (valid since $x \neq -4$):

\[\frac{(x+4)(x-4)}{(x+4)(x-2)} = \frac{x - 4}{x - 2}\]

1. $\dfrac{x+4}{x+2}$: incorrectly factored the denominator as $(x+4)(x+2)$
1. $\dfrac{x-4}{x+2}$: incorrectly factored the denominator and placed the wrong sign
1. $\dfrac{x+4}{x-2}$: canceled $(x-4)$ from numerator and denominator instead of $(x+4)$

Answer: A

Question 20

A circle has a circumference of 20. A central angle of $72°$ cuts off an arc of the circle. What is the length of that arc?

Enter your answer:

Show solution

Arc length equals the fraction of the full circle determined by the central angle, multiplied by the circumference:

\[\text{arc length} = \frac{72°}{360°} \times 20 = \frac{1}{5} \times 20 = 4\]

Answer: 4

Question 21

The graph of $y = f(x)$ is a parabola with vertex at $(3, -4)$. The graph is shifted 2 units to the left and 5 units up to produce a new parabola. What are the coordinates of the vertex of the new parabola?

A) $(1, 1)$
B) $(1, 9)$
C) $(5, 1)$
D) $(5, 9)$

Show solution

A horizontal shift left by 2 decreases the $x$-coordinate of the vertex: $3 - 2 = 1$.

A vertical shift up by 5 increases the $y$-coordinate: $-4 + 5 = 1$.

The new vertex is $(1, 1)$.

1. $(1, 9)$: correctly shifted $x$ left, but treated $-4$ as $+4$ before adding 5, giving $4 + 5 = 9$
1. $(5, 1)$: shifted right 2 instead of left 2 ($3 + 2 = 5$), correctly shifted $y$
1. $(5, 9)$: shifted right instead of left, and used $|-4| + 5 = 9$ for the $y$-coordinate

Answer: A

Question 22

In the figure below, triangle $ABC$ is similar to triangle $DEF$, with $AB$ corresponding to $DE$. If $AB = 6$, $BC = 9$, $AC = 12$, and $DE = 4$, what is the perimeter of triangle $DEF$?

Note: Figure not drawn to scale.

Enter your answer:

Show solution

The scale factor from triangle $ABC$ to triangle $DEF$ is:

\[k = \frac{DE}{AB} = \frac{4}{6} = \frac{2}{3}\]

Perimeter of triangle $ABC = 6 + 9 + 12 = 27$.

Since similar triangles have perimeters in the same ratio as their corresponding sides:

\[\text{Perimeter of } DEF = 27 \times \frac{2}{3} = 18\]

Answer: 18

Module 2

Question 23

If $\dfrac{x}{5} + 3 = 7$, what is the value of $x$?

A) $4$
B) $20$
C) $25$
D) $50$

Show solution

Isolate $x$:

\[\frac{x}{5} + 3 = 7 \Rightarrow \frac{x}{5} = 4 \Rightarrow x = 20\]

1. $4$: reported the value of $\dfrac{x}{5}$ instead of $x$
1. $25$: multiplied both sides of the original equation by 5 before subtracting 3: $x + 15 = 35 \Rightarrow x = 20$… or added $3 + 5 \cdot 4 = 25$ incorrectly
1. $50$: multiplied 7 by 5 without subtracting 3 first

Answer: B

Question 24

A line in the $xy$-plane passes through the points $(0, 6)$ and $(4, 0)$. Which of the following is an equation of this line?

A) $y = -\dfrac{3}{2}x + 6$
B) $y = -\dfrac{2}{3}x + 6$
C) $y = \dfrac{3}{2}x + 6$
D) $y = \dfrac{2}{3}x + 4$

Show solution

The $y$-intercept is $(0, 6)$, so $b = 6$.

Slope: $m = \dfrac{0 - 6}{4 - 0} = \dfrac{-6}{4} = -\dfrac{3}{2}$.

Equation: $y = -\dfrac{3}{2}x + 6$.

1. $y = -\dfrac{2}{3}x + 6$: inverted the slope ($\Delta x / \Delta y$ instead of $\Delta y / \Delta x$)
1. $y = \dfrac{3}{2}x + 6$: correct magnitude of slope but wrong sign (line rises instead of falls)
1. $y = \dfrac{2}{3}x + 4$: used the $x$-intercept as the $y$-intercept and inverted the sign

Answer: A

Question 25

For what value of $k$ does the equation $2x + k = 3x - 5$ have the solution $x = 9$?

A) $-14$
B) $-4$
C) $4$
D) $14$

Show solution

Substitute $x = 9$ into the equation and solve for $k$:

\[2(9) + k = 3(9) - 5\] \[18 + k = 27 - 5 = 22\] \[k = 4\]

1. $-14$: computed $k = 9 - 5 - 18 = -14$ with an arithmetic error
1. $-4$: solved $18 + k = 22$ as $k = 22 - 18$ but got the sign wrong: $k = -4$
1. $14$: added 18 to 22 instead of subtracting: $k = 22 + 18 - 18$… or computed $k = 5 + 9 = 14$

Answer: C

Question 26

The population of a town is modeled by the function $P(t) = 4{,}000 \cdot (1.05)^t$, where $t$ is the number of years after 2010. Which of the following best describes what the value $1.05$ represents in this context?

A) The population increases by $1.05$ people each year.
B) The population in 2010 was $1.05$ thousand people.
C) The population increases by $5\%$ each year.
D) The population doubles every $1.05$ years.

Show solution

In the exponential model $P(t) = P_0 \cdot b^t$, the base $b$ is the growth factor. When $b = 1.05$, each year the population is multiplied by $1.05$, which is an increase of $5\%$ per year ($1.05 - 1 = 0.05 = 5\%$).

1. Confuses multiplicative growth (a percent increase) with additive growth (a fixed amount per year)
1. $4{,}000$ is the initial population, not $1.05$; $P(0) = 4{,}000$
1. The doubling time requires solving $(1.05)^t = 2$, which gives $t \approx 14.2$ years, not $1.05$

Answer: C

Question 27

In the $xy$-plane, the graphs of $y = 2x + 1$ and $y = x^2 - 2$ intersect at two points. What is the $x$-coordinate of the point of intersection where $x > 0$?

Enter your answer:

Show solution

Set the expressions equal:

\[2x + 1 = x^2 - 2\] \[0 = x^2 - 2x - 3\] \[0 = (x - 3)(x + 1)\]

Solutions: $x = 3$ or $x = -1$.

Since $x > 0$, the answer is $x = 3$.

Answer: 3

Question 28

A data set has a mean of 50 and a standard deviation of 8. A new data set is created by adding 10 to every value in the original data set. Which of the following correctly describes the new data set?

A) Mean $= 50$, standard deviation $= 18$
B) Mean $= 60$, standard deviation $= 8$
C) Mean $= 60$, standard deviation $= 18$
D) Mean $= 50$, standard deviation $= 8$

Show solution

Adding a constant to every value shifts all data points by the same amount, so the mean increases by 10: new mean $= 50 + 10 = 60$.

Standard deviation measures spread (distances between values), which does not change when every value shifts by the same constant. New standard deviation $= 8$.

1. Mean unchanged but standard deviation increases — incorrect on both counts
1. Mean increases (correct) and standard deviation also increases — the spread does not change
1. Neither changes — ignores the effect on the mean

Answer: B

Question 29

In the figure below, right triangle $PQR$ has a right angle at $Q$. The length of $PQ$ is $7$ and the length of $PR$ is $25$. What is $\sin(\angle P)$?

Note: Figure not drawn to scale.

A) $\dfrac{7}{25}$
B) $\dfrac{24}{25}$
C) $\dfrac{7}{24}$
D) $\dfrac{24}{7}$

Show solution

First find $QR$ using the Pythagorean theorem:

\[QR = \sqrt{PR^2 - PQ^2} = \sqrt{625 - 49} = \sqrt{576} = 24\]

$\sin(\angle P) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{QR}{PR} = \dfrac{24}{25}$

The side opposite $\angle P$ is $QR$ (not $PQ$, which is adjacent to $\angle P$).

1. $\dfrac{7}{25}$: used $PQ$ (the adjacent leg) as the opposite side; this equals $\cos(\angle P)$
1. $\dfrac{7}{24}$: divided the adjacent leg by the opposite leg; this equals $\tan(\angle R)$
1. $\dfrac{24}{7}$: ratio of opposite to adjacent leg, which is $\tan(\angle P)$ and exceeds 1

Answer: B

Question 30

The system of equations below has infinitely many solutions. What is the value of $c$?

\[3x - 2y = 8\] \[6x - 4y = c\]

A) $4$
B) $8$
C) $16$
D) $32$

Show solution

For a system to have infinitely many solutions, the two equations must represent the same line — one must be a scalar multiple of the other.

The second equation’s left side, $6x - 4y$, equals exactly $2(3x - 2y)$ — a factor of 2 times the first equation’s left side.

For the equations to be identical, the right side must scale by the same factor:

\[c = 2 \times 8 = 16\]

When $c = 16$, both equations are equivalent and every point on the line $3x - 2y = 8$ is a solution.

1. $4$: divided instead of multiplied: $8 \div 2 = 4$
1. $8$: used the same constant as the first equation without scaling
1. $32$: multiplied by 4 instead of 2

Answer: C

Question 31

If $x^2 - 8x + 7 = 0$, what is the sum of all values of $x$ that satisfy the equation?

Enter your answer:

Show solution

Factor: $x^2 - 8x + 7 = (x - 1)(x - 7) = 0$.

Solutions: $x = 1$ and $x = 7$.

Sum $= 1 + 7 = 8$.

Alternatively, by Vieta’s formulas, the sum of roots of $x^2 + bx + c = 0$ is $-b$. Here $b = -8$, so sum $= -(-8) = 8$.

Answer: 8

Question 32

A two-way table shows the results of a survey asking 120 students whether they prefer reading fiction or nonfiction, broken down by grade.

	Fiction	Nonfiction	Total
Grade 11	28	32	60
Grade 12	36	24	60
Total	64	56	120

What is the probability that a randomly selected student is in Grade 12, given that the student prefers fiction?

A) $\dfrac{3}{10}$
B) $\dfrac{9}{16}$
C) $\dfrac{3}{8}$
D) $\dfrac{9}{20}$

Show solution

This is a conditional probability question: $P(\text{Grade 12} \mid \text{Fiction})$.

\[P(\text{Grade 12} \mid \text{Fiction}) = \frac{\text{Grade 12 and Fiction}}{\text{Total Fiction}} = \frac{36}{64} = \frac{9}{16}\]

1. $\dfrac{3}{10}$: divided Grade 12 Fiction (36) by the total number of students (120): $\frac{36}{120} = \frac{3}{10}$ — this is the joint probability, not the conditional
1. $\dfrac{3}{8}$: used Grade 11 Fiction (28) over Grade 12 total (60): incorrect numerator and denominator
1. $\dfrac{9}{20}$: divided Grade 12 Fiction (36) by Grade 12 total (60): $\frac{36}{60} = \frac{3}{5}$… or used wrong denominator of 80

Answer: B

Question 33

In the $xy$-plane, a circle has the equation $(x - 3)^2 + (y + 2)^2 = 49$. Which of the following is the center and radius of the circle?

A) Center $(-3, 2)$, radius $7$
B) Center $(3, -2)$, radius $7$
C) Center $(3, -2)$, radius $49$
D) Center $(-3, 2)$, radius $49$

Show solution

The standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$, with center $(h, k)$ and radius $r$.

Comparing $(x - 3)^2 + (y + 2)^2 = 49$:

$h = 3$, $k = -2$ (note: $y + 2 = y - (-2)$, so $k = -2$)
$r^2 = 49 \Rightarrow r = 7$

Center: $(3, -2)$, radius: $7$.

1. Center $(-3, 2)$: negated both coordinates, confusing the signs in standard form
1. Center correct but radius $= 49$ instead of $\sqrt{49} = 7$; forgot to take the square root of $r^2$
1. Both errors: negated coordinates and used $r = 49$

Answer: B

Question 34

For the polynomial $p(x) = x^3 - 2x^2 - 5x + 6$, which of the following is a factor?

A) $(x + 1)$
B) $(x - 2)$
C) $(x + 2)$
D) $(x + 3)$

Show solution

By the Factor Theorem, $(x - c)$ is a factor of $p(x)$ if and only if $p(c) = 0$. Test each choice:

1. $(x + 1)$: $p(-1) = (-1)^3 - 2(-1)^2 - 5(-1) + 6 = -1 - 2 + 5 + 6 = 8 \neq 0$
1. $(x - 2)$: $p(2) = 8 - 8 - 10 + 6 = -4 \neq 0$
1. $(x + 2)$: $p(-2) = (-8) - 2(4) - 5(-2) + 6 = -8 - 8 + 10 + 6 = 0$ ✓
1. $(x + 3)$: $p(-3) = (-27) - 2(9) - 5(-3) + 6 = -27 - 18 + 15 + 6 = -24 \neq 0$

Only $(x + 2)$ produces $p(-2) = 0$.

For reference, the full factorization is $p(x) = (x - 1)(x - 3)(x + 2)$.

1. $(x+1)$: $p(-1) = 8 \neq 0$; a plausible-looking factor that students may test carelessly
1. $(x-2)$: $p(2) = -4 \neq 0$; close to a root but not one
1. $(x+3)$: $p(-3) = -24 \neq 0$; also not a root

Answer: C

Question 35

The function $f$ is defined by $f(x) = \dfrac{3x + 12}{x - 1}$. For what value of $x$ is $f(x)$ undefined?

Enter your answer:

Show solution

A rational function is undefined when its denominator equals zero:

\[x - 1 = 0 \Rightarrow x = 1\]

Answer: 1

Question 36

The equation $x^2 + bx + 36 = 0$ has exactly one real solution. Which of the following could be the value of $b$?

A) $-6$
B) $6$
C) $12$
D) $18$

Show solution

For exactly one real solution, the discriminant must equal zero:

\[\Delta = b^2 - 4ac = b^2 - 4(1)(36) = b^2 - 144 = 0\]

\[b^2 = 144 \Rightarrow b = \pm 12\]

Among the choices, $b = 12$ satisfies this condition.

Distractor check: - $b = -6$: $\Delta = 36 - 144 = -108 < 0$ → no real solutions - $b = 6$: $\Delta = 36 - 144 = -108 < 0$ → no real solutions - $b = 18$: $\Delta = 324 - 144 = 180 > 0$ → two distinct real solutions

1. $-6$: gives a negative discriminant (no real solutions); student may confuse the sign requirement
1. $6$: same issue as $-6$; both give $\Delta = -108 < 0$
1. $18$: gives positive discriminant (two real solutions); $b$ is too large

Answer: C

Question 37

If $g(x) = 3x - 1$, which of the following defines $g^{-1}(x)$, the inverse of $g$?

A) $g^{-1}(x) = \dfrac{x + 1}{3}$
B) $g^{-1}(x) = \dfrac{x - 1}{3}$
C) $g^{-1}(x) = 3x + 1$
D) $g^{-1}(x) = \dfrac{1}{3x - 1}$

Show solution

To find the inverse, swap $x$ and $y$ then solve for $y$. Starting with $y = 3x - 1$:

\[x = 3y - 1\] \[x + 1 = 3y\] \[y = \frac{x + 1}{3}\]

So $g^{-1}(x) = \dfrac{x + 1}{3}$.

Verify: $g(g^{-1}(x)) = 3\left(\dfrac{x+1}{3}\right) - 1 = (x+1) - 1 = x$ ✓

1. $\dfrac{x-1}{3}$: subtracted 1 instead of adding 1; sign error when isolating $y$
1. $3x + 1$: swapped the operations of multiply/divide and add/subtract without solving properly
1. $\dfrac{1}{3x-1}$: took the reciprocal of $g(x)$ rather than finding the inverse function

Answer: A

Question 38

An investment of $\$1{,}000$ grows according to the model $V(t) = 1{,}000 \cdot (1.08)^t$, where $t$ is the number of years. After how many complete years will the investment first exceed $\$1{,}400$? (Use $\log(1.08) \approx 0.0334$.)

Enter your answer:

Show solution

Solve $1{,}000 \cdot (1.08)^t > 1{,}400$:

\[(1.08)^t > 1.4\]

Taking $\log$ of both sides:

\[t \cdot \log(1.08) > \log(1.4)\]

\[t > \frac{\log(1.4)}{\log(1.08)} \approx \frac{0.1461}{0.0334} \approx 4.37\]

The first complete year after $t = 4.37$ is $t = 5$.

Verify: $V(5) = 1{,}000 \cdot (1.08)^5 \approx 1{,}000 \cdot 1.469 = 1{,}469 > 1{,}400$ ✓

$V(4) = 1{,}000 \cdot (1.08)^4 \approx 1{,}360 < 1{,}400$ ✓

Answer: 5

Question 39

The expression $\dfrac{2x^2 - x - 6}{x - 2}$ is equivalent to which of the following for $x \neq 2$?

A) $2x - 3$
B) $2x + 3$
C) $x - 3$
D) $2x^2 - 3$

Show solution

Factor the numerator: find two numbers that multiply to $2 \cdot (-6) = -12$ and sum to $-1$. Those are $-4$ and $3$.

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

Cancel $(x - 2)$:

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

Verify: $(x-2)(2x+3) = 2x^2 + 3x - 4x - 6 = 2x^2 - x - 6$ ✓

1. $2x - 3$: sign error in factoring — wrote $(x-2)(2x-3) = 2x^2 - 7x + 6$, which doesn’t match
1. $x - 3$: dropped the leading coefficient of 2 from the quotient
1. $2x^2 - 3$: did not perform division; subtracted 3 from the original without dividing

Answer: B

Question 40

A line passes through the point $(4, 7)$ and is perpendicular to the line $y = \dfrac{2}{3}x - 5$. Which of the following is the equation of the line?

A) $y = \dfrac{2}{3}x + \dfrac{13}{3}$
B) $y = -\dfrac{3}{2}x + 13$
C) $y = -\dfrac{2}{3}x + \dfrac{29}{3}$
D) $y = \dfrac{3}{2}x + 1$

Show solution

The slope of the given line is $\dfrac{2}{3}$. A perpendicular line has slope equal to the negative reciprocal: $m_\perp = -\dfrac{3}{2}$.

Using point-slope form with $(4, 7)$:

\[y - 7 = -\frac{3}{2}(x - 4) = -\frac{3}{2}x + 6\]

\[y = -\frac{3}{2}x + 13\]

Verify: $y(4) = -\frac{3}{2}(4) + 13 = -6 + 13 = 7$ ✓

1. Used the original slope $\dfrac{2}{3}$ instead of the perpendicular slope (parallel line, not perpendicular)
1. Negated the slope but did not take the reciprocal: $m = -\dfrac{2}{3}$
1. Took the reciprocal but did not negate: $m = \dfrac{3}{2}$

Answer: B

Question 41

The function $h$ is defined by $h(x) = \dfrac{x^2 - 9}{x^2 - x - 6}$. For which value of $x$ does $h$ have a hole (removable discontinuity)?

A) $x = -2$
B) $x = 2$
C) $x = 3$
D) $x = -3$

Show solution

Factor numerator and denominator:

\[x^2 - 9 = (x-3)(x+3)\] \[x^2 - x - 6 = (x-3)(x+2)\]

The common factor $(x - 3)$ cancels:

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

The canceled factor $(x - 3)$ creates a hole at $x = 3$.

The remaining factor $(x + 2)$ in the denominator creates a vertical asymptote at $x = -2$.

1. $x = -2$: this is a vertical asymptote, not a hole
1. $x = 2$: not a zero of either the numerator or denominator; not a discontinuity
1. $x = -3$: this is a zero of the numerator only, making $h(-3) = 0$, not a discontinuity

Answer: C

Question 42

In the $xy$-plane, the parabola $y = x^2 - 6x + 5$ intersects the $x$-axis at two points. What is the distance between those two points?

Enter your answer:

Show solution

Find the $x$-intercepts by setting $y = 0$:

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

Solutions: $x = 1$ and $x = 5$.

Distance between the two points $= 5 - 1 = 4$.

Answer: 4

Question 43

A circle in the $xy$-plane has center $(2, -1)$ and passes through the point $(2, 6)$. What is the area of the circle?

A) $7\pi$
B) $14\pi$
C) $25\pi$
D) $49\pi$

Show solution

The radius is the distance from the center $(2, -1)$ to the point on the circle $(2, 6)$:

\[r = \sqrt{(2-2)^2 + (6-(-1))^2} = \sqrt{0 + 49} = 7\]

Area $= \pi r^2 = \pi(7)^2 = 49\pi$.

1. $7\pi$: used $r$ instead of $r^2$ in the area formula ($\pi r$ instead of $\pi r^2$)
1. $14\pi$: used the diameter instead of the radius in the area formula: $\pi(14)$… or $\pi(2r) = 14\pi$
1. $25\pi$: computed the distance incorrectly as $r = 5$ (perhaps used $|6 - 1| = 5$ ignoring the negative)

Answer: D

Question 44

The system of equations below has solution $(x, y)$. What is the value of $x + y$?

\[y = x^2 - 4x + 1\] \[y = -x + 5\]

Enter your answer:

Show solution

Set the two expressions equal:

\[x^2 - 4x + 1 = -x + 5\] \[x^2 - 3x - 4 = 0\] \[(x - 4)(x + 1) = 0\]

Solutions: $x = 4$ or $x = -1$.

Find corresponding $y$ values from $y = -x + 5$: - If $x = 4$: $y = -4 + 5 = 1$ → point $(4, 1)$, sum $= 5$ - If $x = -1$: $y = 1 + 5 = 6$ → point $(-1, 6)$, sum $= 5$

Both intersection points give $x + y = 5$.

Answer: 5