Full-Length Practice Test 2

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

Question 1

What is the value of $5 + 3 \times 4 - 2$?

A) $15$
B) $17$
C) $30$
D) $34$

Show solution

Apply the order of operations — multiplication before addition/subtraction:

Multiply first: $3 \times 4 = 12$. Then left to right: $5 + 12 - 2 = 15$.

1. $17$: stopped after $5 + 12 = 17$ without subtracting $2$
1. $30$: computed $(5 + 3) \times 4 - 2 = 32 - 2 = 30$ (added before multiplying)
1. $34$: computed $(5 + 3) \times 4 + 2 = 34$ (added before multiplying and added instead of subtracting)

Answer: A

Question 2

A store sells notebooks for $\$3.50$ each and pens for $\$1.25$ each. If Mia buys 4 notebooks and 3 pens, how much does she spend in total?

F) $\$14.00$
G) $\$17.25$
H) $\$17.75$
J) $\$18.50$

Show solution

Cost of notebooks: $4 \times \$3.50 = \$14.00$.

Cost of pens: $3 \times \$1.25 = \$3.75$.

Total: $\$14.00 + \$3.75 = \$17.75$.

1. $\$14.00$: computed only the notebook cost and forgot the pens
1. $\$17.25$: used $3 \times \$1.00 = \$3.00$ (rounding error on pens), giving $\$14.00 + \$3.25 = \$17.25$
1. $\$18.50$: used $\$1.50$ per pen instead of $\$1.25$: $14.00 + 4.50 = 18.50$

Answer: H

Question 3

Which of the following expressions is equivalent to $x^2 + 7x + 12$?

A) $(x + 3)(x + 4)$
B) $(x - 3)(x - 4)$
C) $(x + 2)(x + 6)$
D) $(x + 1)(x + 12)$

Show solution

Find two numbers that multiply to $+12$ and add to $+7$: those numbers are $+3$ and $+4$.

\[x^2 + 7x + 12 = (x + 3)(x + 4)\]

Check: $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ ✓

1. $(x-3)(x-4)$: both factors have the wrong sign; expands to $x^2 - 7x + 12$
1. $(x+2)(x+6)$: multiplies to $12$ but adds to $8$, not $7$; expands to $x^2 + 8x + 12$
1. $(x+1)(x+12)$: multiplies to $12$ but adds to $13$; expands to $x^2 + 13x + 12$

Answer: A

Question 4

If $4y + 9 = 33$, what is the value of $y$?

F) $4$
G) $6$
H) $8$
J) $10$

Show solution

Subtract $9$ from both sides:

\[4y = 24\]

Divide both sides by $4$:

\[y = 6\]

1. $4$: divided $33$ by $4$ then subtracted (wrong order): $\frac{33}{4} - 9 \approx -0.75$; or divided $33 - 4 = 29$ by something
1. $8$: computed $4y = 33 - 1 = 32$ (arithmetic error on subtraction), giving $y = 8$
1. $10$: added $9$ to both sides instead of subtracting: $4y = 42$, $y = 10.5 \approx 10$

Answer: G

Question 5

Triangle $DEF$ has angles $D$, $E$, and $F$. If $\angle D = 47°$ and $\angle E = 68°$, what is the measure of $\angle F$?

A) $55°$
B) $65°$
C) $75°$
D) $115°$

Show solution

The angles of a triangle sum to $180°$:

\[\angle F = 180° - 47° - 68° = 180° - 115° = 65°\]

1. $55°$: computed $180 - 47 - 68$ with an arithmetic slip: $180 - 115 = 65$, not $55$
1. $75°$: computed $47 + 68 = 105$ (addition error) then $180 - 105 = 75$
1. $115°$: reported the sum of $\angle D + \angle E$ instead of subtracting from $180°$

Answer: B

Question 6

If $g(x) = 3x - 5$, what is $g(0)$?

F) $-8$
G) $-5$
H) $0$
J) $3$

Show solution

Substitute $x = 0$:

\[g(0) = 3(0) - 5 = 0 - 5 = -5\]

1. $-8$: computed $3(0) - 5 = -8$ (arithmetic error; treated $3(0)$ as $3$ rather than $0$, then $3 - 5 - 6$… or subtracted an extra $3$)
1. $0$: set $g(0) = 0$ (found a zero of the function rather than evaluating at $x=0$)
1. $3$: returned the coefficient of $x$ without substituting, or evaluated $g(3)$… or computed $3 - 0 = 3$ (reversed the expression)

Answer: G

Question 7

A circle has a diameter of 14 centimeters. What is the area of the circle, in square centimeters?

A) $14\pi$
B) $28\pi$
C) $49\pi$
D) $196\pi$

Show solution

The diameter is $14$, so the radius is $r = 7$.

\[A = \pi r^2 = \pi (7)^2 = 49\pi\]

1. $14\pi$: used the diameter directly as $r$ in a formula, giving $\pi(14)/1 = 14\pi$; or computed $\pi d/\pi = d$
1. $28\pi$: computed the circumference $\pi d = 14\pi$ and doubled it, or used the circumference formula $2\pi r = 14\pi$ and doubled
1. $196\pi$: used the diameter as the radius: $\pi(14)^2 = 196\pi$

Answer: C

Question 8

A jar contains 6 yellow marbles, 4 red marbles, and 2 white marbles. If one marble is chosen at random, what is the probability that it is NOT yellow?

F) $\dfrac{1}{2}$
G) $\dfrac{1}{3}$
H) $\dfrac{2}{3}$
J) $\dfrac{3}{4}$

Show solution

Total marbles $= 6 + 4 + 2 = 12$.

Non-yellow marbles $= 4 + 2 = 6$.

\[P(\text{not yellow}) = \frac{6}{12} = \frac{1}{2}\]

1. $\frac{1}{3}$: computed $P(\text{yellow}) = \frac{6}{12} = \frac{1}{2}$… or used denominator 6 (yellow count): $\frac{2}{6} = \frac{1}{3}$
1. $\frac{2}{3}$: used only 9 marbles in the denominator (dropped the white marbles): $\frac{6}{9} = \frac{2}{3}$
1. $\frac{3}{4}$: used denominator 8 (non-yellow marbles as denominator): $\frac{6}{8} = \frac{3}{4}$, confusing part for whole

Answer: F

Question 9

What is the perimeter of a rectangle with length $9$ inches and width $5$ inches?

A) $14$ inches
B) $28$ inches
C) $45$ inches
D) $90$ inches

Show solution

Perimeter of a rectangle $= 2(\ell + w)$:

\[P = 2(9 + 5) = 2(14) = 28 \text{ inches}\]

1. $14$: computed $\ell + w$ but forgot to multiply by $2$
1. $45$: computed the area $\ell \times w = 9 \times 5 = 45$ instead of the perimeter
1. $90$: doubled the area ($2 \times 45 = 90$) or computed $2 \times \ell \times w$

Answer: B

Question 10

The table below shows the number of books read by 5 students during the summer.

Student	Books Read
Ali	4
Beth	7
Carlos	3
Dana	9
Evan	2

What is the median number of books read?

F) $3$
G) $4$
H) $5$
J) $7$

Show solution

Arrange the values in order: $2, 3, 4, 7, 9$.

The middle value (3rd of 5) is $4$.

1. $3$: selected the 2nd value instead of the middle value, or sorted incorrectly
1. $5$: computed the mean: $(4+7+3+9+2)/5 = 25/5 = 5$, confusing mean with median
1. $7$: selected the 4th value instead of the middle value

Answer: G

Question 11

What is the slope of the line $3x + 6y = 12$?

A) $-2$
B) $-\dfrac{1}{2}$
C) $2$
D) $3$

Show solution

Solve for $y$ to put the equation in slope-intercept form:

\[6y = -3x + 12 \implies y = -\frac{1}{2}x + 2\]

The slope is $m = -\dfrac{1}{2}$.

1. $-2$: used $-\frac{6}{3} = -2$ (inverted the coefficients)
1. $2$: dropped the negative sign and/or inverted the ratio
1. $3$: read the coefficient of $x$ directly from the standard form without converting

Answer: B

Question 12

In right triangle $RST$, the right angle is at $T$. If $\tan(\angle R) = \dfrac{5}{12}$, what is $\sin(\angle R)$?

F) $\dfrac{5}{12}$
G) $\dfrac{5}{13}$
H) $\dfrac{12}{13}$
J) $\dfrac{13}{12}$

Show solution

$\tan(\angle R) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}$.

So the opposite leg $= 5$ and the adjacent leg $= 12$.

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

\[\sin(\angle R) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}\]

1. $\dfrac{5}{12}$: confused $\sin$ with $\tan$ — reported $\tan(\angle R)$ as the answer
1. $\dfrac{12}{13}$: computed $\cos(\angle R) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$ instead of $\sin$
1. $\dfrac{13}{12}$: inverted the cosine ratio; a value greater than 1 cannot be a sine value

Answer: G

Question 13

What is $40\%$ of $75$?

A) $18.75$
B) $28$
C) $30$
D) $35$

Show solution

\[40\% \times 75 = 0.40 \times 75 = 30\]

1. $18.75$: computed $25\%$ of $75 = 18.75$
1. $28$: computed $40\%$ of $70 = 28$ (used $70$ instead of $75$)
1. $35$: computed $\frac{75}{40} \times 18.67...$, or computed $50\%$ of $70 = 35$; a common off-by-one error

Answer: C

Question 14

The sum of three consecutive integers is 51. What is the largest of the three integers?

F) $16$
G) $17$
H) $18$
J) $19$

Show solution

Let the three consecutive integers be $n$, $n+1$, and $n+2$:

\[n + (n+1) + (n+2) = 51 \implies 3n + 3 = 51 \implies 3n = 48 \implies n = 16\]

The three integers are $16$, $17$, and $18$. The largest is $\mathbf{18}$.

1. $16$: reported the smallest integer $n$ instead of the largest $n+2$
1. $17$: reported the middle integer $n+1$ instead of the largest
1. $19$: computed the largest integer as $n + 3 = 19$ (used $n+3$ instead of $n+2$)

Answer: H

Question 15

A line passes through the points $(0, -3)$ and $(4, 5)$. What is the equation of the line?

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

Show solution

Slope: $m = \dfrac{5 - (-3)}{4 - 0} = \dfrac{8}{4} = 2$.

The line passes through $(0, -3)$, so the $y$-intercept is $b = -3$.

Equation: $y = 2x - 3$.

1. $y = 2x + 3$: correct slope but wrong sign on the $y$-intercept
1. $y = \frac{1}{2}x - 3$: inverted the slope to $\frac{4}{8} = \frac{1}{2}$ instead of $\frac{8}{4} = 2$
1. $y = -2x - 3$: correct $y$-intercept but wrong sign on slope (computed $\frac{-8}{4}$)

Answer: A

Question 16

Two trains leave the same station at the same time and travel in opposite directions. Train A travels at 55 mph and Train B travels at 70 mph. After how many hours will the two trains be 375 miles apart?

F) $2$
G) $3$
H) $4$
J) $5$

Show solution

The trains travel in opposite directions, so their combined speed is $55 + 70 = 125$ mph.

Let $t$ = time in hours:

\[125t = 375 \implies t = 3 \text{ hours}\]

1. $2$: computed $375 / 125 = 3$ but reported $2$; or divided by only one train’s speed: $375 / 70 \approx 5.4$, or used a combined speed of $125/2 \approx 62.5$ and rounded
1. $4$: divided the total distance by only one speed: $375 / 70 \approx 5.4$… or used $375/55 \approx 6.8$; or computed the combined rate as $70 - 55 = 15$, giving $375/15 = 25$ hours — that’s a different error
1. $5$: used only Train B’s speed: $375 / 70 \approx 5.4 \approx 5$

Answer: G

Question 17

What is the value of $x$ in the solution to the system of equations below?

\[\begin{cases} 3x + 2y = 16 \\ x - y = 2 \end{cases}\]

A) $2$
B) $3$
C) $4$
D) $6$

Show solution

From the second equation: $x = y + 2$. Substitute into the first:

\[3(y + 2) + 2y = 16 \implies 3y + 6 + 2y = 16 \implies 5y = 10 \implies y = 2\]

Then $x = y + 2 = 4$.

1. $2$: reported the value of $y$ instead of $x$
1. $3$: substitution error — e.g. $3y + 2 + 2y = 16$, $5y = 14$, $y \approx 2.8$, $x \approx 4.8$; or a different arithmetic error
1. $6$: added the right-hand sides: $x = 16 + 2 - 12 = 6$, an incorrect shortcut

Answer: C

Question 18

The graph of $y = f(x)$ is shown below. Which of the following could be the equation of $f(x)$?

Note: Figure not drawn to scale.

F) $f(x) = (x - 1)^2 - 4$
G) $f(x) = (x + 1)^2 - 4$
H) $f(x) = (x + 1)^2 + 4$
J) $f(x) = (x - 1)^2 + 4$

Show solution

The parabola opens upward and has vertex at $(-1, -4)$.

Vertex form: $y = (x - h)^2 + k$ with vertex $(h, k) = (-1, -4)$:

\[f(x) = (x - (-1))^2 + (-4) = (x + 1)^2 - 4\]

Verify roots: $(x+1)^2 = 4 \implies x + 1 = \pm 2 \implies x = 1$ or $x = -3$ ✓ (matches the graph)

1. $(x-1)^2 - 4$: vertex at $(1, -4)$, not $(-1,-4)$ — sign error on $h$
1. $(x+1)^2 + 4$: vertex at $(-1, +4)$ — vertex is above the $x$-axis, but the graph shows it below
1. $(x-1)^2 + 4$: vertex at $(1, 4)$ — both signs wrong

Answer: G

Question 19

A lamp post that is 12 feet tall casts a shadow that is 9 feet long. At the same time, a nearby tree casts a shadow that is 15 feet long. How tall, in feet, is the tree?

A) $11.25$
B) $16$
C) $18$
D) $20$

Show solution

Similar triangles: the ratio of height to shadow is constant.

\[\frac{\text{lamp height}}{\text{lamp shadow}} = \frac{\text{tree height}}{\text{tree shadow}}\]

\[\frac{12}{9} = \frac{h}{15}\]

Cross-multiply:

\[9h = 180 \implies h = 20 \text{ feet}\]

1. $11.25$: inverted the ratio — $\frac{9}{12} \times 15 = 11.25$ (divided instead of the correct setup)
1. $16$: computed $12 + (15 - 9) = 12 + 6 = 18$… or a different arithmetic error
1. $18$: computed $12 + \frac{15 \cdot 9}{12} = 12 + ?$… or used an additive rather than multiplicative relationship

Answer: D

Question 20

A survey of 120 students asked whether they prefer morning or evening study sessions, broken down by grade.

	Morning	Evening	Total
Grade 9	28	32	60
Grade 10	22	38	60
Total	50	70	120

What is the probability that a randomly selected student prefers evening sessions, given that the student is in Grade 9?

F) $\dfrac{4}{15}$
G) $\dfrac{7}{12}$
H) $\dfrac{8}{15}$
J) $\dfrac{16}{35}$

Show solution

The condition restricts us to Grade 9 students: 60 total.

Of those, 32 prefer evening sessions:

\[P(\text{evening} \mid \text{Grade 9}) = \frac{32}{60} = \frac{8}{15}\]

1. $\dfrac{4}{15}$: used Grade 10 evening students (32… wait, Grade 10 evening = 38) — or divided Grade 9 morning by total: $\frac{32}{120} = \frac{4}{15}$
1. $\dfrac{7}{12}$: divided total evening by total students: $\frac{70}{120} = \frac{7}{12}$ (ignored the Grade 9 condition)
1. $\dfrac{16}{35}$: divided Grade 9 evening by total evening: $\frac{32}{70} = \frac{16}{35}$ (used the wrong denominator)

Answer: H

Question 21

The first term of a geometric sequence is 2, and the common ratio is 3. What is the 5th term of the sequence?

A) $54$
B) $162$
C) $486$
D) $1458$

Show solution

The formula for the $n$th term of a geometric sequence: $a_n = a_1 \cdot r^{n-1}$.

\[a_5 = 2 \cdot 3^{5-1} = 2 \cdot 3^4 = 2 \cdot 81 = 162\]

1. $54$: computed $a_4 = 2 \cdot 3^3 = 54$ (off by one — used $n-2$ instead of $n-1$)
1. $486$: computed $a_6 = 2 \cdot 3^5 = 486$ (used $n$ instead of $n-1$)
1. $1458$: computed $2 \cdot 3^6 = 1458$ (used $n+1$ as the exponent)

Answer: B

Question 22

What is the area of a triangle with vertices at $(0, 0)$, $(8, 0)$, and $(3, 6)$?

F) $12$
G) $24$
H) $36$
J) $48$

Show solution

The base lies along the $x$-axis from $(0,0)$ to $(8,0)$, so $\text{base} = 8$.

The height is the perpendicular distance from $(3,6)$ to the $x$-axis, which is $6$.

\[A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24\]

1. $12$: divided the area by $2$ again, or used only the height: $\frac{1}{2} \times 6 \times 4 = 12$
1. $36$: forgot to halve: $8 \times 6 = 48$… or used base $= 6$ and height $= 12$: $\frac{1}{2}(6)(12) = 36$
1. $48$: computed $\text{base} \times \text{height} = 8 \times 6 = 48$ without halving

Answer: G

Question 23

Which of the following is equivalent to $\sin(90° - \theta)$?

A) $\sin\theta$
B) $-\sin\theta$
C) $\cos\theta$
D) $-\cos\theta$

Show solution

By the complementary angle identity: the sine of an angle equals the cosine of its complement.

\[\sin(90° - \theta) = \cos\theta\]

This can also be verified with a specific value: $\sin(90° - 30°) = \sin(60°) = \frac{\sqrt{3}}{2}$ and $\cos(30°) = \frac{\sqrt{3}}{2}$ ✓

1. $\sin\theta$: confused the identity; $\sin\theta = \cos(90° - \theta)$, not $\sin(90° - \theta)$
1. $-\sin\theta$: applied a reflection identity that doesn’t apply here
1. $-\cos\theta$: the correct function is $\cos\theta$, not $-\cos\theta$

Answer: C

Question 24

A line $\ell$ has equation $y = \dfrac{3}{4}x - 2$. A second line $m$ is perpendicular to $\ell$ and passes through the point $(0, 5)$. What is the equation of line $m$?

F) $y = \dfrac{3}{4}x + 5$
G) $y = -\dfrac{3}{4}x + 5$
H) $y = -\dfrac{4}{3}x + 5$
J) $y = \dfrac{4}{3}x + 5$

Show solution

The slope of $\ell$ is $\dfrac{3}{4}$. A perpendicular line has slope equal to the negative reciprocal:

\[m_\perp = -\frac{4}{3}\]

Line $m$ passes through $(0, 5)$, so the $y$-intercept is $5$:

\[y = -\frac{4}{3}x + 5\]

1. Same slope as $\ell$ — this would be parallel, not perpendicular
1. Negative slope but not the reciprocal: $-\frac{3}{4}$ instead of $-\frac{4}{3}$
1. Reciprocal slope but positive — missed the negative sign

Answer: H

Question 25

A population of bacteria triples every hour. If there are initially 200 bacteria, which function gives the number of bacteria $B$ after $t$ hours?

A) $B(t) = 200 + 3t$
B) $B(t) = 3(200)^t$
C) $B(t) = 200(3)^t$
D) $B(t) = 200(3t)$

Show solution

Exponential growth: $B(t) = B_0 \cdot r^t$, where $B_0$ is the initial amount and $r$ is the growth factor.

Here $B_0 = 200$ and $r = 3$:

\[B(t) = 200 \cdot 3^t\]

Check: at $t=1$, $B = 200 \cdot 3 = 600$ (tripled from 200) ✓.

1. $200 + 3t$: linear model — adds 3 per hour instead of multiplying
1. $3(200)^t$: switched the base and initial value; at $t=1$: $3 \times 200 = 600$ ✓ by coincidence, but at $t=2$: $3 \times 40000 \neq 1800$
1. $200(3t)$: linear in $t$, not exponential; at $t=1$: $600$ ✓ but at $t=2$: $1200 \neq 1800$

Answer: C

Question 26

In the figure below, $\overline{BD}$ is a diameter of the circle with center $O$. If $\angle BAC = 38°$ and $A$ is a point on the circle, what is the measure of arc $BC$?

Note: Figure not drawn to scale.

F) $38°$
G) $52°$
H) $76°$
J) $104°$

Show solution

By the Inscribed Angle Theorem, the intercepted arc is twice the inscribed angle:

\[\text{Arc } BC = 2 \times \angle BAC = 2 \times 38° = 76°\]

1. $38°$: set the arc equal to the inscribed angle (forgot to double)
1. $52°$: computed $90° - 38° = 52°$ — a complementary angle error with no geometric basis here
1. $104°$: doubled correctly to get $76°$, then added to $28°$; or computed the arc $BD$ minus arc $BC$: $180° - 76° = 104°$ (arc $DC$, not arc $BC$)

Answer: H

Question 27

Which of the following values of $x$ satisfies both $x + 3 > 7$ and $2x - 1 < 11$?

A) $2$
B) $4$
C) $5$
D) $8$

Show solution

Solve each inequality:

Inequality 1: $x + 3 > 7 \implies x > 4$

Inequality 2: $2x - 1 < 11 \implies 2x < 12 \implies x < 6$

The solution satisfies $4 < x < 6$.

Of the choices, only $x = 5$ lies in this interval.

1. $2$: satisfies the second inequality ($2(2)-1 = 3 < 11$ ✓) but not the first ($2+3 = 5 \not> 7$)
1. $4$: $x > 4$ is strict, so $x = 4$ does not satisfy the first inequality
1. $8$: satisfies the first inequality ($8 > 4$ ✓) but not the second ($2(8)-1 = 15 \not< 11$)

Answer: C

Question 28

The function $h(x) = -2(x-4)^2 + 8$ represents the height, in feet, of a ball $x$ seconds after it is thrown. What is the maximum height of the ball?

F) $2$ feet
G) $4$ feet
H) $8$ feet
J) $16$ feet

Show solution

The function is in vertex form $h(x) = a(x - h)^2 + k$ with $a = -2$, $h = 4$, $k = 8$.

Since $a < 0$, the parabola opens downward and the vertex is the maximum. The vertex is at $(4, 8)$.

The maximum height is $\mathbf{8}$ feet.

1. $2$: found the leading coefficient’s absolute value instead of reading $k$
1. $4$: read the $x$-coordinate of the vertex ($x = 4$) instead of the height ($k = 8$)
1. $16$: evaluated $h(0) = -2(0-4)^2 + 8 = -32 + 8 = -24$… or computed $-2 \cdot (-4)^2 = -32$, then $+8 = -24$; or doubled the $k$ value

Answer: H

Question 29

How many degrees are in $\dfrac{5\pi}{6}$ radians?

A) $120°$
B) $150°$
C) $210°$
D) $300°$

Show solution

Convert using $180°= \pi$ radians:

\[\frac{5\pi}{6} \times \frac{180°}{\pi} = \frac{5 \times 180°}{6} = \frac{900°}{6} = 150°\]

1. $120°$: converted $\frac{4\pi}{6} = \frac{2\pi}{3}$ or used $\frac{5}{6} \times 144 = 120$; an off-by-fraction error
1. $210°$: converted $\frac{7\pi}{6}$, using $7$ in the numerator instead of $5$
1. $300°$: computed $360° - 60° = 300°$ (complement/supplement confusion) or $\frac{5}{3} \times 180 = 300$

Answer: B

Question 30

A line of best fit is drawn through the scatterplot below, which shows the relationship between advertising spending (in thousands of dollars) and monthly sales (in thousands of units).

Note: Figure not drawn to scale.

According to the line of best fit, what is the predicted monthly sales (in thousands of units) when advertising spending is $\$7{,}000$?

F) $28$
G) $32$
H) $33$
J) $37$

Show solution

The line of best fit passes through approximately $(0, 4.5)$ and $(10, 44.5)$, giving a slope of about $4$ and equation $y \approx 4x + 4.5$.

At $x = 7$ (representing $\$7{,}000$):

\[y \approx 4(7) + 4.5 = 28 + 4.5 = 32.5 \approx 33\]

1. $28$: read the actual data point at $x = 7$ (28 units) rather than the line value, or computed $4 \times 7 = 28$ without adding the intercept
1. $32$: computed $4(7) + 4 = 32$ (used intercept of 4 rather than 4.5)
1. $37$: overestimated the intercept or extrapolated with too steep a slope

Answer: H

Question 31

Which of the following is equivalent to $\dfrac{2x^2 + x - 6}{x^2 - x - 6}$ for all values of $x$ for which the expression is defined?

A) $\dfrac{2x - 3}{x - 3}$
B) $\dfrac{2x + 3}{x + 3}$
C) $\dfrac{2x - 3}{x + 3}$
D) $\dfrac{2x + 3}{x - 3}$

Show solution

Factor the numerator and denominator:

\[\text{Numerator: } 2x^2 + x - 6 = (2x - 3)(x + 2)\]

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

\[\text{Denominator: } x^2 - x - 6 = (x - 3)(x + 2)\]

Cancel the common factor $(x + 2)$:

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

1. $\dfrac{2x+3}{x+3}$: factored the numerator incorrectly as $(2x+3)(x-2)$
1. $\dfrac{2x-3}{x+3}$: factored the denominator incorrectly as $(x+3)(x-2)$
1. $\dfrac{2x+3}{x-3}$: signed errors in both the numerator and denominator factoring

Answer: A

Question 32

If $5x + 2y = 0$ and $y \neq 0$, what is the value of $\dfrac{y}{x}$?

F) $-\dfrac{5}{2}$
G) $-\dfrac{2}{5}$
H) $\dfrac{2}{5}$
J) $\dfrac{5}{2}$

Show solution

Solve for $y$ in terms of $x$:

\[2y = -5x \implies y = -\frac{5}{2}x\]

Therefore:

\[\frac{y}{x} = -\frac{5}{2}\]

1. $-\dfrac{2}{5}$: computed $\dfrac{x}{y}$ instead of $\dfrac{y}{x}$ (inverted the ratio)
1. $\dfrac{2}{5}$: inverted and dropped the negative sign
1. $\dfrac{5}{2}$: read the coefficient of $x$ without accounting for the negative sign from rearranging

Answer: F

Question 33

Given $i = \sqrt{-1}$, what is the value of $(3 + 2i)^2$?

A) $5$
B) $9 + 4i$
C) $5 + 12i$
D) $13$

Show solution

Expand using FOIL:

\[(3 + 2i)^2 = 9 + 12i + 4i^2\]

Since $i^2 = -1$:

\[= 9 + 12i + 4(-1) = 9 + 12i - 4 = 5 + 12i\]

1. $5$: computed the real part $9 - 4 = 5$ but dropped the imaginary part $12i$
1. $9 + 4i$: squared each term separately without the cross term: $(3)^2 + (2i)^2 = 9 - 4 + ?$… or forgot to expand properly, keeping $3^2 = 9$ and $2i$ without squaring
1. $13$: computed $|3 + 2i|^2 = 3^2 + 2^2 = 13$ (the modulus squared, not the square of the complex number)

Answer: C

Question 34

What is the real solution to the equation $\ln(2x - 1) = 3$?

F) $x = \dfrac{e^3 + 1}{2}$
G) $x = e^3 - 1$
H) $x = \dfrac{3}{2} + 1$
J) $x = \dfrac{e^3}{2}$

Show solution

Convert from logarithmic to exponential form:

\[\ln(2x - 1) = 3 \implies 2x - 1 = e^3\]

Solve for $x$:

\[2x = e^3 + 1 \implies x = \frac{e^3 + 1}{2}\]

1. $x = e^3 - 1$: correctly exponentiated but subtracted 1 instead of adding 1 before dividing by 2; i.e. got $2x = e^3$, then $x = e^3/2 - 1$… a careless step
1. $x = \frac{3}{2} + 1 = 2.5$: treated $\ln$ as division by $e$, not exponentiation; solved $\frac{2x-1}{e} = 3$ or similar
1. $x = \dfrac{e^3}{2}$: correctly found $2x - 1 = e^3$ but dropped the $+1$: $2x = e^3$, $x = e^3/2$

Answer: F

Question 35

A function $f$ is defined as:

\[f(x) = \begin{cases} x + 4 & \text{if } x \leq 1 \\ x^2 - 3 & \text{if } x > 1 \end{cases}\]

What is the value of $f(f(-2))$?

A) $-3$
B) $1$
C) $2$
D) $6$

Show solution

Step 1: Find $f(-2)$. Since $-2 \leq 1$, use the first rule:

\[f(-2) = (-2) + 4 = 2\]

Step 2: Find $f(f(-2)) = f(2)$. Since $2 > 1$, use the second rule:

\[f(2) = (2)^2 - 3 = 4 - 3 = 1\]

Therefore $f(f(-2)) = 1$.

1. $-3$: evaluated $f(0) = 0 + 4 = 4$… or applied $x^2 - 3$ to $x = 0$, giving $-3$ (wrong piece for $x = 0$)
1. $2$: stopped after Step 1, reporting $f(-2) = 2$ without computing the outer $f$
1. $6$: applied the second piece to $-2$ directly: $(-2)^2 - 3 + 5 = 6$… or incorrectly used $x^2 - 3$ at $x = 3$: $9 - 3 = 6$

Answer: B

Question 36

Which of the following equations represents an ellipse with a vertical major axis?

F) $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$
G) $\dfrac{x^2}{9} + \dfrac{y^2}{16} = 1$
H) $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$
J) $x^2 + y^2 = 16$

Show solution

For an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$:

If $a^2 > b^2$: major axis is horizontal (along the $x$-axis)
If $b^2 > a^2$: major axis is vertical (along the $y$-axis)

In choice G, $b^2 = 16 > a^2 = 9$, so the major axis is vertical ✓.

1. $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$: $a^2 = 16 > b^2 = 9$ → horizontal major axis (not vertical)
1. $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$: the minus sign makes this a hyperbola, not an ellipse
1. $x^2 + y^2 = 16$: equal coefficients make this a circle, not an ellipse

Answer: G

Question 37

If $f(x) = x^2 - 5x + 4$, which of the following gives $f^{-1}(0)$? (Assume $x \geq \dfrac{5}{2}$.)

A) $0$
B) $1$
C) $4$
D) $5$

Show solution

$f^{-1}(0)$ is the value of $x$ such that $f(x) = 0$.

Solve $x^2 - 5x + 4 = 0$:

\[(x - 1)(x - 4) = 0 \implies x = 1 \text{ or } x = 4\]

Since the restriction is $x \geq \dfrac{5}{2}$, we take $x = 4$.

\[f^{-1}(0) = 4\]

1. $0$: assumed $f^{-1}(0) = f(0)$; computed $f(0) = 4$… or set $f^{-1}(0) = 0$ without solving
1. $1$: found the other root $x = 1$, but this doesn’t satisfy $x \geq 2.5$
1. $5$: off-by-one on the factoring: used $x^2 - 5x + 4 = (x-1)(x-5)$ incorrectly

Answer: C

Question 38

A rectangular garden has a length of $(x + 6)$ meters and a width of $(x - 2)$ meters. If the area of the garden is 48 square meters, what is the value of $x$?

F) $4$
G) $6$
H) $8$
J) $10$

Show solution

Set up the area equation:

\[(x + 6)(x - 2) = 48\]

Expand:

\[x^2 + 4x - 12 = 48 \implies x^2 + 4x - 60 = 0\]

Factor:

\[(x + 10)(x - 6) = 0 \implies x = 6 \text{ or } x = -10\]

Since $x - 2 > 0$ (width must be positive), $x > 2$, so $x = 6$.

Verify: length $= 12$, width $= 4$, area $= 48$ ✓

1. $4$: substituted $x = 4$: $(10)(2) = 20 \neq 48$; found $x = -10 + 6 = -4$, then took the absolute value; or a factoring error
1. $8$: substituted $x = 8$: $(14)(6) = 84 \neq 48$; an arithmetic error in solving the quadratic
1. $10$: found $-x = -10$ from the factored form and reported $x = 10$ (sign error)

Answer: G

Question 39

What is the product of $\begin{bmatrix} 2 & 1 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$?

A) $\begin{bmatrix} 6 \\ 4 \end{bmatrix}$
B) $\begin{bmatrix} 6 & 4 \end{bmatrix}$
C) $[10]$
D) $[14]$

Show solution

This is a $(1 \times 2)$ matrix multiplied by a $(2 \times 1)$ matrix. The result is a $(1 \times 1)$ matrix (a scalar):

\[\begin{bmatrix} 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = 2(3) + 1(4) = 6 + 4 = [10]\]

1. $\begin{bmatrix}6\\4\end{bmatrix}$: multiplied entry-by-entry into a column vector without summing
1. $\begin{bmatrix}6 & 4\end{bmatrix}$: multiplied entry-by-entry into a row vector without summing
1. $[14]$: computed $2(3) + 1(4) = 6 + 4 = 10$… or $2(4) + 1(3) \cdot 2 = 8 + 6 = 14$ (swapped or doubled a term)

Answer: C

Question 40

The polynomial $p(x) = x^3 - 2x^2 - 5x + 6$ has a zero at $x = 1$. Which of the following gives the complete factored form of $p(x)$?

F) $(x - 1)(x - 2)(x + 3)$
G) $(x - 1)(x + 2)(x - 3)$
H) $(x + 1)(x - 2)(x + 3)$
J) $(x - 1)(x^2 - x - 6)$

Show solution

Since $x = 1$ is a zero, $(x - 1)$ is a factor. Divide:

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

Factor $x^2 - x - 6$:

\[(x - 3)(x + 2)\]

So $p(x) = (x - 1)(x - 3)(x + 2) = (x-1)(x+2)(x-3)$.

Verify zeros: $x = 1, x = 3, x = -2$ ✓

1. $(x-1)(x-2)(x+3)$: zeros at $1, 2, -3$; check $p(2) = 8-8-10+6=-4 \neq 0$ ✗
1. $(x+1)(x-2)(x+3)$: includes $x = -1$, but $p(-1) = -1-2+5+6 = 8 \neq 0$ ✗
1. $(x-1)(x^2-x-6)$: partially factored; $x^2-x-6=(x-3)(x+2)$, so this is correct but not fully factored as the question requires “complete factored form”

Answer: G

Question 41

A fence encloses a rectangular field. The length of the field is $3$ times its width. If the perimeter of the field is 160 feet, what is the area of the field, in square feet?

A) $600$
B) $1{,}200$
C) $1{,}500$
D) $2{,}400$

Show solution

Let width $= w$, length $= 3w$.

Perimeter: $2(w + 3w) = 2(4w) = 8w = 160 \implies w = 20$.

Length $= 3(20) = 60$.

Area $= 20 \times 60 = 1{,}200$.

1. $600$: computed $\frac{1}{2} \times 20 \times 60 = 600$ (halved the area as if it were a triangle)
1. $1{,}500$: used $w = 25$: incorrectly solved $8w = 200$ from a perimeter of 200, or misread the problem
1. $2{,}400$: doubled the area: $2 \times 1200 = 2400$, or used width $= 40$ by treating $2w = 160/2 = 80$

Answer: B

Question 42

Elena drives from City A to City B, a distance of 180 miles, then returns from City B to City A. She drives at 60 mph on the way there and 45 mph on the way back. What is her average speed, in miles per hour, for the entire round trip?

F) $50$
G) $51.4$
H) $52.5$
J) $54$

Show solution

Average speed $= \dfrac{\text{total distance}}{\text{total time}}$.

Total distance $= 180 + 180 = 360$ miles.

Time there: $\dfrac{180}{60} = 3$ hours. Time back: $\dfrac{180}{45} = 4$ hours. Total time $= 7$ hours.

\[\text{Average speed} = \frac{360}{7} \approx 51.4 \text{ mph}\]

1. $50$: computed the harmonic mean incorrectly or used $\frac{60+45}{2} - 2.5$
1. $52.5$: computed the arithmetic mean of the two speeds: $\frac{60 + 45}{2} = 52.5$ (incorrect for different distances/times)
1. $54$: weighted average error using time weights instead of the correct formula

Answer: G

Question 43

A box contains 5 red cards and 3 blue cards. Two cards are drawn without replacement. What is the probability that both cards drawn are red?

A) $\dfrac{5}{16}$
B) $\dfrac{25}{64}$
C) $\dfrac{5}{14}$
D) $\dfrac{4}{7}$

Show solution

Without replacement, the draws are dependent events.

\[P(\text{1st red}) = \frac{5}{8}, \quad P(\text{2nd red} \mid \text{1st red}) = \frac{4}{7}\]

\[P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\]

1. $\dfrac{5}{16}$: computed $\frac{5}{8} \times \frac{1}{2}$ or some other product not accounting for conditional probability correctly
1. $\dfrac{25}{64}$: treated draws as independent: $\left(\frac{5}{8}\right)^2 = \frac{25}{64}$ (didn’t reduce the deck after the first draw)
1. $\dfrac{4}{7}$: computed only $P(\text{2nd red} \mid \text{1st red})$ without multiplying by $P(\text{1st red})$

Answer: C

Question 44

A right triangle is inscribed in a semicircle of radius 10. One leg of the triangle lies along the diameter. If the other leg has length 8, what is the length of the hypotenuse of the triangle?

F) $6$
G) $12$
H) $\sqrt{164}$
J) $20$

Show solution

A right triangle inscribed in a semicircle has its hypotenuse as the diameter of the full circle (Thales’ theorem). Any angle inscribed in a semicircle that subtends the diameter is a right angle.

The diameter of the circle $= 2 \times 10 = 20$.

Therefore the hypotenuse $= 20$.

The other leg can be verified: $\sqrt{20^2 - 8^2} = \sqrt{400 - 64} = \sqrt{336} \approx 18.3$ — both legs and the hypotenuse form a valid right triangle ✓

1. $6$: found the leg of an 8-10 right triangle using $\sqrt{10^2-8^2}=6$, confusing the radius with the hypotenuse
1. $12$: used a 6-8-10 triple and doubled the short leg, or computed $\sqrt{10^2-8^2} \times 2 = 12$
1. $\sqrt{164}$: computed $\sqrt{10^2+8^2}=\sqrt{164}$, incorrectly adding legs when the radius is not a leg
1. $6$: found the missing leg of an 8-10 Pythagorean relationship: $\sqrt{10^2-8^2}=6$, confusing the radius with the hypotenuse
1. $12$: computed $\sqrt{10^2 - 8^2} \times 2 = 12$… or a 6-8-10 triple misapplied
1. $\sqrt{164}$: applied the Pythagorean theorem with the radius as one leg: $\sqrt{10^2+8^2}=\sqrt{164}$

Answer: J

Question 45

A store sells two types of gift boxes. A small box costs $\$12$ and a large box costs $\$20$. On a certain day, the store sold a total of 30 boxes and collected $\$456$. How many large boxes were sold?

A) $9$
B) $12$
C) $18$
D) $21$

Show solution

Let $s$ = small boxes, $L$ = large boxes.

\[\begin{cases} s + L = 30 \\ 12s + 20L = 456 \end{cases}\]

From the first equation: $s = 30 - L$. Substitute:

\[12(30 - L) + 20L = 456 \implies 360 - 12L + 20L = 456 \implies 8L = 96 \implies L = 12\]

Verify: $s = 18$; $12(18) + 20(12) = 216 + 240 = 456$ ✓

1. $9$: solved $8L = 72$, an arithmetic error ($456 - 360 = 96$, not $72$)
1. $18$: reported the number of small boxes instead of large boxes
1. $21$: sign error when substituting: $360 + 12L + 20L = 456$, $32L = 96$… or set up $20s + 12L = 456$

Answer: B