Full-Length Practice Test 1

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

Question 1

What is the value of $3x - 7$ when $x = 4$?

A) $-5$
B) $5$
C) $12$
D) $19$

Show solution

Substitute $x = 4$:

\[3(4) - 7 = 12 - 7 = 5\]

1. $-5$: subtracted $12$ from $7$ (reversed the order)
1. $12$: computed $3(4)$ but forgot to subtract $7$
1. $19$: added $7$ instead of subtracting

Answer: B

Question 2

A car travels 240 miles in 4 hours. At the same speed, how many miles will the car travel in 7 hours?

F) $300$
G) $360$
H) $420$
J) $480$

Show solution

Speed $= \dfrac{240}{4} = 60$ miles per hour.

Distance in 7 hours $= 60 \times 7 = 420$ miles.

1. $300$: used 5 hours instead of 7
1. $360$: used 6 hours instead of 7
1. $480$: used 8 hours instead of 7

Answer: H

Question 3

Which of the following expressions is equivalent to $x^2 - 9x + 20$?

A) $(x - 4)(x - 5)$
B) $(x + 4)(x + 5)$
C) $(x - 4)(x + 5)$
D) $(x + 4)(x - 5)$

Show solution

Find two numbers that multiply to $+20$ and add to $-9$: those numbers are $-4$ and $-5$.

\[x^2 - 9x + 20 = (x - 4)(x - 5)\]

Check: $(x-4)(x-5) = x^2 - 5x - 4x + 20 = x^2 - 9x + 20$ ✓

1. $(x+4)(x+5)$: both factors have the wrong sign; expands to $x^2 + 9x + 20$
1. $(x-4)(x+5)$: mixed signs give $x^2 + x - 20$
1. $(x+4)(x-5)$: mixed signs give $x^2 - x - 20$

Answer: A

Question 4

In the figure below, lines $\ell$ and $m$ are parallel and are cut by transversal $t$. What is the value of $x$?

Note: Figure not drawn to scale.

F) $65$
G) $75$
H) $115$
J) $125$

Show solution

The labeled angles are corresponding angles — both on the same side of the transversal, one at each parallel line, in the same relative position. When two parallel lines are cut by a transversal, corresponding angles are equal.

Therefore $x = 65$.

1. $75$: no geometric justification
1. $115$: computed $180 - 65 = 115$, confusing these for co-interior (same-side interior) angles, which are supplementary — but co-interior angles are both located between the two parallel lines, not in the configuration shown
1. $125$: arithmetic error on a supplement calculation

Answer: F

Question 5

The mean of five numbers is 18. Four of the numbers are 12, 15, 22, and 24. What is the fifth number?

A) $14$
B) $17$
C) $18$
D) $22.5$

Show solution

The required total sum $= 18 \times 5 = 90$.

Sum of the four known numbers: $12 + 15 + 22 + 24 = 73$.

Fifth number $= 90 - 73 = 17$.

1. $14$: divided the known sum by 5 instead of subtracting it from the required total: $73 \div 5 = 14.6$, truncated to $14$
1. $18$: assumed the missing number must equal the mean
1. $22.5$: divided the required total by the number of known values instead of subtracting: $90 \div 4 = 22.5$

Answer: B

Question 6

If $f(x) = 2x^2 - 3x + 1$, what is $f(3)$?

F) $10$
G) $14$
H) $16$
J) $28$

Show solution

Substitute $x = 3$:

\[f(3) = 2(3)^2 - 3(3) + 1 = 2(9) - 9 + 1 = 18 - 9 + 1 = 10\]

1. $14$: made two sign slips — evaluated $-3(3)$ as $-3$ and $+1$ as $-1$: $18 - 3 - 1 = 14$
1. $16$: substituted $-3$ instead of $-3(3) = -9$: $18 - 3 + 1 = 16$
1. $28$: applied the exponent to $2x$ instead of $x$: $(2 \cdot 3)^2 - 9 + 1 = 36 - 9 + 1 = 28$

Answer: F

Question 7

In a right triangle, the two legs have lengths 8 and 15. What is the length of the hypotenuse?

A) $7$
B) $\sqrt{161}$
C) $17$
D) $23$

Show solution

Apply the Pythagorean theorem:

\[c^2 = 8^2 + 15^2 = 64 + 225 = 289 \implies c = \sqrt{289} = 17\]

This is the 8-15-17 Pythagorean triple.

1. $7$: subtracted the legs: $15 - 8 = 7$
1. $\sqrt{161}$: computed $15^2 - 8^2 = 225 - 64 = 161$ — subtracted instead of added
1. $23$: added the two legs directly, $8 + 15 = 23$, without squaring

Answer: C

Question 8

A bag contains 4 red marbles, 5 blue marbles, and 3 green marbles. If one marble is selected at random, what is the probability that it is blue?

F) $\dfrac{1}{5}$
G) $\dfrac{5}{12}$
H) $\dfrac{5}{9}$
J) $\dfrac{7}{12}$

Show solution

Total marbles $= 4 + 5 + 3 = 12$.

\[P(\text{blue}) = \frac{5}{12}\]

1. $\frac{1}{5}$: used the number of blue marbles as the denominator instead of the total
1. $\frac{5}{9}$: used only the red and blue marbles as the total: $4 + 5 = 9$
1. $\frac{7}{12}$: computed the complement $P(\text{not blue}) = \frac{4 + 3}{12} = \frac{7}{12}$

Answer: G

Question 9

Solve for $y$: $\dfrac{3y + 6}{3} = 8$

A) $2$
B) $4$
C) $6$
D) $10$

Show solution

Multiply both sides by 3:

\[3y + 6 = 24\]

Subtract 6 from both sides:

\[3y = 18 \implies y = 6\]

1. $2$: simplified $\frac{3y+6}{3}$ as $y + 6$ (incorrectly divided only $6$ by $3$ but not $3y$), giving $y + 6 = 8$, $y = 2$
1. $4$: after reaching $3y + 6 = 24$, divided $24$ by $6$ instead of subtracting $6$: $y = 24 \div 6 = 4$
1. $10$: added $6$ to both sides instead of subtracting: $3y = 24 + 6 = 30$, $y = 10$

Answer: C

Question 10

A rectangle has a length of $2x + 5$ and a width of $x - 1$. Which of the following expressions represents the area of the rectangle?

F) $2x^2 + 3x - 5$
G) $2x^2 - 5$
H) $3x + 4$
J) $6x + 8$

Show solution

Area $=$ length $\times$ width:

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

1. $2x^2 - 5$: dropped the middle terms ($-2x + 5x = 3x$)
1. $3x + 4$: added the two expressions instead of multiplying: $(2x+5)+(x-1) = 3x+4$
1. $6x + 8$: computed the perimeter $2[(2x+5)+(x-1)] = 2(3x+4) = 6x+8$

Answer: F

Question 11

What is $25\%$ of $160$?

A) $32$
B) $40$
C) $48$
D) $64$

Show solution

$25\% = \dfrac{1}{4}$, so $\dfrac{160}{4} = 40$.

1. $32$: computed $20\%$ of $160$
1. $48$: computed $30\%$ of $160$
1. $64$: computed $40\%$ of $160$

Answer: B

Question 12

The table below shows the results of a survey of 50 students about their preferred after-school activity.

Activity	Number of Students
Sports	18
Reading	7
Gaming	15
Art	10

What percent of the students preferred Sports or Gaming?

F) $33\%$
G) $60\%$
H) $66\%$
J) $70\%$

Show solution

Students preferring Sports or Gaming $= 18 + 15 = 33$.

\[\frac{33}{50} \times 100 = 66\%\]

1. $33\%$: used the raw count (33) as the percent without dividing by the total
1. $60\%$: used $30$ instead of $33$ in the numerator ($\frac{30}{50} = 60\%$)
1. $70\%$: used $35$ in the numerator ($\frac{35}{50} = 70\%$) — overcounted by 2 students

Answer: H

Question 13

What is the slope of the line passing through the points $(-2,\, 5)$ and $(4,\, -1)$?

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

Show solution

\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1\]

1. $-2$: computed $\frac{-6}{3}$ by using only $4 - 1 = 3$ in the denominator (didn’t treat $-2$ as negative)
1. $1$: dropped the negative sign from $\frac{-6}{6}$
1. $2$: flipped numerator and denominator and dropped the sign: $\frac{6}{-6} \to 2$ (arithmetic error)

Answer: B

Question 14

In right triangle $ABC$, the right angle is at $C$. If $AB = 10$ and $\angle A = 30°$, what is the length of $BC$?

Note: Figure not drawn to scale.

F) $5$
G) $5\sqrt{2}$
H) $5\sqrt{3}$
J) $10\sqrt{3}$

Show solution

In a 30-60-90 triangle, the side opposite the 30° angle equals half the hypotenuse.

$BC$ is opposite $\angle A = 30°$, and $AB = 10$ is the hypotenuse:

\[BC = \frac{AB}{2} = \frac{10}{2} = 5\]

Using trigonometry: $\sin(30°) = \dfrac{BC}{AB}$, so $BC = 10 \cdot \sin(30°) = 10 \cdot 0.5 = 5$.

1. $5\sqrt{2}$: used a 45-45-90 ratio (legs $= \frac{\text{hyp}}{\sqrt{2}}$) instead of 30-60-90
1. $5\sqrt{3}$: found $AC$ (the leg adjacent to $30°$) instead of $BC$: $AC = 10 \cdot \cos(30°) = 5\sqrt{3}$
1. $10\sqrt{3}$: multiplied the hypotenuse by $\sqrt{3}$ without halving

Answer: F

Question 15

If $2(3x - 1) = 4x + 10$, what is the value of $x$?

A) $2$
B) $4$
C) $6$
D) $12$

Show solution

Distribute on the left:

\[6x - 2 = 4x + 10\]

Subtract $4x$:

\[2x - 2 = 10\]

Add 2:

\[2x = 12 \implies x = 6\]

1. $2$: dropped the $4x$ term after distributing: $6x - 2 = 10$ gives $6x = 12$, $x = 2$
1. $4$: subtracted $2$ instead of adding it: $2x = 10 - 2 = 8$, $x = 4$
1. $12$: solved correctly to $2x = 12$ but reported the value of $2x$ instead of dividing by $2$

Answer: C

Question 16

The first term of an arithmetic sequence is $7$, and the common difference is $-3$. What is the 10th term of the sequence?

F) $-30$
G) $-23$
H) $-20$
J) $34$

Show solution

The formula for the $n$th term of an arithmetic sequence:

\[a_n = a_1 + (n - 1)d\]

With $a_1 = 7$, $d = -3$, $n = 10$:

\[a_{10} = 7 + (10 - 1)(-3) = 7 + 9(-3) = 7 - 27 = -20\]

1. $-30$: computed $10(-3) = -30$ and forgot to add the first term
1. $-23$: used $n = 10$ instead of $n - 1 = 9$: $7 + 10(-3) = 7 - 30 = -23$
1. $34$: used $d = +3$ (ignored the negative sign): $7 + 9(3) = 7 + 27 = 34$

Answer: H

Question 17

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

\[\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}\]

A) $1$
B) $2$
C) $3$
D) $5$

Show solution

Add the two equations to eliminate $y$:

\[(2x + y) + (x - y) = 7 + 2 \implies 3x = 9 \implies x = 3\]

Substitute $x = 3$ into $x - y = 2$:

\[3 - y = 2 \implies y = 1\]

1. $2$: reported the constant from the second equation ($x - y = 2$) as the value of $y$
1. $3$: reported the value of $x$ instead of $y$
1. $5$: substituted into $2x + y = 7$ but solved for $y$ using $x = 1$ instead of $x = 3$: $2(1) + y = 7$, $y = 5$

Answer: A

Question 18

The graph of $f(x) = x^2$ is transformed to produce the graph of $g(x) = (x - 3)^2 + 2$. Which of the following describes the transformation?

F) Shifted left 3 units and up 2 units
G) Shifted right 3 units and up 2 units
H) Shifted right 3 units and down 2 units
J) Shifted left 3 units and down 2 units

Show solution

The vertex form of a parabola is $y = (x - h)^2 + k$, where the vertex is at $(h, k)$.

For $g(x) = (x - 3)^2 + 2$: $h = 3$ and $k = 2$, so the vertex is at $(3, 2)$.

The original $f(x) = x^2$ has vertex at the origin $(0, 0)$.

The graph shifted right 3 units (positive $h$) and up 2 units (positive $k$).

1. Left 3 and up 2: the $-3$ inside the parentheses indicates a rightward shift, not leftward (common sign-direction error)
1. Right 3 and down 2: the $+2$ outside is an upward shift, not downward
1. Left 3 and down 2: both directions wrong

Answer: G

Question 19

In the figure below, triangle $PQR$ is similar to triangle $STU$ with the vertices listed in corresponding order. If $PQ = 6$, $QR = 9$, and $ST = 8$, what is the length of $TU$?

Note: Figure not drawn to scale.

A) $6$
B) $6.75$
C) $11$
D) $12$

Show solution

Since the triangles are similar with vertices in corresponding order, the corresponding sides are proportional:

\[\frac{PQ}{ST} = \frac{QR}{TU}\]

Substituting the known values:

\[\frac{6}{8} = \frac{9}{TU}\]

Cross-multiply:

\[6 \cdot TU = 8 \cdot 9 = 72 \implies TU = \frac{72}{6} = 12\]

1. $6$: matched $TU$ with $PQ = 6$ — paired the wrong corresponding sides
1. $6.75$: inverted the proportion: $\frac{6}{8} = \frac{TU}{9}$ gives $TU = \frac{54}{8} = 6.75$
1. $11$: added the scale difference $2$ to $9$: $9 + 2 = 11$ (used addition instead of proportional reasoning)

Answer: D

Question 20

The two-way table below shows the number of students in two grades who play a sport or do not play a sport.

	Plays a Sport	Does Not Play a Sport	Total
Grade 10	24	16	40
Grade 11	18	22	40
Total	42	38	80

What is the probability that a randomly selected student from this group is in Grade 10, given that the student plays a sport?

F) $\dfrac{3}{10}$
G) $\dfrac{4}{7}$
H) $\dfrac{1}{2}$
J) $\dfrac{3}{5}$

Show solution

This is a conditional probability question: $P(\text{Grade 10} \mid \text{plays a sport})$.

The condition restricts the sample to students who play a sport: 42 total.

Of those, 24 are in Grade 10:

\[P(\text{Grade 10} \mid \text{plays a sport}) = \frac{24}{42} = \frac{4}{7}\]

1. $\frac{3}{10}$: divided $24$ by $80$ (used the total students, not the conditional group): $\frac{24}{80} = \frac{3}{10}$
1. $\frac{1}{2}$: divided $40$ (Grade 10 total) by $80$ (overall total), ignoring the sport condition
1. $\frac{3}{5}$: divided $24$ by $40$ (used the Grade 10 total as denominator instead of the sport-players total)

Answer: G

Question 21

A plumber charges a flat fee of $75 plus $50 per hour for a service call. A customer received a bill of $275. How many hours did the plumber work?

A) $3$
B) $4$
C) $5$
D) $5.5$

Show solution

Let $h$ = number of hours. The total bill is:

\[75 + 50h = 275\]

Subtract 75:

\[50h = 200 \implies h = 4\]

1. $3$: divided the bill by the flat fee instead: $275 \div 75 \approx 3.7$, truncated to $3$
1. $5$: divided the full bill by the hourly rate without subtracting the flat fee: $\frac{275}{50} = 5.5 \approx 5$
1. $5.5$: divided the full amount by the hourly rate: $\frac{275}{50} = 5.5$ (forgot to subtract the flat fee)

Answer: B

Question 22

What is the domain of the function $f(x) = \dfrac{1}{x^2 - 4}$?

F) All real numbers
G) All real numbers except $x = 2$
H) All real numbers except $x = 2$ and $x = -2$
J) All real numbers except $x = 4$

Show solution

The function is undefined when the denominator equals zero. Set $x^2 - 4 = 0$:

\[x^2 = 4 \implies x = 2 \text{ or } x = -2\]

The domain is all real numbers except $x = 2$ and $x = -2$.

1. All real numbers: ignores the restriction from the denominator
1. All real numbers except $x = 2$: forgot that $x = -2$ also makes the denominator zero ($(-2)^2 - 4 = 0$)
1. All real numbers except $x = 4$: set $x - 4 = 0$ instead of $x^2 - 4 = 0$

Answer: H

Question 23

A circle has a radius of 9 inches. What is the length, in inches, of the arc intercepted by a central angle of $\dfrac{2\pi}{3}$ radians?

A) $3\pi$
B) $6\pi$
C) $9\pi$
D) $18\pi$

Show solution

Arc length formula: $s = r\theta$, where $r$ is the radius and $\theta$ is the central angle in radians.

\[s = 9 \cdot \frac{2\pi}{3} = \frac{18\pi}{3} = 6\pi\]

1. $3\pi$: used $\frac{\pi}{3}$ instead of $\frac{2\pi}{3}$ for the angle: $9 \cdot \frac{\pi}{3} = 3\pi$
1. $9\pi$: multiplied only the radius by $\pi$, ignoring the $\frac{2}{3}$ factor
1. $18\pi$: computed $r \cdot 2\pi$ (the full circumference factor) without applying the $\frac{1}{3}$: $9 \cdot 2\pi = 18\pi$

Answer: B

Question 24

A student scored 80 on each of 3 tests and 92 on each of 2 tests. What is the student’s average (arithmetic mean) score across all 5 tests?

F) $84$
G) $84.8$
H) $86$
J) $88$

Show solution

Total score $= 3(80) + 2(92) = 240 + 184 = 424$.

Mean $= \dfrac{424}{5} = 84.8$.

1. $84$: computed the correct mean $84.8$ but truncated the decimal to $84$
1. $86$: computed the simple average of $80$ and $92$: $\frac{80 + 92}{2} = 86$ (didn’t account for the different frequencies)
1. $88$: dropped two of the $80$s and averaged the remaining three scores: $\frac{80 + 92 + 92}{3} = 88$

Answer: G

Question 25

For what value of $k$ does the system of equations below have infinitely many solutions?

\[\begin{cases} 3x - y = 5 \\ 6x - 2y = k \end{cases}\]

A) $5$
B) $7$
C) $10$
D) $15$

Show solution

A system has infinitely many solutions when the two equations represent the same line — the equations are multiples of each other.

Multiply the first equation by 2:

\[2(3x - y) = 2(5) \implies 6x - 2y = 10\]

The second equation is $6x - 2y = k$. For the equations to be identical, $k$ must equal $10$.

1. $5$: that is the constant in the first equation, not the second when doubled
1. $7$: $k = 7$ makes the equations parallel with no solution ($6x - 2y$ can’t equal both $10$ and $7$)
1. $15$: $k = 15$ also gives no solution (parallel lines)

Answer: C

Question 26

Let $f(x) = 2x + 1$ and $g(x) = x^2 - 3$. What is the value of $f(g(2))$?

F) $-1$
G) $1$
H) $3$
J) $9$

Show solution

Evaluate from the inside out. First find $g(2)$:

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

Then find $f(g(2)) = f(1)$:

\[f(1) = 2(1) + 1 = 3\]

1. $-1$: computed $g(2) = 2^2 - 3 = 1$, then evaluated $f$ at $-1$ (sign error)
1. $1$: stopped after computing $g(2) = 1$ without applying $f$
1. $9$: forgot the $-3$ in $g$: $g(2) = 2^2 = 4$, then $f(4) = 2(4) + 1 = 9$

Answer: H

Question 27

In the figure below, points $A$, $B$, and $C$ lie on a circle with center $O$. If $\angle ABC = 35°$, what is the measure of central angle $\angle AOC$?

Note: Figure not drawn to scale.

A) $17.5°$
B) $35°$
C) $55°$
D) $70°$

Show solution

The Inscribed Angle Theorem states that a central angle is twice the inscribed angle that subtends the same arc.

$\angle ABC$ is an inscribed angle and $\angle AOC$ is the central angle, both subtending arc $AC$.

\[\angle AOC = 2 \times \angle ABC = 2 \times 35° = 70°\]

1. $17.5°$: halved the inscribed angle instead of doubling ($35° \div 2$)
1. $35°$: set the central angle equal to the inscribed angle (forgot to apply the theorem)
1. $55°$: computed $90° - 35° = 55°$ — used a complementary angle relationship that doesn’t apply here

Answer: D

Question 28

Which of the following datasets has the largest standard deviation?

F) $\{5,\ 5,\ 5,\ 5\}$
G) $\{4,\ 5,\ 6,\ 7\}$
H) $\{3,\ 5,\ 7,\ 9\}$
J) $\{1,\ 5,\ 9,\ 13\}$

Show solution

Standard deviation measures how spread out values are from the mean. All four datasets have mean $= 5$.

The dataset with values furthest from the mean has the largest standard deviation.

$\{5, 5, 5, 5\}$: all values equal the mean — SD $= 0$
$\{4, 5, 6, 7\}$: values 1 unit from mean at most — small SD
$\{3, 5, 7, 9\}$: values up to 4 units from mean — medium SD
$\{1, 5, 9, 13\}$: values up to 8 units from mean — largest SD ✓
1. $\{5,5,5,5\}$: no spread at all — the minimum possible SD of zero
1. $\{4,5,6,7\}$: small spread around the mean
1. $\{3,5,7,9\}$: more spread than G but less than J

Answer: J

Question 29

A cylinder has a radius of 3 inches and a height of 8 inches. What is the volume of the cylinder, in cubic inches?

A) $24\pi$
B) $48\pi$
C) $72\pi$
D) $288\pi$

Show solution

Volume of a cylinder $= \pi r^2 h$.

\[V = \pi (3)^2 (8) = \pi (9)(8) = 72\pi\]

1. $24\pi$: used $r$ instead of $r^2$: $\pi(3)(8) = 24\pi$
1. $48\pi$: used $r^2 = 6$ (doubled $r$ instead of squaring): $\pi(6)(8) = 48\pi$
1. $288\pi$: used the diameter ($6$) instead of the radius: $\pi(6)^2(8) = 288\pi$

Answer: C

Question 30

The scatterplot below shows the relationship between the number of hours students spent studying and the number of errors they made on a test.

Note: Figure not drawn to scale.

Which of the following best describes the association shown in the scatterplot?

F) Positive linear association
G) No association
H) Negative linear association
J) Positive nonlinear association

Show solution

As hours studied increases (left to right), the number of errors decreases (downward trend). This is a negative association — the two variables move in opposite directions. The points fall close to a straight line, making it linear.

1. Positive linear: would require errors to increase as hours increase — the opposite of what’s shown
1. No association: the data has a clear downward trend, not a random scatter
1. Positive nonlinear: both “positive” and “nonlinear” are wrong — the trend is negative and linear

Answer: H

Question 31

Which of the following is equivalent to $\dfrac{x^2 - 4}{x^2 + x - 6}$ for all values of $x$ for which the expression is defined?

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

Show solution

Factor the numerator and denominator:

\[\text{Numerator: } x^2 - 4 = (x+2)(x-2)\] \[\text{Denominator: } x^2 + x - 6 = (x+3)(x-2)\]

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

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

1. $\frac{x-2}{x-3}$: factored numerator as $(x-2)^2$ and denominator incorrectly
1. $\frac{x-2}{x+3}$: cancelled $(x+2)$ instead of $(x-2)$ from the numerator
1. $\frac{x+2}{x-3}$: factored the denominator incorrectly as $(x-3)(x+2)$

Answer: D

Question 32

If $2x - 3y = 0$ and $x \neq 0$, what is the value of $\dfrac{x}{y}$?

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

Show solution

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

\[2x = 3y \implies x = \frac{3}{2}y\]

Therefore:

\[\frac{x}{y} = \frac{3}{2}\]

1. $\frac{2}{3}$: computed $\frac{y}{x}$ instead of $\frac{x}{y}$ (inverted the ratio)
1. $2$: read the coefficient of $x$ directly without solving for the ratio
1. $3$: read the coefficient of $y$ directly without solving for the ratio

Answer: G

Question 33

A function $f$ is defined as:

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

What is the value of $f(-1) + f(3)$?

A) $6$
B) $8$
C) $10$
D) $12$

Show solution

Since $-1 < 2$, use the first rule: $f(-1) = (-1) + 3 = 2$.

Since $3 \geq 2$, use the second rule: $f(3) = (3)^2 - 1 = 9 - 1 = 8$.

\[f(-1) + f(3) = 2 + 8 = 10\]

1. $6$: applied the first rule to $3$ as well — $f(3) = 3 + 3 = 6$ — and omitted $f(-1)$
1. $8$: used the second rule for $f(-1)$: $(-1)^2 - 1 = 0$, giving $0 + 8 = 8$
1. $12$: computed $3^2 + 1 = 10$ instead of $3^2 - 1$ (sign slip on the constant), then $2 + 10 = 12$

Answer: C

Question 34

If $f(x) = \sqrt[3]{x - 4}$, which of the following gives $f^{-1}(x)$, the inverse of $f$?

F) $\dfrac{1}{\sqrt[3]{x-4}}$
G) $\sqrt[3]{x} + 4$
H) $x^3 - 4$
J) $x^3 + 4$

Show solution

To find the inverse, replace $f(x)$ with $y$, swap $x$ and $y$, then solve for $y$:

\[y = \sqrt[3]{x - 4}\]

Swap $x$ and $y$:

\[x = \sqrt[3]{y - 4}\]

Cube both sides:

\[x^3 = y - 4 \implies y = x^3 + 4\]

Verify: $f(f^{-1}(x)) = \sqrt[3]{(x^3+4)-4} = \sqrt[3]{x^3} = x$ ✓

1. $\frac{1}{\sqrt[3]{x-4}}$: confused “inverse function” with “reciprocal”
1. $\sqrt[3]{x} + 4$: took the cube root of $x$ and added 4, rather than cubing $x$ and adding 4
1. $x^3 - 4$: cubed correctly but subtracted $4$ instead of adding (sign error on the $-4$ when moving it across the equals sign)

Answer: J

Question 35

What is the real solution to the equation $3e^x = 12$?

A) $x = \ln 3$
B) $x = \ln 9$
C) $x = \ln 12$
D) $x = \ln 4$

Show solution

Divide both sides by 3:

\[e^x = \frac{12}{3} = 4\]

Take the natural log of both sides:

\[x = \ln 4\]

1. $\ln 3$: took $\ln$ of the coefficient ($3$) rather than dividing first
1. $\ln 9$: computed $12 - 3 = 9$ (subtracted instead of dividing)
1. $\ln 12$: took $\ln$ of both sides without dividing — $x + \ln 3 = \ln 12$ — then dropped the $\ln 3$ term

Answer: D

Question 36

What are the center and radius of the circle with equation $(x - 3)^2 + (y + 2)^2 = 25$?

F) Center $(-3,\ 2)$, radius $5$
G) Center $(3,\ -2)$, radius $5$
H) Center $(3,\ -2)$, radius $25$
J) Center $(-3,\ 2)$, radius $25$

Show solution

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

Matching $(x - 3)^2 + (y + 2)^2 = 25$:

\[h = 3, \quad k = -2 \quad \text{(since } y + 2 = y - (-2)\text{)}, \quad r^2 = 25 \implies r = 5\]

Center: $(3, -2)$, radius: $5$.

1. $(-3, 2)$: flipped the signs of both $h$ and $k$
1. $(3, -2)$, radius $25$: correct center but used $r^2$ instead of $r$ for the radius
1. $(-3, 2)$, radius $25$: both sign error on center and used $r^2$ for radius

Answer: G

Question 37

Which of the following equations represents a hyperbola?

A) $x^2 + y^2 = 16$
B) $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$
C) $y = 2x^2 - 3$
D) $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$

Show solution

A hyperbola has the standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ (or $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$) — the key feature is the minus sign between the two squared terms.

1. $x^2 + y^2 = 16$: both terms positive and equal coefficients — this is a circle
1. $\frac{x^2}{9} + \frac{y^2}{4} = 1$: both terms positive with different denominators — this is an ellipse
1. $y = 2x^2 - 3$: one variable is squared, one is linear — this is a parabola
1. $\frac{x^2}{9} - \frac{y^2}{4} = 1$: difference of two squared terms equals 1 — this is a hyperbola ✓

Answer: D

Question 38

The equation $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ represents an ellipse. What is the length of the major axis?

F) $3$
G) $5$
H) $6$
J) $10$

Show solution

The standard form of an ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ where $a > b > 0$.

Here $a^2 = 25$, so $a = 5$, and $b^2 = 9$, so $b = 3$.

The major axis runs along the $x$-axis and has length $2a = 2(5) = 10$.

1. $3$: reported $b$ (the semi-minor axis) instead of $2a$
1. $5$: reported $a$ (the semi-major axis) instead of the full major axis length $2a$
1. $6$: computed $2b = 2(3) = 6$ — used the minor axis length instead of major

Answer: J

Question 39

Which of the following is equal to $3\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix}$?

A) $\begin{bmatrix} 5 & 2 \\ 3 & 7 \end{bmatrix}$
B) $\begin{bmatrix} 6 & -1 \\ 0 & 4 \end{bmatrix}$
C) $\begin{bmatrix} 6 & -3 \\ 0 & 12 \end{bmatrix}$
D) $\begin{bmatrix} 6 & 3 \\ 0 & 12 \end{bmatrix}$

Show solution

Scalar multiplication multiplies every entry in the matrix by the scalar:

\[3\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 3(2) & 3(-1) \\ 3(0) & 3(4) \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ 0 & 12 \end{bmatrix}\]

1. $\begin{bmatrix}5&2\\3&7\end{bmatrix}$: added 3 to each entry instead of multiplying
1. $\begin{bmatrix}6&-1\\0&4\end{bmatrix}$: only multiplied the first column by 3
1. $\begin{bmatrix}6&3\\0&12\end{bmatrix}$: correct magnitudes but $3 \times (-1) = -3$, not $+3$ — sign error

Answer: C

Question 40

Given that $i = \sqrt{-1}$, what is the value of $i^{47}$?

F) $1$
G) $i$
H) $-1$
J) $-i$

Show solution

Powers of $i$ cycle with period 4: $i^1 = i,\ i^2 = -1,\ i^3 = -i,\ i^4 = 1$, then repeat.

Divide the exponent by 4 to find the remainder:

\[47 = 4 \times 11 + 3 \implies 47 \equiv 3 \pmod{4}\]

Therefore $i^{47} = i^3 = -i$.

1. $1$: $47 \equiv 0 \pmod 4$ would give $i^4 = 1$, but $47 \div 4$ has remainder $3$, not $0$
1. $i$: corresponds to remainder $1$ (i.e. $i^{45} = i^1 = i$), not remainder $3$
1. $-1$: corresponds to remainder $2$ (i.e. $i^{46} = i^2 = -1$), off by one

Answer: J

Question 41

Keiko invests $\$500$ in an account that earns $8\%$ annual interest compounded semiannually. The amount $A$ in the account after $t$ years is given by $A = 500\left(1 + \dfrac{0.08}{2}\right)^{2t}$. To the nearest cent, what is the amount in the account after 3 years?

A) $\$620.00$
B) $\$629.86$
C) $\$632.66$
D) $\$635.12$

Show solution

Substitute $t = 3$ into the formula:

\[A = 500\left(1 + \frac{0.08}{2}\right)^{2(3)} = 500(1.04)^6\]

Compute $(1.04)^6$:

\[(1.04)^2 = 1.0816, \quad (1.04)^4 = 1.0816^2 = 1.16986, \quad (1.04)^6 = 1.16986 \times 1.0816 \approx 1.26532\]

\[A = 500 \times 1.26532 \approx \$632.66\]

1. $\$620.00$: used simple interest: $500 \times (1 + 0.08 \times 3) = 500 \times 1.24 = 620$
1. $\$629.86$: compounded annually instead of semiannually: $500(1.08)^3 \approx 629.86$
1. $\$635.12$: compounded monthly ($n = 12$) instead of semiannually: $500\left(1 + \frac{0.08}{12}\right)^{36} \approx 635.12$

Answer: C

Question 42

A sequence is defined recursively by $a_1 = 3$ and $a_n = 2a_{n-1} - 1$ for $n \geq 2$. What is the value of $a_6$?

F) $9$
G) $17$
H) $33$
J) $65$

Show solution

Apply the rule $a_n = 2a_{n-1} - 1$ repeatedly:

\[a_1 = 3\] \[a_2 = 2(3) - 1 = 5\] \[a_3 = 2(5) - 1 = 9\] \[a_4 = 2(9) - 1 = 17\] \[a_5 = 2(17) - 1 = 33\] \[a_6 = 2(33) - 1 = 65\]

1. $9$: reported $a_3 = 9$ instead of $a_6$ — stopped three steps early
1. $17$: reported $a_4$ instead of $a_6$
1. $33$: reported $a_5$ instead of $a_6$

Answer: J

Question 43

A game gives the following payouts: a player wins $\$10$ with probability $0.3$, wins $\$5$ with probability $0.4$, and loses $\$8$ with probability $0.3$. What is the expected value of the player’s winnings per game?

A) $\$0.60$
B) $\$2.00$
C) $\$2.60$
D) $\$7.50$

Show solution

Expected value $= \sum (\text{outcome} \times \text{probability})$:

\[E = 10(0.3) + 5(0.4) + (-8)(0.3)\] \[= 3 + 2 - 2.4 = 2.6\]

The expected value is $\$2.60$.

1. $\$0.60$: omitted the $\$5$ outcome: $10(0.3) - 8(0.3) = 3 - 2.4 = 0.60$
1. $\$2.00$: included only the $\$5$ outcome: $5(0.4) = 2.00$
1. $\$7.50$: averaged the two winning payouts, $\frac{10 + 5}{2} = 7.50$, ignoring the probabilities and the loss

Answer: C

Question 44

The polynomial $p(x) = (x - 2)^2(x + 3)$ is graphed in the coordinate plane. Which of the following describes all the $x$-intercepts of the graph?

F) $x = 2$ only
G) $x = -3$ only
H) $x = -2$ and $x = 3$
J) $x = -3$ and $x = 2$

Show solution

The $x$-intercepts occur where $p(x) = 0$. Set each factor equal to zero:

\[(x - 2)^2 = 0 \implies x = 2 \quad \text{(double root — touches but doesn't cross)}\] \[(x + 3) = 0 \implies x = -3\]

The $x$-intercepts are at $x = -3$ and $x = 2$.

1. $x = 2$ only: found one root but missed $x = -3$
1. $x = -3$ only: found one root but missed $x = 2$
1. $x = -2$ and $x = 3$: flipped the signs of both roots (common sign error with factored form)

Answer: J

Question 45

How many integers $n$ satisfy $\dfrac{17}{5} < n < \dfrac{29}{4}$?

A) $3$
B) $4$
C) $5$
D) $6$

Show solution

Convert the bounds to decimals:

\[\frac{17}{5} = 3.4 \qquad \frac{29}{4} = 7.25\]

The integers strictly between $3.4$ and $7.25$ are: $4, 5, 6, 7$ — that is 4 integers.

1. $3$: counted only $5, 6, 7$ — excluded $4$ by treating the lower bound as $n > 4$
1. $5$: also counted $3$ by rounding the lower bound $3.4$ down to $3$
1. $6$: counted $3$ through $8$ inclusive — rounded both bounds outward

Answer: B