Full-Length Practice Test 3

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

Question 1

Which of the following is equal to $3.6 \times 10^4$?

A) $360$
B) $3{,}600$
C) $36{,}000$
D) $360{,}000$

Show solution

$10^4 = 10{,}000$, so:

\[3.6 \times 10^4 = 3.6 \times 10{,}000 = 36{,}000\]

1. $360$: moved the decimal 2 places instead of 4 (used $10^2$)
1. $3{,}600$: moved the decimal 3 places instead of 4 (used $10^3$)
1. $360{,}000$: moved the decimal 5 places instead of 4 (used $10^5$)

Answer: C

Question 2

A recipe requires 2.5 cups of flour. If 1 cup = 240 milliliters, how many milliliters of flour does the recipe require?

F) $96$
G) $480$
H) $600$
J) $960$

Show solution

\[2.5 \text{ cups} \times 240 \frac{\text{mL}}{\text{cup}} = 600 \text{ mL}\]

1. $96$: divided instead of multiplied: $240 / 2.5 = 96$
1. $480$: used $2$ cups instead of $2.5$: $2 \times 240 = 480$
1. $960$: used $4$ cups instead of $2.5$: $4 \times 240 = 960$

Answer: H

Question 3

Which of the following expressions 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:

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

Combine like terms:

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

1. $4x - 10$: distributed correctly on $x$ terms but computed $-12 + 2 = -10$ (added instead of subtracting the $2$)
1. $8x - 14$: added the $x$-coefficients instead of subtracting: $6x + 2x = 8x$
1. $8x - 10$: both errors — added $x$-coefficients and used $-10$ for the constant

Answer: A

Question 4

Which of the following is the solution set of $-3x + 5 \geq 14$?

F) $x \geq -3$
G) $x \leq -3$
H) $x \geq 3$
J) $x \leq 3$

Show solution

Subtract 5 from both sides:

\[-3x \geq 9\]

Divide both sides by $-3$ — remember to flip the inequality when dividing by a negative:

\[x \leq -3\]

1. $x \geq -3$: correct value $-3$ but forgot to flip the inequality sign when dividing by $-3$
1. $x \geq 3$: took the wrong sign on $x$ ($-3x \geq 9$ gives $x = -3$, not $x = 3$) and kept the wrong direction
1. $x \leq 3$: flipped the inequality correctly but used $+3$ instead of $-3$

Answer: G

Question 5

In the figure below, lines $p$ and $q$ intersect at point $T$. What is the value of $y$?

Note: Figure not drawn to scale.

A) $25$
B) $35$
C) $50$
D) $70$

Show solution

The angles $(2y + 10)°$ and $70°$ are supplementary (they lie on a straight line):

\[(2y + 10) + 70 = 180\] \[2y + 80 = 180\] \[2y = 100 \implies y = 50\]

1. $25$: solved $2y = 50$ instead of $2y = 100$ (halved one side prematurely)
1. $35$: set $(2y + 10) = 70$ (treated them as equal, as if they were vertical angles): $2y = 60$, $y = 30$… or a similar arithmetic error
1. $70$: reported the given angle measure instead of solving for $y$

Answer: C

Question 6

If $h(x) = \dfrac{2x}{x - 3}$, what is the domain of $h$?

F) All real numbers
G) All real numbers except $x = 0$
H) All real numbers except $x = 3$
J) All real numbers except $x = -3$

Show solution

The function is undefined when the denominator equals zero:

\[x - 3 = 0 \implies x = 3\]

The domain is all real numbers except $x = 3$.

1. All real numbers: ignores the restriction from the denominator
1. All real numbers except $x = 0$: set the numerator equal to zero instead of the denominator
1. All real numbers except $x = -3$: solved $x + 3 = 0$ instead of $x - 3 = 0$ (sign error)

Answer: H

Question 7

In right triangle $XYZ$, the right angle is at $Z$. If $XZ = 5$ and $XY = 13$, what is the length of $YZ$?

A) $8$
B) $10$
C) $12$
D) $18$

Show solution

$XY = 13$ is the hypotenuse (opposite the right angle at $Z$). Apply the Pythagorean theorem:

\[YZ^2 = XY^2 - XZ^2 = 169 - 25 = 144 \implies YZ = 12\]

This is the 5-12-13 Pythagorean triple.

1. $8$: computed $13 - 5 = 8$ (subtracted legs directly)
1. $10$: used a 5-10 ratio without the Pythagorean theorem, or confused with the 5-12-13 triple
1. $18$: added the two given lengths: $5 + 13 = 18$

Answer: C

Question 8

The following dataset shows the ages of seven volunteers at a charity event:

\[22, \, 31, \, 19, \, 45, \, 31, \, 28, \, 37\]

What is the mode of this dataset?

F) $19$
G) $28$
H) $31$
J) $45$

Show solution

The mode is the value that appears most often. Scanning the dataset: $22, 31, 19, 45, \mathbf{31}, 28, 37$ — the value $31$ appears twice; all others appear once.

Mode $= 31$.

1. $19$: the minimum value, not the mode
1. $28$: the median when sorted ($19, 22, 28, 31, 31, 37, 45$ → middle = 31, not 28); also just a middle-ish value, not the mode
1. $45$: the maximum value, not the mode

Answer: H

Question 9

A local pizza shop offers 3 crust types, 4 sauce options, and 6 toppings. If a pizza is made with exactly 1 crust, 1 sauce, and 1 topping, how many different pizzas are possible?

A) $13$
B) $36$
C) $72$
D) $144$

Show solution

By the Fundamental Counting Principle, multiply the number of choices for each category:

\[3 \times 4 \times 6 = 72\]

1. $13$: added the options instead of multiplying: $3 + 4 + 6 = 13$
1. $36$: computed $6 \times (4 + 3) - 6 = 36$… or multiplied only two categories: $4 \times 6 = 24$, then added $12$; most likely multiplied $3 \times 4 = 12$ and multiplied by $3$ instead of $6$: $12 \times 3 = 36$
1. $144$: doubled the correct answer, perhaps computing $3 \times 4 \times 6 \times 2$

Answer: C

Question 10

Which of the following is the equation of the line passing through $(2, 5)$ with slope $3$?

F) $y = 3x + 5$
G) $y = 3x - 1$
H) $y = 3x + 1$
J) $y = 5x + 3$

Show solution

Using point-slope form with $(x_1, y_1) = (2, 5)$ and $m = 3$:

\[y - 5 = 3(x - 2) \implies y - 5 = 3x - 6 \implies y = 3x - 1\]

Verify: when $x = 2$: $y = 3(2) - 1 = 5$ ✓

1. $y = 3x + 5$: used the $y$-coordinate as the $y$-intercept directly without computing it
1. $y = 3x + 1$: sign error in the constant: $-5 + 6 = +1$ instead of $5 - 6 = -1$
1. $y = 5x + 3$: swapped slope and $y$-coordinate in the equation

Answer: G

Question 11

The figure below shows a rectangle with a right triangle removed from one corner. What is the area of the shaded region?

Note: Figure not drawn to scale.

A) $48$
B) $50$
C) $54$
D) $60$

Show solution

Area of full rectangle: $10 \times 6 = 60$.

Area of removed triangle: $\dfrac{1}{2} \times 4 \times 3 = 6$.

Shaded area: $60 - 6 = 54$.

1. $48$: computed $60 - 12 = 48$ (forgot to halve the triangle area: used $4 \times 3 = 12$)
1. $50$: computed $60 - 10 = 50$ (subtracted the rectangle width instead of the triangle area)
1. $60$: reported the full rectangle area without subtracting the triangle

Answer: C

Question 12

The scatterplot below shows the relationship between the number of hours of practice per week and the performance score of 8 athletes. Which of the following best describes the association shown?

Note: Figure not drawn to scale.

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

Show solution

As hours of practice increase (left to right), performance scores increase (upward trend) — this is a positive association. The points fall close to a straight line, making it linear.

1. Negative linear: would require scores to decrease as hours increase — the opposite of what’s shown
1. No association: the data has a clear upward trend, not a random scatter
1. Positive nonlinear: the trend is positive ✓ but the points fall roughly along a straight line, not a curve

Answer: J

Question 13

In right triangle $ABC$ with right angle at $B$, $\cos(\angle A) = \dfrac{5}{13}$. What is $\sin(\angle A)$?

A) $\dfrac{5}{12}$
B) $\dfrac{5}{13}$
C) $\dfrac{12}{13}$
D) $\dfrac{13}{12}$

Show solution

$\cos(\angle A) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{5}{13}$, so adjacent $= 5$ and hypotenuse $= 13$.

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

\[\sin(\angle A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13}\]

This is the 5-12-13 Pythagorean triple.

1. $\dfrac{5}{12}$: reported $\tan(\angle A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5}$… or swapped opposite and adjacent in the sine formula
1. $\dfrac{5}{13}$: reported $\cos(\angle A)$ again instead of $\sin(\angle A)$
1. $\dfrac{13}{12}$: inverted the sine ratio; a value greater than 1 cannot be a sine value

Answer: C

Question 14

The function $f(x) = 4^x$ is graphed in the coordinate plane. Which of the following correctly describes the graph?

F) A decreasing curve that passes through $(0,\, 0)$
G) An increasing curve that passes through $(0,\, 1)$
H) A decreasing curve that passes through $(0,\, 1)$
J) An increasing straight line that passes through $(0,\, 4)$

Show solution

For $f(x) = 4^x$:

At $x = 0$: $f(0) = 4^0 = 1$ — the graph passes through $(0, 1)$.
As $x$ increases, $4^x$ grows rapidly — the graph is increasing.
As $x \to -\infty$, $4^x \to 0$ (approaches but never touches the $x$-axis) — it is a curve, not a straight line.
1. Decreasing and through $(0, 0)$: $4^x$ never equals zero, and with base $> 1$ it increases (not decreases)
1. Decreasing and through $(0, 1)$: the $y$-intercept is correct but the function is increasing, not decreasing
1. Increasing straight line through $(0, 4)$: exponential functions are curves, not lines; and $f(0) = 1 \neq 4$

Answer: G

Question 15

Solve the system of equations:

\[\begin{cases} 4x + 3y = 23 \\ 2x - 3y = 1 \end{cases}\]

What is the value of $x$?

A) $2$
B) $3$
C) $4$
D) $5$

Show solution

Add the two equations to eliminate $y$:

\[(4x + 3y) + (2x - 3y) = 23 + 1 \implies 6x = 24 \implies x = 4\]

Verify with second equation: $2(4) - 3y = 1 \implies 8 - 3y = 1 \implies y = \frac{7}{3}$. Check in first: $4(4) + 3\cdot\frac{7}{3} = 16 + 7 = 23$ ✓

1. $2$: arithmetic error after adding: $6x = 12$, $x = 2$
1. $3$: halved the sum of right-hand sides: $(23+1)/8 = 3$; or a different setup error
1. $5$: added right-hand sides and divided by the wrong coefficient: $24/5 \approx 5$

Answer: C

Question 16

A jacket originally priced at $\$80$ is on sale for $\$60$. What is the percent decrease in price?

F) $20\%$
G) $25\%$
H) $33\%$
J) $75\%$

Show solution

\[\text{Percent decrease} = \frac{\text{original} - \text{new}}{\text{original}} \times 100 = \frac{80 - 60}{80} \times 100 = \frac{20}{80} \times 100 = 25\%\]

1. $20\%$: computed $\frac{20}{100} = 20\%$ (divided by $100$ instead of the original price $80$)
1. $33\%$: computed $\frac{20}{60} \approx 33\%$ (divided by the new price instead of the original)
1. $75\%$: computed $\frac{60}{80} = 75\%$ (the fraction of the original still remaining, not the decrease)

Answer: G

Question 17

A taxi company charges a $\$2.50$ base fare plus $\$1.75$ per mile. A rideshare app charges a $\$1.00$ base fare plus $\$2.25$ per mile. For how many miles are the total charges of the two services equal?

A) $2$
B) $3$
C) $4$
D) $5$

Show solution

Set the two cost expressions equal:

\[2.50 + 1.75m = 1.00 + 2.25m\]

Subtract $1.75m$ and $1.00$ from both sides:

\[1.50 = 0.50m \implies m = 3 \text{ miles}\]

Verify: taxi $= 2.50 + 1.75(3) = 2.50 + 5.25 = \$7.75$; rideshare $= 1.00 + 2.25(3) = 1.00 + 6.75 = \$7.75$ ✓

1. $2$: error in setup or arithmetic: $1.50 = 0.50m$ but solved $m = 1.50/0.75 = 2$
1. $4$: arithmetic error in dividing $1.50/0.50$: computed $1.50/0.375 = 4$
1. $5$: solved $1.50 = 0.30m$ (used the difference $2.25 - 1.75 - 1.50$ incorrectly)

Answer: B

Question 18

What is the solution to $\dfrac{3}{x} + \dfrac{1}{4} = \dfrac{7}{4}$?

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

Show solution

Subtract $\dfrac{1}{4}$ from both sides:

\[\frac{3}{x} = \frac{7}{4} - \frac{1}{4} = \frac{6}{4} = \frac{3}{2}\]

Since $\dfrac{3}{x} = \dfrac{3}{2}$, cross-multiply:

\[3 \cdot 2 = 3 \cdot x \implies x = 2\]

Verify: $\dfrac{3}{2} + \dfrac{1}{4} = \dfrac{6}{4} + \dfrac{1}{4} = \dfrac{7}{4}$ ✓

1. $\dfrac{4}{3}$: inverted $x$ and $2$ after cross-multiplying: found $\frac{3}{x} = \frac{3}{2}$ but wrote $x = \frac{2}{3}$ then inverted to $\frac{4}{3}$… or solved without subtracting first: $\frac{3}{x} = \frac{7}{4}$, $x = \frac{12}{7}$ and rounded
1. $\dfrac{3}{2}$: reported the value of $\dfrac{3}{x}$ rather than $x$ itself
1. $3$: did not subtract $\frac{1}{4}$ from both sides; set $\frac{3}{x} = \frac{7}{4} - \frac{4}{4} = \frac{3}{4}$… or solved $\frac{3}{x} = 1$, giving $x = 3$

Answer: H

Question 19

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

A) $3x^3 + 10x^2 - 11x + 2$
B) $3x^3 + 10x^2 - 11x - 2$
C) $3x^3 + 14x^2 - 3x + 2$
D) $3x^3 - 10x^2 + 11x + 2$

Show solution

Distribute each term of $(3x - 2)$:

\[3x(x^2 + 4x - 1) - 2(x^2 + 4x - 1)\] \[= 3x^3 + 12x^2 - 3x - 2x^2 - 8x + 2\] \[= 3x^3 + (12-2)x^2 + (-3-8)x + 2\] \[= 3x^3 + 10x^2 - 11x + 2\]

1. $3x^3 + 10x^2 - 11x - 2$: correct except sign error on constant: $-2(-1) = +2$, not $-2$
1. $3x^3 + 14x^2 - 3x + 2$: added $x^2$ coefficients ($12 + 2 = 14$) instead of subtracting ($12 - 2 = 10$)
1. $3x^3 - 10x^2 + 11x + 2$: distributed $-2$ as $+2$, flipping signs on all middle terms

Answer: A

Question 20

The table below shows the prices and quantities sold for two models of headphones at an electronics store.

Model	Price	Units Sold
Standard	$\$45$	7
Premium	$\$120$	3

What was the average (arithmetic mean) revenue per unit sold across both models?

F) $\$65.00$
G) $\$67.50$
H) $\$72.00$
J) $\$82.50$

Show solution

Total revenue $= 45(7) + 120(3) = 315 + 360 = \$675$.

Total units $= 7 + 3 = 10$.

Average revenue per unit $= \dfrac{675}{10} = \$67.50$.

1. $\$65.00$: arithmetic error in the total revenue, giving $650/10 = 65$
1. $\$72.00$: weighted the premium model more heavily, or arithmetic error: $720/10 = 72$
1. $\$82.50$: computed the simple (unweighted) average of the two prices: $\dfrac{45 + 120}{2} = 82.50$, ignoring the different unit counts

Answer: G

Question 21

What is the surface area, in square inches, of a cube with edge length 5 inches?

A) $25$
B) $100$
C) $125$
D) $150$

Show solution

A cube has 6 faces, each with area $s^2 = 5^2 = 25$.

Surface area $= 6 \times 25 = 150$ square inches.

1. $25$: computed the area of one face only ($5^2 = 25$)
1. $100$: multiplied by $4$ instead of $6$: $4 \times 25 = 100$ (forgot two faces)
1. $125$: computed the volume $s^3 = 5^3 = 125$ instead of surface area

Answer: D

Question 22

What is the distance between the points $(-3, 4)$ and $(5, -2)$ in the coordinate plane?

F) $\sqrt{28}$
G) $\sqrt{52}$
H) $10$
J) $14$

Show solution

Apply the distance formula:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(5-(-3))^2 + (-2-4)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\]

1. $\sqrt{28}$: computed $(5-3)^2 + (-2-4)^2 = 4 + 36 = 40$… or used $\Delta x = 2$ and $\Delta y = 6$: $\sqrt{4+36} = \sqrt{40}$
1. $\sqrt{52}$: computed $(5-3)^2 + (-2+4)^2 = 4 + 4 = 8$… or used $\Delta x = 4, \Delta y = 6$: $\sqrt{16+36} = \sqrt{52}$
1. $14$: added $|\Delta x| + |\Delta y| = 8 + 6 = 14$ (Manhattan distance, not Euclidean)

Answer: H

Question 23

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

A) $6$
B) $16$
C) $26$
D) $38$

Show solution

First find $f(1)$:

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

Then find $g(f(1)) = g(5)$:

\[g(5) = 5^2 + 1 = 25 + 1 = 26\]

1. $6$: computed $f(g(1)) = f(2) = 8$… or applied $g$ first: $g(1) = 2$, then stopped and added 4
1. $16$: computed $g(f(1)) = g(5)$ but used $g(x) = x^2 - 9$: $25 - 9 = 16$; or evaluated $f(g(1)) = f(2) = 8$ then doubled
1. $38$: computed $f(g(1))$ using wrong order: $g(1) = 2$, $f(2) = 8$… not 38; or $g(6) = 36+1=37\approx38$

Answer: C

Question 24

The two-way table below shows the results of a survey asking 200 students whether they prefer cats or dogs, broken down by grade level.

	Cats	Dogs	Total
Grade 11	48	52	100
Grade 12	36	64	100
Total	84	116	200

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

F) $\dfrac{9}{50}$
G) $\dfrac{3}{7}$
H) $\dfrac{9}{25}$
J) $\dfrac{1}{2}$

Show solution

The condition restricts to Grade 12 students: 100 total.

Of those, 36 prefer cats:

\[P(\text{cats} \mid \text{Grade 12}) = \frac{36}{100} = \frac{9}{25}\]

1. $\dfrac{9}{50}$: divided Grade 12 cat-lovers by total students: $\frac{36}{200} = \frac{9}{50}$ (used total instead of Grade 12 total)
1. $\dfrac{3}{7}$: divided total cat-lovers by total dog-lovers: $\frac{84}{196}$… or Grade 12 cats divided by total cats: $\frac{36}{84} = \frac{3}{7}$ (wrong denominator)
1. $\dfrac{1}{2}$: assumed equal distribution, or divided Grade 11 cats (48) by Grade 11 total (100) and confused grades

Answer: H

Question 25

The graph of $y = \sin(x)$ is transformed to produce the graph of $y = \sin(x - \pi) + 3$. Which of the following describes the transformation?

A) Shifted left $\pi$ units and up 3 units
B) Shifted right $\pi$ units and down 3 units
C) Shifted left $\pi$ units and down 3 units
D) Shifted right $\pi$ units and up 3 units

Show solution

For $y = \sin(x - h) + k$: - $h > 0$ means a shift right by $h$ units (the $-$ inside moves right) - $k > 0$ means a shift up by $k$ units

Here $h = \pi$ and $k = 3$: the graph shifts right $\pi$ units and up 3 units.

1. Left $\pi$ and up 3: the $-\pi$ inside the function indicates a rightward shift, not leftward
1. Right $\pi$ and down 3: correct horizontal direction but $+3$ outside shifts up, not down
1. Left $\pi$ and down 3: both directions wrong

Answer: D

Question 26

In the figure below, $\overline{PQ}$ is tangent to a circle with center $O$ at point $P$. If $OQ = 15$ and $OP = 9$, what is the length of $\overline{PQ}$?

Note: Figure not drawn to scale.

F) $6$
G) $\sqrt{306}$
H) $12$
J) $\sqrt{144 + 81}$

Show solution

A tangent to a circle is perpendicular to the radius at the point of tangency, so $\angle OPQ = 90°$.

By the Pythagorean theorem in right triangle $OPQ$:

\[PQ^2 = OQ^2 - OP^2 = 15^2 - 9^2 = 225 - 81 = 144 \implies PQ = 12\]

This is the 9-12-15 Pythagorean triple (a scaled 3-4-5 triple).

1. $6$: computed $15 - 9 = 6$ — subtracted directly without squaring
1. $\sqrt{306}$: added instead of subtracted: $\sqrt{15^2 + 9^2} = \sqrt{306}$ — confused which side is the hypotenuse
1. $\sqrt{144 + 81} = \sqrt{225} = 15$: reversed the roles of $PQ$ and $OQ$ — found $OQ$ instead of $PQ$

Answer: H

Question 27

The boxplot below displays the distribution of quiz scores for a class of students.

Note: Figure not drawn to scale.

What is the interquartile range (IQR) of the quiz scores?

A) $12$
B) $22$
C) $32$
D) $44$

Show solution

The IQR is the difference between the third and first quartiles:

\[\text{IQR} = Q3 - Q1 = 84 - 62 = 22\]

1. $12$: computed the distance from the median to $Q1$: $74 - 62 = 12$
1. $32$: computed the range from $Q1$ to the median to $Q3$: $74 - 62 = 12$ … or median minus minimum: $74 - 52 = 22$… or $Q3 - \text{Median} + \text{Median} - Q1 = 10 + 12 = 22$… or $Q3 - \text{Min} = 84 - 52 = 32$ (confused quartile range with partial range)
1. $44$: computed the full range: $\text{Max} - \text{Min} = 96 - 52 = 44$

Answer: B

Question 28

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

F) $(x - 2)^2 + (y + 1)^2 = 5$
G) $(x - 2)^2 + (y + 1)^2 = 25$
H) $(x + 2)^2 + (y - 1)^2 = 25$
J) $(x - 2)^2 + (y + 1)^2 = 50$

Show solution

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

\[r = \sqrt{(6-2)^2 + (2-(-1))^2} = \sqrt{16 + 9} = \sqrt{25} = 5\]

The equation is $(x - 2)^2 + (y - (-1))^2 = 5^2$:

\[(x - 2)^2 + (y + 1)^2 = 25\]

1. $(x-2)^2 + (y+1)^2 = 5$: used $r = 5$ instead of $r^2 = 25$
1. $(x+2)^2 + (y-1)^2 = 25$: flipped both signs of the center coordinates
1. $(x-2)^2 + (y+1)^2 = 50$: computed $r^2 = 4^2 + 3^2 = 16 + 9 = 25$… then doubled to $50$; or computed $r^2 = 4^2 + (3+4)^2 = 16+49 = 65$ — a different error; or added $r^2$ twice

Answer: G

Question 29

Let $f(x) = 3x - 9$. Which of the following gives $f^{-1}(x)$?

A) $\dfrac{x + 9}{3}$
B) $\dfrac{x - 9}{3}$
C) $3x + 9$
D) $\dfrac{1}{3x - 9}$

Show solution

Replace $f(x)$ with $y$, swap $x$ and $y$, then solve:

\[y = 3x - 9 \xrightarrow{\text{swap}} x = 3y - 9 \implies 3y = x + 9 \implies y = \frac{x + 9}{3}\]

Verify: $f\!\left(\frac{x+9}{3}\right) = 3\cdot\frac{x+9}{3} - 9 = x + 9 - 9 = x$ ✓

1. $\dfrac{x-9}{3}$: subtracted 9 instead of adding when isolating $y$: $3y = x - 9$
1. $3x + 9$: multiplied by 3 instead of dividing (confused inverse with the “reverse” arithmetic pattern)
1. $\dfrac{1}{3x-9}$: confused inverse function with reciprocal

Answer: A

Question 30

A line of best fit is shown on the scatterplot below, which displays the relationship between average daily temperature (°F) and daily ice cream sales (in dollars) at a beachside stand.

Note: Figure not drawn to scale.

According to the line of best fit, what is the predicted ice cream sales (in dollars) when the temperature is $85°$F?

F) $\$315$
G) $\$370$
H) $\$440$
J) $\$520$

Show solution

The line of best fit passes through approximately $(62, 80)$ and $(98, 560)$.

Slope $\approx \dfrac{560 - 80}{98 - 62} = \dfrac{480}{36} \approx 13.3$ dollars per °F.

Using the point $(80, 290)$: \[\text{At } 85°\text{F}: \quad 290 + 13.3(85 - 80) = 290 + 66.5 \approx \$357 \approx \$370\]

Reading directly from the line at $x = 85$ gives approximately $\$370$.

1. $\$315$: read the line at a lower temperature (around $78°$F) or underestimated the slope
1. $\$440$: read the actual data point near $90°$F ($\$440$) instead of the line at $85°$F
1. $\$520$: read the line at $95°$F instead of $85°$F

Answer: G

Question 31

Which of the following is equivalent to $\dfrac{x^2 - 9}{2x^2 + 5x - 3}$ for all values of $x$ for which the expression is defined?

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

Show solution

Factor the numerator and denominator:

\[\text{Numerator: } x^2 - 9 = (x+3)(x-3)\]

\[\text{Denominator: } 2x^2 + 5x - 3 = (2x - 1)(x + 3)\]

Check: $(2x-1)(x+3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3$ ✓

Cancel the common factor $(x + 3)$:

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

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

Answer: A

Question 32

For what value of $k$ does the equation $3x^2 - kx + 3 = 0$ have exactly one real solution?

F) $3$
G) $6$
H) $9$
J) $12$

Show solution

A quadratic has exactly one real solution when the discriminant equals zero: $b^2 - 4ac = 0$.

Here $a = 3$, $b = -k$, $c = 3$:

\[(-k)^2 - 4(3)(3) = 0 \implies k^2 - 36 = 0 \implies k^2 = 36 \implies k = 6 \text{ (taking positive value)}\]

1. $3$: set $k^2 = 4ac = 36$ but solved $k = \sqrt{9} = 3$ (computation error)
1. $9$: computed $k = 4ac/b = 36/4 = 9$ (confused the discriminant formula)
1. $12$: computed $k = 4ac = 4(3)(3) = 36$… or $k = 2 \times 6 = 12$ (doubled the correct answer)

Answer: G

Question 33

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

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

Show solution

This is a product of complex conjugates: $(a - bi)(a + bi) = a^2 + b^2$.

\[(2 - 3i)(2 + 3i) = 2^2 + 3^2 = 4 + 9 = 13\]

Alternatively: $(2-3i)(2+3i) = 4 + 6i - 6i - 9i^2 = 4 - 9(-1) = 4 + 9 = 13$.

1. $-5$: computed $2^2 - 3^2 = 4 - 9 = -5$ (subtracted instead of adding, treating $i^2 = +1$)
1. $4 + 9i$: squared only the real parts and imaginary parts separately: $(2)^2 + (3i)^2$… with sign error keeping $9i$ instead of $-9i^2 = +9$
1. $4 - 9i$: same error but kept the imaginary unit; $4 + 9i^2 = 4 - 9$ with an extra $i$ attached

Answer: D

Question 34

What is the real solution to the equation $\log_2(3x + 4) = 5$?

F) $\dfrac{28}{3}$
G) $\dfrac{32}{3}$
H) $12$
J) $\dfrac{2^5 + 4}{3}$

Show solution

Convert from logarithmic to exponential form:

\[\log_2(3x + 4) = 5 \implies 3x + 4 = 2^5 = 32\]

Solve for $x$:

\[3x = 28 \implies x = \frac{28}{3}\]

Verify: $\log_2\!\left(3 \cdot \frac{28}{3} + 4\right) = \log_2(32) = 5$ ✓

1. $\dfrac{32}{3}$: forgot to subtract $4$ before dividing: $3x = 32$, $x = \frac{32}{3}$
1. $12$: solved $3x + 4 = 40$ (computed $2^5 = 40$ instead of $32$) or $3x = 36$, $x = 12$
1. $\dfrac{2^5 + 4}{3} = \dfrac{36}{3} = 12$: added $4$ instead of subtracting when rearranging

Answer: F

Question 35

The function $f(x) = -3\sin(2x) + 1$ has amplitude $A$ and midline $y = k$. What are the values of $A$ and $k$?

A) $A = -3$, $k = 1$
B) $A = 2$, $k = 1$
C) $A = 3$, $k = 0$
D) $A = 3$, $k = 1$

Show solution

For $f(x) = A\sin(Bx) + k$, the amplitude is $|A|$ (always non-negative) and the midline is $y = k$.

Here $A = -3$ (coefficient), $B = 2$, $k = 1$:

Amplitude $= |-3| = 3$
Midline: $y = 1$
1. $A = -3$: amplitude is always the absolute value; $A = |-3| = 3$, not $-3$
1. $A = 2$: confused the amplitude with the coefficient of $x$ (the frequency factor $B$)
1. $A = 3$, $k = 0$: amplitude is correct but the midline is $y = 1$ (the vertical shift), not $y = 0$

Answer: D

Question 36

The equation $\dfrac{y^2}{36} - \dfrac{x^2}{25} = 1$ is the equation of a hyperbola. What are the coordinates of the vertices?

F) $(\pm 5,\ 0)$
G) $(0,\ \pm 5)$
H) $(\pm 6,\ 0)$
J) $(0,\ \pm 6)$

Show solution

The equation is $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ with $a^2 = 36$ and $b^2 = 25$, so $a = 6$ and $b = 5$.

When the $y^2$ term is positive (leading term), the hyperbola opens up and down, and the vertices lie on the $y$-axis at $(0, \pm a) = (0, \pm 6)$.

1. $(\pm 5, 0)$: used $b = 5$ and placed vertices on the $x$-axis (wrong axis and wrong value)
1. $(0, \pm 5)$: correct axis but used $b$ instead of $a$
1. $(\pm 6, 0)$: correct value $a = 6$ but wrong axis — the $y^2$ term being positive means vertices are on the $y$-axis

Answer: J

Question 37

In the figure below, right triangle $PQR$ has a right angle at $R$. If $PR = 7$ and $\angle P = 40°$, what is the length of $QR$, to the nearest tenth?

Note: Figure not drawn to scale.

A) $4.5$
B) $5.4$
C) $5.9$
D) $9.1$

Show solution

In right triangle $PQR$ with right angle at $R$:

$PR$ is adjacent to $\angle P$
$QR$ is opposite $\angle P$

Using tangent:

\[\tan(\angle P) = \frac{QR}{PR} \implies QR = PR \cdot \tan(40°) = 7 \times 0.8391 \approx 5.9\]

1. $4.5$: used $\sin(40°) \approx 0.643$ instead of $\tan$: $7 \times 0.643 \approx 4.5$
1. $5.4$: used $\cos(40°) \approx 0.766$: $7 \times 0.766 \approx 5.4$… that gives the adjacent/hypotenuse ratio; or used $\tan(37°) \approx 0.754 \times 7 = 5.3 \approx 5.4$
1. $9.1$: used $QR = PR / \tan(40°) = 7 / 0.766 \approx 9.1$ (divided instead of multiplied, using cosine in denominator)… or found the hypotenuse $PQ = 7/\cos(40°) = 7/0.766 \approx 9.1$

Answer: C

Question 38

A bacteria culture starts with 500 cells and doubles every 3 hours. The number of cells $N$ after $t$ hours is given by $N(t) = 500 \cdot 2^{t/3}$. How long does it take for the culture to reach 4,000 cells?

F) $6$ hours
G) $9$ hours
H) $12$ hours
J) $16$ hours

Show solution

Set $N(t) = 4{,}000$:

\[500 \cdot 2^{t/3} = 4{,}000 \implies 2^{t/3} = 8 = 2^3 \implies \frac{t}{3} = 3 \implies t = 9 \text{ hours}\]

1. $6$ hours: solved $2^{t/3} = 4 = 2^2$, giving $t/3 = 2$, $t = 6$ (used 4,000/500 = 4 but found $\log_2(4) = 2$ correctly — this means the culture is only at 2,000 after 6 hours)

Actually $500 \cdot 2^{6/3} = 500 \cdot 4 = 2000 \neq 4000$. So F is indeed wrong ✓.

1. $12$ hours: solved $2^{t/3} = 16 = 2^4$, giving $t = 12$ (corresponds to $N = 500 \cdot 16 = 8{,}000$)
1. $16$ hours: computed $t = 3 \times \log_2(8000/500)$ with an error, or added doubling periods incorrectly

Answer: G

Question 39

A chemist mixes $x$ liters of a $20\%$ acid solution with $y$ liters of a $50\%$ acid solution to obtain 12 liters of a $30\%$ acid solution. How many liters of the $20\%$ solution are used?

A) $4$
B) $6$
C) $8$
D) $10$

Show solution

Set up the system:

\[\begin{cases} x + y = 12 \\ 0.20x + 0.50y = 0.30(12) = 3.6 \end{cases}\]

From the first equation: $y = 12 - x$. Substitute:

\[0.20x + 0.50(12 - x) = 3.6 \implies 0.20x + 6 - 0.50x = 3.6 \implies -0.30x = -2.4 \implies x = 8\]

Verify: $x = 8, y = 4$; acid $= 0.20(8) + 0.50(4) = 1.6 + 2.0 = 3.6 = 0.30(12)$ ✓

1. $4$: reported $y$ (the liters of $50\%$ solution) instead of $x$
1. $6$: split 12 liters equally: $x = y = 6$ (ignored the concentration constraint)
1. $10$: arithmetic error in the substitution step: $-0.30x = -3$ → $x = 10$

Answer: C

Question 40

A sequence is defined by $a_1 = 4$ and $a_n = 3a_{n-1} + 2$ for $n \geq 2$. What is $a_4$?

F) $47$
G) $50$
H) $134$
J) $152$

Show solution

Apply the rule $a_n = 3a_{n-1} + 2$ repeatedly:

\[a_1 = 4\] \[a_2 = 3(4) + 2 = 14\] \[a_3 = 3(14) + 2 = 44\] \[a_4 = 3(44) + 2 = 134\]

1. $47$: applied only addition, treating the sequence as arithmetic: $4 + 43 = 47$ or similar
1. $50$: computed $a_3 = 44$ and added $6$ (the rule constant times 3): $44 + 6 = 50$
1. $152$: computed $3(50) + 2 = 152$ using the wrong value of $a_3$

Answer: H

Question 41

A rectangular swimming pool measures 25 meters long, 10 meters wide, and 2 meters deep. It is being filled at a rate of 5 cubic meters per minute. How many minutes will it take to fill the pool completely?

A) $50$
B) $70$
C) $100$
D) $500$

Show solution

Volume of pool $= 25 \times 10 \times 2 = 500$ cubic meters.

Time $= \dfrac{\text{volume}}{\text{rate}} = \dfrac{500}{5} = 100$ minutes.

1. $50$: computed the base area $25 \times 10 = 250$ and divided by 5: $250/5 = 50$ (forgot the depth)
1. $70$: computed $25 + 10 + 2 = 37$ (added dimensions), then $37 \times 5 = 185$… or a different error giving 350/5=70
1. $500$: reported the volume instead of dividing by the fill rate

Answer: C

Question 42

Two events $A$ and $B$ are mutually exclusive. $P(A) = 0.35$ and $P(B) = 0.40$. What is $P(A \text{ or } B)$?

F) $0.14$
G) $0.35$
H) $0.40$
J) $0.75$

Show solution

For mutually exclusive events (they cannot both occur), $P(A \cap B) = 0$.

By the addition rule:

\[P(A \text{ or } B) = P(A) + P(B) - P(A \cap B) = 0.35 + 0.40 - 0 = 0.75\]

1. $0.14$: computed $P(A) \times P(B) = 0.35 \times 0.40 = 0.14$ — the product (appropriate for independent events), not the sum
1. $0.35$: reported $P(A)$ alone
1. $0.40$: reported $P(B)$ alone

Answer: J

Question 43

Two vectors are given: $\vec{u} = \langle 3, -4 \rangle$ and $\vec{v} = \langle -1, 2 \rangle$. What is $\vec{u} + 2\vec{v}$?

A) $\langle 1, 0 \rangle$
B) $\langle 1, -8 \rangle$
C) $\langle 5, -6 \rangle$
D) $\langle 2, -2 \rangle$

Show solution

First compute $2\vec{v} = 2\langle -1, 2 \rangle = \langle -2, 4 \rangle$.

Then add:

\[\vec{u} + 2\vec{v} = \langle 3, -4 \rangle + \langle -2, 4 \rangle = \langle 3 + (-2),\; -4 + 4 \rangle = \langle 1, 0 \rangle\]

1. $\langle 1, -8 \rangle$: computed $\vec{u} - 2\vec{v}$: $\langle 3-(-2), -4-4 \rangle = \langle 5, -8 \rangle$… or a sign error on one component
1. $\langle 5, -6 \rangle$: added $\vec{u} + \vec{v}$ without doubling $\vec{v}$: $\langle 3+(-1), -4+2 \rangle$… then scaled wrong component
1. $\langle 2, -2 \rangle$: computed $\vec{u} + \vec{v}$ (without the factor of 2): $\langle 3-1, -4+2 \rangle = \langle 2, -2 \rangle$

Answer: A

Question 44

A spinner has 5 equal sections labeled 1, 2, 3, 4, and 5. A player wins $\$6$ if the spinner lands on 5, wins $\$2$ if it lands on an even number (2 or 4), and loses $\$4$ otherwise (1 or 3). What is the expected value of the player’s winnings per spin?

F) $-\$0.60$
G) $\$0.40$
H) $\$1.00$
J) $\$2.20$

Show solution

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

The outcomes and probabilities are:

Land on 5 (win $\$6$): probability $\dfrac{1}{5}$
Land on 2 or 4 (win $\$2$): probability $\dfrac{2}{5}$
Land on 1 or 3 (lose $\$4$): probability $\dfrac{2}{5}$

\[E = 6 \cdot \frac{1}{5} + 2 \cdot \frac{2}{5} + (-4) \cdot \frac{2}{5} = \frac{6}{5} + \frac{4}{5} - \frac{8}{5} = \frac{2}{5} = \$0.40\]

1. $-\$0.60$: computed only the expected loss $(-4)(2/5) = -1.6$ and the win from 5: $6/5 = 1.2$; total $= -0.4$… or used wrong probabilities
1. $\$1.00$: computed the unweighted average $(6 + 2 - 4)/3 = 4/3 \approx 1.33 \approx 1.00$, ignoring probabilities
1. $\$2.20$: computed $6(1/5) + 2(2/5) + 4(2/5) = 1.2 + 0.8 + 1.6 = 3.6$… or treated the loss as a gain: $6/5 + 4/5 + 8/5 = 18/5 = 3.60$; or otherwise summed without the negative

Answer: G

Question 45

A 10-foot ladder leans against a vertical wall. The base of the ladder is 4 feet from the wall, and the ladder makes an angle of $\theta$ with the ground. A painter standing on the ladder drops a tool that falls straight down 7.5 feet to the ground. How high above the ground was the painter when the tool was dropped? (Use the Pythagorean theorem, not trigonometry.)

A) $\sqrt{59}$ feet
B) $\sqrt{6.25}$ feet
C) $7.5$ feet
D) $\sqrt{43.75}$ feet

Show solution

The ladder forms a right triangle with base $= 4$ ft and hypotenuse $= 10$ ft.

The height of the top of the ladder: $h = \sqrt{10^2 - 4^2} = \sqrt{84} \approx 9.17$ ft.

The tool falls $7.5$ feet straight down to the ground, so the painter was exactly $7.5$ feet above the ground. Since $7.5 < 9.17$, the painter is at a valid position on the ladder ✓.

1. $\sqrt{59}$: incorrectly applied the Pythagorean theorem using $7.5$ as a hypotenuse and $4$ as a leg: $\sqrt{7.5^2 - 4^2} = \sqrt{56.25 - 16} = \sqrt{40.25}$; not $\sqrt{59}$, but a similar misapplication
1. $\sqrt{6.25}$: computed $\sqrt{4^2 - (10-7.5)^2} = \sqrt{16 - 6.25} = \sqrt{9.75}$; a different misapplication
1. $\sqrt{43.75}$: computed the horizontal distance along the ground at the painter’s height: $\sqrt{10^2 - 7.5^2} = \sqrt{56.25} = 7.5$… or used $\sqrt{7.5^2 - 4^2} = \sqrt{43.75}$ (swapped the roles of the sides)

Answer: C