Full-Length Practice Test 1

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

Module 1

Question 1

If $5x - 3 = 17$, what is the value of $x$?

A) 2
B) 4
C) 7
D) 14

Show solution

Add 3 to both sides: $5x = 20$. Divide both sides by 5: $x = 4$.

Choice A (2): arithmetic slip — divided $14$ by $5$ after an incorrect subtraction. Choice C (7): added 3 to 17 first to get $5x = 20$, then computed $20 - 13 = 7$ instead of dividing. Choice D (14): added 3 to 17 to get 20, then did not divide by the coefficient.

Answer: B

Question 2

If $f(x) = 3x + 8$, what is the value of $f(-2)$?

A) $-14$
B) $2$
C) $6$
D) $14$

Show solution

Substitute $x = -2$: $f(-2) = 3(-2) + 8 = -6 + 8 = 2$.

Choice A ($-14$): computed $3(-2) - 8 = -14$ — subtracted instead of added the constant. Choice C (6): used $|{-2}| = 2$ — ignored the negative sign on the input. Choice D (14): computed $3(2) + 8 = 14$ — used $+2$ instead of $-2$.

Answer: B

Question 3

Two angles are supplementary. One angle measures $(2x + 15)°$ and the other measures $(3x - 10)°$. What is the measure, in degrees, of the larger angle?

A) 35°
B) 85°
C) 95°
D) 180°

Show solution

Supplementary angles sum to 180°: \[(2x + 15) + (3x - 10) = 180\] \[5x + 5 = 180 \implies 5x = 175 \implies x = 35\]

The two angle measures are: - $2(35) + 15 = 85°$ - $3(35) - 10 = 95°$

The larger angle is $95°$.

Choice A (35°): the value of $x$, not an angle measure. Choice B (85°): the smaller of the two angles. Choice D (180°): the total, not either individual angle.

Answer: C

Question 4

A store is having a sale. All items are marked down 20% from the original price. Jada buys a jacket whose original price is $p$ dollars. Which expression represents the sale price of the jacket?

A) $p - 20$
B) $0.20p$
C) $0.80p$
D) $1.20p$

Show solution

A 20% markdown means the customer pays $100\% - 20\% = 80\%$ of the original price. As a decimal, $80\% = 0.80$, so the sale price is $0.80p$.

Choice A ($p - 20$): subtracts a fixed $20 instead of 20% — confuses a flat discount with a percent discount. Choice B ($0.20p$): gives the amount saved (the discount), not the amount paid. Choice D ($1.20p$): adds 20% instead of subtracting — this would be a 20% markup, not a markdown.

Answer: C

Question 5

If $\dfrac{x}{4} + 7 = 13$, what is the value of $x$?

Enter your answer:

Show solution

Subtract 7 from both sides: $\dfrac{x}{4} = 6$. Multiply both sides by 4: $x = 24$.

Answer: 24

Question 6

The function $g$ is defined by $g(x) = x^2 - 5x + 6$. What is the value of $g(3)$?

A) $0$
B) $2$
C) $6$
D) $12$

Show solution

Substitute $x = 3$: \[g(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0\]

Choice B (2): arithmetic slip — computed $9 - 5 + 6 - 6 = 4$, then divided by 2. Choice C (6): read only the constant term, or computed $3^2 - 3 = 6$ ignoring the rest of the expression. Choice D (12): added the coefficient of $x$ rather than multiplying: $9 + 5 - 6 - 12$… likely a sign error chain.

Answer: A

Question 7

The bar chart below shows the number of books read by five students during the summer.

According to the bar chart, how many more books did Carla read than Ben?

A) 3
B) 5
C) 7
D) 10

Show solution

From the bar chart: Carla read 10 books and Ben read 5 books. The difference is $10 - 5 = 5$.

Choice A (3): the difference between Diego (3) and the origin — misread Diego’s bar as the answer. Choice C (7): difference between Carla (10) and Diego (3) — read Diego’s bar instead of Ben’s. Choice D (10): Carla’s total, not the difference between Carla and Ben.

Answer: B

Question 8

A rectangle has a length of $(2x + 3)$ and a width of $4$. If the area of the rectangle is $44$ square units, what is the value of $x$?

A) 4
B) 5
C) 5.5
D) 9.25

Show solution

Area = length × width: \[4(2x + 3) = 44\] \[8x + 12 = 44\] \[8x = 32\] \[x = 4\]

Choice B (5): results from $2x + 3 = 44$ (forgot to multiply by the width 4) giving $2x = 41$, then rounded — not exact. Choice C (5.5): results from dividing $44$ by $8$ directly without subtracting $12$ first: $44 / 8 = 5.5$. Choice D (9.25): results from treating the problem as $(2x + 3) = 44$ (ignoring the width factor) and solving: $2x = 41$, $x = 20.5$… or from $4(2x) + 3 = 44 \Rightarrow 8x = 41 \Rightarrow x \approx 5.1$ — forgot to distribute the 4 to the constant term.

Answer: A

Question 9

The equation $x^2 - 7x + 12 = 0$ has two solutions. What is the greater of the two solutions?

Enter your answer:

Show solution

Factor the quadratic: $x^2 - 7x + 12 = (x - 3)(x - 4) = 0$.

By the zero product property: $x = 3$ or $x = 4$.

The greater solution is $4$.

Answer: 4

Question 10

A rideshare company charges a flat fee of $\$2.50$ plus $\$1.75$ per mile. Which equation gives the total cost $C$, in dollars, for a ride of $m$ miles?

A) $C = 1.75 + 2.50m$
B) $C = 2.50m + 1.75$
C) $C = 1.75m + 2.50$
D) $C = 4.25m$

Show solution

The flat fee is a one-time constant (the $y$-intercept), and the per-mile charge is the rate (the slope). Total cost: \[C = 1.75m + 2.50\]

Choice A ($C = 1.75 + 2.50m$): swaps the roles of the two numbers — makes $2.50$ the per-mile rate and $1.75$ the flat fee. Choice B ($C = 2.50m + 1.75$): same error — $2.50$ is the per-mile rate here, which is incorrect. Choice D ($C = 4.25m$): adds both values into a single rate, eliminating the flat-fee structure.

Answer: C

Question 11

The dot plot below shows the number of hours five students spent studying last week.

What is the mean number of hours studied by the five students?

Enter your answer:

Show solution

Reading the dot plot, the five values are 2, 3, 5, 5, 5.

\[\text{Mean} = \frac{2 + 3 + 5 + 5 + 5}{5} = \frac{20}{5} = 4\]

Answer: 4

Question 12

Which of the following is equivalent to $\dfrac{x^5 \cdot x^3}{x^4}$, where $x \neq 0$?

A) $x^2$
B) $x^4$
C) $x^{12}$
D) $x^{60}$

Show solution

Multiply the exponents in the numerator, then divide by subtracting: \[\frac{x^5 \cdot x^3}{x^4} = \frac{x^{5+3}}{x^4} = \frac{x^8}{x^4} = x^{8-4} = x^4\]

Choice A ($x^2$): subtracted the numerator exponents instead of adding — computed $5 - 3 = 2$. Choice C ($x^{12}$): added all three exponents $5 + 3 + 4 = 12$ — did not account for the division step. Choice D ($x^{60}$): multiplied all three exponents $5 \times 3 \times 4 = 60$ — applied a multiplication rule that doesn’t exist.

Answer: B

Question 13

A system of equations is shown below.

\[\begin{aligned} 2x + y &= 11 \\ x - y &= 1 \end{aligned}\]

What is the value of $x$?

A) 2
B) 4
C) 5
D) 7

Show solution

Add the two equations to eliminate $y$: \[(2x + y) + (x - y) = 11 + 1 \implies 3x = 12 \implies x = 4\]

To verify: $x = 4 \Rightarrow y = 4 - 1 = 3$. Check first equation: $2(4) + 3 = 11$ ✓

Choice A (2): solved for $y$ — substituting $x = 4$ back gives $y = 3$, not 2; likely a setup error. Choice C (5): arithmetic error in the addition step — subtracted the right sides as $11 - 1 = 10$ instead of adding, giving $3x = 10$. Choice D (7): added $x$ values from both equations directly without solving the system.

Answer: B

Question 14

In the figure below, lines $\ell$ and $m$ are parallel, and line $t$ is a transversal. The measure of angle 1 is $(3x + 20)°$ and the measure of angle 2 is $(5x - 40)°$.

Note: Figure not drawn to scale.

What is the measure, in degrees, of angle 1?

A) 90°
B) 110°
C) 120°
D) 130°

Show solution

Angles 1 and 2 are alternate interior angles formed by parallel lines $\ell$ and $m$ cut by transversal $t$. Alternate interior angles are equal: \[(3x + 20) = (5x - 40)\] \[60 = 2x \implies x = 30\]

Angle 1: $3(30) + 20 = 90 + 20 = 110°$.

Verify angle 2: $5(30) - 40 = 150 - 40 = 110°$ ✓

Choice A (90°): set the angle expression equal to 90° without using the parallel-line relationship. Choice C (120°): arithmetic slip in solving — likely computed $3x = 100$ at some step. Choice D (130°): computed $5(30) = 150$ and subtracted only 20 instead of 40.

Answer: B

Question 15

The function $f$ is defined by $f(x) = 2x^2 - 16x + 35$. Which of the following is an equivalent form of $f(x)$ that reveals the minimum value of $f$?

A) $f(x) = 2(x - 8)^2 - 93$
B) $f(x) = 2(x - 4)^2 + 3$
C) $f(x) = 2(x - 4)^2 - 3$
D) $f(x) = (2x - 8)(x - 4) + 3$

Show solution

Factor out the leading coefficient from the variable terms: \[f(x) = 2(x^2 - 8x) + 35\]

Complete the square: half of $-8$ is $-4$, and $(-4)^2 = 16$. Add and subtract 16 inside: \[f(x) = 2(x^2 - 8x + 16 - 16) + 35 = 2\bigl((x-4)^2 - 16\bigr) + 35\] \[= 2(x-4)^2 - 32 + 35 = 2(x-4)^2 + 3\]

The vertex form is $2(x-4)^2 + 3$, so the minimum value of $f$ is $\mathbf{3}$, occurring at $x = 4$.

Choice A ($2(x-8)^2 - 93$): used $-8$ as the value of $h$ without dividing by 2 — did not account for the leading coefficient when halving $b$. Choice C ($2(x-4)^2 - 3$): sign error in the final constant — subtracted $32$ from $35$ with the wrong sign. Choice D: partial factoring that does not cleanly reveal the vertex or minimum value.

Answer: B

Question 16

The scatterplot below shows the relationship between the number of hours a student studied and their score on a history exam. A line of best fit is shown.

Which of the following is the best interpretation of the slope of the line of best fit?

A) For each additional hour studied, the predicted exam score decreases by about 5 points.
B) A student who studies 0 hours is predicted to score about 49 points.
C) For each additional hour studied, the predicted exam score increases by about 5 points.
D) A student who studies 10 hours is predicted to score about 100 points.

Show solution

The slope of a line of best fit represents the rate of change: for each one-unit increase in the $x$-variable (hours studied), the $y$-variable (exam score) changes by approximately that amount on average.

The line rises from roughly $(1, 54)$ to $(10, 98)$, giving an approximate slope of $\dfrac{98 - 54}{10 - 1} \approx \dfrac{44}{9} \approx 5$ points per hour. The positive slope means each additional hour predicts an increase of about 5 points.

Choice A: describes a negative slope — the scatterplot clearly shows a positive association. Choice B: describes the $y$-intercept (predicted score when hours = 0), not the slope. Choice D: describes a specific predicted value for a single input, not the rate of change.

Answer: C

Question 17

The equation $x^2 - 6x + k = 0$ has exactly one real solution. What is the value of $k$?

Enter your answer:

Show solution

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

Here $a = 1$, $b = -6$, $c = k$: \[(-6)^2 - 4(1)(k) = 0 \implies 36 - 4k = 0 \implies k = 9\]

Answer: 9

Question 18

Maya earns $\$12$ per hour at a part-time job. She wants to earn at least $\$150$ in a week. She has already worked 5 hours this week. Which inequality represents the number of additional hours $h$ she must work to meet her goal?

A) $12h \geq 150$
B) $12h + 60 \geq 150$
C) $12h + 60 \leq 150$
D) $12(h + 5) \leq 150$

Show solution

Maya has already earned $12 \times 5 = \$60$. From $h$ more hours she will earn $12h$. Total earnings: $12h + 60$.

She wants to earn at least $\$150$: \[12h + 60 \geq 150\]

Choice A ($12h \geq 150$): ignores the $\$60$ already earned — doesn’t add the earnings from the 5 hours worked. Choice C ($12h + 60 \leq 150$): reverses the inequality — this says she wants to earn at most $\$150$. Choice D ($12(h+5) \leq 150$): treats total hours as $(h+5)$ inside the product, and uses the wrong inequality direction.

Answer: B

Question 19

The table below shows the results of a survey of 200 students about whether they prefer to study alone or in a group, and whether they prefer morning or evening study sessions.

	Morning	Evening	Total
Alone	54	66	120
Group	36	44	80
Total	90	110	200

A student is selected at random from those who prefer to study in the evening. What is the probability that this student prefers to study alone?

A) $\dfrac{33}{100}$
B) $\dfrac{3}{5}$
C) $\dfrac{11}{20}$
D) $\dfrac{3}{4}$

Show solution

The student is selected from those who prefer the evening — this restricts the sample space to the Evening column: 110 students total. Of those, 66 prefer to study alone.

\[P(\text{alone} \mid \text{evening}) = \frac{66}{110} = \frac{3}{5}\]

Choice A ($\frac{33}{100}$): used the total 200 as the denominator and divided 66 by 200, then halved — failed to condition on evening. Choice C ($\frac{11}{20}$): computed $\frac{110}{200} = \frac{11}{20}$ — the probability of being an evening student, not the conditional probability of being alone given evening. Choice D ($\frac{3}{4}$): likely computed $\frac{66}{90}$ — used the Morning total (90) as the denominator instead of the Evening total (110).

Answer: B

Question 20

Line $\ell$ passes through the points $(1, 9)$ and $(3, 13)$. Which of the following is an equation of line $\ell$?

A) $y = x + 2$
B) $y = x + 8$
C) $y = 2x + 7$
D) $y = 2x - 1$

Show solution

Find the slope: \[m = \frac{13 - 9}{3 - 1} = \frac{4}{2} = 2\]

Use point-slope form with $(1, 9)$: \[y - 9 = 2(x - 1) \implies y = 2x - 2 + 9 = 2x + 7\]

Verify with $(3, 13)$: $2(3) + 7 = 13$ ✓

Choice A ($y = x + 2$): used slope 1 — computed $\frac{4}{2}$ as 4 − 2 = 2, then misread as the slope. Choice B ($y = x + 8$): used slope 1 and computed the intercept using $9 - 1 = 8$ — two errors. Choice D ($y = 2x - 1$): correct slope but wrong intercept — likely used $9 - 2(3) = 3$ instead of $9 - 2(1) = 7$.

Answer: C

Question 21

The polynomial $p(x) = x^3 - 2x^2 + ax - 24$ has a factor of $(x - 3)$. What is the value of $a$?

Enter your answer:

Show solution

By the Factor Theorem, if $(x - 3)$ is a factor of $p(x)$, then $p(3) = 0$: \[p(3) = (3)^3 - 2(3)^2 + a(3) - 24 = 0\] \[27 - 18 + 3a - 24 = 0\] \[3a - 15 = 0\] \[a = 5\]

Answer: 5

Question 22

A right triangle has legs of length $x$ and $x + 2$ and a hypotenuse of length $x + 4$. What is the area of the triangle?

A) 12
B) 24
C) 30
D) 60

Show solution

Apply the Pythagorean theorem: (leg₁)² + (leg₂)² = (hypotenuse)²: \[x^2 + (x+2)^2 = (x+4)^2\] \[x^2 + x^2 + 4x + 4 = x^2 + 8x + 16\] \[2x^2 + 4x + 4 = x^2 + 8x + 16\] \[x^2 - 4x - 12 = 0\] \[(x - 6)(x + 2) = 0\]

Since lengths must be positive, $x = 6$. The legs are $6$ and $8$, and the hypotenuse is $10$ — a 6–8–10 right triangle.

\[\text{Area} = \frac{1}{2} \times 6 \times 8 = 24\]

Choice A (12): computed $\frac{1}{2} \times 6 \times 4$ — used $x$ and $x - 2$ instead of $x$ and $x + 2$. Choice C (30): computed $\frac{1}{2} \times 6 \times 10$ — multiplied a leg by the hypotenuse instead of the two legs together. Choice D (60): computed $6 \times 10$ — used a leg times the hypotenuse and forgot the $\frac{1}{2}$ factor.

Answer: B

Module 1

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Module 2

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44