ACT Math Diagnostic Quiz

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

Question 1

Which of the following is equal to $\dfrac{2^8}{2^5}$?

A) $2^3$
B) $2^{13}$
C) $4^3$
D) $1^3$

Show solution

When dividing powers with the same base, subtract exponents:

\[\frac{2^8}{2^5} = 2^{8-5} = 2^3 = 8\]

1. $2^{13}$: added exponents instead of subtracting ($8 + 5 = 13$)
1. $4^3$: $4^3 = 64 \neq 8$; confused $2^3$ with $(2^2)^3$
1. $1^3$: divided the bases instead of subtracting exponents ($2/2 = 1$)

Answer: A

Question 2

Solve for $x$: $5(x - 2) = 3x + 8$

F) $3$
G) $7$
H) $9$
J) $11$

Show solution

Distribute on the left:

\[5x - 10 = 3x + 8\]

Subtract $3x$, then add $10$:

\[2x = 18 \implies x = 9\]

Verify: $5(9-2) = 5(7) = 35$ and $3(9)+8 = 35$ ✓

1. $3$: arithmetic error after collecting terms: $2x = 6$, $x = 3$
1. $7$: solved $5x - 10 = 3x + 4$ (halved the right-side constant): $2x = 14$, $x = 7$
1. $11$: added $10$ to both sides instead of subtracting, then made a further error: $2x = 22$, $x = 11$

Answer: H

Question 3

If $f(x) = x^2 - 2x + 5$, what is $f(-1)$?

A) $2$
B) $4$
C) $8$
D) $10$

Show solution

Substitute $x = -1$:

\[f(-1) = (-1)^2 - 2(-1) + 5 = 1 + 2 + 5 = 8\]

1. $2$: computed $(-1)^2 - 2(-1) + 5 = 1 - 2 + 5 - 2 = 2$… or evaluated $f(1) = 1 - 2 + 5 = 4$ and halved
1. $4$: evaluated $f(1) = 1 - 2 + 5 = 4$ (used $x = +1$ instead of $x = -1$)
1. $10$: computed $(-1)^2 + 2(-1)(-1) + 5 = 1 + 2(1) + ... = 8$; or treated $-2(-1) = -2$ instead of $+2$: $1 - 2 + … $; or added an extra term

Answer: C

Question 4

In the figure below, $\triangle ABC$ and $\triangle DEF$ are similar with $\angle A \cong \angle D$. If $AB = 6$, $BC = 8$, and $DE = 9$, what is the length of $EF$?

Note: Figure not drawn to scale.

F) $10$
G) $11$
H) $12$
J) $15$

Show solution

Since the triangles are similar with $\angle A \cong \angle D$, corresponding sides are proportional:

\[\frac{AB}{DE} = \frac{BC}{EF} \implies \frac{6}{9} = \frac{8}{EF}\]

Cross-multiply:

\[6 \cdot EF = 9 \cdot 8 = 72 \implies EF = 12\]

1. $10$: added the scale difference $9 - 6 = 3$ to $BC$: $8 + 3 = 11$… or used wrong proportion
1. $11$: $8 + (9-6) = 11$ (added the difference of corresponding sides instead of using proportional reasoning)
1. $15$: added the two given sides: $6 + 9 = 15$, an unrelated calculation

Answer: H

Question 5

A data set has values $\{3, 7, 7, 9, 12, 14, 18\}$. What is the range of the data set?

A) $7$
B) $9$
C) $10$
D) $15$

Show solution

Range $= \text{maximum} - \text{minimum} = 18 - 3 = 15$.

1. $7$: reported the mode (the most frequent value) instead of the range
1. $9$: reported the median (the middle value of 7 values) instead of the range
1. $10$: computed $18 - 8 = 10$ (subtracted the wrong minimum) or $12 - 3 + 1 = 10$

Answer: D

Question 6

A car’s value depreciates by $15\%$ each year. If the car is currently worth $\$24{,}000$, which expression gives its value after $t$ years?

F) $24{,}000(0.15)^t$
G) $24{,}000(0.85)^t$
H) $24{,}000(1.15)^t$
J) $24{,}000 - 0.15t$

Show solution

A $15\%$ decrease each year means the car retains $100\% - 15\% = 85\% = 0.85$ of its value each year.

\[V(t) = 24{,}000 \times (0.85)^t\]

1. $24{,}000(0.15)^t$: used the decay rate $0.15$ as the base instead of the retention rate $0.85$
1. $24{,}000(1.15)^t$: used $1 + 0.15 = 1.15$, modeling a $15\%$ increase (appreciation) instead of a decrease
1. $24{,}000 - 0.15t$: modeled linear decay (subtracting a fixed dollar amount) instead of exponential decay

Answer: G

Question 7

For what value of $k$ does the equation $2x^2 + kx + 8 = 0$ have exactly one real solution?

A) $4$
B) $6$
C) $7$
D) $8$

Show solution

Exactly one real solution when the discriminant $= 0$: $b^2 - 4ac = 0$.

Here $a = 2$, $b = k$, $c = 8$:

\[k^2 - 4(2)(8) = 0 \implies k^2 = 64 \implies k = 8 \text{ (positive value)}\]

1. $4$: computed $k = \sqrt{4ac/4} = \sqrt{ac} = \sqrt{16} = 4$ (incorrect simplification)
1. $6$: computed $k = \sqrt{4(2)(8) - 16} = \sqrt{48} \approx 6.9 \approx 6$… or set $k^2 = 36$
1. $7$: set $k^2 = 4(2)(8) = 56$… or confused $4ac$ with $4a + c = 4(2) + 8 = 16$… or $k = \sqrt{56} \approx 7.5 \approx 7$

Answer: D

Question 8

Let $f(x) = 4x - 1$ and $g(x) = \dfrac{x+1}{4}$. What is $f(g(x))$?

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

Show solution

Substitute $g(x)$ into $f$:

\[f(g(x)) = f\!\left(\frac{x+1}{4}\right) = 4\left(\frac{x+1}{4}\right) - 1 = (x + 1) - 1 = x\]

Note: $f$ and $g$ are inverse functions of each other, so $f(g(x)) = x$.

1. $x - 1$: subtracted $1$ but forgot to cancel it from the expansion: $4 \cdot \frac{x+1}{4} = x + 1$, then $-1$ gives $x$ — this error would give $x - 1 - 1 = x - 2$, not $x - 1$
1. $x + 1$: computed $4 \cdot \frac{x+1}{4} = x + 1$ but forgot to subtract the $-1$ in $f$
1. $4x^2 - 1$: multiplied $f(x) \cdot g(x)$ instead of composing: $(4x-1)\cdot\frac{x+1}{4} \neq f(g(x))$

Answer: G

Question 9

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

Note: Figure not drawn to scale.

A) $25$
B) $29$
C) $31$
D) $35$

Show solution

Co-interior (same-side interior) angles are supplementary when lines are parallel:

\[(3x + 15) + (2x + 10) = 180\] \[5x + 25 = 180\] \[5x = 155 \implies x = 31\]

Verify: $3(31)+15=108°$, $2(31)+10=72°$; $108+72=180°$ ✓

1. $25$: solved $5x + 25 = 150$: $5x = 125$, $x = 25$ (used $150°$ instead of $180°$ as the sum)
1. $29$: arithmetic error: $5x = 145$, $x = 29$
1. $35$: set the two angles equal as if they were alternate interior angles: $3x+15 = 2x+10$ gives $x = -5$… or solved $5x = 175$, $x = 35$

Answer: C

Question 10

The table below shows scores from two classes on a quiz.

Class	Mean	Number of Students
Class A	78	20
Class B	86	30

What is the combined mean score for all 50 students?

F) $82$
G) $82.8$
H) $83.2$
J) $84$

Show solution

Total score $= 78(20) + 86(30) = 1{,}560 + 2{,}580 = 4{,}140$.

Combined mean $= \dfrac{4{,}140}{50} = 82.8$.

1. $82$: computed the simple average of the two means: $\frac{78 + 86}{2} = 82$ (ignored the different class sizes)
1. $83.2$: arithmetic error in total: $78(20) + 86(30) = 1560 + 2580 = 4140$; $4160/50 = 83.2$ (used $1580$ for Class A)
1. $84$: weighted Class A more heavily, or used incorrect class sizes

Answer: G

Question 11

Which of the following is equivalent to $\dfrac{x^2 - 16}{x^2 - x - 12}$ for all values of $x$ for which the expression is defined?

A) $\dfrac{x - 4}{x - 4}$
B) $\dfrac{x - 4}{x + 3}$
C) $\dfrac{x + 4}{x - 4}$
D) $\dfrac{x + 4}{x + 3}$

Show solution

Factor numerator and denominator:

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

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

Cancel the common factor $(x - 4)$:

\[\frac{(x+4)(x-4)}{(x-4)(x+3)} = \frac{x + 4}{x + 3}\]

1. $\dfrac{x-4}{x-4} = 1$: cancelled $(x+4)$ from the numerator and $(x+3)$ from the denominator — neither is a common factor
1. $\dfrac{x-4}{x+3}$: cancelled $(x+4)$ from the numerator instead of $(x-4)$
1. $\dfrac{x+4}{x-4}$: factored the denominator incorrectly as $(x-4)(x-3)$ or $(x+4)(x-4)$

Answer: D

Question 12

What is the real solution to $5e^{2x} = 20$?

F) $x = \dfrac{\ln 4}{2}$
G) $x = \ln 2$
H) $x = \ln 4$
J) $x = \dfrac{\ln 20}{5}$

Show solution

Divide both sides by 5:

\[e^{2x} = 4\]

Take the natural log of both sides:

\[2x = \ln 4 \implies x = \frac{\ln 4}{2}\]

Since $\ln 4 = 2\ln 2$, this simplifies to $x = \ln 2$.

1. $\dfrac{\ln 4}{2}$: this is equivalent to $\ln 2$ (an unsimplified form of the same value — not wrong, just not fully simplified)
1. $\ln 4$: forgot to divide by $2$: stopped at $2x = \ln 4$ and wrote $x = \ln 4$
1. $\dfrac{\ln 20}{5}$: took $\ln$ of both sides before dividing by $5$, then divided by $10$ incorrectly: $\ln(5e^{2x}) = \ln 20$ and got confused

Answer: G

Question 13

In right triangle $MNP$, the right angle is at $N$. If $\angle M = 50°$ and $MN = 10$, what is the length of $MP$, to the nearest tenth?

Note: Figure not drawn to scale.

A) $7.7$
B) $11.9$
C) $13.1$
D) $15.6$

Show solution

In right triangle $MNP$ with right angle at $N$:

$MN = 10$ is adjacent to $\angle M$
$MP$ is the hypotenuse

Using cosine:

\[\cos(\angle M) = \frac{MN}{MP} \implies MP = \frac{MN}{\cos(50°)} = \frac{10}{0.6428} \approx 15.6\]

1. $7.7$: computed $MP = MN \cdot \cos(50°) = 10 \times 0.766 \approx 7.66 \approx 7.7$ (multiplied instead of divided, treating MN as the hypotenuse)
1. $11.9$: found $NP = MN \cdot \tan(50°) = 10 \times 1.192 \approx 11.9$ (the other leg, not the hypotenuse)
1. $13.1$: computed $MP = MN / \sin(50°) = 10 / 0.766 \approx 13.1$ (used sine instead of cosine — found the hypotenuse using the wrong trig ratio for the given side)

Answer: D

Question 14

A train travels 120 miles in the first 2 hours at a constant speed, then slows down and travels 90 miles in the next 3 hours at a new constant speed. What is the train’s average speed, in miles per hour, for the entire 5-hour trip?

F) $42$
G) $45$
H) $52.5$
J) $55$

Show solution

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

Total distance $= 120 + 90 = 210$ miles. Total time $= 2 + 3 = 5$ hours.

\[\text{Average speed} = \frac{210}{5} = 42 \text{ mph}\]

1. $45$: computed the arithmetic mean of the two speeds $\left(\frac{60 + 30}{2} = 45\right)$, ignoring that the train spent different amounts of time at each speed
1. $52.5$: computed the simple average of the two speeds using the two leg distances as weights: $\frac{120(60) + 90(30)}{120 + 90}$… or averaged $60$ and $45$ (the per-hour rates for each leg)
1. $55$: overcounted the faster leg’s contribution or used the wrong total time

Answer: F

Question 15

A chemist needs to make 10 liters of a $45\%$ saline solution by mixing a $30\%$ solution with a $60\%$ solution. How many liters of the $60\%$ solution are needed?

A) $4$
B) $5$
C) $6$
D) $7$

Show solution

Let $x$ = liters of $60\%$ solution; then $10 - x$ = liters of $30\%$ solution.

Set up the salt equation:

\[0.60x + 0.30(10 - x) = 0.45(10)\] \[0.60x + 3 - 0.30x = 4.5\] \[0.30x = 1.5 \implies x = 5\]

Verify: $0.60(5) + 0.30(5) = 3 + 1.5 = 4.5 = 0.45(10)$ ✓

1. $4$: arithmetic error — solved $0.30x = 1.2$, giving $x = 4$
1. $6$: solved for the $30\%$ solution volume and reported it as the $60\%$ solution
1. $7$: set up the equation with reversed percentages or made a sign error: $0.30x = 2.1$

Answer: B