Final review · Mock placement test 1

Mock placement test 1

32 questions · about 75-90 minutes · optional 90-minute timer

Please read: This is a practice test modeled on the MATH 1010 catalog topics (systems of equations; polynomial, rational and radical expressions; quadratic, rational, radical, exponential and log equations; graphing functions). It is not the real SLCC placement test, whose questions are not public.

About the real test: According to SLCC Testing Services, you take it online on the Möbius website, in one sitting, with a 90-minute time limit (SLCC estimates about 30 minutes), and you can't go back to earlier questions. The same page says notes, books, websites and help from anyone are not allowed; it doesn't say whether a calculator is allowed for math. So practice like the real thing: answer the questions in order, and don't go back to change one once you've moved on.

Answer every question, then press Grade my answers at the bottom. You'll get your score, a breakdown by unit, a worked solution for each question, and links to the lessons worth revisiting.

Timer: optional. The real test also has a 90-minute limit, but SLCC doesn't say how many questions it has, so use the timer only to practice working at a steady pace.
Calculator: work without one unless a question says "Calculator allowed" (only question 28 does). SLCC doesn't say whether one is allowed, so practicing without it is the safe choice. Pencil and paper are encouraged.
Typing answers: use ^ for powers (3x^2), / for fractions (3/2), sqrt( ) for square roots (2sqrt(5)), i for complex numbers (-1+2i). For two solutions, separate them with a comma (4, -3). Intervals look like (-inf, 3]; join pieces with U. Points look like (2, -1). Type none if there is no solution.
Tip: when you have a solution, plug it back into the original problem. It takes seconds and catches most slips, including extraneous solutions.

This is practice, not a test of you. A low score just shows where to look next.

Evaluate $b^2 - 4ac$ when $a = 2$, $b = -3$, and $c = -5$.

Substitute with parentheses: $(-3)^2 - 4(2)(-5)$.

$(-3)^2 = 9$ and $4(2)(-5) = -40$.

$9 - (-40) = 9 + 40 = 49$.

Answer: $49$.

Write $3x - 4y = 12$ in slope-intercept form, $y = mx + b$.

Subtract $3x$ from both sides: $-4y = -3x + 12$.

Divide every term by $-4$: $y = \dfrac{-3}{-4}x + \dfrac{12}{-4}$.

Answer: $y = \tfrac34 x - 3$.

Which is equal to $(3x - 5)^2$?

$9x^2 - 15x + 25$
$9x^2 + 30x + 25$
$9x^2 - 30x + 25$
$9x^2 - 30x - 25$

Use $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 3x$ and $b = 5$.

$a^2 = 9x^2$, $2ab = 2(3x)(5) = 30x$, $b^2 = 25$.

So $(3x - 5)^2 = 9x^2 - 30x + 25$. (Squaring a binomial always gives a middle term; it is never just $9x^2 + 25$.)

Answer: (c).

Solve $\dfrac{x}{3} + \dfrac{x - 1}{2} = 7$.

Multiply every term by the LCD, 6: $2x + 3(x - 1) = 42$.

Distribute: $2x + 3x - 3 = 42$, so $5x - 3 = 42$.

$5x = 45$, so $x = 9$.

Check: $\dfrac93 + \dfrac82 = 3 + 4 = 7$. Answer: $x = 9$.

The graph of $g(x) = (x + 2)^2 - 3$ is the graph of $y = x^2$ moved:

left 2 units and down 3 units
right 2 units and down 3 units
left 2 units and up 3 units
right 2 units and up 3 units

A number added inside with $x$ moves the graph sideways, opposite to its sign: $x + 2$ is zero at $x = -2$, so the graph moves left 2.

A number added outside moves it up or down with its sign: $-3$ moves it down 3.

The vertex moves from $(0, 0)$ to $(-2, -3)$. Answer: (a).

Simplify $5(x - 3) - 2(3x - 4)$.

Distribute the 5: $5x - 15$. Distribute the $-2$ to both terms: $-6x + 8$.

Combine: $5x - 6x - 15 + 8 = -x - 7$.

Answer: $-x - 7$.

Which is the complete factorization of $3x^3 - 75x$?

$3x(x^2 - 25)$
$3x(x - 5)(x + 5)$
$3x(x - 5)^2$
$3(x - 5)(x + 5)$

GCF first: both terms share $3x$, so $3x^3 - 75x = 3x(x^2 - 25)$.

$x^2 - 25$ is a difference of squares: $x^2 - 25 = (x - 5)(x + 5)$. So (a) is not finished.

Complete factorization: $3x(x - 5)(x + 5)$. Answer: (b).

Solve the system.

\[\begin{cases} 3x + 2y = 4 \\ 5x - 3y = 13 \end{cases}\]

Make the $y$ terms opposites: multiply the first equation by 3 and the second by 2.

$9x + 6y = 12$ and $10x - 6y = 26$. Add them: $19x = 38$, so $x = 2$.

Substitute into the first equation: $3(2) + 2y = 4$, so $2y = -2$ and $y = -1$.

Check in the second: $5(2) - 3(-1) = 10 + 3 = 13$. Answer: $(2, -1)$.

Simplify $\dfrac{x^2 - 9}{x^2 + x - 6}$.

Factor the top: $x^2 - 9 = (x - 3)(x + 3)$.

Factor the bottom: $x^2 + x - 6 = (x + 3)(x - 2)$ ($3 \cdot (-2) = -6$, $3 + (-2) = 1$).

Cancel the common factor $x + 3$: $\dfrac{(x - 3)\cancel{(x + 3)}}{\cancel{(x + 3)}(x - 2)} = \dfrac{x - 3}{x - 2}$, for $x \ne -3, 2$.

Answer: $\dfrac{x - 3}{x - 2}$.

Simplify $\dfrac{\left(x^{2/3}\,y^{1/2}\right)^6}{x\,y^2}$. Assume $x$ and $y$ are positive.

Power rule: multiply exponents. $\left(x^{2/3}\right)^6 = x^{\frac23 \cdot 6} = x^4$ and $\left(y^{1/2}\right)^6 = y^{3}$.

Now divide: $\dfrac{x^4y^3}{x^1y^2} = x^{4-1}y^{3-2}$.

Answer: $x^3y$.

Solve $(x - 3)^2 = 20$. Give exact answers.

Square root property (don't forget ±): $x - 3 = \pm\sqrt{20}$.

Simplify: $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt5$.

Add 3: $x = 3 \pm 2\sqrt5$.

Answer: $x = 3 + 2\sqrt5$ or $x = 3 - 2\sqrt5$.

Solve $|2x - 3| \le 7$. Write the answer in interval notation.

"Less than or equal" means "between": $-7 \le 2x - 3 \le 7$.

Add 3 to all three parts: $-4 \le 2x \le 10$.

Divide by 2: $-2 \le x \le 5$.

Answer: $[-2, 5]$ (square brackets, because the endpoints are included).

Find the domain of $f(x) = \sqrt{6 - 2x}$. Write it in interval notation.

The number under a square root can't be negative: $6 - 2x \ge 0$.

Subtract 6: $-2x \ge -6$. Divide by $-2$ and flip the sign: $x \le 3$.

Answer: $(-\infty, 3]$.

Simplify $\dfrac{(2x^3y^{-2})^3}{4x^{-1}y^5}$. Write the answer with positive exponents only.

Numerator: cube each factor. $2^3 = 8$, $(x^3)^3 = x^9$, $(y^{-2})^3 = y^{-6}$. So the top is $8x^9y^{-6}$.

Divide: $\dfrac84 = 2$, $x^{9 - (-1)} = x^{10}$, $y^{-6 - 5} = y^{-11}$.

Move $y^{-11}$ below the bar to make the exponent positive: $\dfrac{2x^{10}}{y^{11}}$.

Answer: $\dfrac{2x^{10}}{y^{11}}$.

Which is equal to $2\log x + \log y - 3\log z$, written as a single logarithm?

$\log \dfrac{2xy}{3z}$
$\log \dfrac{x^2}{yz^3}$
$\log (x^2y - z^3)$
$\log \dfrac{x^2y}{z^3}$

Power rule first: $2\log x = \log x^2$ and $3\log z = \log z^3$.

Product rule (adding logs multiplies): $\log x^2 + \log y = \log(x^2y)$.

Quotient rule (subtracting logs divides): $\log(x^2y) - \log z^3 = \log\dfrac{x^2y}{z^3}$.

Answer: (d).

Factor $6x^2 - 7x - 20$.

ac method: $a \cdot c = 6 \cdot (-20) = -120$. Find two numbers that multiply to $-120$ and add to $-7$: $8$ and $-15$.

Split the middle term: $6x^2 + 8x - 15x - 20$.

Group: $2x(3x + 4) - 5(3x + 4) = (3x + 4)(2x - 5)$.

Check: $(2x - 5)(3x + 4) = 6x^2 + 8x - 15x - 20 = 6x^2 - 7x - 20$. Answer: $(2x - 5)(3x + 4)$.

Find the equation of the line through $(8, -3)$ that is perpendicular to $y = 4x - 1$. Write it as $y = mx + b$.

The given slope is 4. A perpendicular slope is the negative reciprocal: $m = -\tfrac14$.

Point-slope form: $y - (-3) = -\tfrac14(x - 8)$, so $y + 3 = -\tfrac14 x + 2$.

Subtract 3: $y = -\tfrac14 x - 1$.

Check the point: $-\tfrac14(8) - 1 = -2 - 1 = -3$. Answer: $y = -\tfrac14 x - 1$.

Solve $2x^2 - 6x + 3 = 0$. Give exact answers.

It doesn't factor, so use the quadratic formula with $a = 2$, $b = -6$, $c = 3$.

Discriminant: $b^2 - 4ac = 36 - 24 = 12$, and $\sqrt{12} = 2\sqrt3$.

$x = \dfrac{6 \pm 2\sqrt3}{4}$. Divide every term by 2: $x = \dfrac{3 \pm \sqrt3}{2}$.

Answer: $x = \dfrac{3 + \sqrt3}{2}$ or $x = \dfrac{3 - \sqrt3}{2}$.

A school play sold 40 tickets for a total of $395. Adult tickets cost $12 and child tickets cost $7. How many adult tickets were sold?

Let $a$ = adult tickets and $c$ = child tickets. Then $a + c = 40$ and $12a + 7c = 395$.

From the first equation, $c = 40 - a$. Substitute: $12a + 7(40 - a) = 395$.

$12a + 280 - 7a = 395$, so $5a = 115$ and $a = 23$. Then $c = 17$.

Check: $12(23) + 7(17) = 276 + 119 = 395$. Answer: 23 adult tickets.

Subtract. Which is equal to $\dfrac{4}{x - 2} - \dfrac{3}{x + 1}$, written as one fraction?

$\dfrac{x + 2}{(x - 2)(x + 1)}$
$\dfrac{x + 10}{(x - 2)(x + 1)}$
$\dfrac{7x - 2}{(x - 2)(x + 1)}$
$\dfrac{x - 11}{(x - 2)(x + 1)}$

The LCD is $(x - 2)(x + 1)$. Rewrite each fraction over it: $\dfrac{4(x + 1)}{(x - 2)(x + 1)} - \dfrac{3(x - 2)}{(x - 2)(x + 1)}$.

Subtract the numerators, keeping parentheses: $4(x + 1) - 3(x - 2) = 4x + 4 - 3x + 6 = x + 10$. (The $-3$ multiplies both terms, so $-3 \cdot (-2) = +6$.)

So the result is $\dfrac{x + 10}{(x - 2)(x + 1)}$. Answer: (b).

Rationalize the denominator. Which is equal to $\dfrac{6}{3 - \sqrt5}$?

$\dfrac{9 - 3\sqrt5}{2}$
$\dfrac{9 + 3\sqrt5}{7}$
$\dfrac{9 + 3\sqrt5}{2}$
$\dfrac{3 + \sqrt5}{4}$

Multiply top and bottom by the conjugate $3 + \sqrt5$.

Bottom: $(3 - \sqrt5)(3 + \sqrt5) = 9 - 5 = 4$. Top: $6(3 + \sqrt5) = 18 + 6\sqrt5$.

$\dfrac{18 + 6\sqrt5}{4}$. Divide every term by 2: $\dfrac{9 + 3\sqrt5}{2}$.

Answer: (c).

Solve $2x^2 + 5x = 12$.

Get 0 on one side: $2x^2 + 5x - 12 = 0$.

Factor: $(2x - 3)(x + 4) = 0$. (Check: $2x^2 + 8x - 3x - 12$.)

Zero product property: $2x - 3 = 0$ or $x + 4 = 0$, so $x = \tfrac32$ or $x = -4$.

Answer: $x = \tfrac32$ or $x = -4$.

Find the inverse function of $f(x) = \dfrac{2x - 5}{3}$.

Write $y = \dfrac{2x - 5}{3}$, then swap $x$ and $y$: $x = \dfrac{2y - 5}{3}$.

Solve for $y$: $3x = 2y - 5$, so $2y = 3x + 5$ and $y = \dfrac{3x + 5}{2}$.

Check: $f\!\left(\dfrac{3x + 5}{2}\right) = \dfrac{(3x + 5) - 5}{3} = x$.

Answer: $f^{-1}(x) = \dfrac{3x + 5}{2}$.

Find the vertex of the parabola $f(x) = -2x^2 + 8x - 3$.

The vertex's $x$-value is $x = -\dfrac{b}{2a} = -\dfrac{8}{2(-2)} = 2$.

Its $y$-value: $f(2) = -2(4) + 16 - 3 = -8 + 16 - 3 = 5$.

Answer: $(2, 5)$. (The parabola opens down, so this is its highest point.)

Divide: $(6x^3 - 7x^2 - 7x + 6) \div (2x - 3)$.

Long division. $6x^3 \div 2x = 3x^2$. Multiply: $3x^2(2x - 3) = 6x^3 - 9x^2$. Subtract: $-7x^2 - (-9x^2) = 2x^2$. Bring down: $2x^2 - 7x$.

$2x^2 \div 2x = x$. Multiply: $x(2x - 3) = 2x^2 - 3x$. Subtract: $-7x - (-3x) = -4x$. Bring down: $-4x + 6$.

$-4x \div 2x = -2$. Multiply: $-2(2x - 3) = -4x + 6$. Subtract: remainder 0.

Answer: $3x^2 + x - 2$.

Solve $x^2 - x - 12 \lt 0$. Write the answer in interval notation.

Find the zeros: $x^2 - x - 12 = (x - 4)(x + 3) = 0$, so $x = 4$ or $x = -3$.

The parabola opens up, so it is below zero between the zeros. Test $x = 0$: $-12 \lt 0$. True.

The inequality is strict ($\lt$), so the endpoints are not included.

Answer: $(-3, 4)$.

Solve $\sqrt{2x + 7} = x + 2$.

The radical is already alone. Square both sides: $2x + 7 = (x + 2)^2 = x^2 + 4x + 4$.

Get 0 on one side: $0 = x^2 + 2x - 3 = (x + 3)(x - 1)$, so $x = -3$ or $x = 1$.

Check both. $x = 1$: $\sqrt{9} = 3$ and $1 + 2 = 3$. Works. $x = -3$: $\sqrt{1} = 1$ but $-3 + 2 = -1$. Extraneous.

Answer: $x = 1$.

Calculator allowed. An account starts with $5,000 and earns 3% interest compounded continuously, so its balance is $A = 5000e^{0.03t}$ after $t$ years. After how many years will it hold $8,000? Round to the nearest tenth of a year.

Set up: $5000e^{0.03t} = 8000$. Divide by 5000: $e^{0.03t} = 1.6$.

Take the natural log of both sides: $0.03t = \ln 1.6$.

$t = \dfrac{\ln 1.6}{0.03} \approx \dfrac{0.4700}{0.03} \approx 15.67$.

Answer: about 15.7 years.

Solve the system.

\[\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}\]

Eliminate $z$. Add equations 1 and 3: $2x + 3y = 8$. Add equations 2 and 3: $3x + y = 5$.

From $3x + y = 5$: $y = 5 - 3x$. Substitute: $2x + 3(5 - 3x) = 8$, so $2x + 15 - 9x = 8$, $-7x = -7$, $x = 1$.

Then $y = 5 - 3 = 2$, and from equation 1, $z = 6 - 1 - 2 = 3$.

Check equation 2: $2(1) - 2 + 3 = 3$. Answer: $(1, 2, 3)$.

Solve $x^2 + 2x + 5 = 0$. Write complex solutions in the form $a + bi$.

Quadratic formula with $a = 1$, $b = 2$, $c = 5$. Discriminant: $4 - 20 = -16$, which is negative, so the solutions are complex.

$\sqrt{-16} = 4i$, so $x = \dfrac{-2 \pm 4i}{2}$.

Divide each term by 2: $x = -1 \pm 2i$.

Answer: $x = -1 + 2i$ or $x = -1 - 2i$.

Solve $\dfrac{x}{x - 2} - \dfrac{8}{x^2 - 4} = \dfrac{1}{x + 2}$.

$x^2 - 4 = (x - 2)(x + 2)$, so the LCD is $(x - 2)(x + 2)$. Note $x \ne 2$ and $x \ne -2$.

Multiply every term by the LCD: $x(x + 2) - 8 = x - 2$.

$x^2 + 2x - 8 = x - 2$, so $x^2 + x - 6 = 0$, which factors as $(x + 3)(x - 2) = 0$. So $x = -3$ or $x = 2$.

$x = 2$ makes a denominator 0, so throw it out. Check $x = -3$: $\dfrac{-3}{-5} - \dfrac{8}{5} = \dfrac{3}{5} - \dfrac{8}{5} = -1$, and $\dfrac{1}{-1} = -1$. Answer: $x = -3$.

Solve $\log_2 x + \log_2(x - 2) = 3$.

Combine with the product rule: $\log_2\big(x(x - 2)\big) = 3$.

Rewrite in exponential form: $x(x - 2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$, which factors as $(x - 4)(x + 2) = 0$.

So $x = 4$ or $x = -2$. But $\log_2(-2)$ is not defined, so $x = -2$ is extraneous.

Check $x = 4$: $\log_2 4 + \log_2 2 = 2 + 1 = 3$. Answer: $x = 4$.

After the mock

Look at the By topic table in your results. Any unit where you missed 2 or more questions is your next study block: open the lessons listed under "Lessons worth revisiting", redo their practice problems, then retake that unit's checkpoint quiz.

For each miss, read the worked solution and decide which kind it was: a slip (sign, arithmetic, forgot to check), or a method you don't have yet. Slips get fixed by checking your answers; methods get fixed by the lesson.

Aim for 80% or more (26 of 32) on both mocks before the real test. If you're not there yet, that's what the remaining study days are for. Your plan for the test, and what to do if the score comes in under 400, is in the schedule's retake plan. Mock test 2 is on the course home; try it after a few days of review, not right away.

Want to look something up first? Glossary · Quadratics cheat sheet · Exponentials and logs cheat sheet

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