Final review · Mock placement test 2

Mock placement test 2

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 $-x^2 + 3xy$ when $x = -3$ and $y = 2$.

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

The exponent comes first, and the minus sign in front stays: $-(-3)^2 = -(9) = -9$.

$3(-3)(2) = -18$. So $-9 + (-18) = -27$.

Answer: $-27$.

Write $5x + 2y = 8$ in slope-intercept form, $y = mx + b$.

Subtract $5x$ from both sides: $2y = -5x + 8$.

Divide every term by 2: $y = -\tfrac52 x + 4$.

Answer: $y = -\tfrac52 x + 4$.

Which is equal to $(4x + 3)^2$?

$16x^2 + 12x + 9$
$16x^2 + 24x + 9$
$16x^2 - 24x + 9$
$4x^2 + 24x + 9$

Use $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 4x$ and $b = 3$.

$a^2 = (4x)^2 = 16x^2$, $2ab = 2(4x)(3) = 24x$, $b^2 = 9$.

So $(4x + 3)^2 = 16x^2 + 24x + 9$. (It is never just $16x^2 + 9$: the middle term matters.)

Answer: (b).

Solve $\dfrac{x + 2}{4} - \dfrac{x}{6} = 2$.

Multiply every term by the LCD, 12: $3(x + 2) - 2x = 24$.

Distribute: $3x + 6 - 2x = 24$, so $x + 6 = 24$.

$x = 18$.

Check: $\dfrac{20}{4} - \dfrac{18}{6} = 5 - 3 = 2$. Answer: $x = 18$.

Starting from the graph of $y = |x|$, which steps give the graph of $g(x) = -|x - 4| + 1$?

flip over the x-axis, move left 4, move up 1
flip over the x-axis, move right 4, move down 1
flip over the x-axis, move left 4, move down 1
flip over the x-axis, move right 4, move up 1

The minus sign in front of the absolute value flips the V upside down (over the x-axis).

Inside, $x - 4$ is zero at $x = 4$, so the graph moves right 4. Outside, $+1$ moves it up 1.

The tip of the upside-down V is at $(4, 1)$. Answer: (d).

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

Distribute the $-3$ to both terms: $-6x + 12$. Distribute the 2: $2x + 10$.

Now you have $-6x + 12 + 2x + 10 - x$.

Combine like terms: $-6x + 2x - x = -5x$ and $12 + 10 = 22$.

Answer: $-5x + 22$.

Which is the complete factorization of $x^3 + 3x^2 - 4x - 12$?

$(x + 3)(x^2 - 4)$
$(x - 3)(x - 2)(x + 2)$
$(x + 3)(x - 2)(x + 2)$
$(x + 3)(x - 2)^2$

Four terms, so try grouping: $(x^3 + 3x^2) + (-4x - 12) = x^2(x + 3) - 4(x + 3)$.

Pull out the common factor $x + 3$: $(x + 3)(x^2 - 4)$.

$x^2 - 4$ is a difference of squares, so keep going: $(x + 3)(x - 2)(x + 2)$. Choice (a) is not finished.

Answer: (c).

Solve the system.

\[\begin{cases} 4x - 3y = 18 \\ 3x + 2y = 5 \end{cases}\]

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

$8x - 6y = 36$ and $9x + 6y = 15$. Add them: $17x = 51$, so $x = 3$.

Substitute into the second equation: $3(3) + 2y = 5$, so $2y = -4$ and $y = -2$.

Check in the first: $4(3) - 3(-2) = 12 + 6 = 18$. Answer: $(3, -2)$.

Simplify $\dfrac{x^2 - 5x + 6}{6 - 3x}$.

Factor the top: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Factor the bottom, taking out $-3$ so the binomial matches: $6 - 3x = -3(x - 2)$.

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

Answer: $\dfrac{3 - x}{3}$ (the same as $-\dfrac{x - 3}{3}$).

Simplify $\dfrac{(27x^9)^{2/3}}{x^4}$. Assume $x$ is positive.

Raise each factor to the $\tfrac23$ power. $27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$.

$\left(x^9\right)^{2/3} = x^{9 \cdot \frac23} = x^6$.

Divide: $\dfrac{9x^6}{x^4} = 9x^{6-4} = 9x^2$.

Answer: $9x^2$.

Solve $3(x + 1)^2 = 24$. Give exact answers.

Divide by 3 first: $(x + 1)^2 = 8$.

Square root property (don't forget ±): $x + 1 = \pm\sqrt8 = \pm 2\sqrt2$.

Subtract 1: $x = -1 \pm 2\sqrt2$.

Answer: $x = -1 + 2\sqrt2$ or $x = -1 - 2\sqrt2$.

Solve $|3x + 1| \gt 8$. Write the answer in interval notation.

"Greater than" means "or": $3x + 1 \gt 8$ or $3x + 1 \lt -8$.

First piece: $3x \gt 7$, so $x \gt \tfrac73$. Second piece: $3x \lt -9$, so $x \lt -3$.

Answer: $(-\infty, -3) \cup \left(\tfrac73, \infty\right)$ (round brackets, because the inequality is strict).

Find the domain of $f(x) = \dfrac{\sqrt{x + 4}}{x - 1}$. Write it in interval notation.

Two rules. Under the square root: $x + 4 \ge 0$, so $x \ge -4$.

The denominator can't be 0: $x - 1 \ne 0$, so $x \ne 1$.

Start at $-4$ (included) and skip over 1.

Answer: $[-4, 1) \cup (1, \infty)$.

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

Numerator: raise each factor to the $-2$. $2^{-2} = \tfrac14$, $(x^{-3})^{-2} = x^6$, $(y^2)^{-2} = y^{-4}$. So the top is $\tfrac14 x^6y^{-4}$.

Divide: $x^{6 - 4} = x^2$ and $y^{-4 - (-1)} = y^{-3}$. That gives $\tfrac14 x^2y^{-3}$.

Move $y^{-3}$ below the bar: $\dfrac{x^2}{4y^3}$.

Answer: $\dfrac{x^2}{4y^3}$.

Which is equal to $\ln\dfrac{x^3}{y\sqrt z}$? (Assume $x, y, z$ are positive.)

$3\ln x - \ln y - \tfrac12\ln z$
$3\ln x - \ln y + \tfrac12\ln z$
$\ln 3x - \ln y - \tfrac12\ln z$
$3\ln x - \ln y - 2\ln z$

Quotient rule: $\ln x^3 - \ln\big(y\sqrt z\big)$.

Product rule on the bottom: $\ln(y\sqrt z) = \ln y + \ln\sqrt z$. Subtracting it subtracts both parts.

Power rule: $\ln x^3 = 3\ln x$ and $\ln\sqrt z = \ln z^{1/2} = \tfrac12\ln z$.

Result: $3\ln x - \ln y - \tfrac12\ln z$. Answer: (a).

Factor $10x^2 - 11x - 6$.

ac method: $a \cdot c = 10 \cdot (-6) = -60$. Find two numbers that multiply to $-60$ and add to $-11$: $4$ and $-15$.

Split the middle term: $10x^2 + 4x - 15x - 6$.

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

Check: $(5x + 2)(2x - 3) = 10x^2 - 15x + 4x - 6 = 10x^2 - 11x - 6$. Answer: $(5x + 2)(2x - 3)$.

Find the equation of the line through $(5, -1)$ that is parallel to $2x + 5y = 10$. Write it as $y = mx + b$.

Find the given line's slope: $5y = -2x + 10$, so $y = -\tfrac25 x + 2$. Parallel lines have the same slope: $m = -\tfrac25$.

Point-slope form: $y - (-1) = -\tfrac25(x - 5)$, so $y + 1 = -\tfrac25 x + 2$.

Subtract 1: $y = -\tfrac25 x + 1$.

Check the point: $-\tfrac25(5) + 1 = -2 + 1 = -1$. Answer: $y = -\tfrac25 x + 1$.

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

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

Discriminant: $b^2 - 4ac = 4 - 4(3)(-2) = 4 + 24 = 28$, and $\sqrt{28} = 2\sqrt7$.

$x = \dfrac{-2 \pm 2\sqrt7}{6}$. Divide every term by 2: $x = \dfrac{-1 \pm \sqrt7}{3}$.

Answer: $x = \dfrac{-1 + \sqrt7}{3}$ or $x = \dfrac{-1 - \sqrt7}{3}$.

A boat travels 36 miles downstream in 2 hours. The return trip upstream takes 3 hours. What is the boat's speed in still water, in miles per hour?

Let $b$ = boat speed in still water and $c$ = current speed. Downstream the current helps; upstream it slows the boat.

Downstream: $b + c = \dfrac{36}{2} = 18$. Upstream: $b - c = \dfrac{36}{3} = 12$.

Add the equations: $2b = 30$, so $b = 15$. Then $c = 3$.

Check: $2(15 + 3) = 36$ and $3(15 - 3) = 36$. Answer: 15 miles per hour.

Subtract and simplify: $\dfrac{x}{x - 3} - \dfrac{18}{x^2 - 9}$.

$x^2 - 9 = (x - 3)(x + 3)$, so the LCD is $(x - 3)(x + 3)$. Rewrite: $\dfrac{x(x + 3)}{(x - 3)(x + 3)} - \dfrac{18}{(x - 3)(x + 3)}$.

Subtract the numerators: $x^2 + 3x - 18$. It factors: $(x + 6)(x - 3)$.

$\dfrac{(x + 6)(x - 3)}{(x - 3)(x + 3)}$. Cancel the common factor $x - 3$.

Answer: $\dfrac{x + 6}{x + 3}$.

Rationalize the denominator. Which is equal to $\dfrac{\sqrt3 + 1}{\sqrt3 - 1}$?

$\dfrac{4 + 2\sqrt3}{4}$
$\dfrac{4 + 2\sqrt3}{2}$
$\dfrac{4 - 2\sqrt3}{2}$
$\dfrac{3 + 2\sqrt3}{2}$

Multiply top and bottom by the conjugate of the bottom, $\sqrt3 + 1$.

Bottom: $(\sqrt3 - 1)(\sqrt3 + 1) = 3 - 1 = 2$.

Top: $(\sqrt3 + 1)^2 = 3 + 2\sqrt3 + 1 = 4 + 2\sqrt3$. So the result is $\dfrac{4 + 2\sqrt3}{2}$, which simplifies to $2 + \sqrt3$.

Answer: (b).

A rectangle's length is 3 feet more than twice its width. Its area is 65 square feet. What is its width, in feet?

Let $w$ = width. Then length = $2w + 3$, and area: $w(2w + 3) = 65$.

Standard form: $2w^2 + 3w - 65 = 0$. Factor: $(2w + 13)(w - 5) = 0$. (Check: $2w^2 - 10w + 13w - 65$.)

So $w = 5$ or $w = -\tfrac{13}{2}$. A width can't be negative.

Check: $5 \cdot 13 = 65$. Answer: 5 feet.

Find the inverse function of $f(x) = x^3 - 6$.

Write $y = x^3 - 6$, then swap $x$ and $y$: $x = y^3 - 6$.

Solve for $y$: $y^3 = x + 6$, so $y = \sqrt[3]{x + 6}$.

Check: $f\!\left(\sqrt[3]{x + 6}\right) = (x + 6) - 6 = x$.

Answer: $f^{-1}(x) = \sqrt[3]{x + 6}$.

A ball is thrown upward. Its height after $t$ seconds is $h(t) = -16t^2 + 48t + 6$ feet. What is its maximum height, in feet?

The maximum is at the vertex: $t = -\dfrac{b}{2a} = -\dfrac{48}{2(-16)} = 1.5$ seconds.

Height then: $h(1.5) = -16(2.25) + 48(1.5) + 6 = -36 + 72 + 6 = 42$.

Answer: 42 feet.

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

There is no $x^2$ term, so write the dividend as $x^3 + 0x^2 - 7x + 6$.

$x^3 \div x = x^2$. Multiply: $x^2(x + 3) = x^3 + 3x^2$. Subtract: $0x^2 - 3x^2 = -3x^2$. Bring down: $-3x^2 - 7x$.

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

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

Solve $2x^2 + 5x - 3 \ge 0$. Write the answer in interval notation.

Find the zeros: $2x^2 + 5x - 3 = (2x - 1)(x + 3) = 0$, so $x = \tfrac12$ or $x = -3$.

The parabola opens up, so it is at or above zero outside the zeros. Test $x = 0$: $-3 \ge 0$ is false, so the middle is not part of the answer.

The inequality includes "equal", so the endpoints are included.

Answer: $(-\infty, -3] \cup \left[\tfrac12, \infty\right)$.

Solve $\sqrt{x + 7} + 5 = x$.

Get the radical alone first: $\sqrt{x + 7} = x - 5$.

Square both sides: $x + 7 = (x - 5)^2 = x^2 - 10x + 25$. So $0 = x^2 - 11x + 18 = (x - 9)(x - 2)$, and $x = 9$ or $x = 2$.

Check both. $x = 9$: $\sqrt{16} + 5 = 4 + 5 = 9$. Works. $x = 2$: $\sqrt9 + 5 = 8$, not 2. Extraneous.

Answer: $x = 9$.

Calculator allowed. A car bought for $24,000 loses 15% of its value each year, so its value after $t$ years is $V = 24000(0.85)^t$. After how many years is it worth $10,000? Round to the nearest tenth of a year.

Set up: $24000(0.85)^t = 10000$. Divide by 24000: $(0.85)^t = \dfrac{10000}{24000} = \dfrac{5}{12}$.

Take the log of both sides and use the power rule: $t \ln 0.85 = \ln\dfrac{5}{12}$.

$t = \dfrac{\ln(5/12)}{\ln 0.85} \approx \dfrac{-0.8755}{-0.1625} \approx 5.39$.

Answer: about 5.4 years.

Solve the system.

\[\begin{cases} x + y + z = 2 \\ 2x + y - z = 3 \\ x - 2y + 3z = 10 \end{cases}\]

Eliminate $z$. Add equations 1 and 2: $3x + 2y = 5$. Multiply equation 2 by 3 and add equation 3: $(6x + 3y - 3z) + (x - 2y + 3z) = 9 + 10$, so $7x + y = 19$.

From $7x + y = 19$: $y = 19 - 7x$. Substitute: $3x + 2(19 - 7x) = 5$, so $3x + 38 - 14x = 5$, $-11x = -33$, $x = 3$.

Then $y = 19 - 21 = -2$, and from equation 1, $z = 2 - 3 - (-2) = 1$.

Check equation 3: $3 - 2(-2) + 3(1) = 3 + 4 + 3 = 10$. Answer: $(3, -2, 1)$.

Solve $x^2 + 4x + 13 = 0$. Write complex solutions in the form $a + bi$.

Quadratic formula with $a = 1$, $b = 4$, $c = 13$. Discriminant: $16 - 52 = -36$, which is negative, so the solutions are complex.

$\sqrt{-36} = 6i$, so $x = \dfrac{-4 \pm 6i}{2}$.

Divide each term by 2: $x = -2 \pm 3i$.

Answer: $x = -2 + 3i$ or $x = -2 - 3i$.

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

$x^2 - 1 = (x - 1)(x + 1)$, so the LCD is $(x - 1)(x + 1)$. Note $x \ne 1$ and $x \ne -1$.

Multiply every term by the LCD: $x(x + 1) - 2(x - 1) = 4$.

$x^2 + x - 2x + 2 = 4$, so $x^2 - x - 2 = 0$, which factors as $(x - 2)(x + 1) = 0$. So $x = 2$ or $x = -1$.

$x = -1$ makes a denominator 0, so throw it out. Check $x = 2$: the left side is $\dfrac21 - \dfrac23 = \dfrac43$, and the right side is $\dfrac{4}{2^2 - 1} = \dfrac43$. Answer: $x = 2$.

Solve $\log_3 x + \log_3(x + 6) = 3$.

Combine with the product rule: $\log_3\big(x(x + 6)\big) = 3$.

Rewrite in exponential form: $x(x + 6) = 3^3 = 27$, so $x^2 + 6x - 27 = 0$, which factors as $(x + 9)(x - 3) = 0$.

So $x = 3$ or $x = -9$. But $\log_3(-9)$ is not defined, so $x = -9$ is extraneous.

Check $x = 3$: $\log_3 3 + \log_3 9 = 1 + 2 = 3$. Answer: $x = 3$.

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 1 is on the course home. If you already took it, compare the two breakdowns: a unit that is weak on both is the one to fix first.

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

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