SBAC Practice Questions (2026) — Free Sample Questions with Answers

14 free SBAC questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

Concepts and Procedures (5)Communicating Reasoning (3)Reading (3)Writing (1)Speaking/Listening (1)Research (1)

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Concepts and Procedures

Question 1

In the figure, triangle $A B C$ has a right angle at $C$ , and $\overline{C D}$ is perpendicular to $\overline{A B}$ . The measure of angle $A$ is 32° and $A C = 10$ units. Select the segment whose length equals $10 sin 3 2^{\circ}$ .

Click the correct area of the figure.

Right triangle ABC with altitude CD

Figure not drawn to scale.

Show answer & explanation

Answer: D

$10 sin 3 2^{\circ}$ is (hypotenuse) × sin(angle), so look for a right triangle with hypotenuse $A C = 10$ that contains angle $A$ .
Triangle $A D C$ has its right angle at $D$ , so $A C$ is its hypotenuse; the side opposite $A$ is $C D$ .
Therefore $C D = 10 sin 3 2^{\circ} \approx 5.30$ , while $A D = 10 cos 3 2^{\circ} \approx 8.48$ .
In triangle $A B C$ , $A C$ is a leg: $B C = 10 tan 3 2^{\circ} \approx 6.25$ ; and $B D = B C sin 3 2^{\circ} \approx 3.31$ .
Check: $A D^{2} + C D^{2} \approx 71.9 + 28.1 = 100 = A C^{2}$ . ✓

Segment AD: In right triangle $A D C$ , $A D$ is adjacent to angle $A$ , so $A D = 10 cos 3 2^{\circ} \approx 8.48$ .

Segment BC: In triangle $A B C$ , $B C$ is opposite angle $A$ , but there $A C$ is a leg, not the hypotenuse: $B C = 10 tan 3 2^{\circ} \approx 6.25$ .

Segment BD: $B D$ is not a side of a right triangle with hypotenuse $A C$ ; $B D = B C sin 3 2^{\circ} \approx 3.31$ .

Segment CD: Correct. In right triangle $A D C$ (right angle at $D$ ), $A C = 10$ is the hypotenuse and $C D$ is opposite angle $A$ , so $C D = 10 sin 3 2^{\circ} \approx 5.30$ .

A sine ratio uses the hypotenuse of the right triangle that contains the angle; when an altitude creates several right triangles, first find the one whose hypotenuse is the given side.

Question 2

For the function $f (x) = 2 x^{2} - 3 x$ , which expression is equivalent to $f (a + 1) - f (a)$ ?

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Answer: B

$f (a + 1) = 2 (a + 1)^{2} - 3 (a + 1) = 2 a^{2} + 4 a + 2 - 3 a - 3 = 2 a^{2} + a - 1$ .
$f (a) = 2 a^{2} - 3 a$ .
$f (a + 1) - f (a) = (2 a^{2} + a - 1) - (2 a^{2} - 3 a) = 4 a - 1$ .
Check with $a = 1$ : $f (2) - f (1) = 2 - (- 1) = 3$ and $4 (1) - 1 = 3$ . ✓

$- 1$ : Expands $(a + 1)^{2}$ as $a^{2} + 1$ , which drops the $2 a$ term and makes the $a$ -terms cancel.

$4 a - 1$ : Correct. $f (a + 1) = 2 (a^{2} + 2 a + 1) - 3 a - 3 = 2 a^{2} + a - 1$ , and subtracting $f (a) = 2 a^{2} - 3 a$ leaves $4 a - 1$ .

$4 a + 5$ : Writes $- 3 (a + 1)$ as $- 3 a + 3$ instead of $- 3 a - 3$ .

$2 a^{2} + a - 1$ : This is $f (a + 1)$ alone; $f (a)$ was never subtracted.

To evaluate $f$ at an expression, replace every $x$ with the whole expression in parentheses, then expand carefully.

Question 3

The lines $y = 2 x - 1$ (dashed) and $y = - x + 5$ (solid) divide the coordinate plane into four regions. Select the region that contains all the solutions of the system $y > 2 x - 1$ and $y \leq - x + 5$ .

Click the correct area of the figure.

Two lines in the coordinate plane

The lines intersect at (2, 3). The dashed line is y = 2x − 1 and the solid line is y = −x + 5.

Show answer & explanation

Answer: D

$y > 2 x - 1$ is the half-plane above the dashed line; $y \leq - x + 5$ is the half-plane on or below the solid line.
Test one point in each region: Top $(2, 6)$ fails the second inequality; Right $(6, 3)$ fails the first; Bottom $(2, 0)$ fails the first.
Left $(- 2, 0)$ : $0 > - 5$ ✓ and $0 \leq 7$ ✓, so the Left region is the solution set.

Top region: Test $(2, 6)$ : $6 > 3$ is true, but $6 \leq 3$ is false — this region is above both lines.

Right region: Test $(6, 3)$ : $3 > 11$ is false — this region is below the dashed line.

Bottom region: Test $(2, 0)$ : $0 > 3$ is false — this region is below both lines.

Left region: Correct. Test $(- 2, 0)$ : $0 > - 5$ and $0 \leq 7$ are both true, so this region is above the dashed line and on or below the solid line.

The solution set of a system of inequalities is the overlap of the half-planes; a test point in each region decides which one it is.

Question 4

What are all the solutions of $x + 7 = x - 5$ ?

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Answer: A

Square both sides: $x + 7 = (x - 5)^{2} = x^{2} - 10 x + 25$ .
Rearrange: $x^{2} - 11 x + 18 = 0 \Rightarrow (x - 2) (x - 9) = 0$ .
Candidates: $x = 2$ and $x = 9$ .
Check $x = 9$ : $16 = 4$ and $9 - 5 = 4$ . ✓
Check $x = 2$ : $9 = 3$ but $2 - 5 = - 3$ . ✗ (extraneous)

$x = 9$ only: Correct. Squaring gives $x^{2} - 11 x + 18 = 0$ , so $x = 2$ or $x = 9$ ; only $x = 9$ checks, because $9 = 3$ but $2 - 5 = - 3$ .

$x = 2$ only: Keeps the extraneous root and rejects the real one: at $x = 9$ , $16 = 4$ and $9 - 5 = 4$ .

$x = 2$ and $x = 9$ : Accepts both roots of the squared equation without checking; $x = 2$ makes the right side negative.

$x = - 9$ and $x = - 2$ : Factors $x^{2} - 11 x + 18$ as $(x + 2) (x + 9)$ , which has the wrong signs.

Squaring both sides can create extraneous solutions; always substitute each candidate into the original radical equation.

Question 5

The amount of money in a savings account $t$ years after it is opened is modeled by $A (t) = 500 (1.02)^{4 t}$ . Which statement best interprets the base 1.02 in this model?

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Answer: A

Rewrite the exponent: $(1.02)^{4 t}$ applies the factor 1.02 once for each $\frac{1}{4}$ year.
A factor of 1.02 is a 2% increase, so the amount grows 2% per quarter year.
Check: $A (\frac{1}{4}) = 500 (1.02)^{1} = 510$ , which is 2% more than $500. ✓

The amount increases by 2% every quarter of a year.: Correct. The exponent $4 t$ increases by 1 each time $t$ increases by $\frac{1}{4}$ , so the amount is multiplied by 1.02 (a 2% increase) every quarter year.

The amount increases by 2% every year.: Ignores the 4 in the exponent; in one year the factor 1.02 is applied four times.

The amount increases by 8% every year.: Multiplies 2% by 4; the yearly factor is $1.0 2^{4} \approx 1.0824$ , an increase of about 8.24%, and the base itself describes a quarter-year change.

The amount increases by 2% every four years.: Reads $4 t$ as "every 4 years"; the factor is applied 4 times per year, not once every 4 years.

In $a (b)^{k t}$ , the factor $b$ is applied $k$ times per unit of $t$ , so it describes the change over $\frac{1}{k}$ of a unit.

Communicating Reasoning

Question 6

Diego solves $∣ x - 3∣ = 2 x$ by cases. Case 1: $x - 3 = 2 x$ , so $x = - 3$ . Case 2: $x - 3 = - 2 x$ , so $x = 1$ . He concludes that the solutions are $- 3$ and $1$ . Which statement about Diego's work is true?

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Answer: D

Case 1 ( $x \geq 3$ ): $x - 3 = 2 x \Rightarrow x = - 3$ , which is not $\geq 3$ → reject.
Case 2 ( $x < 3$ ): $- (x - 3) = 2 x \Rightarrow x = 1$ , which is $< 3$ → keep.
Check: $∣1 - 3∣ = 2 = 2 (1)$ ✓; $∣ - 3 - 3∣ = 6 \neq = - 6$ ✗.

Diego is correct, because each case gives a solution of its own equation.: Each case equation has a solution, but a case result counts only if it meets that case's condition; $x = - 3$ fails.

Diego should reject $x = 1$ , because it makes $x - 3$ negative.: Case 2 is the case where $x - 3 < 0$ , so $x = 1$ fits it: $∣1 - 3∣ = 2 = 2 (1)$ .

Diego should have used only Case 1, because the right side of the equation, 2x, must be positive.: Both cases are needed; the right side is not known to be positive until $x$ is found.

Diego should reject $x = - 3$ , because it fails the Case 1 condition $x \geq 3$ .: Correct. Case 1 assumes $x \geq 3$ ; also $∣ - 3 - 3∣ = 6$ but $2 (- 3) = - 6$ . The only solution is $x = 1$ .

When an argument is broken into cases, each result must satisfy the condition that defines its case.

Question 7

Let $a$ , $b$ and $c$ be nonzero real numbers. Determine whether each statement is always true, sometimes true, or never true. Select one answer for each statement.

	Always true	Sometimes true	Never true
If $a > b$ , then $a^{2} > b^{2}$ .
If $a > b$ , then $a + c > b + c$ .
If $a > b$ and $c < 0$ , then $a c > b c$ .
If $a > b$ , then $\frac{1}{a} < \frac{1}{b}$ .

Select one option in each row.

Show answer & explanation

Answer: If a > b, then a^2 > b^2.: Sometimes true; If a > b, then a + c > b + c.: Always true; If a > b and c < 0, then ac > bc.: Never true; If a > b, then 1/a < 1/b.: Sometimes true

Test each statement with numbers of the same sign and of opposite signs.
Squares: $3 > 2$ gives $9 > 4$ ; $1 > - 2$ gives $1 < 4$ → sometimes.
Addition preserves order → always. Multiplying by $c < 0$ reverses order → never.
Reciprocals: order reverses when $a$ and $b$ have the same sign; when $a > 0 > b$ , $\frac{1}{a} > 0 > \frac{1}{b}$ → sometimes.

To decide always / sometimes / never, look for one example and one counterexample; a general property (such as multiplying by a negative) can make a statement never true.

Question 8

Ellis says, "A square encloses the most area for a given perimeter, so the dog park should be a square, 80 feet on each side." Which statement best evaluates Ellis's reasoning?

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Answer: D

Ellis's square: $w = 80$ , parallel side $240 - 160 = 80$ , area $6, 400$ square feet.
The model $A (w) = w (240 - 2 w)$ is greatest at $w = 60$ : $60 \times 120 = 7, 200$ square feet.
The square rule applies to a closed rectangle whose whole perimeter is fixed; the wall changes the constraint.

Ellis is correct, because a square encloses the most area for any fixed perimeter.: The rule is about a fixed perimeter of all four sides; here only three sides use the 240 feet.

Ellis is correct, because an 80-foot square uses exactly 240 feet of fencing.: True that 80 + 80 + 80 = 240, but using all the fencing does not make the area greatest: 6,400 < 7,200 square feet.

Ellis is incorrect, because 240 feet of fencing cannot form a square park with the wall as one side.: An 80-by-80-foot square is possible (three sides of 80 feet); the problem is that it is not the largest.

Ellis is incorrect, because the wall is not fenced; a 60-by-120-foot park encloses more area.: Correct. The square rule assumes all four sides come from the perimeter. With the wall as one side, $w = 60$ gives 7,200 square feet, more than the square's 6,400.

Before using a known result, check that its conditions match the situation; here the fixed-perimeter rule does not apply because one side is free.

Reading

Question 9

Which two statements best describe how the ideas in Text 1 and Text 2 relate to each other? Select two statements.

0 of 2 selected

Show answer & explanation

Answer: A, C

Both authors treat the Elwha River as evidence that rivers can recover after dams are removed.: Correct. Text 1 describes salmon returning (paragraph 3); Text 2 grants that the recovery "deserves the attention it has received" and that rivers "can recover remarkably" (paragraphs 6 and 10).

Both authors argue that dams producing electricity should be the first ones removed.: Neither says this; Text 2 treats power production as a reason for caution.

Text 1 focuses on dams that no longer serve a purpose, while Text 2 warns that a dam may serve purposes that are not obvious.: Correct. Text 1 targets "old and unneeded dams" (paragraph 5); Text 2 says a dam "may still be doing a job no one designed it for" (paragraph 8).

Text 2 rejects the claim in Text 1 that old dams can be dangerous to people nearby.: Text 2 never addresses the safety hazard described in paragraph 4.

Text 1 argues that every dam should be removed, while Text 2 argues that none should be.: Both positions are misstated: Text 1 says "None of this means every dam should disappear," and Text 2 allows removal when conditions are right.

Text 2 claims that the problem of sediment makes removing a dam impossible.: Overstates paragraph 9: sediment adds time and cost but can "rebuild a river" if released carefully.

When comparing paired texts, find both the common ground and the point where the authors' emphasis differs; reject options that overstate either position.

Question 10

Click on the sentence in paragraph 2 in which Captain Ward sets up a familiar expectation that she then rejects.

Thank you, and good morning. I'm told that commencement speakers are supposed to tell graduates that the world is changing faster than ever. I'm not going to say that. Partly that's because every commencement speaker since the invention of commencements has said it. Mostly it's because, at sea, the important things change very slowly.

Click the sentence to select it.

Show answer & explanation

Answer: "I'm told that commencement speakers are supposed to tell graduates that the world is changing faster than ever."

Keyed: "I'm told that commencement speakers are supposed to tell graduates that the world is changing faster than ever." It states the expected message, which the next sentences refuse. Text structure: an expectation-and-reversal opening.
"Thank you, and good morning." is a greeting.
"I'm not going to say that." is the rejection itself, not the expectation it rejects.
The "Partly" sentence gives a humorous reason for the rejection.
The "Mostly" sentence gives the real reason and previews her theme.

In a set-up-and-reversal opening, separate the sentence that states the expectation from the sentences that reject or explain it.

Question 11

Read the article. Which statement best expresses the central idea of the article?

Borrowed Tools, Shared Knowledge

¹On a Saturday morning outside a converted storefront, a line forms before the doors open. The people in it are not waiting for concert tickets or a sale on sneakers. They have come to borrow a tile saw, a pressure washer, a set of pipe wrenches, a ladder long enough to reach a second-story gutter. A woman near the front holds a list written on the back of an envelope. By noon, most of the shelves will have gaps where the most popular items used to be.

²The storefront is a tool library, and it works much the way a book library does. Members sign up, often for a small annual fee or for free, and check out equipment for a few days at a time. Return it late, and someone else's weekend project waits. Return it broken, and a volunteer will try to fix it. Tool libraries have existed in some American cities for decades, but in recent years they have spread to suburbs and small towns, often run by public libraries, neighborhood associations, or a handful of determined volunteers.

³The usual argument for them is economic. Most households own tools they rarely touch: the hedge trimmer used twice a summer, the drill that comes out of the closet to hang a picture and goes back in. A widely repeated claim holds that the average power drill is used for only thirteen minutes in its entire lifetime. The figure appears in books, speeches, and advertisements for sharing services, yet no one seems able to trace it to an actual study. It may be roughly true; it may be invented. Either way, the argument for tool libraries does not depend on it. Anyone who has bought a specialized tool for a single job and then stored it for a decade understands the basic math.

⁴But spending less is only part of what members describe when they explain why they keep coming back. In the Saturday line, a retired machinist named Walter Nakamura is explaining to a nervous first-time homeowner how to keep a circular saw from binding. Nobody assigned him this job. He simply knows how, and the person next to him does not.

⁵This is the part of the tool library that rarely shows up in a budget. A tool on a shelf is only potential; it becomes useful when someone knows what to do with it. Many tool libraries now offer free workshops on basic repairs—patching drywall, replacing a faucet washer, sharpening a mower blade—and volunteers report that the classes fill faster than the most popular equipment is checked out. Members who arrive to borrow a tool often leave having borrowed a skill as well.

⁶There is an older idea underneath all of this. For most of history, expensive equipment was shared out of necessity. Farm communities pooled threshing machines; neighbors passed a single ladder down the street. The modern habit of every household owning its own complete set of tools is, in historical terms, fairly new, made possible by cheap manufacturing and large garages. The tool library does not so much invent a new model as recover an old one, adjusted for a world of online catalogs and barcode scanners.

⁷The model has limits. Tool libraries depend on volunteers, and volunteers burn out. Insurance can be expensive, especially for power tools that could injure an inexperienced user. Items disappear or come back damaged beyond repair. Some tool libraries have closed after a few years when the founding group moved on and no one replaced them. The most stable ones tend to be those connected to an existing institution, such as a public library system, that can provide space, staff, and a budget line that does not vanish when a volunteer does.

⁸Still, the idea keeps spreading, and it has begun to reach beyond tools. Some public libraries now lend sewing machines, telescopes, cake pans, and camping gear, and at least one rural library lends fishing rods to families who want to try the sport before buying anything. The logic is the same: an object that sits unused in one closet could be useful in another, and the person who knows how to use it is often only a few doors away.

⁹Back at the storefront, Nakamura is helping a member return a borrowed ladder to its rack. Asked whether he worries that the library takes business away from hardware stores, he shrugs. "People who learn to fix one thing," he says, "usually end up buying the tools to fix the next one." He has no study to prove it. But the line outside, he points out, is longer than it was last year.

Show answer & explanation

Answer: D

Tool libraries save households money by reducing the number of tools each family must buy.: Paragraph 3 makes this point, but paragraph 4 calls saving money "only part" of the story; this is a supporting idea, not the central one.

Tool libraries often struggle to survive because they depend on unpaid volunteers.: Paragraph 7 discusses this limit, but the article ends with the idea spreading; a single paragraph's concern is not the central idea.

Tool libraries are a new kind of community service made possible by online catalogs and barcode scanners.: Contradicts paragraph 6, which says tool libraries "recover an old" model; the technology is only an adjustment.

Tool libraries offer savings, but their deeper value lies in sharing skills and resources in a community.: Correct. The article moves from the economic case (paragraph 3) to shared knowledge (paragraphs 4–5) and neighbors sharing objects (paragraph 8). Central idea: an idea developed across the whole text.

The central idea must cover the whole article; options that match one paragraph are supporting ideas.

Writing

Question 12

Read the paragraph from the student's article. Click on the sentence that contains an error in subject-verb agreement.

A student is writing an article for the school website about a new community garden.

The list of volunteers who signed up for the new community garden are posted outside the main office. Each of the twelve plots in the garden has been assigned to a different club. Neither the art club nor the science students want the plot nearest the fence, because it gets the least sun. To help every club get started, the tools that the hardware store donated are stored in the shed beside the plots. Once the seedlings are in, watering the beds during summer break is the job of whoever lives closest to the garden.

Click the sentence to select it.

Show answer & explanation

Answer: "The list of volunteers who signed up for the new community garden are posted outside the main office."

Keyed: "The list ... are posted." The subject is "list" (singular); "of volunteers who signed up" is an intervening phrase, so the verb should be "is." Edit/clarify: subject-verb agreement.
"Each ... has" is correct: "each" is singular even when followed by "of the twelve plots."
With "neither ... nor," the verb agrees with the nearer subject, "students," so "want" is correct.
"The tools ... are" is correct; "tools" is plural.
"Watering ... is" is correct; a gerund phrase used as a subject is singular.

Find the true subject by setting aside phrases between it and the verb.

Speaking/Listening

Question 13

Decide whether each statement is supported by information in the interview. Select Supported or Not Supported for each.

	Supported	Not Supported
Magnetic tape can become sticky or brittle as its binder breaks down.
The lab digitizes recordings in the order in which it receives them.
Old tape players are hard to keep working because parts and repair experts are scarce.
Families should play their old tapes regularly to make sure they still work.

Select one option in each row.

Show answer & explanation

Answer: Magnetic tape can become sticky or brittle as its binder breaks down.: Supported; The lab digitizes recordings in the order in which it receives them.: Not Supported; Old tape players are hard to keep working because parts and repair experts are scarce.: Supported; Families should play their old tapes regularly to make sure they still work.: Not Supported

Check each statement against what the speaker actually says; watch for statements that reverse her advice.

Research

Question 14

Which two statements best describe how the information in Source 1 and Source 2 fits together to answer the student's research question? Select two statements.

0 of 2 selected

Show answer & explanation

Answer: C, F

Both sources show that noise has no lasting effect on birds that adjust their songs.: Neither source says this; Source 1 says scientists are still studying the effects on birds that adjust.

Source 1 proves that city birds raise fewer young than birds in quiet places.: Overstates Source 1, which says scientists "are still working out" whether this is true.

Source 1 shows how some birds adapt to noise, while Source 2 shows that noise can drive other birds away.: Correct. Source 1 describes changes in song; Source 2 shows fewer birds when the traffic sound played.

Source 2 contradicts Source 1 by showing that birds never change their behavior because of noise.: Source 2 shows behavior changing (leaving, watching for predators), so it does not contradict Source 1.

Source 2 relies on the same kind of observation from a single location as Source 3.: Source 2 is a controlled experiment that switched the noise on and off, unlike the single backyard in Source 3.

Both sources suggest that noise may have costs for birds even when they do not leave.: Correct. Source 1 says adjusting songs "may come at a cost"; Source 2 says birds that stayed "gained weight more slowly."

Integrating sources means noticing both what each adds (adaptation versus avoidance) and where they agree (hidden costs).

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