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

Answer: D

The waggle dance is so unreliable that most of the bees in a hive simply ignore it when they search for food.: Overstates: "not perfectly precise" does not mean unreliable, and bees actively follow the runs.

Recruits find flowers by scent alone and do not use the dance.: This is the challenge some scientists raised (paragraph 5); paragraph 4 says scent is used in addition to the dance.

Dancers repeat their runs because the first run is always wrong.: Paragraph 4 says each run "varies slightly," not that the first run is always wrong.

One run gives only rough directions, so recruits do better by combining several runs and other clues.: Correct. Paragraph 4 says one run "may search in the wrong place," but following "several runs" gives "a more accurate average," and recruits also use "the scent of the flowers."

A supported conclusion stays within the text's degree ("not perfectly precise") and does not turn a limitation into a total failure.

Answer: A

I saw that the projector was already showing the first slide.: Correct. The opening phrase "Walking into the lecture hall" must describe the subject right after the comma; "I" is the one walking, so the modifier is no longer dangling.

the projector that I noticed was already showing the first slide.: Adds a person only inside a clause; the subject right after the comma is still "the projector," so the sentence says the projector was walking.

the projector was already showing the first slide.: Dangling modifier: the sentence says the projector was walking into the hall.

my eyes went straight to the first slide on the projector.: Still dangling: the subject is now "my eyes," and eyes do not walk into a lecture hall; the walker is "I."

An introductory -ing phrase must be followed directly by the person or thing doing that action.

Answer: C

Asphalt roads are the only reason that cities are hotter than the countryside.: Too narrow and contradicted: paragraph 1 says the effect "has no single cause," and paragraph 3 adds buildings and waste heat.

Air conditioners make cities hotter by releasing waste heat into the streets.: A supporting detail from paragraph 3, not the idea of the whole passage.

Several features of city life make cities warmer than nearby areas, and cities can reduce that heat.: Correct. Paragraph 1 says "several features of city life combine to trap and produce heat," paragraphs 2–4 explain the causes and costs, and paragraph 5 says "Cities are not powerless." The statement covers the whole passage.

Cities should replace their pavement with forests, because only forests can fully end the urban heat island effect.: Goes beyond the passage; paragraph 5 says no step "can make a city as cool as a forest" and never proposes replacing pavement with forests.

The main idea must cover the whole passage (causes, effects and responses), not one paragraph's detail.

Answer: D

1. The LCD of 4 and 6 is 12.
2. Rename: $\frac{3}{4}=\frac{9}{12}$ and $\frac{5}{6}=\frac{10}{12}$.
3. Add the numerators: $\frac{9+10}{12}=\frac{19}{12}$.
4. Write as a mixed number: $19÷12=1$ remainder $7$, so $1\frac{7}{12}$.

$\frac{5}{8}$: Multiplies the fractions instead of adding them.

$\frac{2}{3}$: Puts $3+5$ over the LCD 12 without renaming the numerators, then simplifies $\frac{8}{12}$.

$\frac{4}{5}$: Adds the numerators and the denominators ($\frac{8}{10}$), then simplifies.

$1\frac{7}{12}$: Correct. Rename with the LCD 12: $\frac{9}{12}+\frac{10}{12}=\frac{19}{12}=1\frac{7}{12}$.

Add fractions only after renaming them with a common denominator; add the numerators and keep the denominator.

Answer: D

1. Read the graph: housing 35%, savings 15%.
2. Add the percents: 35% + 15% = 50%.
3. Find 50% of $3,200: $0.5\times 3,200=1,600$.

$160: Writes 50% as 0.05 instead of 0.5 (decimal-point error).

$480: Finds the amount for savings only (15%).

$1,120: Finds the amount for housing only (35%).

$1,600: Correct. Housing (35%) + savings (15%) = 50%, and $0.5\times 3,200=1,600$ dollars.

When a graph gives percents of a total, add the percents first, then multiply by the total.

Answer: A

1. $-3^2$ means $-(3\cdot 3)=-9$.
2. $(-2{)}^3=(-2)(-2)(-2)=-8$ (an odd number of negative factors is negative).
3. Add: $-9+(-8)=-17$.

$-17$: Correct. $-3^2=-(3^2)=-9$ because the exponent applies only to 3, and $(-2{)}^3=-8$; so $-9+(-8)=-17$.

$-1$: Evaluates $-3^2$ correctly as $-9$ but treats $(-2{)}^3$ as $+8$.

$1$: Treats $-3^2$ as $(-3{)}^2=9$; without parentheses the negative is not squared.

$17$: Makes both powers positive, ignoring the sign rules for both terms.

An exponent applies only to the number directly below it: $-a^2=-(a^2)$, while $(-a{)}^2$ squares the negative number.

Answer: 120 loaves

1. Cost: $C=420+1.25x$. Revenue: $R=4.75x$.
2. Break even when $R=C$: $4.75x=420+1.25x$.
3. Subtract $1.25x$: $3.5x=420$.
4. Divide: $x=120$ loaves.

Break-even means revenue equals total cost; set the two expressions equal and solve.

Answer: D

1. Isolate the absolute value: subtract 6 from both sides to get $|x+4|=-4$.
2. An absolute value is a distance, so it cannot equal a negative number.
3. The equation has no solution.

$x=-12$ or $x=4$: Adds 6 to both sides instead of subtracting, solving $|x+4|=8$.

$x=-8$ or $x=0$: Drops the negative sign, solving $|x+4|=4$ instead of $|x+4|=-4$.

$x=-8$: Removes the absolute-value bars and solves $x+4+6=2$ as an ordinary equation.

The equation has no solution.: Correct. Isolating gives $|x+4|=-4$; an absolute value is never negative, so there is no solution.

Always isolate the absolute value first; if it equals a negative number, the equation has no solution.

Answer: 13

1. Let $s$ = boxes of small and $L$ = boxes of large: $s+L=30$ and $8s+11L=279$.
2. Substitute $s=30-L$: $8(30-L)+11L=279$.
3. $240-8L+11L=279\Rightarrow 3L=39$.
4. $L=13$ (and $s=17$).
5. Check: $8(17)+11(13)=136+143=279$. ✓

Write one equation for the count and one for the cost, then solve by substitution or elimination and check both equations.

Answer: B

1. Slope from $(0,-1)$ to $(1,1)$: $m=\frac{2}{1}=2$; $b=-1$. Boundary: $y=2x-1$.
2. Dashed line → strict inequality ($<$ or $>$).
3. Test $(0,0)$, which is shaded: $0?2(0)-1$ gives $0>-1$, so the symbol is $>$.
4. Answer: $y>2x-1$.

$y<2x-1$: Describes the region below the line; the shaded side contains $(0,0)$, and $0<-1$ is false.

$y>2x-1$: Correct. The boundary has slope $\frac{1-(-1)}{1-0}=2$ and $y$-intercept $-1$. It is dashed, so the symbol is strict, and the shaded side contains $(0,0)$, where $0>-1$ is true.

$y\ge 2x-1$: Includes the boundary line; that would be drawn solid, but this one is dashed.

$y>-2x-1$: Uses the wrong sign for the slope; the line rises from left to right.

Find the boundary line's equation, use a dashed line for $<$ or $>$ and a solid line for $\le$ or $\ge$, then test a point in the shaded region to choose the direction.

Answer: C

1. Factor the second denominator: $x^2-4=(x-2)(x+2)$. LCD $=(x-2)(x+2)$.
2. Rewrite: $\frac{x(x+2)}{(x-2)(x+2)}-\frac{8}{(x-2)(x+2)}$.
3. Numerator: $x^2+2x-8=(x+4)(x-2)$.
4. $\frac{(x+4)(x-2)}{(x-2)(x+2)}=\frac{x+4}{x+2}$.

$\frac{x-8}{x^2-4}$: Subtracts the numerators without first rewriting $\frac{x}{x-2}$ over the common denominator $x^2-4$.

$\frac{x-4}{x-2}$: Factors $x^2+2x-8$ as $(x-4)(x+2)$; the correct factors are $(x+4)(x-2)$.

$\frac{x+4}{x+2}$: Correct. Over the LCD $(x-2)(x+2)$ the numerator is $x(x+2)-8=x^2+2x-8=(x+4)(x-2)$; cancel $(x-2)$.

$\frac{x^2-8}{x^2-4}$: Multiplies the numerator $x$ by $x$ only instead of by the whole factor $(x+2)$.

Factor denominators to find the LCD, combine the numerators, then factor the result and cancel any common factor.

Answer: 3

1. Square both sides: $2x+3=x^2$.
2. Standard form: $x^2-2x-3=0$, so $(x-3)(x+1)=0$.
3. Candidates: $x=3$ or $x=-1$.
4. Check $x=3$: $\sqrt{9}=3$. ✓ Check $x=-1$: $\sqrt{1}=1\ne -1$, so $-1$ is extraneous.
5. Solution: $x=3$.

Squaring both sides can create extraneous roots, so check every candidate in the original equation; a principal square root cannot equal a negative number.

Answer: A

1. The vertex is $(1,4)$ and the parabola opens downward.
2. The maximum output is $y=4$, and the outputs decrease without bound.
3. Range: $(-\infty ,4]$.

$(-\infty ,4]$: Correct. The parabola opens downward, so its highest $y$-value is the vertex's, 4, and it takes every $y$-value below that: $(-\infty ,4]$.

$(-\infty ,1]$: Uses the $x$-coordinate of the vertex instead of its $y$-coordinate.

$[-1,3]$: Gives the $x$-values between the $x$-intercepts; range describes $y$-values.

$[4,\infty )$: Points the wrong way; that range belongs to a parabola that opens upward.

The range of a parabola runs from the vertex's $y$-value: down to $-\infty$ if it opens downward, up to $\infty$ if it opens upward.