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

Answer: C

1. Compute $f(2k)=(2k{)}^2-4(2k)=4k^2-8k$.
2. Set equal: $k^2-4k=4k^2-8k$, so $3k^2-4k=0$ and $k(3k-4)=0$.
3. $k>0$, so $k=\frac{4}{3}$.
4. Faster: $f$ is symmetric about $x=2$, so two different inputs with equal outputs add to 4: $k+2k=4$, giving $k=\frac{4}{3}$.

$\frac{2}{3}$: Sets $k+2k$ equal to the vertex, 2, instead of twice the vertex: inputs with equal outputs are symmetric about $x=2$, so their sum is 4.

$1$: Makes $2k$ equal to the vertex, 2; but $f(1)=-3$ and $f(2)=-4$.

$\frac{4}{3}$: Correct. $k^2-4k=4k^2-8k$ gives $3k^2-4k=0$, so $k=\frac{4}{3}$ (for $k>0$). Both $f\left(\frac{4}{3}\right)$ and $f\left(\frac{8}{3}\right)$ equal $-\frac{32}{9}$.

$2$: This is the vertex of the parabola; $f(2)=-4$ but $f(4)=0$.

$4$: Comes from writing $f(2k)=2k^2-8k$ (squaring $k$ but not 2); $f(4)=0$ but $f(8)=32$.

Substitute the whole expression for $x$, using parentheses; for a parabola, two different inputs with the same output are equally far from the vertex.

Answer: C

1. An absolute value is never negative, so a solution needs $2x-6\ge 0$, that is, $x\ge 3$.
2. Case 1: $x+3=2x-6$ gives $x=9$. Since $9\ge 3$, it is valid: $|12|=12=2(9)-6$.
3. Case 2: $-(x+3)=2x-6$ gives $-x-3=2x-6$, so $x=1$. Since $1<3$, it is rejected: $|4|=4$, but $2(1)-6=-4$.
4. The sum of all solutions is 9.

1: Keeps only the extraneous value from the negative case; at $x=1$ the right side is $-4$, but an absolute value cannot be negative.

3: Comes from a sign slip in the negative case, $-x+3=2x-6$; at $x=3$, $|6|\ne 0$.

9: Correct. The case $x+3=2x-6$ gives $x=9$, which checks ($|12|=12$). The case $-(x+3)=2x-6$ gives $x=1$, which fails because $2(1)-6<0$. The only solution is 9.

10: Adds both case results, 9 and 1, without checking them; $x=1$ makes the right side negative.

12: Adds 9 and the value 3 from a sign slip in the negative case.

When a variable appears outside the absolute value, solve both cases and check each result in the original equation; one case often gives an extraneous root.

Answer: C

1. Let $c$ = child tickets, so $420-c$ = adult tickets. Revenue: $15(420-c)+9c=5,220$.
2. Expand: $6,300-15c+9c=5,220$, so $6c=1,080$ and $c=180$.
3. Check: 240 adult tickets bring $3,600 and 180 child tickets bring $1,620; $3,600 + $1,620 = $5,220.
4. Faster: if all 420 were adult tickets, revenue would be $6,300, which is $1,080 too much; each child ticket lowers revenue by $6, so there are 1,080 ÷ 6 = 180 child tickets.

60: This is the difference between the adult and child counts (240 − 180), not the number of child tickets.

120: Divides the $1,080 gap between the all-adult revenue and the actual revenue by $9 instead of by the $6 price difference.

180: Correct. With $c$ child tickets, $15(420-c)+9c=5,220$ gives $6,300-6c=5,220$, so $c=180$.

210: Assumes the tickets were split evenly (420 ÷ 2) and ignores the revenue.

240: This is the number of adult tickets; it answers for the wrong variable.

For a two-item total problem, write one equation in one variable, or assume every item is one type and divide the shortfall by the price difference.

Answer: E

1. Number of equally likely pairs: $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$.
2. The sum is odd exactly when one number is odd (1, 3, 5) and the other is even (2, 4): $3\times 2=6$ pairs.
3. Probability $=\frac{6}{10}=\frac{3}{5}$.
4. Check by sequence: $P(\text{odd, then even})+P(\text{even, then odd})=\frac{3}{5}\cdot \frac{2}{4}+\frac{2}{5}\cdot \frac{3}{4}=\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$.

$\frac{3}{10}$: Counts only one order (an odd number first, then an even one: $\frac{3}{5}\cdot \frac{2}{4}$); "even, then odd" is equally likely and must be added.

$\frac{2}{5}$: Gives the probability that the sum is even (both odd or both even), the complement of what is asked.

$\frac{12}{25}$: Treats the draws as independent with replacement ($2\cdot \frac{3}{5}\cdot \frac{2}{5}$); the 2 proposals must be different.

$\frac{1}{2}$: Assumes odd and even sums are equally likely; there are 3 odd numbers and only 2 even ones.

$\frac{3}{5}$: Correct. An odd sum needs one odd and one even number: $3\times 2=6$ of the $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ equally likely pairs.

Count unordered outcomes with combinations, or, if you work in sequence, include every order that produces the event.

Answer: A

1. Sum of the 7 integers: $7\times 20=140$. In order, the 4th is the median, 18.
2. To maximize the largest, minimize the other six: the three smallest are 1, 2, 3; the 5th and 6th must exceed 18 and differ, so they are 19 and 20.
3. Largest $=140-(1+2+3+18+19+20)=140-63=77$.
4. Check: 1, 2, 3, 18, 19, 20, 77 has sum 140, mean 20 and median 18.

77: Correct. The sum is 140. Make the other six as small as allowed: 1, 2, 3 below the median, then 18, then 19 and 20; $140-63=77$.

80: Uses 1, 2, 3 but repeats 18 as the fifth and sixth values; the integers must all be different, so they must be at least 19 and 20.

83: Ignores the "different" condition entirely, using 1, 1, 1, 18, 18, 18.

95: Leaves the median out of the subtraction: $140-(1+2+3+19+20)$.

122: Subtracts only the median from the sum and forgets the other five integers.

To maximize one value under a fixed sum, minimize every other value subject to all constraints (distinctness, positivity, the median position).

Answer: C

1. Rates: P fills $\frac{1}{4}$ tank per hour; Q fills $\frac{1}{6}$; together $\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$ tank per hour.
2. After P's first hour, $\frac{1}{4}$ is full and $\frac{3}{4}$ remains.
3. Time together: $\frac{3}{4}÷\frac{5}{12}=\frac{3}{4}\cdot \frac{12}{5}=\frac{9}{5}=1.8$ hours.
4. Total: $1+1.8=2.8$ hours $=2$ hours 48 minutes. Check: in 2.8 hours P fills $\frac{2.8}{4}=0.7$ and Q, running 1.8 hours, fills $\frac{1.8}{6}=0.3$; $0.7+0.3=1$ tank.

1 hour 48 minutes: Gives only the time the two pumps run together and leaves out pump P's first hour.

2 hours 24 minutes: Has both pumps running from the start ($\frac{12}{5}$ hours), ignoring that P ran alone for the first hour.

2 hours 48 minutes: Correct. P fills $\frac{1}{4}$ in the first hour; the remaining $\frac{3}{4}$ at the combined rate $\frac{5}{12}$ per hour takes $\frac{9}{5}$ hours, for $2.8$ hours in total.

3 hours 24 minutes: Adds the first hour to the full combined fill time ($1+2.4$), as if the tank were still empty after P's first hour.

5 hours: Averages the two fill times; combined rates add, so the times do not average.

Add rates (jobs per hour), not times. When machines start at different times, subtract the work already done before applying the combined rate.

Answer: E

Brightway's main competitors did not lower the prices of their frozen dinners last year.: Explains the jump in unit sales, not the fall in profit.

Frozen dinners account for less than one-fifth of Brightway's total sales.: Irrelevant: the share of total sales does not explain lower profit on frozen dinners themselves.

Shoppers who bought Brightway's frozen dinners last year also bought more of the company's frozen desserts than before.: Irrelevant to the puzzle, which concerns profit from frozen dinners only; extra dessert sales would not lower it.

Industry-wide sales of frozen dinners rose by about 5 percent last year.: Does not explain: a general sales increase says nothing about Brightway's costs.

To meet the higher demand, Brightway ran its plants on overtime, which raised its cost of producing each dinner.: Correct. Puzzle: revenue rose (0.85 × 1.20 = 1.02, about a 2 percent rise) yet profit fell. Profit is revenue minus costs, so costs must have grown faster than revenue. Overtime that raised the cost of every dinner, on more dinners, explains this: total costs rose by more than 20 percent while revenue rose by only about 2 percent.

When revenue rises but profit falls, look for a choice that raises costs; changes in volume or price alone cannot resolve it.

Answer: E

A cup of green tea contains less caffeine, on average, than a cup of brewed coffee.: Irrelevant: caffeine content compared with coffee does not bear on whether tea prevented diagnoses in this study.

Some participants who drank green tea daily were nonetheless diagnosed with high blood pressure.: Does not weaken: the claim is a 30 percent lower rate, not total protection.

The overall rate of high blood pressure among participants was close to the national rate.: Irrelevant: whether the whole sample is typical says nothing about why the two groups differ.

The tea drinkers and the non-drinkers in the study had nearly identical age distributions.: Strengthens: it rules out age as an alternative explanation for the difference.

Before the study began, physicians commonly advised patients whose blood pressure was already creeping upward to avoid caffeinated drinks, including green tea.: Correct. Conclusion: green tea prevents high blood pressure. Premise: tea drinkers were diagnosed less often. Gap: the correlation is assumed to run from tea to health. If people already heading toward high blood pressure had been told to stop drinking tea, the non-drinkers group was loaded with at-risk people, so the difference can be explained by the reverse direction of influence.

In a correlation-to-causation argument, a strong weakener can show that the supposed effect (or its early signs) influenced the supposed cause.

Answer: D

vote less reliably when local elections coincide with national ones: Contradicted: their members "vote reliably whenever elections are held."

generally favor proposals to hold local elections on the same dates as state and national elections: Reversed: since on-cycle timing dilutes their influence, the passage suggests they may be among those resisting such proposals.

are largely indifferent to decisions about zoning and school budgets: Opposite: they have "an organized stake in local policy."

have relatively more influence over local elections held separately from national contests: Correct. Paragraph 3 says low-turnout elections "tend to be dominated" by such groups, whose members vote reliably whenever elections are held, and that moving elections on-cycle "dilutes" their influence. Off-cycle elections therefore give them proportionally more weight.

make up a majority of the voters in national elections: Unsupported: the passage describes their share of low-turnout local electorates only.

An inference combines stated pieces: reliable turnout plus diluted influence when turnout rises implies outsized influence when turnout is low.

Answer: E

arguing that all local elections should be moved to coincide with national elections: Not argued: the author reports research and debate but makes no recommendation.

explaining why municipal employees vote more reliably than other citizens: Too narrow: reliable voting by organized groups is one point in paragraph 3, and its cause is not explained.

showing that voters regard local government as more important than national government: Unsupported: the author claims only that indifference "cannot be the whole story," not that voters rank local government higher.

surveying the factors that political scientists have identified as influencing voter turnout in elections at every level of government: Too broad: the passage examines two explanations of low turnout in local elections only, not every factor affecting turnout at every level.

arguing that election timing, more than voter indifference, explains low local turnout, and noting who gains from that timing: Correct. Paragraph 1 gives the indifference explanation, paragraph 2 argues that holding local elections apart from national ones better explains the turnout gap, and paragraph 3 notes that low-turnout elections favor organized groups such as municipal employees, which may explain resistance to changing election dates.

Include every paragraph's job in the purpose: here a challenge (paragraph 2) and an implication (paragraph 3).

Answer: B

explaining why large mammals tend to become smaller after colonizing small islands: Too narrow: dwarfing of large species is mentioned only in paragraph 1; the passage centers on why small species grow.

presenting an alternative to a widely accepted explanation of a pattern in island mammals: Correct. Paragraph 1 describes the size pattern and "the most widely cited explanation" (food); paragraph 2 presents researchers who argue the account "neglects the role of predators" and reports evidence for their view.

demonstrating that the availability of food has no effect on the body size of island mammals: Too extreme: the researchers say the food account "neglects" predators, not that food has no effect.

describing how small mammals on the mainland use burrows to escape predators: Too narrow: burrowing is one step in the predator argument, not the subject of the passage.

comparing the body sizes of mammals on islands with and without abundant vegetation: Too narrow: sparse vegetation appears only in the final piece of evidence.

Two-paragraph passages often follow "standard view, then challenge." The primary purpose names the challenge without overstating it.

Answer: A

They may play some role in price rigidity but are probably not its main cause.: Correct. Paragraph 2 argues that if physical repricing costs "were the main source of rigidity," prices online would be far more flexible than they are, so the author rejects them as the main cause. The author does not deny that they exist or matter at all, which is why the modest wording fits.

They are higher for online retailers than for traditional stores.: Reversed: paragraph 2 says repricing online requires little more than "changing a number in a database."

They vary according to whether customers can observe the reason for a price change.: Contradicted: paragraph 4 says menu costs "are the same whatever the reason a price changes."

They are entirely irrelevant to the decisions firms make about when to change prices.: Too extreme: the author attacks menu costs only as "the main source" of rigidity, not as a factor of no importance.

They explain why firms raise prices more readily after cost increases that have been widely publicized.: Contradicted: paragraph 4 calls this a pattern the menu-cost hypothesis "leaves unexplained."

Watch the scope of the author's criticism: rejecting something as the main cause is not the same as calling it irrelevant. The best inference matches that degree.

Answer: C

describing how online retailers have nearly eliminated the costs associated with changing prices: Too narrow: the online example in paragraph 2 is evidence against the menu-cost view, not the passage's subject.

arguing that firms would earn greater profits if they adjusted their prices more frequently: Not argued: the passage explains why firms keep prices steady and suggests that doing so preserves goodwill; it gives no advice to change more often.

arguing that price rigidity reflects firms' concern for customers' sense of fairness more than the cost of repricing: Correct. Paragraph 1 sets out the menu-cost hypothesis as "the standard explanation" for price rigidity, paragraph 2 shows that it fails for online sellers, whose repricing costs are near zero, and paragraphs 3–4 argue that firms hold prices steady to avoid seeming unfair to customers, which also explains why prices rise more readily after visible cost increases.

tracing the development of economic theories of pricing from their origins to the present day: Too broad: the passage discusses only two explanations of one phenomenon, not a history of pricing theory.

reconciling two competing explanations of a pricing pattern by showing that each one applies to a different kind of firm: Wrong structure: the author favors one explanation over the other rather than dividing them between online and traditional firms.

Map the paragraphs (standard view, problem for it, alternative, extra support) and pick the option that names that whole arc in neutral terms.

Answer: (1) 750, (2) 1.4 times

On a cumulative graph, the amount for a period is the difference between the readings at its end and at the end of the previous period.

Answer: The program with the highest admission rate also had the lowest yield.: Yes; More women enrolled in Economics than in Biology.: Yes; Across the eight programs combined, women made up more than half of all enrolled students.: No

A percentage column must be converted to counts before comparing groups of different sizes or combining groups; an overall percent is a weighted, not simple, average.

Answer: (1) For a decade before the increase, restaurant employment had grown at nearly the same rate in the region as in the neighboring regions., (2) For a decade before the increase, restaurant employment had grown at nearly the same rate in the region as in the neighboring regions.

When two arguments disagree over the same evidence, one finding can weaken one and strengthen the other; here it validates the comparison group. The two parts can share a choice.

Answer: A

1. Rephrase: $\frac{8^n}{2^{n+6}}=\frac{2^{3n}}{2^{n+6}}=2^{2n-6}$, which is $>1$ exactly when $2n-6>0$, i.e. $n>3$.
2. Statement (1): positive integers with $n^2>10$ are $n=4,5,6,\dots$ ($3^2=9$ fails). All are greater than 3, so always yes. Sufficient. Keep A and D.
3. Statement (2): $n$ = 1, 2, 3 or 4. Case 1: $n=4$ gives $2^2=4>1$ (yes). Case 2: $n=3$ gives $2^0=1$ (no, not greater than 1). NOT sufficient. Eliminate D.
4. Answer A.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.: Correct. $\frac{8^n}{2^{n+6}}=2^{3n-n-6}=2^{2n-6}$, which is greater than 1 exactly when $n>3$. For a positive integer, $n^2>10$ forces $n\ge 4$, so the answer is always yes.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.: Statement (2) alone is not sufficient: $n=4$ gives $2^2=4>1$ (yes), but $n=3$ gives $2^0=1$, which is not greater than 1 (no).

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.: "Neither alone" is false, because statement (1) alone is already sufficient: for a positive integer, $n^2>10$ means $n\ge 4$, and every $n\ge 4$ makes the expression greater than 1.

EACH statement ALONE is sufficient.: Statement (2) alone is not sufficient: $n=4$ gives $2^2=4>1$ (yes), but $n=3$ gives $2^0=1$, which is not greater than 1 (no).

Statements (1) and (2) TOGETHER are NOT sufficient.: Statement (1) alone is sufficient: for a positive integer, $n^2>10$ means $n\ge 4$, and every $n\ge 4$ makes the expression greater than 1.

Rewrite exponent questions with a common base to reduce them to a simple inequality, and use stated integer constraints when testing statements.

Answer: B

1. Statement (1): Case 1: 6, 8, 10, 12, 14 (sum 50, mean 10) has median 10. Case 2: 1, 2, 3, 4, 40 (sum 50, mean 10) has median 3. Two different medians, so NOT sufficient. Eliminate A and D.
2. Statement (2): there are exactly five integers from 8 through 12 (8, 9, 10, 11, 12). Five different integers in that range must be all of them, so the median is 10. Sufficient.
3. Answer B.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.: Statement (1) alone is not sufficient: the list 6, 8, 10, 12, 14 has mean 10 and median 10, but the list 1, 2, 3, 4, 40 has mean 10 and median 3.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.: Correct. Five different integers from 8 through 12 can only be 8, 9, 10, 11, 12, so the median must be 10. A fixed mean, by contrast, allows very different medians.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.: "Neither alone" is false, because statement (2) alone is already sufficient: five different integers from 8 to 12 must be exactly 8, 9, 10, 11, 12, so the median is 10.

EACH statement ALONE is sufficient.: Statement (1) alone is not sufficient: the list 6, 8, 10, 12, 14 has mean 10 and median 10, but the list 1, 2, 3, 4, 40 has mean 10 and median 3.

Statements (1) and (2) TOGETHER are NOT sufficient.: Statement (2) alone is sufficient: five different integers from 8 to 12 must be exactly 8, 9, 10, 11, 12, so the median is 10.

Read every constraint in the stem: "different integers" plus a narrow range can force the whole list, while a mean alone never fixes a median.