The hardest GRE math questions, with every step worked

What makes a GRE math question hard?

Hidden cases, not hard content. The Quantitative Reasoning measure covers arithmetic, algebra, geometry and data analysis at a level you met in high school, and that ceiling does not move for the difficult questions. What changes is how much the question leaves for you to notice.

Four things account for nearly all of the difficulty:

• More than one case. A variable that could be negative, zero or a fraction. A number that could be 6 or 12. The hard question is the one where the cases give different answers, and where the first case you try looks conclusive.
• A constraint that is easy to read past.“Distinct”, “positive”, “integer”, “without replacement”. One word changes the answer, and it is rarely the word in bold.
• A structure you have to see before you can compute. A difference of squares to factor, a remainder that cycles, a process that multiplies instead of subtracting. The arithmetic afterwards is short. Finding the structure is the question.
• A wrong answer built from the natural mistake. Hard multiple-choice questions list the result of the obvious error among the choices, so getting an answer that appears is not evidence that it is right.

One measured detail from our own bank, which is written to GRE style: of the questions we rate at the top difficulty level, about a third are number properties (divisibility, primes, remainders), ahead of algebra, then probability and data interpretation. Geometry is a small share. That is our editorial rating, not an ETS statistic, but it matches where most people report losing points, and it is why four of the ten questions below are about integers.

Where the hardest questions show up

In the second Quant section, and only if you earned it. The GRE is adaptive by section: how you do on the first Quant section decides whether your second one is the easier, the medium or the harder version. The hardest questions on the test live in that harder second section.

This matters for how you read your own practice. If you never see questions like the ones below, it may be because your first section is not yet routing you to them, and the highest scaled scores are only reachable through the harder section. The mechanics are explained in how the section-adaptive GRE works, and what those top scores are worth is on how to get a 330.

10 hard GRE math questions with solutions

Six are quantitative comparison, three are multiple choice and one is numeric entry. Try each one under roughly two minutes before opening the solution, since that is about what the section allows per question. These are original questions written to GRE style, not ETS items, and the difficulty ratings are our own.

1. Question 1Quantitative comparison · Hardest · Inequalities and sign reasoning

   $a$ and $b$ are nonzero numbers with $a<b$.

   Quantity A

   $\frac{1}{a}$

   Quantity B

   $\frac{1}{b}$
   Show the answer and worked solution

   Answer: D. The relationship cannot be determined from the information given.

   We are told only that $a$ and $b$ are nonzero with $a<b$. Taking reciprocals does not always preserve or reverse an inequality, because the result depends on the signs of $a$ and $b$, so we test cases.

   Case 1: let $a=2$ and $b=4$ (both positive, $a<b$). Then Quantity A is $\frac{1}{2}=0.5$ and Quantity B is $\frac{1}{4}=0.25$. Here Quantity A is greater.

   Case 2: let $a=-3$ and $b=2$ (opposite signs, $a<b$). Then Quantity A is $\frac{1}{-3}\approx -0.33$ and Quantity B is $\frac{1}{2}=0.5$. Here Quantity B is greater.

   Because one allowed case makes A greater and another makes B greater, the relationship cannot be determined.

   The trap is to assume that $a<b$ forces $\frac{1}{a}>\frac{1}{b}$; that holds only when $a$ and $b$ share the same sign. When they straddle zero, the conclusion flips.

   Do

   Test both a same-sign case and an opposite-sign case, since reciprocation behaves differently across zero.

   Avoid

   Do not blindly flip the inequality when taking reciprocals without checking the signs of $a$ and $b$.

2. Question 2Quantitative comparison · Hardest · Coordinate geometry

   Compare the two quantities.

   Quantity A

   The shortest distance from the point $(8,6)$ to the circle $x^2+y^2=25$

   Quantity B

   $5$
   Show the answer and worked solution

   Answer: C. The two quantities are equal.

   The circle $x^2+y^2=25$ is centered at the origin with radius $5$.

   The distance from the origin to $(8,6)$ is $\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$.

   Since the point lies outside the circle ($10>5$), the shortest distance from the point to the circle equals the center distance minus the radius: $10-5=5$.

   So Quantity A $=5$, which equals Quantity B.

   Answer: (C) The two quantities are equal. Choices A and B fail because the center distance is exactly $10$ and the radius is exactly $5$, giving exactly $5$.

   Do

   Find the distance to the center, then subtract the radius since the point is outside.

   Avoid

   Don't subtract the radius from a single coordinate; use the full center-to-point distance.

3. Question 3Quantitative comparison · Hardest · Number properties

   $n$ is a positive integer such that $n^2$ is divisible by $12$.

   Quantity A

   $n$

   Quantity B

   $6$
   Show the answer and worked solution

   Answer: D. The relationship cannot be determined from the information given.

   $12=2^2\cdot 3$. For $3\mid n^2$, since $3$ is prime, $n$ must be divisible by $3$. For $4\mid n^2$, $n$ must be divisible by $2$. So $n$ must be a multiple of $6$, and the smallest such value is $n=6$. But $n$ could also equal $12$, $18$, and so on. If $n=6$, Quantity A equals Quantity B. If $n=12$, Quantity A is greater than Quantity B. Since both cases satisfy the given condition and yield different relationships, the relationship cannot be determined.

   Do

   Determine the minimal divisibility requirement on $n$, then test at least two admissible values (such as $6$ and $12$) to check whether the relationship stays fixed.

   Avoid

   Don't assume $n$ must equal exactly $6$; any positive multiple of $6$ also makes $n^2$ divisible by $12$.

4. Question 4Quantitative comparison · Hardest · Statistics

   The list $S$ consists of the five numbers $2,4,6,8,$ and $x$, where $x$ is a real number.

   Quantity A

   the standard deviation of the five numbers in $S$

   Quantity B

   the standard deviation of the four numbers $2,4,6,8$
   Show the answer and worked solution

   Answer: D. The relationship cannot be determined from the information given.

   The four numbers $2,4,6,8$ have mean $5$ and a fixed spread. Adding a fifth value $x$ can either tighten or widen the spread, depending on where $x$ falls.

   Case $x=5$ (right at the mean): the new point sits at the center, which pulls the standard deviation down. So A $<$ B.

   Case $x=20$ (far from the mean): the new point is an outlier that increases the spread sharply, so A $>$ B.

   Because a value at the center decreases the standard deviation while a distant value increases it, the relationship cannot be determined.

   Do

   Recall standard deviation measures spread about the mean; test a central $x$ and an extreme $x$.

   Avoid

   Assume adding any point increases the standard deviation; a point at the mean lowers it.

5. Question 5Quantitative comparison · Hardest · Number properties

   $x$ and $y$ are positive integers such that $x^2-y^2=45$.

   Quantity A

   $x-y$

   Quantity B

   $5$
   Show the answer and worked solution

   Answer: D. The relationship cannot be determined from the information given.

   Factor the left side: $x^2-y^2=(x-y)(x+y)=45$. Since $(x-y)+(x+y)=2x$ is even, the two factors must have the same parity, and since their product $45$ is odd, both factors must be odd. The odd factor pairs of $45$ with the smaller factor equal to $x-y$ are $(1,45)$, $(3,15)$, and $(5,9)$, giving $x-y=1$, $3$, or $5$ (each yields a valid positive integer solution for $x$ and $y$, for instance $x-y=1$ gives $x=23,y=22$; $x-y=3$ gives $x=9,y=6$; $x-y=5$ gives $x=7,y=2$). When $x-y=1$ or $3$, Quantity A is less than Quantity B; when $x-y=5$, Quantity A equals Quantity B. Since both a strict inequality and equality are possible, the relationship cannot be determined. Answer: D

   Do

   Factor the difference of squares and enumerate the odd factor pairs of 45 to find every admissible value of $x-y$.

   Avoid

   Don't assume the unique 'nicest' factor pair (5,9) is forced; smaller factor pairs also produce valid positive integer solutions and change the comparison.

6. Question 6Multiple choice · Hardest · Data analysis

   List $A$ consists of 9 distinct integers arranged in increasing order. The median of List $A$ is $20$. What is the least possible value of the sum of the four greatest integers in the list?

   • (A)84
   • (B)86
   • (C)88
   • (D)90
   • (E)92
   Show the answer and worked solution

   Answer: (D) 90

   With 9 distinct integers in increasing order, the median is the 5th term, so the four greatest integers occupy the 6th through 9th positions and must each be strictly greater than 20, with all four distinct and increasing. To minimize their sum, choose the smallest possible such integers: 21, 22, 23, 24, which sum to $21+22+23+24=90$. A sum of 84, 86, or 88 would require at least one of these four terms to be at most 20, contradicting that each exceeds the median; 92 is not the minimum.

   Do

   Recognize that the four greatest terms must be distinct integers strictly greater than the median, then minimize them as consecutive integers starting just above the median.

   Avoid

   Don't simply multiply the median by 4 (getting 80) or allow repeated or non-increasing values among the four greatest terms.

7. Question 7Multiple choice · Hardest · Arithmetic

   A tank holds $20$ liters of pure acid. $4$ liters are drained off and replaced with water, and the mixture is stirred. Then $4$ liters of the mixture are drained off and again replaced with water. How many liters of acid remain in the tank?

   • (A)$9.6$
   • (B)$12$
   • (C)$12.8$
   • (D)$14.4$
   • (E)$16$
   Show the answer and worked solution

   Answer: (C) $12.8$

   Each draw-and-replace step removes a fixed fraction of whatever acid is currently present. Draining $4$ of $20$ liters removes $\frac{4}{20}=\frac{1}{5}$ of the contents, leaving the fraction $1-\frac{1}{5}=\frac{4}{5}$ of the acid. The key insight: each step multiplies the remaining acid by $\frac{4}{5}$.

   After the first replacement: acid $=20\times \frac{4}{5}=16$ liters.

   After the second replacement: acid $=16\times \frac{4}{5}=12.8$ liters.

   Equivalently, acid $=20{\left(\frac{4}{5}\right)}^2=20\times \frac{16}{25}=12.8$ liters.

   Answer: (C) $12.8$. Choice E ($16$) stops after one step; choice B ($12$) wrongly removes $4$ liters of acid each time as if the mixture stayed pure ($20-4-4$); choice D ($14.4$) and choice A ($9.6$) come from other partial-step errors.

   Do

   Multiply the acid by the same surviving fraction $\frac{4}{5}$ for each replacement: $20{\left(\frac{4}{5}\right)}^2$.

   Avoid

   Don't subtract $4$ liters of acid on the second step; after stirring, the drained $4$ liters is mostly water.

8. Question 8Numeric entry · Hardest · Rates/ratios/percents

   Machine A can complete a job alone in 6 hours, and Machine B can complete the same job alone in 8 hours. Machine A works alone for 2 hours, after which Machine B joins and the two work together until the job is finished. What is the total number of hours, measured from when Machine A began working, required to complete the job? Give your answer as a fraction.

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{30}{7}$ hours, about 4.29

   Machine A's rate is $\frac{1}{6}$ job per hour, and Machine B's rate is $\frac{1}{8}$ job per hour. In the first 2 hours, Machine A alone completes $2\times \frac{1}{6}=\frac{1}{3}$ of the job, leaving $\frac{2}{3}$ remaining. Working together, their combined rate is $\frac{1}{6}+\frac{1}{8}=\frac{4}{24}+\frac{3}{24}=\frac{7}{24}$ job per hour. The time to finish the remaining $\frac{2}{3}$ of the job together is $\frac{2/3}{7/24}=\frac{2}{3}\times \frac{24}{7}=\frac{16}{7}$ hours. The total time from when Machine A started is $2+\frac{16}{7}=\frac{14}{7}+\frac{16}{7}=\frac{30}{7}$ hours.

   Do

   Track the fraction of the job completed during the solo phase, then apply the combined rate only to the remaining fraction of the job before adding back the solo time.

   Avoid

   Do not apply the combined rate to the entire job from time zero; Machine A already completed part of the job alone before Machine B joined, so only the leftover portion should use the combined rate.

9. Question 9Quantitative comparison · Hardest · Number properties

   Compare the two quantities.

   Quantity A

   The remainder when $7^{83}$ is divided by $5$

   Quantity B

   The remainder when $3^{83}$ is divided by $5$
   Show the answer and worked solution

   Answer: A. Quantity A is greater.

   Reduce each base mod $5$ and find the period of the remainder cycle.

   For $7$: $7\equiv 2(\mathrm{mod}5)$, so $7^n$ behaves like $2^n$. The cycle of $2^n\mathrm{mod}5$ is $2,4,3,1$ (then repeats), with period $4$. Since $83=4\cdot 20+3$, the exponent lands on the $3$rd term of the cycle, which is $3$. So $7^{83}\equiv 3(\mathrm{mod}5)$, giving Quantity A $=3$.

   For $3$: the cycle of $3^n\mathrm{mod}5$ is $3,4,2,1$ (then repeats), with period $4$. With $83\equiv 3(\mathrm{mod}4)$, the exponent lands on the $3$rd term, which is $2$. So $3^{83}\equiv 2(\mathrm{mod}5)$, giving Quantity B $=2$.

   Since $3>2$, Quantity A is greater.

   Answer: (A) Quantity A is greater. The trap is assuming equal exponents force equal remainders; choice C fails because $3\ne 2$, and B fails because Quantity A is the larger remainder.

   Do

   Reduce each base mod $5$, find the period-4 cycle, then use $83\mathrm{mod}4=3$ to pick the term.

   Avoid

   Don't assume identical exponents give identical remainders for different bases.

10. Question 10Multiple choice · Hardest · Number properties

    For every prime number $p$ greater than $3$, the value of $p^2-1$ is divisible by $24$. What is the greatest positive integer that is guaranteed to divide $p^2-1$ for every such prime $p$?

    • (A)$8$
    • (B)$12$
    • (C)$16$
    • (D)$24$
    • (E)$48$
    Show the answer and worked solution

    Answer: (D) $24$

    Since $p>3$ is prime, $p$ is odd and not a multiple of $3$. Write $p^2-1=(p-1)(p+1)$. Because $p-1$ and $p+1$ are consecutive even integers, one of them is divisible by $4$, so their product is divisible by $8$. Also, among the three consecutive integers $p-1$, $p$, $p+1$, since $p$ is not a multiple of $3$, exactly one of $p-1$ or $p+1$ must be, contributing a factor of $3$. Together this gives divisibility by $8\times 3=24$ for every such prime. This bound is exact: for $p=5$, $p^2-1=24$, which is not divisible by $48$, so $48$ cannot be guaranteed. Thus the greatest integer guaranteed to divide $p^2-1$ for every prime $p>3$ is $24$.

    Do

    Factor $p^2-1=(p-1)(p+1)$ and combine the guaranteed factor of $8$ (consecutive even integers) with the guaranteed factor of $3$ (one of three consecutive integers), then test the smallest case ($p=5$) to confirm no larger bound holds.

    Avoid

    Don't assume a larger multiple of $24$, such as $48$, is also guaranteed just because it worked for a larger prime like $p=7$; the smallest qualifying prime, $p=5$, disproves that.

Four habits that solve hard GRE math questions

Each of these maps to questions above, so you can see the habit do its work.

• Test a negative, zero and a fraction before you commit. In quantitative comparison, one extra case is the difference between a confident wrong answer and the right one. Question 1 falls to a single negative number. Question 4 falls to trying a value at the mean and a value far from it.
• Factor before you calculate. Questions 5 and 10 look like they need large numbers. Both collapse the moment you write the expression as a product. If a hard question seems to demand heavy arithmetic, you have probably missed a factorization.
• Look for the cycle. Remainders of powers repeat with a short period. Nobody computes a number raised to the 83rd power. In question 9 you find a cycle of four and read off the third term.
• Ask what the process multiplies by. Repeated replacement, successive percent changes and compound growth all scale by a fixed factor each step. Question 7 is two multiplications, and the tempting wrong answer comes from subtracting instead.

The same list of traps, with one worked example each, is in the 12 GRE Quant traps.

Should you spend time on the hardest questions?

It depends on your target, and for many people the honest answer is not yet. Every question in a section is worth the same, and there is no penalty for a wrong answer. So a medium question you drop through carelessness costs exactly as much as a hard one you never solve, and it is far cheaper to fix.

• Aiming for the mid 150s in Quant: your points are in accuracy on easy and medium questions. Use the mark-and-review feature to skip anything like the questions above, guess, and come back only if time is left.
• Aiming for 160 to 165: you need most medium questions and some hard ones. Learn the four habits, and practice until quantitative comparison with cases feels routine.
• Aiming for 166 and above: these questions are the score. There is very little room for misses at the top of the Quant scale, where a 170 is only around the 89th percentile.

You can check what any Quant score is worth with the GRE percentile calculator. To practice at this level by topic, start with our GRE probability questions and GRE word problems, then drill the format that produced six of the ten above in quantitative comparison practice. When you want them mixed and timed, the free mock tests are section-adaptive like the real exam.

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