Hard GMAT math questions, with every solution worked through

Answer: D. 19

• A. 3 is the value of y that goes with the largest x. The question asks for x.
• B. 5 is the x that pairs with the largest admissible y, y = 11. Maximising y minimises x.
• C. 12 is the next solution down, pairing with y = 7. There is a larger one at y = 3.
• D. Correct. y = 3 gives 4x = 76 and x = 19, and no smaller positive y makes 97 − 7y a multiple of 4.
• E. This takes y = 1, gets 4x = 90 and rounds 22.5 down. x must be an integer exactly, so y = 1 is not admissible.

Why it is hard: A linear equation in two integers where the answer depends on finding the smallest admissible y.

Answer: B. 107

• A. 18 = 6 + 5 + 7 adds the three pair products. The numbers themselves are products of two pair products each, 42, 30 and 35, so their sum is 107.
• B. Correct. 42, 30 and 35 have a least common multiple of 210 in every pair, and no other grouping of the primes gives a smaller sum.
• C. This groups 2 with 5 instead of with 3, so the pair products are 10, 3 and 7 and the numbers are 70, 30 and 21. That is a valid triple, but its pair products are further apart than 6, 5 and 7.
• D. This groups 2 with 7, so the pair products are 14, 3 and 5 and the numbers are 70, 42 and 15. It meets the condition but not with the smallest sum.
• E. This groups 3 with 7, so the pair products are 21, 2 and 5 and the numbers are 105, 42 and 10.

Why it is hard: The condition has to be recast as each prime of 210 dividing at least two of the numbers, and the sum then becomes a trade-off: the choice is how to group four primes into three pair products, and the first valid grouping a solver finds is usually not the smallest.

Answer: C. 6

• A. This finds only the numbers with three different prime factors, 60, 84 and 90, and adds N = 1. It misses 72 = 2³ × 3² and 96 = 2⁵ × 3, which also have 12 divisors.
• B. This counts the five numbers with 12 divisors but leaves out N = 1, whose divisor product, 1, equals 1⁶.
• C. Correct. 60, 72, 84, 90 and 96 have 12 divisors each, and N = 1 also meets the condition.
• D. 12 is the number of divisors N must have, not the number of values of N.
• E. This takes the product of the divisors to be N raised to the number of divisors, so it counts the 16 numbers below 100 with exactly 6 divisors. Each pair of divisors contributes only one factor of N.

Why it is hard: The product of the divisors has to be rewritten as N raised to half the number of divisors, the numbers below 100 with exactly 12 divisors have to be found across three factorization shapes, and N = 1, which meets the condition without having 12 divisors, is easy to miss.

Answer: E. 4,950

• A. 3,150 is the combined value of the two bonuses removed, not the average of what remains.
• B. This divides the reduced total of 34,650 by the original 9 consultants rather than by the 7 who are left.
• C. 4,200 is the average over all nine. Dropping the two smallest bonuses must raise the average of the rest.
• D. This removes only the 1,200 bonus and averages the other eight: 36,600 ÷ 8. Both small bonuses have to go.
• E. Correct. 34,650 ÷ 7 = 4,950.

Why it is hard: Requires reconstructing a total from a mean, removing two members, and re-dividing by the new count.

Answer: D. 33

• A. This solves (72 + n)/4 = 21, dividing by the four listed numbers and forgetting that n is itself one of the five values being averaged.
• B. This solves (72 + n)/5 = n, setting the mean equal to n rather than to the median.
• C. 21 is the median of the list, the target the mean must reach, not the missing value.
• D. Correct. With n = 33 the five numbers total 105, so the mean is 21, matching the median.
• E. This takes the median to be 24, the largest of the four listed numbers, and solves (72 + n)/5 = 24. With n > 24 the middle value is 21.

Why it is hard: The median must be identified from the constraint on n before the mean equation can be written.

Answer: D. 34

• A. 9 counts only sets of four equally spaced numbers, such as 2, 4, 6, 8. Equal spacing is enough for the mean to equal the median but not necessary: 1, 2, 5, 6 has mean and median 3.5.
• B. 16 is the number of sets of 3 numbers from 1 to 9 whose mean equals their median, the equally spaced triples. The question asks about sets of 4.
• C. This allows, for each sum, one fewer set than the number of pairs with that sum, as if a pair could only be matched with its neighbor. Any two pairs with the same sum can be matched, so a sum with 4 pairs gives 6 sets, not 3.
• D. Correct. Adding the number of ways to choose 2 pairs for each sum gives 34.
• E. This counts each set twice, once for each choice of which pair is on the outside. The outer pair is always the one holding the smallest and largest numbers, so each set is counted once.

Why it is hard: The condition has to be translated into a + d = b + c before counting is possible, and the count then needs a second idea, grouping pairs by their sum and choosing two pairs per sum, instead of a search through 126 sets; equally spaced sets are the tempting partial answer.

Answer: E. 71

• A. This reads the scores as all different: 1, 2, 3, 4, 25, 26, 27, 28 total 116, leaving 64. Nothing requires different scores, and in that list 26 appears only once, so it is not the most common score.
• B. This lets 26 appear only twice, so the low scores must all differ (1, 2, 3, 4) and the eighth score must be 27: 180 − 114 = 66. Letting 26 appear three times allows 1, 1, 2, 2 and frees 5 more points.
• C. This uses 26 three times but still keeps the low scores different: 1, 2, 3, 4, 25, 26, 26, 26 total 113, leaving 67. With 26 appearing three times, the low scores may repeat twice each.
• D. This uses 1, 1, 2, 2, 25, 26, 26, 27, which totals 110 and leaves 70. But then 1, 2 and 26 each appear twice, so 26 is not given by more judges than every other score.
• E. Correct. The scores 1, 1, 2, 2, 25, 26, 26, 26 and 71 total 180 and have median 25, and 26 appears three times while no other score appears more than twice.

Why it is hard: Maximizing one score means minimizing the other eight, but the mode condition fights that: repeating a low score saves points only while 26 stays strictly the most common score, so the solver has to weigh how often 26 appears against how often low scores may repeat, and several near-miss lists break a condition.

Answer: A. 115

• A. Correct. 100 panels with exactly three economists plus 15 with four gives 115.
• B. This is C(6,3) × C(8,1) = 20 × 8, choosing three economists and then a fourth speaker from everyone left. That counts each four-economist panel four times over.
• C. This is 100 + C(6,4) × C(5,1) = 100 + 75, attaching a journalist to the all-economist case as well and so building panels of five.
• D. This is C(10,4), which uses a pool of 10 people and drops the restriction on economists entirely.
• E. This is C(11,4), every panel of 4 from the 11 candidates, before the at-least-3-economists condition is imposed.

Why it is hard: An at-least restriction that has to be split into disjoint cases before the combinations are added.

Answer: C. 8/15

• A. This counts only the orders in which the fourth defective bulb is the ninth tested: C(8,3)/C(10,4) = 56/210. The orders in which the sixth good bulb is the ninth tested also make her test 9 bulbs, and they add another 4/15.
• B. This is (4 × 3 + 6 × 5)/90 = 42/90, the probability that the last two bulbs are of the same kind. That is when she finishes within 8 tests, the opposite of the case asked about.
• C. Correct. She needs a ninth test only when the last two bulbs are one defective and one good, which happens with probability 48/90 = 8/15.
• D. This is 1 − (6 × 5)/90, which rules out only the case of two good bulbs at the end. Two defective bulbs at the end also let her stop after 8 tests, since she has then found all 6 good ones.
• E. This is 1 − (4 × 3)/90, which rules out only the case of two defective bulbs at the end. Two good bulbs at the end also let her stop after 8 tests, since she has then found all 4 defective ones.

Why it is hard: The stopping rule has two exits, all defective bulbs found or all good bulbs found, and both have to be seen; the clean route is to ignore the order of the tests and look only at the last two bulbs, while counting just the defective-bulb exit gives exactly half the answer.

Answer: B. 242

• A. This is 320 − 81: strings with no two adjacent letters alike, less all 3⁴ strings of B, C and D. It never applies the rule about the first and last letters, and it subtracts vowel-free strings that were never counted.
• B. Correct. 260 codes meet both rules about repeated letters, and 18 of them use no vowel.
• C. 260 codes meet the rules about adjacent and first-and-last letters, but 18 of them contain no vowel.
• D. This is 320 − 24, which handles the vowel rule correctly within codes with no two adjacent letters alike but never applies the rule that the first and last letters differ.
• E. 320 = 5 × 4 × 4 × 4 applies only the rule about adjacent letters.

Why it is hard: The rule that the first and last letters differ makes the count depend on whether the third letter repeats the first, a case split most solvers miss, and the vowel rule then has to be handled by a complement taken inside the same restricted count, not inside all strings.

Answer: B. 18

• A. 17 is half of 34, the total of the six requirements, on the idea that each agent counts toward two blocks. That bound is real but cannot be reached here, because blocks 1, 3 and 5 alone need 18 different agents.
• B. Correct. Blocks 1, 3 and 5 force at least 18 agents, and starting 4, 4, 6 and 4 agents in blocks 1, 2, 3 and 5 meets every requirement.
• C. This fixes the shifts as blocks 2 and 3, 4 and 5, and 6 and 1, so each pair needs as many agents as its busier block: 10 + 6 + 4 = 20. Letting shifts start in different blocks saves 2 agents.
• D. This fixes the shifts as blocks 1 and 2, 3 and 4, and 5 and 6, so it needs 8 + 10 + 4 = 22 agents.
• E. 34 adds the six requirements, which counts every agent twice, once for each block in the shift.

Why it is hard: No standard method applies: the solver has to find a lower bound by noticing that every shift covers exactly one of the alternate blocks, check that the other alternate blocks give a weaker bound, and then build a schedule that reaches it, while fixed pairings of blocks and halving the total both miss.

Answer: C. 342

• A. 312 = 24 × 13 pays Chloe for the whole job, which takes her 24 hours, well over the 15-hour limit.
• B. 324 = 18 × 18 pays Ben for the whole job, which takes him 18 hours, over the limit.
• C. Correct. Ana for 6 hours and Ben for 9 hours finishes the job in 15 hours for 180 + 162 = $342.
• D. This gives the spare hours to Chloe, the cheapest per job: Chloe works 6 hours (1/4 of the job) and Ana 9 hours, for 78 + 270 = $348. Chloe saves only $4 per extra hour, against Ben's $6.
• E. 360 has Ana do the whole job in 12 hours. That meets the limit, but the 3 unused hours can replace part of her work with cheaper work.

Why it is hard: The obvious move, giving as much work as possible to the cheapest worker, is wrong here; the solver has to compare how much money each substitution saves per hour of the limited machine time, and see that the faster, less cheap worker is the better use of the spare hours.

Answer: E. 6

• A. This is 5.4 × 0.80, applying a 20 percent fall to the 2024 figure. The 20 percent was a rise, and it applied to the 2023 figure.
• B. This is 5.4 ÷ 1.20, which reverses only the rise and leaves the earlier 25 percent fall undone.
• C. This is 5.4 × 0.90, applying the net factor forward again instead of dividing by it to work backwards.
• D. This assumes a 25 percent fall and a 20 percent rise cancel. They do not: the rise is taken on a smaller base.
• E. Correct. 5.4 ÷ 0.90 = 6, and 6 × 0.75 × 1.20 returns 5.4.

Why it is hard: Two compounded percent changes must be reversed together to recover a starting value.