Maximum and minimum questions in CAT LR: bound it, then build it

Many CAT sets do not have a single solution. They leave some values open and ask for the extremes: the least possible top score, the most units one shop could have sold, the fewest people who could have chosen all three options. These questions punish guessing, because the obvious extreme often breaks a condition you forgot.

The method has two halves. Bound: use the conditions to show the value cannot go past some number. Build: write down one complete case that reaches that number and satisfies every condition. If you have both, you have the answer. If you have only one, you have a guess.

The core moves

When values must be distinct, "as equal as possible" means consecutive whole numbers. When there is an order ("P sold more than Q"), the pull goes only as far as the order allows.

Set 1: five distinct scores

Five students scored a total of 400 marks in a test marked out of 100. Every score is a different whole number from 0 to 100.

Q1. What is the least possible top score?

• Bound. Make the scores as close as possible. If the top score is m, the five scores are at most m, m − 1, m − 2, m − 3 and m − 4, a total of 5m − 10. That must reach 400, so 5m ≥ 410 and m ≥ 82.
• Build. 82, 81, 80, 79, 78 adds to 400.
• Answer: 82.

Q2. What is the least possible lowest score?

• Bound. Make the other four as large as possible: 100, 99, 98 and 97, a total of 394. The lowest is at least 400 − 394 = 6.
• Build. 100, 99, 98, 97, 6.
• Answer: 6.

Q3. What is the greatest possible lowest score?

• Bound. If the lowest is m, the others are at least m + 1, m + 2, m + 3, m + 4, so 5m + 10 ≤ 400 and m ≤ 78.
• Build. 78, 79, 80, 81, 82.
• Answer: 78.

Q4. What is the least possible second-highest score?

This one needs care. To make the second-highest (call it s) small, push the top score to its limit, 100, and push the three below s as close to s as possible: s − 1, s − 2, s − 3.

• Bound. 100 + s + (s − 1) + (s − 2) + (s − 3) must reach 400. So 4s + 94 ≥ 400, 4s ≥ 306 and s ≥ 76.5. As a whole number, s ≥ 77.
• Build. With s = 77, the maximum total is 100 + 77 + 76 + 75 + 74 = 402, which is 2 too many. Lower the smallest score by 2: 100, 77, 76, 75, 72. Total 400, all distinct.
• Answer: 77.

A search through every set of five distinct scores from 0 to 100 that total 400 confirms all four answers.

Set 2: four sellers and a hidden condition

Four sellers, P, Q, R and S, sold 100 copies of a book in a week. Each sold at least 10 copies, and no two sold the same number. P sold the most, and Q sold more than R, who sold more than S. P sold fewer copies than Q and R together.

The last sentence, P < Q + R, is what makes this set more than arithmetic. Watch how it changes the answers.

Q1. What is the maximum number P could have sold?

• Without the last condition, P would be 100 − 10 − 11 − 12 = 67.
• Bound with it. P + Q + R = 100 − S ≤ 90, and Q + R > P, so 2P < P + Q + R ≤ 90 and P < 45. So P ≤ 44.
• Build. S = 10, P = 44, Q + R = 46: take Q = 24, R = 22. Check: 44 < 46, 44 > 24 > 22 > 10, total 100.
• Answer: 44.

Q2. What is the minimum number P could have sold?

• Bound. Four distinct numbers totalling 100, with P the largest: P + (P − 1) + (P − 2) + (P − 3) ≥ 100, so 4P ≥ 106 and P ≥ 27.
• Build. 27, 26, 25, 22. P < Q + R holds easily.
• Answer: 27.

Q3. What is the maximum number S could have sold?

• Bound. S + (S + 1) + (S + 2) + (S + 3) ≤ 100, so 4S ≤ 94 and S ≤ 23.
• Build. 27, 26, 24, 23. Total 100, and 27 < 50.
• Answer: 23.

Q4. What is the maximum number R could have sold?

• Bound. S ≥ 10, Q ≥ R + 1, P ≥ R + 2. So R + (R + 1) + (R + 2) + 10 ≤ 100, 3R ≤ 87, R ≤ 29.
• Build. 31, 30, 29, 10. Check P < Q + R: 31 < 59.
• Answer: 29.

Q5. What is the minimum number Q could have sold? Here the hidden condition bites from the other side.

• If Q = 20, then R ≤ 19 and S ≤ 18, so P ≥ 100 − 20 − 19 − 18 = 43. But P < Q + R ≤ 39. Impossible.
• If Q = 21, then R = 20, S = 19 gives P = 40, and 40 < 41 holds.
• Answer: 21.

An exhaustive search of all 905 valid sales patterns confirms all five answers.

Common mistakes

• Only bounding. "P cannot exceed 44" is not an answer until you show a case with P = 44.
• Only building. Finding a case with S = 23 does not prove 24 is impossible.
• Forgetting distinctness. The equal-split answer (25 each) is not allowed when values must differ.
• Rounding a bound the wrong way.
• Dropping a condition while building. Check every condition against your final case, as in Q1 of set 2.

Practice set

1. In set 2, suppose S sold exactly 20 copies. What is the maximum P could have sold?
2. Seven different positive whole numbers add up to 50. What is the largest value the biggest of them can take?
3. For the same seven numbers, what is the smallest value the biggest of them can take?

Answers:

1. 37. Now Q > R > 20, so Q ≥ 22 and R ≥ 21, and P = 80 − Q − R ≤ 37. Build: 37, 22, 21, 20, with 37 < 43. Note that the binding limit here is distinctness, not P < Q + R, which would allow up to 39.
2. 29. The other six are at least 1 + 2 + 3 + 4 + 5 + 6 = 21, so the biggest is at most 29. Build: 1, 2, 3, 4, 5, 6, 29.
3. 11. If the biggest is m, the total is at most m + (m − 1) + … + (m − 6) = 7m − 21, so 7m ≥ 71 and m ≥ 10.14, that is m ≥ 11. Build: 11, 10, 9, 8, 5, 4, 3, which adds to 50.

What to do next

• For every extreme-value question this week, write "Bound:" and "Build:" as two separate lines.
• Redo set 2 with the condition P < Q + R removed and see which answers change.
• Apply the same method to tournaments in games and tournaments and to overlaps in Venn diagram sets.