"30 Students Borrowed Books From the Library — Maximize One Student" — GMAT® Worked Solution

From Episode 42 of Real GMAT® Problems (The GMAT® Strategy Podcast). For the full strategy behind GMAT® statistics questions, read: GMAT® Statistics: The Average, the Median, and the Shortcut That Saves You Time.

The Problem

Source: Official Guide for GMAT® Review, 11th Edition

In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest of the students each borrowed at least 3 books. If the average (arithmetic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?

(A) 3

(B) 5

(C) 8

(D) 13

(E) 15

Setting Up the Table

Use a simple two-column layout — one column for the number of students, one for the books they borrowed in total.

The first three rows are given directly. The fourth row covers the students who borrowed at least 3 books. We do not know how many of them there are or how many books they borrowed in total — yet.

Step 1: Find the Number of "At Least 3" Students

The first three groups have 2 + 12 + 10 = 24 students.

The class has 30 students.

30 − 24 = 6 students borrowed at least 3 books each.

Step 2: Find the Class Total From the Average

The class average is 2 books per student. Use the average formula:

sum / 30 = 2

sum = 2 × 30 = 60

The class borrowed 60 books in total.

Step 3: Find the Books for the "At Least 3" Group

The first three groups account for 0 + 12 + 20 = 32 books.

60 − 32 = 28 books were borrowed by the 6 students in the last group.

Step 4: Maximize One Student in That Group

Now apply the optimization logic. Six students borrowed 28 books in total. To push one of them as high as possible, push the other five as low as possible.

The constraint is that each of these students borrowed at least 3 books. So the smallest the other five can be is 3 books each.

5 × 3 = 15 books for the other five students.

28 − 15 = 13 books for the maximum student.

The answer is (D) 13.

Why This Problem Matters

This question misses at a much higher rate than the warm-up — about three times higher. The reason is the number of steps, not the difficulty of any single step.

The setup integrates two ideas from earlier in the episode. First, the average formula is the bridge to the total. Without converting the average of 2 books per student into a fixed total of 60 books, the question has no upper limit and "infinity" would be a valid answer.

Second, the optimization logic from the wood problem applies directly. Maximize one value by minimizing the others within the constraints. Here the binding constraint is "at least 3 books" — that is the floor for the other five students in the final group.

Two places people lose this question.

The first is missing the unused information move. If you get stuck halfway through, the prompt to ask is: what given information have I not used yet? On this problem, the unused information is the class average. Once you connect that to a fixed total, the rest of the steps fall into place.

The second is computational. Long subtraction (60 − 32, 28 − 15) and small multiplication (5 × 3, 2 × 30) come up repeatedly. None of it is hard math. Under time pressure, though, careless arithmetic is the most common reason people land on (C) 8 or (B) 5 instead of (D) 13.

A note for Episode 43 on word problems: the same library books problem shows up again there, framed as a word-problem organization question. The math is identical. The discussion focuses on how to translate the prose into the table structure, rather than on the statistics shortcut.

Back to the strategy article: GMAT® Statistics: The Average, the Median, and the Shortcut That Saves You Time

Episode page: Real GMAT® Problems — Ep. 42 — Statistics