2025 AMC 10B Problem 1

P1-10~1 minPrint

Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed

The instructions on a $350$ -gram bag of coffee beans say that proper brewing of a large mug of pour-over coffee requires $20$ grams of coffee beans. What is the greatest number of properly brewed large mugs of coffee that can be made from the coffee beans in that bag?

Show solution without answering

Key insight

Each mug needs a full 20 grams. Seventeen mugs use 340 grams, and the 10 grams left over cannot make another proper mug.

Solution

Each proper mug takes $20$ grams. Count how many full groups of $20$ grams fit into $350$ grams.
Try a round number first: $10$ mugs use $200$ grams, which leaves $150$ grams.
Seven more mugs use $140$ grams, which leaves $10$ grams. So far that is $10 + 7 = 17$ mugs using $340$ grams.
The last $10$ grams are only half of what a mug needs. An eighteenth mug would not be properly brewed, so it does not count.

So the most mugs is $17$ .

The answer is $\boxed{\textbf{(B)}\ 17}$ .

Why this works

Dividing gives $350 \div 20 = 17.5$ , but you can only make whole mugs, so you drop the half. Whenever a question asks how many complete things you can make from a fixed supply, divide and then round down, even if the decimal part is $.5$ or more.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Every properly brewed mug uses exactly $20$ grams, so $n$ mugs use $20n$ grams and the requirement is $20n \le 350$ . Divide:
$350 = 20 \cdot 17 + 10.$
Seventeen mugs use $340$ grams and leave $10$ grams, which is not enough for an eighteenth mug brewed properly. Hence the greatest number is $17$ .

The answer is $\boxed{\textbf{(B)}\ 17}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)