AMC 10 · 2011 · #3
Grade 5 arithmeticPick an answer.
The question asks for the MINIMUM number of bottles, and the whole trap lives at the boundary: dividing 500 by 35 gives about 14.3, which sits between two answer choices. Tool #14 (Extreme Principle) is exactly the min/max lens — we look for the smallest whole number of bottles that first pushes the total to 500 ml or more. Tool #8 (Analyze the Units) keeps the bookkeeping honest: each bottle adds 35 ml, so n bottles give 35n ml, and we compare that to 500 ml. Tool #6 (Guess and Check) then confirms the boundary by testing the two neighboring counts, 14 and 15.
Write the demand as an inequality
The demand is one plain inequality.
The goal is the smallest count of bottles whose shampoo first covers all 500 ml.
4.OA.A.3Extreme PrincipleLook at the remainder
Dividing leaves a nonzero remainder.
Division tells you how many whole 35-ml pours fit inside 500 ml, and 10 ml is left uncovered.
5.NBT.B.6Analyze The UnitsWhy it must round up
So the count must round up.
A nonzero remainder always forces one extra bottle, because a partial fill still leaves the bottle unfilled.
A nonzero remainder always forces one extra bottle, because a partial fill still leaves the bottle unfilled.
▸ Why?
Dividing leaves a remainder smaller than one bottle, and that leftover still has to be covered.
▸ Why?
One fewer bottle would fall short of the target, so the next whole number is the smallest that works.
Check the neighbours
Checking neighbours gives 15, choice (E).
Testing 14 and 15 pins the exact boundary: 15 is the first count that reaches 500 ml.
4.NBT.B.5Guess And CheckWhen a real-world division has a leftover, round UP — 14 bottles fall 10 ml short, so you need a 15th to truly fill the big bottle.