AMC 10 · 2011 · #2
Grade 5 arithmeticA small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A small shampoo bottle holds $35$ ml and a large bottle holds $500$ ml. Find the fewest small bottles Jasmine must buy so their combined shampoo completely fills the large bottle.
Givens: Small bottle capacity: $35$ ml; Large bottle capacity: $500$ ml; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Unknowns: The minimum number of small bottles whose total volume reaches at least $500$ ml
Understand
Restated: A small shampoo bottle holds $35$ ml and a large bottle holds $500$ ml. Find the fewest small bottles Jasmine must buy so their combined shampoo completely fills the large bottle.
Givens: Small bottle capacity: $35$ ml; Large bottle capacity: $500$ ml; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Plan
Primary tool: #14 Extreme Principle
Secondary: #8 Analyze the Units, #6 Guess and Check
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 $15$.
Execute — Answer: E
4.OA.A.3 Step 1 - State what "completely fill" demands.
- Each small bottle pours in $35$ ml, so $n$ bottles pour in $35 \times n$ ml.
- To fill the large bottle, this total must reach at least $500$ ml.
💡 The goal is the smallest count of bottles whose shampoo first covers all $500$ ml.
5.NBT.B.6 Step 2 - Divide $500$ ml by the $35$ ml each bottle supplies to see how many full bottles fit.
- Because $35 \times 14 = 490$ and $35 \times 15 = 525$, the quotient is $14$ with a remainder.
💡 Division tells you how many whole $35$-ml pours fit inside $500$ ml, and $10$ ml is left uncovered.
4.OA.A.3 Step 3 - Interpret that leftover.
- With $14$ bottles you have only $490$ ml — that is $10$ ml short, so the large bottle is not yet full.
- To cover the remaining $10$ ml you need one more bottle, rounding the count up to $15$.
💡 A nonzero remainder always forces one extra bottle, because a partial fill still leaves the bottle unfilled.
4.NBT.B.5 Step 4 - Check the two boundary counts directly.
- $14$ bottles give $490$ ml, which is too little; $15$ bottles give $525$ ml, which fills the $500$ ml (with $25$ ml to spare).
- So the minimum that works is $15$ bottles.
💡 Testing $14$ and $15$ pins the exact boundary: $15$ is the first count that reaches $500$ ml.
4.OA.A.3 State what "completely fill" demands. Each small bottle pours in $35$ ml, so $n$ 5.NBT.B.6 Divide $500$ ml by the $35$ ml each bottle supplies to see how many full bottles 4.OA.A.3 Interpret that leftover. With $14$ bottles you have only $490$ ml — that is $10$ 4.NBT.B.5 Check the two boundary counts directly. $14$ bottles give $490$ ml, which is too Review
Reasonableness: Estimate: $500 \div 35 \approx 14.3$, so the answer must be just above $14$. The choices $(D)\ 14$ and $(E)\ 15$ are the two candidates, and $14$ is the trap for anyone who rounds down. Since $14$ bottles give $490$ ml — $10$ ml short — the bottle would not be full, so $15$ is the smallest count that works. That matches $\textbf{(E)}$.
Alternative: Skip the remainder reasoning and just multiply the two nearby choices (Tool #6, Guess and Check): $14 \times 35 = 490$ ml is not enough, while $15 \times 35 = 525$ ml is enough. The first count that reaches $500$ ml is $15$, confirming $\textbf{(E)}$.
CCSS standards used (min grade 5)
5.NBT.B.6Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors (Dividing $500$ by the two-digit divisor $35$ to get $14$ with a remainder of $10$.)4.OA.A.3Solve multistep word problems, including problems in which remainders must be interpreted (Reading the problem as "reach at least $500$ ml" and interpreting the remainder of $10$ to round the count up from $14$ to $15$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Checking $14 \times 35 = 490$ and $15 \times 35 = 525$ to confirm $15$ is the smallest count that fills $500$ ml.)
⭐ When a real-world division has a leftover, round UP — $14$ bottles fall $10$ ml short, so you need a $15$th to truly fill the big bottle.
⭐ When a real-world division has a leftover, round UP — $14$ bottles fall $10$ ml short, so you need a $15$th to truly fill the big bottle.
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