AMC 10 · 2011 · #3
Easy mode Grade 5A small bottle holds 35 milliliters of shampoo. A large bottle holds 500 milliliters. Jasmine pours small bottles into the large one until it is full. What is the fewest small bottles she needs to buy?
Pick an answer.
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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