AMC 10 · 2012 · #3
Easy mode Grade 5A box is 2 centimeters tall, 3 centimeters wide, and 5 centimeters long. Packed full of clay, it holds 40 grams. A second box is twice as tall as the first, three times as wide, and the same length. It is packed full of the same clay. How many grams of clay does the second box hold?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A rectangular box measuring $2$ cm high, $3$ cm wide, and $5$ cm long is packed full of clay and holds $40$ grams. A second box is built twice as high, three times as wide, and the same length. Find $n$, the number of grams of the same clay the second box holds.
Givens: First box: height $2$ cm, width $3$ cm, length $5$ cm.; Filled up, the first box holds $40$ grams of clay.; Second box: twice the height, three times the width, same length as the first box.; Both boxes are packed with the same clay.; Answer choices: (A) $120$, (B) $160$, (C) $200$, (D) $240$, (E) $280$.
Unknowns: $n$, the grams of clay the second box holds when it is packed full.
Understand
Restated: A rectangular box measuring $2$ cm high, $3$ cm wide, and $5$ cm long is packed full of clay and holds $40$ grams. A second box is built twice as high, three times as wide, and the same length. Find $n$, the number of grams of the same clay the second box holds.
Givens: First box: height $2$ cm, width $3$ cm, length $5$ cm.; Filled up, the first box holds $40$ grams of clay.; Second box: twice the height, three times the width, same length as the first box.; Both boxes are packed with the same clay.; Answer choices: (A) $120$, (B) $160$, (C) $200$, (D) $240$, (E) $280$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #17 Visualize Spatial Relationships, #1 Draw a Diagram, #8 Analyze the Units
The second box is not just "bigger" — its dimensions are whole-number multiples of the first box's, and that is worth much more than a size comparison. Tool #17 makes the payoff visible: because the height doubles and the width triples, the second box can be sliced cleanly into blocks that are exact copies of the first box. Tool #7 then turns the big box into a problem already solved, since each copy's capacity is the $40$ grams handed to us. Tool #1 keeps the two boxes and their labelled edges straight while slicing, and Tool #8 does the first job of all: notice that the problem states grams but describes centimetres, and pin down what lets one be read off the other.
Execute — Answer: D
5.MD.C.3 Step 1 Grams are decided by space
- The problem gives a weight in grams but describes the boxes in centimetres, so something has to connect the two.
- That something is the clay: both boxes are packed with the same clay, so one cubic centimetre of it weighs the same wherever it sits.
- The grams a box holds therefore depend on exactly one thing — how much space is inside.
- Two boxes enclosing the same space hold the same grams, and a box enclosing $6$ times the space holds $6$ times the grams.
- Every step below leans on this sentence.
💡 Same clay everywhere means the box's only job is to count how much room it offers.
4.OA.A.1 Step 2 Size up the second box
- Read the description one dimension at a time.
- "Twice the height" turns $2$ into $4$.
- "Three times the width" turns $3$ into $9$.
- "The same length" leaves $5$ alone, which is multiplying by $1$.
- Sketch the two boxes side by side with those edge labels on them.
💡 "Twice" and "three times" are multiplications applied to one edge each, and "same" is multiplication by $1$.
5.MD.C.5 Step 3 Slice it into copies
- Because $4 = 2 \times 2$ and $9 = 3 \times 3$, the second box cuts up perfectly.
- One cut halfway up the height splits it into $2$ layers each $2$ cm high.
- Two cuts across the width split it into $3$ columns each $3$ cm wide.
- The length $5$ is never cut.
- That leaves $2 \times 3 = 6$ blocks, and every single block measures $2$ by $3$ by $5$ — an exact copy of the first box, with nothing left over and nothing overlapping.
💡 Doubling and tripling whole edges means the big box is a tidy stack of small boxes, not merely something larger.
4.OA.A.2 Step 4 Six copies, six times the clay
- Now pack the second box to the top.
- Those imaginary cuts split the clay inside into $6$ regions, and each region exactly fills a $2$-by-$3$-by-$5$ box, so each region weighs the $40$ grams the problem already gave us.
- The regions do not overlap and they leave no gap, so their weights simply add up to the whole.
- That makes $n$ six lots of $40$ grams.
💡 If the big box is six small boxes pushed together, its clay is six small boxes' worth of clay.
5.MD.C.3 The problem gives a weight in grams but describes the boxes in centimetres, so s 4.OA.A.1 Read the description one dimension at a time. "Twice the height" turns $2$ into 5.MD.C.5 Because $4 = 2 \times 2$ and $9 = 3 \times 3$, the second box cuts up perfectly. 4.OA.A.2 Now pack the second box to the top. Those imaginary cuts split the clay inside i Review
Reasonableness: The answer is only as good as the slicing picture, so check that picture against raw volumes. The first box encloses $2 \times 3 \times 5 = 30$ cubic centimetres and the second encloses $4 \times 9 \times 5 = 180$, and $180 = 6 \times 30$ exactly — six copies, no remainder, just as the cuts claimed. The weights agree too: $\frac{240}{180} = \frac{40}{30} = \frac{4}{3}$ grams per cubic centimetre in both boxes, which is the one promise the problem made about the clay. Two of the wrong choices are readable: $120 = 3 \times 40$ is what counting only the tripled width gives, and $280 = 7 \times 40$ is what "six times bigger" gives if it is read as six copies added on top of the original instead of six copies in total. One coincidence here is worth distrusting: the scale factors $2$, $3$, $1$ multiply to $6$ but they also add to $6$, so a student who adds them lands on $240$ and never finds out the rule is wrong. With twice the height and four times the width, adding would give $7$ where the truth is $8$. The factors multiply, because each edge carries its own factor into the product of the three edges.
Alternative: Tool #4 (Introduce a Variable) with Tool #8 (Analyze the Units) gets there without cutting anything. Let $d$ be the grams of clay in one cubic centimetre. The first box says $30d = 40$. The second box's space is $(2\cdot 2)(3\cdot 3)(5) = 180$ cubic centimetres, so $n = 180d = 6 \times (30d) = 6 \times 40 = 240$ — and $d$ itself, $\frac{4}{3}$ grams per cubic centimetre, never has to be worked out. The two routes trade off honestly: this one survives scale factors that are not whole numbers, such as $1.5$ times the height, where no clean slicing exists; the slicing route needs the whole-number factors but assumes nothing at all about the clay being evenly packed.
CCSS standards used (min grade 5)
5.MD.C.3Recognize volume as an attribute of solid figures (Establishing that the grams a box holds are decided by the space inside it, since the same clay fills both boxes.)4.OA.A.1Interpret a multiplication equation as a comparison (Turning "twice the height" and "three times the width" into the edges $4$ cm and $9$ cm.)5.MD.C.5Relate volume to the operations of multiplication and addition (Cutting the $4 \times 9 \times 5$ box into $6$ blocks of $2 \times 3 \times 5$ and knowing the pieces fill the whole with no gap or overlap.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Combining $6$ blocks of $40$ grams each into $n = 6 \times 40 = 240$.)
⭐ Doubling one edge and tripling another turns the box into six copies of the original, so it holds six times as much clay: $6 \times 40 = 240$ grams.
⭐ Doubling one edge and tripling another turns the box into six copies of the original, so it holds six times as much clay: $6 \times 40 = 240$ grams.
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