AMC 10 · 2025 · #19
Grade 7 geometry-3dA container has a 1×1 square bottom, a 3×3 open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes 35 minutes to fill the container up to the midline of the trapezoids.
How many more minutes will it take to fill the remainder of the container?
Pick an answer.
AMC 10 2025 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 tank has a $1\times1$ square bottom, a $3\times3$ open square top, and four matching slanted sides. Water fills it at a steady rate. Reaching the midline of the slanted sides (the halfway height) takes $35$ minutes. Find how many more minutes are needed to finish filling the tank.
Givens: The bottom is a $1\times1$ square and the open top is a $3\times3$ square; The four trapezoidal sides are congruent, so a horizontal slice is always a square; Water enters at a constant rate; Filling up to the midline of the trapezoids (half the height) takes $35$ minutes; Answer choices: (A) $70$, (B) $85$, (C) $90$, (D) $95$, (E) $105$
Unknowns: The extra time to fill from the midline up to the $3\times3$ top
Understand
Restated: A tank has a $1\times1$ square bottom, a $3\times3$ open square top, and four matching slanted sides. Water fills it at a steady rate. Reaching the midline of the slanted sides (the halfway height) takes $35$ minutes. Find how many more minutes are needed to finish filling the tank.
Givens: The bottom is a $1\times1$ square and the open top is a $3\times3$ square; The four trapezoidal sides are congruent, so a horizontal slice is always a square; Water enters at a constant rate; Filling up to the midline of the trapezoids (half the height) takes $35$ minutes; Answer choices: (A) $70$, (B) $85$, (C) $90$, (D) $95$, (E) $105$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships, #8 Analyze the Units
The tank is a frustum (a pyramid with its tip sliced off), and there is no clean K-8 formula for its volume. Tool #9 (Solve an Easier Related Problem) fixes this: extend the four slanted sides downward until they meet at a point, turning the awkward frustum into a full square pyramid. Then the shape is made of three nested pyramids whose square bases have sides $1$, $2$, and $3$. Tool #17 (Visualize Spatial Relationships) sees that these three pyramids are scaled copies, so their volumes compare as cubes. Tool #1 (Draw a Diagram) tracks the growing square slice, and Tool #8 (Analyze the Units) turns the volume ratio into a time by using the constant fill rate.
Execute — Answer: D
7.G.A.3 Step 1 Complete the tip into a full pyramid
- The four slanted sides all lean inward the same way, so if you extend them downward they meet at a single point below the tank.
- That makes a full square pyramid.
- The real tank is this big pyramid with its small pointed tip cut off.
- Working with a whole pyramid is far easier than the cut-off shape.
💡 Filling in the missing tip turns a hard frustum into an easy pyramid you can measure by scaling.
7.RP.A.2 Step 2 Slices are squares of side 1, 2, 3
- Every horizontal cut through the tank is a square.
- At the bottom its side is $1$, at the top its side is $3$, and exactly halfway up its side is the average, $2$.
- Measuring from the pyramid's tip, the square's side is proportional to the height, so the tip reaches side $1$ at one unit of height, side $2$ at two units, and side $3$ at three units.
- That gives three nested pyramids with heights in the ratio $1:2:3$.
💡 In a cone or pyramid the cross-section grows in step with height, so the side length and the height climb together.
6.EE.A.1 Step 3 Volumes scale as the cube: 1, 8, 27
- The three pyramids (to the $1\times1$, $2\times2$, and $3\times3$ squares) are the same shape at scales $1$, $2$, and $3$.
- Scaling a solid by a factor multiplies its volume by that factor three times, once for each dimension.
- So the volumes are in the ratio $1^3 : 2^3 : 3^3 = 1 : 8 : 27$.
💡 Doubling every side doubles length, width, and height, so volume grows by $2\times2\times2$.
6.RP.A.3 Step 4 Filled part is 7, remaining part is 19
- The water below the midline fills the region between the pyramid of size $1$ and the pyramid of size $2$: that volume is $8 - 1 = 7$ units.
- The part still to fill is between size $2$ and size $3$: that volume is $27 - 8 = 19$ units.
- So the already-filled portion and the leftover portion are in the ratio $7 : 19$.
💡 Each layer of the tank is the gap between two nested pyramids, found by subtracting their volumes.
6.RP.A.3 Step 5 Convert volume to time
- The rate is constant, so time is proportional to volume.
- The lower $7$ units took $35$ minutes, meaning $1$ unit of volume takes $35 \div 7 = 5$ minutes.
- The remaining $19$ units therefore take $19 \times 5 = 95$ minutes.
- The answer is $(\text{D})\ 95$.
💡 At a steady pour, twice the water needs twice the time, so minutes track volume unit for unit.
7.G.A.3 The four slanted sides all lean inward the same way, so if you extend them downw 7.RP.A.2 Every horizontal cut through the tank is a square. At the bottom its side is $1$ 6.EE.A.1 The three pyramids (to the $1\times1$, $2\times2$, and $3\times3$ squares) are t 6.RP.A.3 The water below the midline fills the region between the pyramid of size $1$ and 6.RP.A.3 The rate is constant, so time is proportional to volume. The lower $7$ units too Review
Reasonableness: The upper half of the tank is much wider than the lower half (a $2$-to-$3$ square band versus a $1$-to-$2$ band), so it should hold far more water and take much longer than $35$ minutes. The ratio $19:7$ gives $35 \times \tfrac{19}{7} = 95$ minutes, comfortably more than $35$, which fits. A quick cross-check with slice areas: the filled volume behaves like $2^3 - 1^3 = 7$ and the remaining like $3^3 - 2^3 = 19$, the same $7:19$ split, confirming (D). Choices $70$ and $85$ are too small for how much wider the top is.
Alternative: Use the frustum (truncated-pyramid) volume formula $V = \tfrac{1}{3}h\,(A_1 + A_2 + \sqrt{A_1 A_2})$. The lower frustum has base areas $1$ and $4$, giving $\tfrac{1}{3}h(1 + 4 + 2) = \tfrac{7}{3}h$; the upper frustum has areas $4$ and $9$, giving $\tfrac{1}{3}h(4 + 9 + 6) = \tfrac{19}{3}h$. Their ratio is $7:19$, so the remaining time is $35 \times \tfrac{19}{7} = 95$ minutes.
CCSS standards used (min grade 7)
7.G.A.3Describe the two-dimensional figures that result from slicing three-dimensional figures (Recognizing that every horizontal slice of the tank is a square and completing the frustum into a full pyramid.)7.RP.A.2Recognize and represent proportional relationships between quantities (Seeing that the square's side is proportional to height, giving nested pyramids with heights in ratio $1:2:3$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Cubing the scale factors so the pyramid volumes come out as $1:8:27$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Subtracting to get the $7:19$ volume split and using the constant fill rate to turn it into $95$ minutes.)
⭐ Fill in a cut-off pyramid's missing tip, then compare scaled copies by cubing the scale — volume grows as the cube of the size, so a slightly wider top can hold a lot more water.
⭐ Fill in a cut-off pyramid's missing tip, then compare scaled copies by cubing the scale — volume grows as the cube of the size, so a slightly wider top can hold a lot more water.
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