AMC 10 · 2010 · #12
Grade 8 geometry-3dLogan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?
Pick an answer.
AMC 10 2010 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 real water tower is $40$ meters tall, and its spherical top holds $100{,}000$ liters. A scale model of the same tower holds only $0.1$ liter in its matching sphere. Find the height, in meters, of the model.
Givens: Real tower height: $40$ m; Real sphere volume: $100{,}000$ liters; Model sphere volume: $0.1$ liter; The model is a scaled copy of the real tower (same shape, all lengths shrunk by the same factor); Answer choices (in m): (A) $0.04$, (B) $\dfrac{0.4}{\pi}$, (C) $0.4$, (D) $\dfrac{4}{\pi}$, (E) $4$
Unknowns: The height of the model tower, in meters
Understand
Restated: A real water tower is $40$ meters tall, and its spherical top holds $100{,}000$ liters. A scale model of the same tower holds only $0.1$ liter in its matching sphere. Find the height, in meters, of the model.
Givens: Real tower height: $40$ m; Real sphere volume: $100{,}000$ liters; Model sphere volume: $0.1$ liter; The model is a scaled copy of the real tower (same shape, all lengths shrunk by the same factor); Answer choices (in m): (A) $0.04$, (B) $\dfrac{0.4}{\pi}$, (C) $0.4$, (D) $\dfrac{4}{\pi}$, (E) $4$
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #8 (Analyze the Units): the two given amounts are volumes (liters), but the answer is a length (meters). Volume measures space in three directions, so it grows like length cubed. That single fact links the volume ratio to the length ratio and is the whole key. Tool #7 (Subproblems): the work splits cleanly into three small jobs — find the volume ratio, take its cube root to get the length ratio, then shrink the height. Tool #3 (Eliminate): a quick reverse check that $100^3$ really is a million confirms one choice and rules out the decoys built from $\pi$.
Execute — Answer: C
6.RP.A.1 Step 1 Compare the two volumes
- First see how many times more water the real sphere holds than the model.
- Divide the real volume by the model volume.
💡 The volume ratio just asks how many tiny models it would take to fill the real sphere.
6.EE.A.1 Step 2 Volume grows like length cubed
- A liter measures space in three directions at once, so if every length is scaled by a factor $k$, the volume is scaled by $k \times k \times k = k^3$.
- So the volume ratio equals the length ratio cubed.
💡 Stretching a box by $k$ in each of its three directions multiplies the space inside by $k$ three times.
8.EE.A.2 Step 3 Undo the cube
- Solve for the length scale $k$ by taking the cube root of a million.
- Since $100 \times 100 \times 100 = 1{,}000{,}000$, the cube root is $100$.
- So every real length is $100$ times its model length.
💡 Volume shrinks fast, so a huge million-to-one volume gap comes from only a hundred-to-one gap in length.
5.NBT.B.7 Step 4 Shrink the height
- The height is a length, so it scales by the same factor $100$.
- Divide the real height by $100$ to get the model height: $40 \div 100 = 0.4$ meter.
- That is choice (C).
💡 Once you know lengths shrink by $100$, every length — including the height — divides by $100$.
6.RP.A.3 Step 5 Check and rule out the rest
- Verify by scaling back up: a $0.1$-liter sphere blown up by $100$ in every direction gains $100^3 = 1{,}000{,}000$ in volume, giving $0.1 \times 1{,}000{,}000 = 100{,}000$ liters — exactly the real sphere.
- The choices with $\pi$ are traps: the $\pi$ in the sphere-volume formula appears on both towers and cancels in the ratio, so no $\pi$ can survive.
- That leaves only (C).
💡 A correct scale must rebuild the original volume when you cube it back, and shared factors like $\pi$ never survive a ratio.
6.RP.A.1 First see how many times more water the real sphere holds than the model. Divide 6.EE.A.1 A liter measures space in three directions at once, so if every length is scaled 8.EE.A.2 Solve for the length scale $k$ by taking the cube root of a million. Since $100 5.NBT.B.7 The height is a length, so it scales by the same factor $100$. Divide the real h 6.RP.A.3 Verify by scaling back up: a $0.1$-liter sphere blown up by $100$ in every direc Review
Reasonableness: The length scale is $100$ because $100^3 = 1{,}000{,}000$ matches the volume ratio, and reversing it gives back $100{,}000$ liters, so the scale is right. A model that fits on a desk being about $0.4$ m tall next to a $40$ m tower is sensible. Choice (A) $0.04$ mistakenly divides by $1000$ (treating the million as $10^3$ wrong); (E) $4$ divides by only $10$; and (B), (D) wrongly keep a $\pi$ that cancels in the ratio. Only (C) $0.4$ survives.
Alternative: Skip the ratio and work with actual radii. From $V = \tfrac{4}{3}\pi r^3$, the real and model radii satisfy $\dfrac{r_{\text{real}}^3}{r_{\text{model}}^3} = \dfrac{100{,}000}{0.1} = 10^6$, so $\dfrac{r_{\text{real}}}{r_{\text{model}}} = 10^2 = 100$. Every length shares this ratio, so the model height is $\dfrac{40}{100} = 0.4$ m — the same answer, with $\pi$ cancelling on its own.
CCSS standards used (min grade 8)
6.RP.A.1Understand the concept of a ratio and use ratio language (Forming the volume ratio $100{,}000 : 0.1 = 1{,}000{,}000$ between the real and model spheres.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Recognizing that scaling every length by $k$ scales volume by $k^3$, so the volume ratio is the length ratio cubed.)8.EE.A.2Use square root and cube root symbols to represent solutions (Taking the cube root of $1{,}000{,}000$ to find the length scale factor $100$.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Dividing the real height $40$ by $100$ to get the model height $0.4$ m.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Checking the scale by rebuilding $100{,}000$ liters and reasoning that shared factors like $\pi$ cancel.)
⭐ Volume grows by length cubed, so a million-times-bigger volume means only a hundred-times-bigger length: $40 \div 100 = 0.4$ m.
⭐ Volume grows by length cubed, so a million-times-bigger volume means only a hundred-times-bigger length: $40 \div 100 = 0.4$ m.
More like this
Same archetype — closest grade level first.