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2010 AMC 10A Problem 12

2010 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 2 of 5 · about 2 min · position P11-15

Logan 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?

A)
$0.04$
B)
$\frac{0.4}{\pi}$
C)
$0.4$
D)
$\frac{4}{\pi}$
E)
$4$

Answer

Key insight

Volume scales as the cube of length: a volume ratio of one to a million means a length ratio of one to 100, so 40/100 = 0.4.

Solution

The model is a scaled copy of the real tower, so every length is multiplied by the same factor $k$ , and every volume is multiplied by $k^3$ .

The spheres' volumes are in the ratio
$\frac{0.1}{100{,}000} = \frac{1}{1{,}000{,}000} = \frac{1}{100^3},$
so $k^3 = \dfrac{1}{100^3}$ and $k = \dfrac{1}{100}$ .

The model's height is therefore $40 \cdot \dfrac{1}{100} = 0.4$ meters.

The answer is $\boxed{\textbf{(C)}\ 0.4}$ .

Why this works

For similar figures, lengths scale by $k$ , areas by $k^2$ , volumes by $k^3$ . Given a volume ratio, take the cube root to recover the length ratio; the shape of the tank (sphere, cylinder, whatever) is irrelevant, which is why $\pi$ never enters and choices (B) and (D) are decoys.

The trap

Using the volume ratio directly on the height (0.00004 m) or taking a square root instead of a cube root (0.04 m, choice A).

Common mistakes

Using the volume ratio directly on the height (0.00004 m) or taking a square root instead of a cube root (0.04 m, choice A).
Trying to compute the sphere's radius from its volume with $\frac43 \pi r^3$ , which introduces $\pi$ and wastes time.

Techniques

Set up the equation/formula and compute; no special trick needed