AMC 10 · 2011 · #14
Grade 7 probabilityA pair of standard 6-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?
Pick an answer.
AMC 10 2011 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: Two standard $6$-sided dice are rolled once. Their sum is used as the diameter of a circle. Compare the plain number that is the circle's area with the plain number that is its circumference, and find the probability that the area value is smaller than the circumference value.
Givens: Two fair $6$-sided dice are rolled, so the sum ranges from $2$ to $12$; The sum becomes the diameter $d$ of a circle; Area of the circle is $\pi (d/2)^2$ and circumference is $\pi d$, both taken as plain numbers; Answer choices: (A) $\frac{1}{36}$, (B) $\frac{1}{12}$, (C) $\frac{1}{6}$, (D) $\frac{1}{4}$, (E) $\frac{5}{18}$
Unknowns: The probability that the area value is less than the circumference value
Understand
Restated: Two standard $6$-sided dice are rolled once. Their sum is used as the diameter of a circle. Compare the plain number that is the circle's area with the plain number that is its circumference, and find the probability that the area value is smaller than the circumference value.
Givens: Two fair $6$-sided dice are rolled, so the sum ranges from $2$ to $12$; The sum becomes the diameter $d$ of a circle; Area of the circle is $\pi (d/2)^2$ and circumference is $\pi d$, both taken as plain numbers; Answer choices: (A) $\frac{1}{36}$, (B) $\frac{1}{12}$, (C) $\frac{1}{6}$, (D) $\frac{1}{4}$, (E) $\frac{5}{18}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
The diameter is the only quantity that changes, so Tool #4 (Introduce a Variable) is the natural anchor: call it $d$, write area and circumference as formulas in $d$, and turn the vague word comparison into a clean inequality. Tool #7 (Identify Subproblems) splits the work into two independent pieces — first a pure geometry/algebra step that finds which diameters satisfy the condition, then a pure probability step that counts how often the dice produce those diameters. Tool #2 (Make a Systematic List) handles that counting cleanly by listing the ordered dice results out of $36$.
Execute — Answer: B
7.G.B.4 Step 1 Write both quantities in terms of $d$
- Let $d$ be the diameter, which equals the dice sum.
- The radius is $d/2$, so the area is $\pi (d/2)^2 = \frac{\pi d^2}{4}$ and the circumference is $\pi d$.
- Both are expressed with the same variable $d$, so they can be compared directly.
💡 Naming the diameter $d$ lets one letter carry the whole comparison, so the geometry becomes a single inequality.
7.EE.B.4 Step 2 Turn the comparison into $d < 4$
- The area value must be less than the circumference value: $\frac{\pi d^2}{4} < \pi d$.
- Since $d > 0$ and $\pi > 0$, divide both sides by $\pi d$ (a positive number, so the inequality direction is unchanged) to get $\frac{d}{4} < 1$, which means $d < 4$.
- So the condition holds exactly when the diameter is less than $4$.
💡 Dividing by the shared positive factor $\pi d$ strips away the clutter and leaves a simple size test on $d$.
7.SP.C.8 Step 3 Find the qualifying dice sums
- The diameter is a dice sum, so it is a whole number at least $2$.
- The condition $d < 4$ leaves only $d = 2$ and $d = 3$.
- List the ordered rolls that make each: $d = 2$ comes only from $(1,1)$, one outcome; $d = 3$ comes from $(1,2)$ and $(2,1)$, two outcomes.
💡 Only the two smallest sums squeeze under the boundary, and each can be built in just a handful of ways.
7.SP.C.8 Step 4 Compute the probability
- There are $1 + 2 = 3$ favorable ordered outcomes out of the $36$ equally likely rolls.
- The probability is $\frac{3}{36} = \frac{1}{12}$.
- This matches choice (B).
💡 Favorable outcomes over all $36$ equally likely rolls gives the probability straight away.
7.G.B.4 Let $d$ be the diameter, which equals the dice sum. The radius is $d/2$, so the 7.EE.B.4 The area value must be less than the circumference value: $\frac{\pi d^2}{4} < \ 7.SP.C.8 The diameter is a dice sum, so it is a whole number at least $2$. The condition 7.SP.C.8 There are $1 + 2 = 3$ favorable ordered outcomes out of the $36$ equally likely Review
Reasonableness: The boundary $d = 4$ gives equal area and circumference ($\frac{\pi\cdot16}{4} = 4\pi = \pi\cdot4$), so only diameters below $4$ work — a small set, so a small probability is expected. Only the sums $2$ and $3$ qualify, and these are the rarest sums, worth $\frac{1}{36}$ and $\frac{2}{36}$. Their total $\frac{3}{36} = \frac{1}{12}$ is choice (B). The trap answers line up with partial work: (A) $\frac{1}{36}$ counts only $d=2$, (C) $\frac{1}{6}$ would wrongly include $d=4$ as well, and (E) $\frac{5}{18}$ ignores the inequality entirely.
Alternative: Skip algebra and test each small sum directly (Tool #6, Guess and Check). For $d=2$: area $=\pi$, circumference $=2\pi$, area smaller — qualifies. For $d=3$: area $=\frac{9\pi}{4}=2.25\pi$, circumference $=3\pi$, still smaller — qualifies. For $d=4$: area $=4\pi$, circumference $=4\pi$, equal — fails. Larger $d$ only makes the area grow faster, so nothing else works. Counting $(1,1),(1,2),(2,1)$ gives $\frac{3}{36} = \frac{1}{12}$, confirming (B).
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for area and circumference of a circle (Writing the area $\frac{\pi d^2}{4}$ and circumference $\pi d$ from the diameter $d$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Reducing the inequality $\frac{\pi d^2}{4} < \pi d$ to the simple test $d < 4$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the ordered dice rolls that give $d = 2$ or $d = 3$ and computing $\frac{3}{36} = \frac{1}{12}$.)
⭐ Turn a wordy compare-two-things question into one inequality in a single variable, then just count the dice rolls that pass.
⭐ Turn a wordy compare-two-things question into one inequality in a single variable, then just count the dice rolls that pass.
More like this
Same archetype — closest grade level first.