AMC 8 · 2001 · #17

Grade 6 rate-ratiopattern

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

percentagefraction-arithmeticratio-proportionsequences-geometric systematic-enumerationidentify-subproblems ↑ Prerequisites: percentagefraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

For the game show Who Wants To Be A Millionaire?, the dollar values of each question are shown in the following table (where K = 1000).

$\begin{tabular}{rccccccccccccccc} \text{Question} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \text{Value} & 100 & 200 & 300 & 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} & 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K} \end{tabular}$

Between which two questions is the percent increase of the value the smallest?

Pick an answer.

(A)

From 1 to 2

(B)

From 2 to 3

(C)

From 3 to 4

(D)

From 11 to 12

(E)

From 14 to 15

AMC 8 2001 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: On Who Wants To Be A Millionaire?, the prize for each of $15$ questions doubles or steps up by an irregular amount. Among five listed pairs of consecutive questions, find the pair whose prize jump is the smallest percent increase.

Givens: Prize values: $100, 200, 300, 500, 1\text{K}, 2\text{K}, 4\text{K}, 8\text{K}, 16\text{K}, 32\text{K}, 64\text{K}, 125\text{K}, 250\text{K}, 500\text{K}, 1000\text{K}$ for questions $1$ through $15$; $1\text{K} = 1000$; Answer choices: (A) From $1$ to $2$, (B) From $2$ to $3$, (C) From $3$ to $4$, (D) From $11$ to $12$, (E) From $14$ to $15$

Unknowns: Which of the five listed question-pairs has the smallest percent increase in prize value

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #6 Guess and Check

The question has five separate candidates, so Tool #7 (Identify Subproblems) splits the work cleanly: compute one percent increase per candidate, then compare. Tool #6 (Guess and Check) is the right second tool because we are literally testing each choice against the formula and keeping the smallest. No algebra is needed — just the percent-increase formula five times and a head-to-head comparison.

Execute — Answer: B

#7 Identify Subproblems 6.RP.A.3 Step 1

Write down the percent-increase formula in the form we will reuse for every pair.
The base of the percent is always the older (smaller-numbered) question.

$$\text{Percent increase} = \dfrac{V_{\text{new}} - V_{\text{old}}}{V_{\text{old}}} \times 100\%$$

💡 Treating the older value as the "whole" (the base of $100\%$) is the Grade 6 percent-change convention.

#6 Guess and Check 6.RP.A.3 Step 2

Apply the formula to the five listed pairs.
Read the prize values straight off the table and reduce each fraction.

$\text{(A)}\;\dfrac{200-100}{100} = 1 = 100\%$ $\text{(B)}\;\dfrac{300-200}{200} = \dfrac{1}{2} = 50\%$ $\text{(C)}\;\dfrac{500-300}{300} = \dfrac{2}{3} \approx 66.7\%$ $\text{(D)}\;\dfrac{125-64}{64} = \dfrac{61}{64} \approx 95.3\%$ $\text{(E)}\;\dfrac{1000-500}{500} = 1 = 100\%$

💡 For (D), the values are in thousands but the units cancel in the ratio, so $\tfrac{125-64}{64}$ uses the same arithmetic as cents or dollars.

#7 Identify Subproblems 6.NS.C.7 Step 3

Line up the five percents and pick the smallest.
The list is $100\%,\;50\%,\;66.7\%,\;95.3\%,\;100\%$, and $50\%$ is clearly the minimum.

$$\min(100\%,\;50\%,\;66.7\%,\;95.3\%,\;100\%) = 50\% \;\Rightarrow\; \textbf{(B)}$$

💡 Ordering five rational numbers and picking the smallest is a Grade 6 number-line comparison.

[1] #7 6.RP.A.3 Write down the percent-increase formula in the form we will reuse for every pair

[2] #6 6.RP.A.3 Apply the formula to the five listed pairs. Read the prize values straight off t

[3] #7 6.NS.C.7 Line up the five percents and pick the smallest. The list is $100\%,\;50\%,\;66.

Review

Reasonableness: The added dollar amount is not what decides the winner — the base matters. (A) adds only $\$100$ but on a base of $\$100$, so the percent is huge ($100\%$). (B) also adds $\$100$, but the base is $\$200$, cutting the percent in half to $50\%$. Every other listed pair at least doubles ($100\%$) or comes close to doubling ($95.3\%$), so $50\%$ stands out as the only "sub-doubling" jump. This matches answer (B).

Alternative: Tool #5 (Look for a Pattern): most of the prize table doubles each question ($100 \to 200 \to 400$ would be perfect doubling), so the percent increase is usually about $100\%$. The only place the table breaks the doubling pattern is the soft step $200 \to 300$ (only $+50\%$) and the slightly soft step $64\text{K} \to 125\text{K}$ (about $+95\%$). Of those two off-pattern jumps, the bigger break — and therefore the smaller percent increase — is from question $2$ to $3$, giving (B).

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including percent problems (Applying the percent-increase formula $\tfrac{V_{\text{new}}-V_{\text{old}}}{V_{\text{old}}}\times 100\%$ to each of the five listed pairs.)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Comparing the five computed percents ($100\%, 50\%, 66.7\%, 95.3\%, 100\%$) to pick the smallest.)

⭐ When a prize jumps by the same dollars, the percent change shrinks as the starting value grows — so $\$200 \to \$300$ is a $50\%$ jump, the smallest of the five pairs.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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