AMC 8 · 2007 · #6

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

percentagefraction-decimal-conversionestimation identify-subproblems ↑ Prerequisites: percentagefraction-arithmetic

📏 Short solution 💡 1 insight

Problem

The average cost of a long-distance call in the USA in $1985$ was
$41$ cents per minute, and the average cost of a long-distance
call in the USA in $2005$ was $7$ cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2007 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: A long-distance call in the USA cost $41$ cents per minute in $1985$ and $7$ cents per minute in $2005$. Find the approximate percent decrease in the cost per minute.

Givens: $1985$ cost: $41$ cents per minute; $2005$ cost: $7$ cents per minute; Answer choices: (A) $7$, (B) $17$, (C) $34$, (D) $41$, (E) $80$

Unknowns: The approximate percent decrease from $41$ cents to $7$ cents

Understand

Restated: A long-distance call in the USA cost $41$ cents per minute in $1985$ and $7$ cents per minute in $2005$. Find the approximate percent decrease in the cost per minute.

Givens: $1985$ cost: $41$ cents per minute; $2005$ cost: $7$ cents per minute; Answer choices: (A) $7$, (B) $17$, (C) $34$, (D) $41$, (E) $80$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #6 Guess and Check

Tool #7 (Identify Subproblems) breaks the question into three small steps: (a) find the amount of decrease, (b) form the ratio of decrease to the original price, and (c) convert that ratio to a percent. Each piece is a single arithmetic move. Tool #6 (Guess and Check) handles the word "approximate" — the answer choices are widely spaced ($7,\,17,\,34,\,41,\,80$), so a quick estimate of $\tfrac{34}{41}$ is enough to pick the only choice close to it. We avoid Tool #13 (Algebra) because no variable is needed.

Execute — Answer: E

#7 Identify Subproblems 4.NBT.B.4 Step 1

Find the amount of decrease.
The price fell from $41$ cents to $7$ cents, so the decrease is the difference of the two prices.

$$\text{decrease} = 41 - 7 = 34 \text{ cents}$$

💡 "How much did it go down?" is just subtraction of the two prices.

#7 Identify Subproblems 6.RP.A.3 Step 2

Form the percent-decrease ratio.
The percent decrease compares the drop to the original $1985$ price, not the new price, so the denominator is $41$.

$$\text{percent decrease} = \dfrac{\text{decrease}}{\text{original}} = \dfrac{34}{41}$$

💡 Percent change always uses the starting amount as the whole, so $41$ goes on the bottom.

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

Estimate $\dfrac{34}{41}$ as a percent.
Since $34$ is close to $41$ but a bit less, the ratio is a little under $1$.
Compare to the easier benchmark $\tfrac{34}{40} = 0.85 = 85\%$; the true ratio is slightly less because the denominator is $41$, not $40$.

$$\dfrac{34}{41} \approx 0.83 = 83\%$$

💡 Dividing by something close to the numerator gives a value close to (but under) $100\%$ — a useful sanity check.

#6 Guess and Check 6.RP.A.3 Step 4

Pick the closest answer choice.
The choices are $7,\,17,\,34,\,41,\,80$ — all far apart.
Only $80$ is anywhere near $83\%$; every other choice is off by $40$ points or more.

$$83\% \;\text{closest to}\; 80\% \;\Rightarrow\; \textbf{(E)}$$

💡 When choices are spread out, a rough estimate is enough — no exact decimal needed.

[1] #7 4.NBT.B.4 Find the amount of decrease. The price fell from $41$ cents to $7$ cents, so the

[2] #7 6.RP.A.3 Form the percent-decrease ratio. The percent decrease compares the drop to the o

[3] #7 6.RP.A.3 Estimate $\dfrac{34}{41}$ as a percent. Since $34$ is close to $41$ but a bit le

[4] #6 6.RP.A.3 Pick the closest answer choice. The choices are $7,\,17,\,34,\,41,\,80$ — all fa

Review

Reasonableness: Double-check with the winning percent. If the $1985$ price drops by $80\%$, the remaining price is $20\%$ of $41$, or $0.20 \times 41 = 8.2$ cents. The actual $2005$ price is $7$ cents — very close to $8.2$, off by only about a cent. The other choices are not close: a $41\%$ drop would leave $24.2$ cents, and a $17\%$ drop would leave $34$ cents — both far above the real $7$ cents. So $80\%$ is the only sensible match.

Alternative: Tool #6 (Guess and Check) directly on the answer choices: for each percent $p$, the predicted $2005$ price is $41 \times (1 - p/100)$. Try (A) $7\%$: $41 \times 0.93 \approx 38$. Try (C) $34\%$: $41 \times 0.66 \approx 27$. Try (D) $41\%$: $41 \times 0.59 \approx 24$. Try (E) $80\%$: $41 \times 0.20 = 8.2$ — the only one near the actual $7$ cents. Answer (E).

CCSS standards used (min grade 6)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the amount of decrease $41 - 7 = 34$ cents.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including percent problems (Forming the percent-change ratio $\tfrac{34}{41}$ over the original price and converting it to roughly $83\%$ to match the closest answer choice.)

⭐ Percent decrease is just "how much it dropped" divided by "what it started at" — one subtraction and one division get you the answer.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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