AMC 8 · 2017 · #4

Easy mode Grade 5

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Problem

Picture multiplying these two numbers together: $0.000315$ and $7{,}928{,}564$ .

The first number is a small decimal. The second number is close to $8$ million.

You don't need the exact answer. You just need to know which of the choices is closest to the real product.

Which one of the choices below is closest?

$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$

Pick an answer.

(A)

210

(B)

240

(C)

2100

(D)

2400

(E)

24000

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Toolkit + CCSS Solution

Understand

Restated: Multiply the tiny decimal $0.000315$ by the large whole number $7{,}928{,}564$ and decide which of the five answer choices the product is closest to. Because the choices ($210, 240, 2100, 2400, 24000$) are spread across very different orders of magnitude, an exact long multiplication is unnecessary — a careful estimate will pick out the right one.

Givens: First factor: $0.000315$ (a small decimal between $0$ and $1$); Second factor: $7{,}928{,}564$ (a $7$-digit whole number, close to $8$ million); Answer choices: (A) $210$, (B) $240$, (C) $2100$, (D) $2400$, (E) $24000$

Unknowns: Which answer choice is closest to the product $0.000315 \times 7{,}928{,}564$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #3 Eliminate Possibilities

The exact product $0.000315 \times 7{,}928{,}564$ is painful to compute by hand, but the problem only asks which choice it is closest to. Tool #9 (Easier Related Problem) says: swap the ugly numbers for friendly nearby ones — $0.000315 \to 0.0003$ and $7{,}928{,}564 \to 8{,}000{,}000$ — and solve that easier version instead. Tool #3 (Eliminate Possibilities) then takes our estimate and matches it against the five widely-spaced choices; because (A)$/$(B) are around $200$, (C)$/$(D) around $2000$, and (E) around $24{,}000$, a single-digit estimate is enough to land on exactly one choice.

Execute — Answer: D

#9 Solve an Easier Related Problem 5.NBT.A.4 Step 1

Replace each factor with a nearby number that has only one non-zero digit.
$0.000315$ is very close to $0.0003$ (we drop the last two digits, $15$, which barely change the value).
$7{,}928{,}564$ is very close to $8{,}000{,}000$ (we round up by about $1\%$).

$$0.000315 \approx 0.0003 \qquad 7{,}928{,}564 \approx 8{,}000{,}000$$

💡 Rounding to a single non-zero digit by place value is exactly the Grade 5 "round decimals to any place" idea, applied to both the tiny decimal and the big whole number.

#9 Solve an Easier Related Problem 5.NBT.A.2 Step 2

Multiply the easy version by separating the leading digits from the zeros.
$0.0003 = 3 \times \tfrac{1}{10{,}000}$ and $8{,}000{,}000 = 8 \times 1{,}000{,}000$.
So the product is $(3 \times 8) \times \tfrac{1{,}000{,}000}{10{,}000}$.

$$0.0003 \times 8{,}000{,}000 = (3 \times 8) \times \dfrac{1{,}000{,}000}{10{,}000} = 24 \times 100 = 2400$$

💡 Multiplying by powers of $10$ just shifts the decimal point — the Grade 5 "patterns in zeros and the decimal point" standard turns the calculation into $3 \times 8 = 24$ with a $\times 100$ rescue.

#3 Eliminate Possibilities 4.NBT.A.2 Step 3

Match the estimate $2400$ against the five choices and eliminate the ones that are in the wrong order of magnitude.
(A) $210$ and (B) $240$ are roughly $10$ times too small; (E) $24{,}000$ is $10$ times too big; (C) $2100$ is in the right ballpark but our estimate of $2400$ is closer to (D) than to (C).
The honest rounding (one factor up by $\sim 1\%$, the other down by $\sim 5\%$) is small enough that (D) is the answer.

$$\text{estimate} = 2400 \;\Rightarrow\; \textbf{(D)}$$

💡 Comparing multi-digit numbers by place value — Grade 4 — is all that is needed to throw out the choices that are off by a factor of $10$ and pick the one that matches the estimate.

[1] #9 5.NBT.A.4 Replace each factor with a nearby number that has only one non-zero digit. $0.00

[2] #9 5.NBT.A.2 Multiply the easy version by separating the leading digits from the zeros. $0.00

[3] #3 4.NBT.A.2 Match the estimate $2400$ against the five choices and eliminate the ones that a

Review

Reasonableness: A quick sanity check: $3 \times 8 = 24$, and the place-value bookkeeping gives $10^{-4} \times 10^{6} = 10^{2} = 100$, so the product is around $24 \times 100 = 2400$. The exact product is $0.000315 \times 7{,}928{,}564 \approx 2497.5$, which rounds to $2500$ — much closer to (D) $2400$ than to (C) $2100$ or (E) $24{,}000$. The answer (D) is consistent with both the rough estimate and the true value.

Alternative: Tool #8 (Analyze the Units) via scientific notation: write $0.000315 = 3.15 \times 10^{-4}$ and $7{,}928{,}564 = 7.93 \times 10^{6}$. Then the product is $(3.15 \times 7.93) \times 10^{-4+6} = (\text{about } 25) \times 10^{2} \approx 2500$. Same answer (D), reached by tracking exponents instead of swapping in easier numbers.

CCSS standards used (min grade 5)

5.NBT.A.4 Round decimals to any place (Rounding $0.000315$ down to $0.0003$ and $7{,}928{,}564$ up to $8{,}000{,}000$ — the place-value rounding move that turns the messy product into a one-digit-by-one-digit estimate.)
5.NBT.A.2 Explain patterns in number of zeros and placement of decimal point (Reducing $0.0003 \times 8{,}000{,}000$ to $3 \times 8 = 24$ followed by a $\times 100$ shift, using the rule that multiplying or dividing by powers of $10$ just moves the decimal point.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Comparing the estimate $2400$ against the five answer choices to eliminate those that are off by a factor of $10$ and pick (D).)

⭐ This AMC 8 problem only needs Grade 5 place-value rounding and the powers-of-$10$ shortcut you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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