AMC 8 · 2017 · #4
Grade 5 arithmeticProblem
When 0.000315 is multiplied by 7,928,564 the product is closest to which of the following?
Pick an answer.
AMC 8 2017 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: 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
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) $24000$
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
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 $1\%$).
💡 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.
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 $(3 \times 8) \times \tfrac{1{,}000{,}000}{10{,}000}$.
💡 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.
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.
💡 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.
5.NBT.A.4 Replace each factor with a nearby number that has only one non-zero digit. $0.00 5.NBT.A.2 Multiply the easy version by separating the leading digits from the zeros. $0.00 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.4Round 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.2Explain 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.2Read 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!
⭐ This AMC 8 problem only needs Grade 5 place-value rounding and the powers-of-$10$ shortcut you already know!
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