AMC 8 · 2024 · #10

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition

ratemulti-digit-arithmeticfraction-decimal-conversion dimensional-analysis ↑ Prerequisites: multi-digit-arithmeticfraction-decimal-conversion

📏 Medium solution 💡 3 insights

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Problem

In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.

Pick an answer.

(A)

399

(B)

414

(C)

420

(D)

444

(E)

459

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Toolkit + CCSS Solution

Understand

Restated: In January 1980 the CO$_2$ level at Mauna Loa was $338$ ppm, and it has gone up by about $1.515$ ppm every year since. We need to estimate the CO$_2$ level in January 2030 and round to the nearest whole number.

Givens: Starting year: January 1980; Starting CO$_2$ level: $338$ ppm; Annual increase (rate): $1.515$ ppm per year; Target year: January 2030; Five answer choices: (A) 399, (B) 414, (C) 420, (D) 444, (E) 459

Unknowns: The expected CO$_2$ level (in ppm) in January 2030, rounded to the nearest integer

Understand

Plan

Primary tool: #8 Analyze the Units

Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

This is a rate problem: ppm per year times years gives ppm of increase, which then adds to a starting ppm. Tool #8 (Analyze Units) keeps the recipe honest — "years × (ppm/year) = ppm," and "ppm + ppm = ppm." The only tricky arithmetic is $1.515 \times 50$, where Tool #9 (Easier Related Problem) lets us replace it with the friendlier $1.5 \times 50 = 75$ and then add a small correction. Finally, since the question is multiple-choice, Tool #3 (Eliminate Possibilities) confirms the answer against the five choices.

Execute — Answer: B

#8 Analyze the Units 4.NBT.B.4 Step 1

First find how many years have passed from January 1980 to January 2030.
Subtract the years: $2030 - 1980 = 50$ years.
The units track: years $-$ years $=$ years, so 50 is the number of years over which CO$_2$ has been climbing.

$$2030 - 1980 = 50 \text{ years}$$

💡 Subtracting four-digit years to find a span of time is the standard multi-digit subtraction skill from Grade 4.

#9 Solve an Easier Related Problem 5.NBT.B.7 Step 2

Now apply the rate to the time.
Multiply $1.515 \text{ ppm/year} \times 50 \text{ years}$.
To make the arithmetic friendlier, first solve the easier related problem $1.5 \times 50 = 75$.
Then notice the real rate is $0.015$ more than $1.5$ per year, so over 50 years that adds $0.015 \times 50 = 0.75$ more.
Total increase: $75 + 0.75 = 75.75$ ppm.
The units cancel correctly: $\tfrac{\text{ppm}}{\text{year}} \times \text{year} = \text{ppm}$.

$$1.515 \times 50 = (1.5 \times 50) + (0.015 \times 50) = 75 + 0.75 = 75.75 \text{ ppm}$$

💡 Multiplying a decimal by a whole number — and breaking the decimal into easier pieces — is exactly the decimal-arithmetic move from Grade 5.

#8 Analyze the Units 5.NBT.B.7 Step 3

Add the total increase to the 1980 starting value to get the expected 2030 level.
Both quantities are in ppm, so the units agree: ppm $+$ ppm $=$ ppm.
The expected 2030 level is $338 + 75.75 = 413.75$ ppm.

$$338 + 75.75 = 413.75 \text{ ppm}$$

💡 Adding a whole number and a decimal by lining up the decimal point is a standard Grade 5 decimal-addition skill.

#3 Eliminate Possibilities 5.NBT.A.4 Step 4

The problem asks for the answer rounded to the nearest integer.
Since the tenths digit of $413.75$ is $7$ (which is $\geq 5$), we round up: $413.75 \approx 414$.
Cross-check against the choices: $414$ matches (B).
Choice (A) 399 would mean the level barely grew; (D) 444 and (E) 459 require a much bigger rate; (C) 420 is too high by about $6$.
Only (B) is consistent with the calculation, so we eliminate the rest.

$$413.75 \approx 414 \;\Rightarrow\; \textbf{(B)}$$

💡 Rounding a decimal to the nearest whole number by looking at the tenths digit is the rounding-decimals skill introduced in Grade 5.

[1] #8 4.NBT.B.4 First find how many years have passed from January 1980 to January 2030. Subtrac

[2] #9 5.NBT.B.7 Now apply the rate to the time. Multiply $1.515 \text{ ppm/year} \times 50 \text

[3] #8 5.NBT.B.7 Add the total increase to the 1980 starting value to get the expected 2030 level

[4] #3 5.NBT.A.4 The problem asks for the answer rounded to the nearest integer. Since the tenths

Review

Reasonableness: Sanity check the size: in 50 years, CO$_2$ goes up by about $1.5$ ppm/year $\times 50 = 75$ ppm, so the 2030 level should be roughly $338 + 75 \approx 413$ ppm. Our answer $414$ matches that ballpark almost exactly, and the small extra $0.75$ from the $0.015$ correction explains the rounding up to $414$. Units are right (ppm), and the answer sits squarely between the too-low (A) 399 and too-high (C) 420 options.

Alternative: An alternative is Tool #13 (Convert to Algebra): write the level $L(t) = 338 + 1.515 t$, where $t$ is years after 1980. Plug in $t = 50$: $L(50) = 338 + 75.75 = 413.75 \approx 414$. Same answer, but for an elementary student the rate-and-units approach is more intuitive and easier to check.

CCSS standards used (min grade 5)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Subtracting the four-digit years $2030 - 1980 = 50$ to find the time span.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Multiplying the decimal rate $1.515 \times 50$ and adding $338 + 75.75$ to combine the initial level with the total increase.)
5.NBT.A.4 Round decimals to any place (Rounding the final decimal level $413.75$ to the nearest whole number to match an answer choice.)

⭐ This AMC 8 problem only needs Grade 5 decimal arithmetic and rounding you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 5)

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