AMC 10 · 2010 · #8
Grade 6 arithmeticTony works 2 hours a day and is paid $0.50 per hour for each full year of his age. During a six month period Tony worked 50 days and earned $630. How old was Tony at the end of the six month period?
Pick an answer.
AMC 10 2010 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: Tony works $2$ hours each day. For every full year of his age he is paid $\$0.50$ per hour. Over a six-month period he worked $50$ days and earned $\$630$ in all. Find how old Tony was at the end of those six months.
Givens: Tony works $2$ hours per day.; His pay is $\$0.50$ per hour for each full year of his age.; In the six-month period he worked $50$ days and earned $\$630$.; Answer choices: (A) $9$, (B) $11$, (C) $12$, (D) $13$, (E) $14$.
Unknowns: Tony's age (in full years) at the end of the six-month period.
Understand
Restated: Tony works $2$ hours each day. For every full year of his age he is paid $\$0.50$ per hour. Over a six-month period he worked $50$ days and earned $\$630$ in all. Find how old Tony was at the end of those six months.
Givens: Tony works $2$ hours per day.; His pay is $\$0.50$ per hour for each full year of his age.; In the six-month period he worked $50$ days and earned $\$630$.; Answer choices: (A) $9$, (B) $11$, (C) $12$, (D) $13$, (E) $14$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities
First untangle the pay rule by tracking units (Tool #8, Analyze the Units): $2$ hours times $\$0.50$ per hour per year of age gives $\$1$ per year of age each day, so his daily pay in dollars is simply his age in years. Then test the two nearby fixed ages (Tool #3, Eliminate Possibilities): no single unchanging age can produce $\$630$, which forces a birthday partway through. Finally name the number of older-age days with a letter (Tool #4, Introduce a Variable) and solve a one-step equation; a positive whole-number answer confirms the birthday, so the ending age is the older one.
Execute — Answer: D
6.RP.A.3 Step 1 Turn the pay rule into daily pay
- Follow the units in the pay rule.
- Each hour Tony earns $\$0.50$ for every full year of his age. He works $2$ hours a day, so his daily pay is $2 \times \$0.50 \times (\text{age in years})$.
- Since $2 \times 0.50 = 1$, that is $\$1$ for each year of his age every day.
- So his daily pay in dollars equals his age in years.
💡 Chasing the units collapses a messy rate into one clean fact: each day pays his age in dollars.
4.NBT.B.5 Step 2 Test a fixed age
- Suppose Tony's age never changed.
- Then his total is $50$ days times his fixed daily pay.
- At age $12$ that is $50 \times 12 = 600$ dollars; at age $13$ it is $50 \times 13 = 650$ dollars.
- His actual total, $\$630$, sits strictly between $\$600$ and $\$650$. No single fixed age gives $\$630$, so his age must have risen from $12$ to $13$ during the period — he had a birthday.
💡 A total trapped between two whole-age totals means the age itself changed partway through.
6.EE.B.7 Step 3 Count the days and read off the age
- Let $d$ be the number of days Tony worked while he was $13$, so he worked $50 - d$ days while he was $12$.
- His earnings are $12(50-d) + 13d = 600 - 12d + 13d = 600 + d$.
- Setting this equal to $630$ gives $600 + d = 630$, so $d = 30$.
- That leaves $50 - 30 = 20$ days at age $12$.
- Both counts are positive whole numbers, so the birthday truly fell inside the period.
- His age only goes up, so the $20$ days at $12$ came first and the $30$ days at $13$ came after; at the end of the six months he was $13$.
- The answer is $\textbf{(D)}\ 13$.
💡 Naming the older-age days turns the split into a one-step equation that must land on a whole number.
6.RP.A.3 Follow the units in the pay rule. Each hour Tony earns $\$0.50$ for every full y 4.NBT.B.5 Suppose Tony's age never changed. Then his total is $50$ days times his fixed da 6.EE.B.7 Let $d$ be the number of days Tony worked while he was $13$, so he worked $50 - Review
Reasonableness: Put the split back in: $20$ days at $\$12$ is $\$240$, and $30$ days at $\$13$ is $\$390$, and $240 + 390 = 630$, matching the total exactly. The day counts $20$ and $30$ are both positive and add to $50$, so the story holds together. The two impossible fixed-age totals, $\$600$ and $\$650$, bracket $\$630$, confirming the age had to cross from $12$ to $13$. So the ending age is $13$, choice (D).
Alternative: Skip the variable and reason with the overshoot. If Tony had been $12$ for all $50$ days he would have earned $\$600$, which is $\$30$ short of $\$630$. Each day he was actually $13$ instead of $12$ adds exactly $\$1$. To make up the $\$30$ he needs $30$ such days, so $30$ days at age $13$ and $20$ days at age $12$. Since he ends at the older age, he was $13$ at the end.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Combining $2$ hours per day with $\$0.50$ per hour per year of age to find that daily pay in dollars equals his age in years.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $50 \times 12 = 600$ and $50 \times 13 = 650$ to bracket the $\$630$ total between two fixed ages.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Setting up $12(50-d) + 13d = 630$, simplifying to $600 + d = 630$, and solving $d = 30$ to split the days.)
⭐ His pay each day is just his age in dollars, so a total stuck between $50\times12$ and $50\times13$ means he turned $13$ partway through — and $13$ is his age at the end.
⭐ His pay each day is just his age in dollars, so a total stuck between $50\times12$ and $50\times13$ means he turned $13$ partway through — and $13$ is his age at the end.
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