AMC 10 · 2007 · #12
Grade 8 arithmeticTom's age is T years, which is also the sum of the ages of his three children. His age N years ago was twice the sum of their ages then. What is T/N?
Pick an answer.
AMC 10 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.
Toolkit + CCSS Solution
Understand
Restated: Tom's current age is $T$ years, and $T$ also equals the sum of the ages of his three children right now. Exactly $N$ years ago, Tom's age at that time was twice the sum of his three children's ages at that same time. Find the value of the ratio $T/N$.
Givens: Tom's age now is $T$; The three children's ages now add up to $T$ (the same $T$); $N$ years ago, Tom's age then was twice the sum of the three children's ages then; Answer choices: (A) 2, (B) 3, (C) 4, (D) 5, (E) 6
Unknowns: The ratio $T/N$ of Tom's age to the number of years ago
Understand
Restated: Tom's current age is $T$ years, and $T$ also equals the sum of the ages of his three children right now. Exactly $N$ years ago, Tom's age at that time was twice the sum of his three children's ages at that same time. Find the value of the ratio $T/N$.
Givens: Tom's age now is $T$; The three children's ages now add up to $T$ (the same $T$); $N$ years ago, Tom's age then was twice the sum of the three children's ages then; Answer choices: (A) 2, (B) 3, (C) 4, (D) 5, (E) 6
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra
The problem already hands us two letters, $T$ and $N$, so naming the remaining quantities in terms of them (Tool #4) is the natural move. The key that this tool unlocks is that '$N$ years ago' shifts each of the three children back by $N$, so their combined age drops by $3N$, not by $N$. Once every age is written in $T$ and $N$, the single sentence 'Tom's age then was twice their sum then' becomes one equation (Tool #13, Convert to Algebra), which we solve for the ratio $T/N$.
Execute — Answer: D
6.EE.A.2 Step 1 Write every age in terms of $T$ and $N$
- Tom's age now is $T$, and the three children's ages now add up to $T$.
- Go back $N$ years.
- Tom's age then was $T-N$.
- Each of the three children was also $N$ years younger, so the three of them together were $3N$ years younger than now.
- Their combined age $N$ years ago was therefore $T-3N$, not $T-N$ — this is the step that trips people up, because there are three children losing $N$ years each.
💡 Three kids each go back $N$ years, so their ages together drop by $3N$, three times as fast as one person's.
7.EE.A.1 Step 2 Turn the sentence into an equation
- The sentence says Tom's age $N$ years ago was twice the sum of the children's ages $N$ years ago.
- Tom's age then is $T-N$; twice the children's sum then is $2(T-3N)$.
- Setting them equal gives the equation, and expanding the right side using the distributive property spreads the $2$ across both terms.
💡 'Was twice' means one side equals two copies of the other, which is exactly an equals sign with a factor of $2$.
8.EE.C.7 Step 3 Solve for the ratio $T/N$
- Collect the $T$ terms on one side and the $N$ terms on the other.
- From $T-N=2T-6N$, subtract $T$ from both sides to get $-N=T-6N$, then add $6N$ to both sides to get $5N=T$.
- So $T=5N$, which means dividing both sides by $N$ gives $T/N=5$.
- That matches choice (D).
💡 Gathering the $T$'s and $N$'s onto opposite sides leaves a clean relationship between the two, which is all a ratio needs.
6.EE.A.2 Tom's age now is $T$, and the three children's ages now add up to $T$. Go back $ 7.EE.A.1 The sentence says Tom's age $N$ years ago was twice the sum of the children's ag 8.EE.C.7 Collect the $T$ terms on one side and the $N$ terms on the other. From $T-N=2T-6 Review
Reasonableness: Pick numbers that fit $T=5N$. Let $N=6$, so $T=30$: Tom is $30$ and his three children's ages sum to $30$ now. Six years ago Tom was $24$, and the children summed to $30-3(6)=12$. Indeed $24=2\times 12$, so the condition holds and $T/N=30/6=5$. Trying the tempting wrong answer that forgets the three children — using $T-N=2(T-N)$ — would force $T=N$, i.e. Tom's whole age passed in $N$ years, which is impossible, confirming the $3N$ drop is essential.
Alternative: Work with the drop directly. Over $N$ years, Tom gains $N$ and the children's total gains $3N$. Now Tom's age equals the children's sum ($T=T$); back then Tom's age was double it. So going forward from 'then' to 'now,' Tom's lead over the children's sum shrank from (double $-$ single $=$ one children's-sum-then) to zero. In numbers, then-sum $=T-3N$, and Tom's then-lead $T-N-(T-3N)=2N$ must equal that then-sum, giving $T-3N=2N$, so $T=5N$ and $T/N=5$.
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing each age $N$ years ago in terms of $T$ and $N$, including the children's sum as $T-3N$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding $2(T-3N)$ to $2T-6N$ when translating 'twice their sum' into an equation.)8.EE.C.7Solve linear equations in one variable (Collecting like terms across both sides of $T-N=2T-6N$ to get $T=5N$ and hence $T/N=5$.)
⭐ When you rewind time in an age problem, every person loses those years, so three children lose three times as many — count all of them before writing the equation.
⭐ When you rewind time in an age problem, every person loses those years, so three children lose three times as many — count all of them before writing the equation.
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