AMC 10 · 2007 · #8

Grade 8 algebra
linear-equations-two-varsystems-of-equations convert-to-algebra ↑ Prerequisites: linear-equations-one-var
📏 Medium solution 💡 2 insights
Problem
A father's age equals the total of his three children's ages, and some years ago it was twice their total. Find the ratio of his age to that number of years.

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
5
(E)
6
How to solve
Strategy Introduce a Variable

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.

1STEP 1

Write every age in terms of T and N

The children's total drops three times as fast going back.

Tom then=T-N, children's sum then=T-3N
2STEP 2

Turn the sentence into an equation

The past sentence becomes one equation.

T-N=2(T-3N)=2T-6N
3STEP 3

Solve for the ratio T/N

Solving gives the ratio 5, choice (D).

T-N=2T-6N → 5N=T → T/N=5
Answer
5
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 × 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.
💡Key takeaway

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.

  • Write every age in terms of T and N
  • Turn the sentence into an equation
  • Solve for the ratio T/N