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2007 AMC 10B Problem 12

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2007 AMC 10B Problem 12AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 2 of 5 · about 2 min · position P11-15

Tom's age is TT years, which is also the sum of the ages of his three children. His age NN years ago was twice the sum of their ages then. What is T/NT/N ?

  1. A)

    22

  2. B)

    33

  3. C)

    44

  4. D)

    55

  5. E)

    66

Answer

D

Key insight

N years ago Tom was T - N and the three children together were T - 3N; setting T - N = 2(T - 3N) gives T = 5N.

Solution

Right now Tom is TT and the children's ages add to TT.

Go back NN years. Tom was T−NT - N. Each of the three children was NN years younger, so their total was T−3NT - 3N. The condition "Tom's age was twice their sum" reads
T−N=2(T−3N)=2T−6N.T - N = 2(T - 3N) = 2T - 6N.
Rearranging, 5N=T5N = T, so
TN=5.\frac{T}{N} = 5.

The answer is (D) 5\boxed{\textbf{(D)}\ 5}.

Why this works

Age problems are linear because everyone ages at the same rate; the only subtlety is that a sum of kk ages drops by kNkN over NN years. Setting up both "then" quantities carefully turns the problem into one line of algebra.

Alternative approach

Pick numbers: try N=1N = 1. Then T−1=2(T−3)T - 1 = 2(T - 3) gives T=5T = 5, so T/N=5T/N = 5. A concrete check like this also guards against the "T−NT - N for the children" slip.

The trap

Subtracting N only once from the children's total, forgetting that each of the three children was N years younger.

Common mistakes

  • Subtracting N only once from the children's total, forgetting that each of the three children was N years younger; this gives T=NT = N and no valid choice.
  • Writing the "twice" on the wrong side: 2(T−N)=T−3N2(T - N) = T - 3N leads to a negative ratio.

Techniques

Set up the equation/formula and compute; no special trick needed