1996 AMC 12 Problem 7

Attempt Problem 7 of the 1996 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

A father takes his twins and a younger child out to dinner on the twins’ birthday. The restaurant charges $4.95\$4.95 for the father and $0.45\$0.45 for each year of a child’s age, where age is defined as the age at the most recent birthday. If the bill is $9.45,\$9.45, which of the following could be the age of the youngest child?
11
22
33
44
55
Answer: B
Concepts:linear Diophantine equationsage problems
Small hint:
Subtract the father’s charge and divide the remainder by the price per year
Big hint:
If the twins are tt years old and the younger child is yy, impose both 2t+y=102t+y=10 and y<ty\lt t
Solution:
The children account for 9.45−4.95=4.509.45-4.95=4.50 dollars, so their ages total 4.500.45=10.\frac{4.50}{0.45}=10. If each twin is tt and the younger child is y,y, then 2t+y=102t+y=10 with y<t.y\lt t. Of the choices, y=2y=2 gives t=4,t=4, which works. Thus the correct answer is B.
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