1996 AHSME Problems/Problem 7

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Problem

A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$, which of the following could be the age of the youngest child?

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution

The bill for the three children is $9.45 - 4.95 = 4.50$. Since the charge is $0.45$ per year for the children, they must have $\frac{4.50}{0.45} = 10$ years among the three of them.

The twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when the first twin was born and the second twin was born). If we let the twins be $5$ years old, that leaves $10 - 2\cdot 5 = 0$ years leftover for the youngest child. But if the twins are each $4$ years old, then the youngest child could be $10 - 2\cdot 4 = 2$ years old, which is choice $\boxed{B}$.

Of note, this is the only possibility for the ages of the children. If the twins were $3$ years old, the "younger" child would be $10 - 2\cdot 3 = 4$ years old.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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