2001 CEMC Gauss (Grade 8) Problems/Problem 17

Problem

Daniel’s age is one-ninth of his father’s age. One year from now, Daniel’s father’s age will be seven times Daniel’s age. The difference between their ages is

$\text{ (A) }\ 24 \qquad\text{ (B) }\ 25 \qquad\text{ (C) }\ 26 \qquad\text{ (D) }\ 27 \qquad\text{ (E) }\ 28$

Solution 1

Let $a$ be Daniel's current age. Using the first sentence of the problem, we then see that his father's age is $9a$.

We also know that Daniel's age $1$ year from now is $a + 1$, and his father's age is $9a + 1$. From the second sentence, we also know his father's age in $1$ year will be $7 \times (a + 1)$.

Equating the two, we get:

$9a + 1 = 7(a + 1)$

$9a + 1 = 7a + 7$

$2a = 6$

$a = 3$

The difference between their ages would be $9a - a = 8a = 8 \times 3 = \boxed {\textbf {(A) } 24}$.

~anabel.disher

2001 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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CEMC Gauss (Grade 8)