During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

2012 CEMC Gauss (Grade 8) Problems/Problem 13

Revision as of 20:59, 18 October 2025 by Anabel.disher (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Three numbers have a mean (average) of $7$. The mode of these three numbers is $9$. What is the smallest of these three numbers?

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

Solution

Since the mode of these three numbers is $9$, we can infer that at least two of the numbers are $9$.

Let $x$ be the number that we don't know. Setting up an equation involving the average, we get:

$\frac{x + 9 + 9}{3} = 7$

$\frac{x + 18}{3} = 7$

$x + 18 = 21$

$x = 3$

$3 < 9 = 9$, so this is the smallest number. Thus, the answer is $\boxed {\textbf {(C) } 3}$.

2012 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
CEMC Gauss (Grade 8)