Difference between revisions of "2012 CEMC Gauss (Grade 8) Problems/Problem 13"

Latest revision as of 20:59, 18 October 2025

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

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

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) (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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CEMC Gauss (Grade 8)

Revision as of 16:00, 17 June 2025 (view source) Anabel.disher (talk \| contribs) (Created page with "==Problem== Three numbers have a mean (average) of <math>7</math>. The mode of these three numbers is <math>9</math>. What is the smallest of these three numbers? <math> \tex...")		Latest revision as of 20:59, 18 October 2025 (view source) Anabel.disher (talk \| contribs)
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	<math>3 < 9 = 9</math>, so this is the smallest number. Thus, the answer is <math>\boxed {\textbf {(C) } 3}</math>.		<math>3 < 9 = 9</math>, so this is the smallest number. Thus, the answer is <math>\boxed {\textbf {(C) } 3}</math>.
		+	{{CEMC box\|year=2012\|competition=Gauss (Grade 8)\|num-b=12\|num-a=14}}