Difference between revisions of "2013 CEMC Gauss (Grade 8) Problems/Problem 5"

Latest revision as of 20:47, 18 October 2025

Jarek multiplies a number by $3$ and gets an answer of $90$ . If instead, he divides the number by $3$ , what answer does he get?

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 270$

Let $x$ be the number that Jarek multiplied by $3$ . Using the first sentence, we have:

$3x = 90$

$x = 30$

We are then looking for $x \div 3$ , which is $30 \div 3 = \boxed {\textbf {(B) } 10}$

~anabel.disher

Without finding the number that Jarek multiplied by $3$ , we can note that $x \div 3 = 3x \div 9$ . Thus, we can divide the product by $9$ :

$90 \div 9 = \boxed {\textbf {(B) } 10}$

~anabel.disher

2013 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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CEMC Gauss (Grade 8)

Revision as of 13:27, 18 September 2025 (view source) Anabel.disher (talk \| contribs) (Created page with "== Problem== Jarek multiplies a number by <math>3</math> and gets an answer of <math>90</math>. If instead, he divides the number by <math>3</math>, what answer does he get?...")		Latest revision as of 20:47, 18 October 2025 (view source) Anabel.disher (talk \| contribs)
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	~anabel.disher		~anabel.disher
		+	{{CEMC box\|year=2013\|competition=Gauss (Grade 8)\|num-b=4\|num-a=6}}