Difference between revisions of "2011 CEMC Gauss (Grade 8) Problems/Problem 6"

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{{CEMC box|year=2011|competition=Gauss (Grade 8)|num-b=5|num-a=7}}

Latest revision as of 15:01, 19 October 2025

Problem

If Clara doubles a number and then adds $3$, the result is $23$. The original number is

$\text{ (A) }\ 13 \qquad\text{ (B) }\ 10 \qquad\text{ (C) }\ 49 \qquad\text{ (D) }\ 17 \qquad\text{ (E) }\ 20$

Solution 1

Let $n$ be the original number that Clara chose. Setting up an equation, we get:

$2n + 3 = 23$

Solving for n gives:

$2n = 20$

$n = \boxed {\textbf {(B) } 10}$

~anabel.disher

Solution 2

We can work backwards, and subtract $3$ from the final result, and then halve the result to get the original number. This gives:

$\frac{23 - 3}{2} = \frac{20}{2} = \boxed {\textbf {(B) } 10}$

~anabel.disher

2011 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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CEMC Gauss (Grade 8)