2011 CEMC Gauss (Grade 8) Problems/Problem 6

Problem

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

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

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

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) (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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CEMC Gauss (Grade 8)