2009 Grade 8 CEMC Gauss Problems/Problem 10

Problem

If two numbers differ by $2$ and their sum is $20$ , the larger number is

$\text{ (A) }\ 11 \qquad\text{ (B) }\ 10 \qquad\text{ (C) }\ 9 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 8$

Let $x$ be the larger number. Since the numbers differ by $2$ , the smaller number must be $x - 2$ . Thus, we can express the sum in terms of $x$ :

$x + x - 2 = 20$

$2x - 2 = 20$

$2x = 22$

$x = \boxed {\textbf {(A) } 11}$

~anabel.disher

Let $x$ be the smaller number. We can use similar logic to solution 1 to conclude that $x + 2$ is the larger number, and we can find that $x = 9$ .

The larger number is therefore $x + 2 = 9 + 2 = \boxed {\textbf {(A) } 11}$ .

~anabel.disher

We can notice that the number must be greater than $10$ because $\frac{20}{2} = 10$ , as the number wouldn't be the larger number otherwise.

Trying answer choice A, we have:

$20 - 11 = 9$ for the other number

These differ by $2$ , so the answer is $\boxed {\textbf {(A) } 11}$

~anabel.disher