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2015 CEMC Gauss (Grade 8) Problems/Problem 15

Revision as of 18:52, 13 October 2025 by Anabel.disher (talk | contribs) (Created page with "== Problem== Two puppies, Walter and Stanley, are growing at different but constant rates. Walter’s mass is <math>12</math> kg and he is growing at a rate of <math>2</math>...")
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Problem

Two puppies, Walter and Stanley, are growing at different but constant rates. Walter’s mass is $12$ kg and he is growing at a rate of $2$ kg/month. Stanley’s mass is $6$ kg and he is growing at a rate of $2.5$ kg/month. What will Stanley’s mass be when it is equal to Walter’s?

$\textbf{(A)}\ 24 \text { kg} \qquad\textbf{(B)}\ 28 \text { kg} \qquad\textbf{(C)}\ 32 \text { kg} \qquad\textbf{(D)}\ 36 \text { kg} \qquad\textbf{(E)}\ 42 \text { kg}$

Solution 1

We can create functions. Let $s(x)$ represent Stanley's mass after $x$ months and $w(x)$ represent Walter's mass after $x$ months.

Since Stanley is originally $6$ kg and is growing at a rate of $2.5$ kg per month, we have:

$s(x) = 2.5x + 6$

$w(x) = 2x + 12$

We can now see at what $x$ value these intersect at by setting the equations equal to each other:

$s(x) = w(x)$

$2.5x + 6 = 2x + 12$

$0.5x = 6$

$x = 6 \times 2 = 12 \text{ months}$

We can now plug-in $x = 12$ into one of the equations (either one would work) to see what weight Stanley would have, as both would be equal:

$s(12) = 2.5 \times 12 + 6 = 30 + 6 = \boxed {\textbf {(D) } 36 \text{ kg}}$

$w(12) = 2 \times 12 + 12 = 24 + 12 = \boxed {\textbf {(D) } 36 \text{ kg}}$

~anabel.disher