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Algebra Difficulty 2.1 Junior Prove it Canada

IMG0 In January, Rebecca measured the
temperature in Yellowknife every day at 11:00 a.m. The average of these
$31$ temperatures was $\text{-20 C}$ . The average of the temperatures from the first $21$ days was $\text{-15 C}$ . What was the average of the temperatures from the last $10$ days? McKayla runs to her grandmother's house and then runs home along the same straight road. The route from McKayla's house, $M$ , to her grandmother's house, $G$ , is on flat ground from $M$ to $H$ , and then uphill from $H$ to $G$ , as shown in the cross-section below. The distance from $M$ to $H$ to $G$ is $10$ km. (That is, $MH+HG=10$ km.)McKayla runs on flat ground at $12$ km/h, uphill at $10$ km/h, and downhill at $15$ km/h. It takes $54$ minutes for her to run from $M$ to $H$ to $G$ . Determine the number of minutes that it takes for her to run from $G$ to $H$ to $M$ .

Figure for this problem

Solution

Since the average of $31$ temperatures was $-20\degree\text{C}$ , then the sum of these $31$ temperatures was $31 \cdot (-20\degree\text{C}) = -620\degree\text{C}$ .

Since the average of $21$ of these temperatures was $-15\degree\text{C}$ , then the sum of these $21$ temperatures was $21 \cdot (-15\degree\text{C}) = -315\degree\text{C}$ .

This means that the sum of the other $10$ temperatures was $-620\degree\text{C} - (-315\degree\text{C}) = -305\degree\text{C}$ , and so the average of these other $10$ temperatures was $\dfrac{-305\degree\text{C}}{10}$ or $-30.5\degree\text{C}$ . Suppose that $MH = $x\text{}$
km} $. This means that$ HG = $(10-x)\text{}$ km}$.

Since McKayla runs on flat ground at $12 \text{ km/h}$ , then the time that it takes her to run from $M$ to $H$ is $\dfrac{x\text{ km}}{12\text{ km/h}}$ or $\dfrac{x}{12}\text{ h}$ . Since McKayla runs uphill at $10 $\text{}$
km/h}, then the time that it takes her to run from

Figure for this problem

Figure for this problem H $to$ G $is$ $\dfrac{(10-x)\text{ km}}{10\text{}}$ km/h}} $or$ $\dfrac{10-x}{10}\text{}$ h} $. Since it takes her$ 54 $minutes to run from$ M $to$ H $to$ G $, and 54 minutes is the same as$ $\dfrac{9}{10}\text{}$ h} $, then$ $\dfrac{x}{12} + \dfrac{10-x}{10} = \dfrac{9}{10}\$.$

Multiplying both sides of this equation by $120$ , we obtain $10x + 12(10-x) = 9 \cdot 12$ and so $2x = 12$ or $x = 6$ .

Therefore, to run from $G$ to $H$ to $M$ , it takes McKayla $\dfrac{4\text{ km}}{15\text{ km/h}} + \dfrac{6\text{ km}}{12\text{ km/h}} = \dfrac{16}{60}\text{ h} + \dfrac{30}{60}\text{ h} = \dfrac{46}{60}\text{ h}$ or $46$ minutes.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.