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A bag of 75 balls has a ratio of red balls to green balls is 2:7. If y, which is the number of red balls is graphed as a function of x, which is the number of green balls, which set of points will the line pass through? a. (0,0)and (2,7) b. (0,0) and (7,2) c. (0,75) and (2,7) d. (0,75)and (7,2)

I think the answer might be c but need clarification. I get that 2 out of every 9 balls would be red and 7 would be green. If there are 75 balls in total, there would be 16 red balls and 56 green balls. Would the equation be Number of red balls= Total number of balls- number of green balls? So that would be y=75-x? or y=9-x (if we are using the ratios?) I think this might be wrong though! I feel like it might be c, but how would this relate to the function/equation?

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  • $\begingroup$ A bag of 75 balls can't have a ratio of red balls to green balls of 2:7 (unless some of the 75 balls are neither red nor green). But here's how to think of it: if there are 7 green balls, how many red balls? That will give you one point on the graph. If there are zero green balls, then how many red? That gives you another point. What contest is this for, please? $\endgroup$ Commented May 19, 2019 at 3:27
  • $\begingroup$ This is actually a question my friend showed me and I wasn't sure how to answer it. She said it was from a contest page, but I didn't ask :) I just like solving math problems in my spare and like to seek clarification when I can't understand it fully. So if there are 7 green balls then there would be 2 red balls right? If there are 0 green balls, then wouldn't the rest of the 9 balls just be red? $\endgroup$
    – Star S
    Commented May 19, 2019 at 3:41
  • $\begingroup$ 7 green, 2 red is right. Zero green, 9 red is not in the ratio 2:7. $\endgroup$ Commented May 19, 2019 at 5:46

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You have $x$ the number of green balls, and $y$ the number of red balls. You know that $y:x=2:7$. So if you would have $2$ red balls, you would have $7$ green ones. So for $x=2$ you have $y=7$. Therefore the graph will go through $(2,7)$, so answer is either a or c. Now if you have $14$ green balls, you would have $4$ red balls, and so on. Notice that we can write the equation as $$y=\frac 27 x$$ For $x=0$, you have $y=0$.

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