A quiz has twenty questions with 7 points awarded for each correct answer, 2 points deducted for each wrong answer and 0 for each question omitted. Jack scored 87 points. How many questions did he omit? My math teacher gave us this as an optional challenge problem for 8th grade math. I got stuck on it, so if someone could give me some hints or at least tell me what I should be doing, I’d appreciate it. Like how the heck do I know which ones are correct vs which ones are omitted? Is it a multiple answer problem?

(he got 2 wrong answers, and 13 correct answers)

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Explanation:

c = number of correct answers

w = number of wrong answers

m = number of missing answers (ie, omitted answers)

7c = number of points awarded from just the correct answers

-2w = number of points awarded from the wrong answers. Note the negative to indicate we lose points here.

0m = 0 = points from the missing answers. We don't have to worry about this portion

In total, Jack would earn 7c-2w points. Since he actually earned 87 points, we can write the equation 7c-2w = 87. We'll come back to this later.

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The quiz has 20 questions, so c+w+m = 20 is another equation we'll be working with. Solve for c to get c = -w-m+20. Plug this into the equation 7c-2w = 87 to get...

7c-2w = 87

7(-w-m+20)-2w = 87 ... replace c with (-w-m+20)

-7w-7m+140-2w = 87 ... distribute

-9w-7m+140 = 87

-9w-7m = 87-140

-9w-7m = -53

9w+7m = 53

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From here, you use trial and error. We don't have enough information to finish solving the system. We would need 3 equations to solve for the 3 unknowns.

What we can do is make a guess of what w could be and use that to find m. Luckily there is a small list of possible values for w.

So let's say w = 0

9w+7m = 53

9(0)+7m = 53 ... plug in w = 0

7m = 53

m = 53/7 ... divide both sides by 7

m = 7.57

We got a non-whole number, so there is no way w = 0 since we can't have 0.57 of a question missing. Jack either answers the question, or he skips it. He can't have both happen.

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Let's repeat for w = 1

9w+7m = 53

9(1)+7m = 53 ... replace w with 1

9+7m = 53

7m = 53-9 ... subtract 9 from both sides

7m = 44

m = 44/7 ... divide both sides by 7

m = 6.28

So we can see that w = 1 isn't possible either.

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However, w = 2 is possible

9w+7m = 53

9(2)+7m = 53

18+7m = 53

7m = 53-18

7m = 35

m = 35/7

m = 5

Meaning that Jack got w = 2 wrong answers and skipped over m = 5 questions. The number of correct answers is c = 20-m-w = 20-5-2 = 13

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An alternative to using trial-and-error is to use a graph.

Replace w and m with x and y respectively. So we go from 9w+7m = 53 to 9x+7y = 53

Use a graphing tool, or plot the graph on graph paper, to get what you see in the diagram below. I used the program GeoGebra to graph.

Then notice how the graph goes through the point (2,5) which means x = 2 and y = 5. Replace x and y with w and m to find (w,m) = (2,5)

note: the point (5,1) looks like it also lies on the graph, but it is not a solution. It's just a point that is very close to the line. I recommend you try out (w,m) = (5,1) and you'll see it doesn't work.

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.

just took the test and my answers down below!

step-by-step explanation:

check the pic :