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.

35-60 students are taking math.

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =