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bennett2968
30.05.2021 •
Mathematics
There are 20 questions in a quiz. Each correct answer scores 7 points, each wrong answer scores -4 points and each question left blank scores 0 points. Eric took the quiz and scored 100 points. How many questions did he left blank?.
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Ответ:
He answered 16 questions correctly, and got 3 wrong answers.
Explanation:
Let
x = number of questions that are correct y = number of wrong answers z = number of questions left blankx,y,z are nonnegative whole numbers.
Since there are 20 questions total, this means the first equation to set up is:
x+y+z = 20
Solving for y leads to
x+y+z = 20
y+z = 20-x
y = 20-x-z
We'll use this later.
Another equation to set up is 7x-4y = 100 because Eric earns 7 points per correct answer and loses 4 points for each incorrect answer, and all that leads to 100 points total which was his quiz score. We'll ignore the questions he left blank since they add 0 points.
Let's plug the equation in which we isolated y
7x-4y = 100
7x-4(20-x-z) = 100
7x-80+4x+4z = 100
7x+4x+4z = 100+80
11x+4z = 180
Now we can guess and check to see which pair of x and z values will make that last equation above true. I suggest starting with the smallest possible value of x and using that x value to solve for z.
If x = 0, then,
11x+4z = 180
11(0)+4z = 180
4z = 180
z = 180/4
z = 45
So if Eric got 0 correct answers, then he left 45 questions blank. But that isn't possible because there are only 20 questions total. So we'll ignore the case that x = 0.
If we increase x by 4, and decrease z by 11, then we get another ordered pair solution to this equation
So another solution is (x,z) = (4,34)
Note that
11x+4y = 180
11(4) + 4(34) = 180
But like before, z = 34 isn't possible since 20 is the max.
Increase x by 4 again, and drop z by 11 to get (x,z) = (8,23). Again we run into the same issue as before.
Increase x by 4 again, and decrease z by 11 to get (x,z) = (12, 12). Now we have both x and z smaller than 20, but note how x+z = 12+12 = 24 which exceeds the total number of questions. So we rule this case out as well.
Do another round of "increase x by 4, decrease z by 11" to get to (x,z) = (16, 1). This is the only case left because anything beyond this, z will be negative.
Luckily, this final case does work. If Eric answers x = 16 questions correctly, then he left z = 1 of them blank. That must mean y = 20-x-z = 20-16-1 = 3 questions were incorrect.
We can see that:
7x-4y = 7(16)-4(3) = 112-12= 100
meaning that (x,y) = (16,3) is a solution to 7x-4y = 100.
To summarize, we found that the only possible solution is (x,y,z) = (16, 3, 1)
Meaning x = 16 questions were correct, y = 3 were wrong, and z = 1 question was left blank.
Ответ: