1. ## Need Algebra help.

Hi, I'm new here just signed up. I got a problem for homework that's been troubling me so I did a quick google for math help forums and came up with this place. Any help you give me is MUCH appreciated, I also hope I posted this in the right area.

Anyway here is the question:

Peter entered a competition in which he got 5 points for every correct answer, but lost 3 points for every wrong answer. At the end of the competition he had no points at all and had answered 16 questions.

That was basically the first statement. Then we had to work out this problem:

Work out how many questions Peter got right from the 16 questions he answered.

I did this correctly, I think.

X questions wrong @ -3 points ea.
16 - X questions right @ 5 points ea.

-3x + 5 (16 - x) = 0

-3x + 80 - 5x = 0

-8x = -80

x = 10.

I'm fairly sure that I got that question right, but further down the track came this question;

What numbers of questions could you answer to end up with no points?

I thought that I could use the same formula in a way as the first question, so I used this:

Let there be N questions where 0 > n > 20
Let the number of correct answers = X @ 5 points ea.
Therefore, let the number wrong = n - x. @ -3 points ea.

5x - 3 (n - x) = 0

Now I'm a bit confused. If I keep working it out I'll get something like this:

5x - 3n - 3x = 0

2x - 3n = 0

Now I guess I should take one of these to the other side of the equation.

2x = 3n

Now I'm totally stumped. The teacher said you'd have to use some guesswork to get the answer, but I have no idea how you could do that either.

Am I missing out on something? How do you do this?

Thanks for the help.

2. Originally Posted by Beasticly
Hi, I'm new here just signed up. I got a problem for homework that's been troubling me so I did a quick google for math help forums and came up with this place. Any help you give me is MUCH appreciated, I also hope I posted this in the right area.

Anyway here is the question:

Peter entered a competition in which he got 5 points for every correct answer, but lost 3 points for every wrong answer. At the end of the competition he had no points at all and had answered 16 questions.

That was basically the first statement. Then we had to work out this problem:

Work out how many questions Peter got right from the 16 questions he answered.

I did this correctly, I think.

X questions wrong @ -3 points ea.
16 - X questions right @ 5 points ea.

-3x + 5 (16 - x) = 0

-3x + 80 - 5x = 0

-8x = -80

x = 10.

I'm fairly sure that I got that question right, but further down the track came this question;

What numbers of questions could you answer to end up with no points?

I thought that I could use the same formula in a way as the first question, so I used this:

Let there be N questions where 0 > n > 20
Let the number of correct answers = X @ 5 points ea.
Therefore, let the number wrong = n - x. @ -3 points ea.

5x - 3 (n - x) = 0

Now I'm a bit confused. If I keep working it out I'll get something like this:

5x - 3n - 3x = 0

2x - 3n = 0

Now I guess I should take one of these to the other side of the equation.

2x = 3n

Now I'm totally stumped. The teacher said you'd have to use some guesswork to get the answer, but I have no idea how you could do that either.

Am I missing out on something? How do you do this?

Thanks for the help.
Umm, you are doing great!

Let me just comment on your choosing x as the questions wrong---when the problem was asking for the questions right.
So you got x=10. Is that your answer? You did not show that the questions right are 6 actually. (16 -10 = 6).
Problem-solving will be easier if you know what you are supposed to find or what the problem asks you to find. In this particular word problem, you were not asked to find the questions wrong.
You can say you can get the asked-for questions right by way of the questions wrong. Yes, but it is awkward.
You should have let x be the questions right.
......my comment only......

--------------------------------------
For the second part of the problem, you are on the correct path.

[Except for that "Let there be N questions where 0 > n > 20" thing. What is that? You mean, 0 < n < 20, I know, but why should 20 be the upper bound for n? The problem told you that? Your teacher? You just guessed the 20? Why have a limit at all? ]

So you said 5x -3(n-x) = 0
Very good.
But your expansion of that was not good. It should have been like this:
5x -3n +3x = 0 ---------negative 3, times negative x, equals +3x.
8x -3n = 0
8x = 3n ---------not 2x = 3n. (2x=3n gives unreasonable numbers.)

Now, like your teacher suggested, proceed by doing some guessworks. Trial and error, in other words.
The idea here is to find numbers, whole numbers only, for n and x to make the equation true.

Say, you assign values for n to get x.
Wait, to make things easier, simplify further the equation into
x = 3n/8 ---------***

If n = 2, x will be 6/8---not whole number, so, not okay.
If n = 3; x = 9/8 ----not okay.
n = 4; x = 12/8 ---no.
For shortcut, get the 3n be multiple of 8,
n = 8, x = 24/8 = 3 --------whole number! [ :-) ]

"n=8, x=3"
That means when there are a total of 8 questions, 3 of those could be right. So, 5 would be wrong. Then check.
5*3 -3*5 =? 0
15 -15 =? 0
Yes, so, OK.
Therefore, 8 questions in total could get you zero points.

Another set of questions would be when n=16.
Then x = 3*16/8 = 6
Meaning, 6 correct, 10 wrong.
5*6 -3*10 =? 0
30 -30 =? 0
Yes, so, Ok.

Another set would be when n=24
Then x = 3*24/8 = 9
That would be 9 correct, and 15 wrong.
5*9 -3*15 =? 0
45 -45 =? 0
Yes, so, Ok.

Etc. (n multiple of 8)

Umm, so if 0 < n < 20, then n = 8 or 16 only. -----------answer.