The Student Room Group

Maths - Simultaneous Equations

In a maths challenge there are 45 questions
A correct answer is awarded 5 points and incorrect answer is awarded -2 points.
Ben answered all questions and is awarded 113 points in the competition.
Use these equations to find how many questions ben answered correctly.

Help please!
Set up an equation yourself using the information given, then start to eliminate or substitute like a normal simultaneous equation or so.
Reply 2
Iv got 5x+(-2y)=113 but I can't think of a second?
Original post by ross2000
Iv got 5x+(-2y)=113 but I can't think of a second?


Have you do considered making an equation on the total amount of questions?
Reply 4
Original post by ross2000
Iv got 5x+(-2y)=113 but I can't think of a second?


So if we take x as being the no of correct questions and y being the number of wrong questions and there are 45 questions in total. Could you could up with an equations to express that?
Reply 5
Original post by Slowbro93
Have you do considered making an equation on the total amount of questions?


I have, just no idea which numbers to use for this question or the next 3...
Reply 6
Original post by KaylaB
So if we take x as being the no of correct questions and y being the number of wrong questions and there are 45 questions in total. Could you could up with an equations to express that?


Thanks so much, I got 29 correct 16 wrong which seems right.

Now onto the last 3...
Reply 7
Original post by ross2000
Thanks so much, I got 29 correct 16 wrong which seems right.

Now onto the last 3...


No problem, happy to help:h:
Reply 8
How could I put (12,6) with y intercept of 2 into an equation to prove its equation (straight line)

Very confused about this :frown:
Reply 9
Original post by ross2000
How could I put (12,6) with y intercept of 2 into an equation to prove its equation (straight line)

Very confused about this :frown:


If you know the y intercept is 2, what would that be as a co-ordinate?
Reply 10
I love simultaneous equations :colondollar:
Don't take my word but looking at this, you could only assume the second set of coordinates are 0,2 right? If the intersection of y is at 2. So using y = mx +c start to sub in values. Y = mx + 2. Now use both coordinates to find the gradient which is given as the change in y divided by the change in x... So 6-2/12 so 4 divided by 12. And then plug that into the MX part so it would be y = 4/12x + 2

Hopefully someone will give you the full right answer this is just an assumption
Reply 12
Original post by KaylaB
If you know the y intercept is 2, what would that be as a co-ordinate?


I see, so I have 3y-x=6 and 3y=1/3x + 2 (where 1/3 is gradient for y=mx+c)

Think i've got it...
Reply 13
Original post by ross2000
I see, so I have 3y-x=6 and 3y=1/3x + 2 (where 1/3 is gradient for y=mx+c)

Think i've got it...


If you mean 3y-x=6 and y=1/3x +2 then yes, there both the same.
Reply 14
Original post by KaylaB
If you mean 3y-x=6 and y=1/3x +2 then yes, there both the same.


Thanks! It's far too late for maths :u:
Reply 15
Original post by ross2000
Thanks! It's far too late for maths :u:


No problem! But it's never too late for maths!:wink:

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