[stateofdreaming]

Ah, I'm doing OCR MEI AS Pure Mathematics and we recently started but I don't understand how to do the basic algebra worded questions. I'm fine with algebra but it's difficult to do the worded questions and put them into formulas and expressions.

e.g. In multiple-choice examination of 25 questions, four marks are given for each correct answer and two marks are deducted for each wrong answer. One mark is deducted for any question which is not attempted. A candidate attempts q questions and gets c correct.

i) write an expression for the candidate's total mark in terms of q and c.
ii) James attempts 22 questions and scores 55 marks. Write down and solve an equation for the number of questions which James gets right.

Ah, I know the answers but I honestly don't understand how to make an expression and solve from them. Could you please help me in terms of understanding worded algebraic questions????

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[the bear]

Start off by choosing values for q & c

eg he does 20 questions and gets 14 correct... how would you find his score ?
Once you can see how to do that you can rework it putting letters in instead of the numbers

[justinawe]

(Original post by stateofdreaming)
Ah, I'm doing OCR MEI AS Pure Mathematics and we recently started but I don't understand how to do the basic algebra worded questions. I'm fine with algebra but it's difficult to do the worded questions and put them into formulas and expressions.

e.g. In multiple-choice examination of 25 questions, four marks are given for each correct answer and two marks are deducted for each wrong answer. One mark is deducted for any question which is not attempted. A candidate attempts q questions and gets c correct.

i) write an expression for the candidate's total mark in terms of q and c.
ii) James attempts 22 questions and scores 55 marks. Write down and solve an equation for the number of questions which James gets right.

Ah, I know the answers but I honestly don't understand how to make an expression and solve from them. Could you please help me in terms of understanding worded algebraic questions????

I'll try and explain this as clearly as I can.

So, you get 4 marks for a correct answer, lose 2 marks for a wrong answer and lose 1 mark for an unattempted question, yes? To find the total mark, you would take the number of correct answers multiplied by 4 and subtract the number of wrong answers multiplied by 2 as well as the number of unattempted questions.

The question says that a candidate gets number of questions correct. So, to find out how many marks the candidate gets from correct answers, we multiply this by 4, giving .

Now, we know that the candidate attempts questions and gets correct. If we subtract the number of correctly answered questions from the total number of attempted question, we would get the number of questions the candidate answered wrong. So, we can say that the candidate got questions wrong. To find out how many marks were deducted due to wrong answers, we multiply this by 2. Giving .

There are 25 questions in the examination, the candidate attempted questions. So, there would be unattempted questions, and also marks deducted as it's 1 mark deducted for every unattempted question.

Subtracting and from gives us:



Simplify this, and you have your expression!

Since you say you're fine with algebra, I assume you'll be able to do part (ii) on your own now. Quote me of you've got any problems.

[WaterWheel]

You need a tutor to go through this with you. Find a tutor. You are looking for the general method of converting words into numbers , that is what the tutor needs to talk to you about. Then any worded question will become easy.

[danny111]

(Original post by stateofdreaming)
Ah, I'm doing OCR MEI AS Pure Mathematics and we recently started but I don't understand how to do the basic algebra worded questions. I'm fine with algebra but it's difficult to do the worded questions and put them into formulas and expressions.

e.g. In multiple-choice examination of 25 questions, four marks are given for each correct answer and two marks are deducted for each wrong answer. One mark is deducted for any question which is not attempted. A candidate attempts q questions and gets c correct.

i) write an expression for the candidate's total mark in terms of q and c.
ii) James attempts 22 questions and scores 55 marks. Write down and solve an equation for the number of questions which James gets right.

Ah, I know the answers but I honestly don't understand how to make an expression and solve from them. Could you please help me in terms of understanding worded algebraic questions????

Start by writing down everything you have been told:

25 questions
correct = 4 marks
incorrect = -2 marks
not attempted = -1 marks
q attempted
i.e. 25-q not attempted => q-25 marks or if you want -(25-q)
c correct => 4c marks
i.e. q-c incorrect => -2(q-c) marks

hence marks are 6c - q -25

[the bear]

(Original post by justinawe)
I'll try and explain this as clearly as I can.

So, you get 4 marks for a correct answer and you lose 2 marks for a wrong answer, yes? To find the total mark, you would take the number of correct answers multiplied by 4 and subtract the number of wrong answers multiplied by 2.

The question says that a candidate gets number of questions correct. So, to find out how many marks the candidate gets from correct answers, we multiply this by 4, giving .

Now, we know that the candidate attempts questions and gets correct. If we subtract the number of correctly answered questions from the total number of attempted question, we would get the number of questions the candidate answered wrong. So, we can say that the candidate got questions wrong. To find out how many marks were deducted due to wrong answers, we multiply this by 2. Giving .

Subtracting from gives us:



There's your expression!

Since you say you're fine with algebra, I assume you'll be able to do part (ii) on your own now. Quote me of you've got any problems.

You missed out something... they deduct marks for non-attempted questions

[Jack--]

(Original post by the bear)
You missed out something... they deduct marks for non-attempted questions

It only asks for it in terms of c and q though, that's just trying to throw you of the trail (I think), you also do t need the number of non attempted questions for part 2.


This was posted from The Student Room's iPhone 4s app.

[danny111]

(Original post by Jack--)
It only asks for it in terms of c and q though, that's just trying to throw you of the trail (I think), you also do t need the number of non attempted questions for part 2.


This was posted from The Student Room's iPhone 4s app.

25-q not attempted means -(25-q) points. answer in terms of q.

[justinawe]

(Original post by the bear)
You missed out something... they deduct marks for non-attempted questions

yeah, you're right, I probably read the OP too quickly. I'll edit the post.

[stateofdreaming]

(Original post by justinawe)
I'll try and explain this as clearly as I can.

So, you get 4 marks for a correct answer, lose 2 marks for a wrong answer and lose 1 mark for an unattempted question, yes? To find the total mark, you would take the number of correct answers multiplied by 4 and subtract the number of wrong answers multiplied by 2 as well as the number of unattempted questions.

The question says that a candidate gets number of questions correct. So, to find out how many marks the candidate gets from correct answers, we multiply this by 4, giving .

Now, we know that the candidate attempts questions and gets correct. If we subtract the number of correctly answered questions from the total number of attempted question, we would get the number of questions the candidate answered wrong. So, we can say that the candidate got questions wrong. To find out how many marks were deducted due to wrong answers, we multiply this by 2. Giving .

There are 25 questions in the examination, the candidate attempted questions. So, there would be unattempted questions, and also marks deducted as it's 1 mark deducted for every unattempted question.

Subtracting and from gives us:



Simplify this, and you have your expression!

Since you say you're fine with algebra, I assume you'll be able to do part (ii) on your own now. Quote me of you've got any problems.

Thank you! That was a perfect explanation but I now understand that I was writing in the wrong number, hence why my answers were wrong but thank you! Also, I went through this with a teacher so it's perfect, once again, thanks!