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    (Original post by Chittesh14)
    No problem. I'm sorry I couldn't help much - I just forgot about this thread.


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    There is one thing I struggle the most with in maths: the equations for algebra. I was wondering if anyone could explain this question to me and also give me some advice on solving future questions like this one:

    7) In a 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 down 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.

    Now, I understand part 'ii', it's just part 'i' that I don't and struggle for every question that asks to do that! However, once I get the answer I normally understand it, except this one. The answer for part 'i' is 6c-q-25 - I just have no idea where this comes from, especially the number 6.
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    (Original post by RDKGames)
    First part for ya

    Attachment 565230


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    Hello, thanks soo much for your help!! But I don't get it still I'm sooo sorry - please could you explain it to me pleasee - I really want to get it but I can't!
    Also can you give me some advice on future questions - I mean I just don't get how you'd work it out and what you think of first as soon as you see a question like this as my brain goes completely blank and I have no idea what to do or where to even start thinking when I come across questions like these for some reason!
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    (Original post by Ak786454)
    Hello, thanks soo much for your help!! But I don't get it still I'm sooo sorry - please could you explain it to me pleasee - I really want to get it but I can't!
    Also can you give me some advice on future questions - I mean I just don't get how you'd work it out and what you think of first as soon as you see a question like this as my brain goes completely blank and I have no idea what to do or where to even start thinking when I come across questions like these for some reason!
    I'll walk you through it. So let's start with the info we are given. The first few sentences just lay out the mark gains/penalties which we can use later on.

    The candidate attempts q amount of questions, therefore he does not attempt 25-q questions as there are 25 of them in total for the candidate to attempt. There is no gain for attempting questions, however we are told there is a penalty for not attempting some questions, one point per question, therefore we can see that the candidate loses (25-q)marks.

    Next we are told the candidate gets c questions correct. If he attempts q questions and gets c correct out of those, then that means he gets q-c incorrect. Now there is a gain for getting the right answer, AND a penalty for getting the wrong answer. Since he got c correct, and you get 4 marks per correct answer, that means the candidate gains 4c marks. As he got (q-c) incorrect, and you lose 2 marks per incorrect answer, the candidate loses 2(q-c) marks.

    Now to find the amount of marks the candidate gains in total, we will begin with 0 and add on the gains, while subtracting the losses. As we found, the only gain is 4c while the two losses are (25-q) and 2(q-c).

    Therefore if you let Tm equal the total marks, you would get the following:
    T_m=0+4c-(25-q)-2(q-c)=6c-q-25

    which gives the required result!
    Of course you can ignore the 0 but I just used it to emphasise the fact that total marks start from 0 then begin to change once you start to add and subtract.

    As for future questions like these, I'd say it is important to recognise that you presented two numerical quantities; such as the marks and the questions, and you must convert between the two by means of the information given to you within the context. Once the question gives you a specific case (such as the candidate there), then use this information that there are finite amount of questions to do, so you can already form expressions such as in my working out. Remember what you need to work out, and always consider which numbers increase the wanted value, and which ones decrease it; thus add/subtract respectively. It's mostly just being able to follow the scenario they give you carefully while taking everything into account because you need to consider the opposite of what is being told in the text as well; as seen from my working out where I work with the amount of unattempted questions. It's hard to explain how to go about these questions but I'd just say think hard about the situations in order to consider everything, and it should be straightforward from there.
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    (Original post by RDKGames)
    I'll walk you through it. So let's start with the info we are given. The first few sentences just lay out the mark gains/penalties which we can use later on.

    The candidate attempts q amount of questions, therefore he does not attempt 25-q questions as there are 25 of them in total for the candidate to attempt. There is no gain for attempting questions, however we are told there is a penalty for not attempting some questions, one point per question, therefore we can see that the candidate loses (25-q)marks.

    Next we are told the candidate gets c questions correct. If he attempts q questions and gets c correct out of those, then that means he gets q-c incorrect. Now there is a gain for getting the right answer, AND a penalty for getting the wrong answer. Since he got c correct, and you get 4 marks per correct answer, that means the candidate gains 4c marks. As he got (q-c) incorrect, and you lose 2 marks per incorrect answer, the candidate loses 2(q-c) marks.

    Now to find the amount of marks the candidate gains in total, we will begin with 0 and add on the gains, while subtracting the losses. As we found, the only gain is 4c while the two losses are (25-q) and 2(q-c).

    Therefore if you let Tm equal the total marks, you would get the following:
    T_m=0+4c-(25-q)-2(q-c)=6c-q-25

    which gives the required result!
    Of course you can ignore the 0 but I just used it to emphasise the fact that total marks start from 0 then begin to change once you start to add and subtract.

    As for future questions like these, I'd say it is important to recognise that you presented two numerical quantities; such as the marks and the questions, and you must convert between the two by means of the information given to you within the context. Once the question gives you a specific case (such as the candidate there), then use this information that there are finite amount of questions to do, so you can already form expressions such as in my working out. Remember what you need to work out, and always consider which numbers increase the wanted value, and which ones decrease it; thus add/subtract respectively. It's mostly just being able to follow the scenario they give you carefully while taking everything into account because you need to consider the opposite of what is being told in the text as well; as seen from my working out where I work with the amount of unattempted questions. It's hard to explain how to go about these questions but I'd just say think hard about the situations in order to consider everything, and it should be straightforward from there.
    That made soo much more sense, thanks for all your help, it's really appreciated! But it really is just soo difficult to think about it all and I also find it really hard to work out where I should start! I really do hope it just comes with practice though
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    Helloo everyone can anyone help me with these questions (the topic is called changing the subject of an equation):

    1) Make b the subject of the formula: h= *Square root everything here* a^2 + b^2.

    The answer for this is b = +/- *square root everything here* h^2 - a^2, however I got this answer, but without the +/- sign, would anyone be able to explain where this +/- sign comes from by any chance?

    2) Make R the subject of the formula: 1/R = 1/R1 + 1/R2.

    The answer is R = R1R2 / R1+R2, however I got 1 / 1/R1 + 1/R2, how do you get to that answer?

    3) Make u the subject of the formula: f = uv/u+v.

    The answer is u = fv/v-f, however I got -fv/f-v, I have no idea why there isn't '-' sign in the answer as I worked it out and I got a '-' sign.
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    (Original post by Ak786454)
    Helloo everyone can anyone help me with these questions (the topic is called changing the subject of an equation):

    1) Make b the subject of the formula: h= *Square root everything here* a^2 + b^2.

    The answer for this is b = +/- *square root everything here* h^2 - a^2, however I got this answer, but without the +/- sign, would anyone be able to explain where this +/- sign comes from by any chance?

    2) Make R the subject of the formula: 1/R = 1/R1 + 1/R2.

    The answer is R = R1R2 / R1+R2, however I got 1 / 1/R1 + 1/R2, how do you get to that answer?

    3) Make u the subject of the formula: f = uv/u+v.

    The answer is u = fv/v-f, however I got -fv/f-v, I have no idea why there isn't '-' sign in the answer as I worked it out and I got a '-' sign.
    For question 1, when you have an equation of the form x^2 = some number, x could be the positive or negative square root. For example, x^2 = 4 -> x = +/- sqrt 4 = +/- 2, since squaring a negative number gives the same result as squaring the positive (remember minus times minus makes plus!).

    For question 2, Multiply the numerator and denominator of your answer by R1*R2 to see that it is equivalent to the required result. Remember that if you multiply/divide both the numerator and denominator of a fraction by the same thing, the value of the fraction stays the same.

    For question 3, just multiply the numerator ad denominator of your result by -1 to again see that it is equivalent.
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    (Original post by Ak786454)
    Helloo everyone can anyone help me with these questions (the topic is called changing the subject of an equation):

    1) Make b the subject of the formula: h= *Square root everything here* a^2 + b^2.

    The answer for this is b = +/- *square root everything here* h^2 - a^2, however I got this answer, but without the +/- sign, would anyone be able to explain where this +/- sign comes from by any chance?

    2) Make R the subject of the formula: 1/R = 1/R1 + 1/R2.

    The answer is R = R1R2 / R1+R2, however I got 1 / 1/R1 + 1/R2, how do you get to that answer?

    3) Make u the subject of the formula: f = uv/u+v.

    The answer is u = fv/v-f, however I got -fv/f-v, I have no idea why there isn't '-' sign in the answer as I worked it out and I got a '-' sign.
    1.) When you square root something, the answer is always \pm of whatever you get because if you take the positive and square it, obviously you will get the positive, but if you square the negative, then you will ALSO get the positive due to the 2 negatives multiplying by each other. Hence there are 2 solutions for square roots.

    Example: \sqrt4=\pm2 because 2^2=4 and (-2)^2

    2.) Your answer is exactly the same. On the denominator, from the two fractions you can make one fraction under the same denominator by the following rule: \frac{1}{a}+\frac{1}{b}=\frac{b}  {ab}+\frac{a}{ba}=\frac{a+b}{ab}

    Once you have one fraction as the denominator, you can multiply the top and bottom of the overall fraction by the denominator of the smaller fraction if that makes sense: \frac{1}{\frac{a+b}{ab}}=\frac{a  b}{a+b} and this should help you arrive at the answer they want.

    3.) Again, exactly the same answer. Multiply the top and bottom by -1.
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    (Original post by RDKGames)
    1.) When you square root something, the answer is always \pm of whatever you get because if you take the positive and square it, obviously you will get the positive, but if you square the negative, then you will ALSO get the positive due to the 2 negatives multiplying by each other. Hence there are 2 solutions for square roots.

    Example: \sqrt4=\pm2 because 2^2=4 and (-2)^2
    No, this is plainly wrong. \sqrt{4} = +2 only since \sqrt{.} \, \colon \, \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}. Hapax gives the correct explanation.
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    (Original post by Zacken)
    No, this is plainly wrong. \sqrt{4} = +2 only since \sqrt{.} \, \colon \, \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}. Hapax gives the correct explanation.
    To me it seems we have literally written the same thing. Maybe I just haven't explained it "by the book"
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    (Original post by RDKGames)
    To me it seems we have literally written the same thing. Maybe I just haven't explained it "by the book"
    sqrt(4) = 2, but x^2 = 4 -> x = +/- sqrt(4) = +/- 2.
    If what you said in your post was correct, the +/- wouldn't be needed in the quadratic formula as the square root would already include it.
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    (Original post by HapaxOromenon3)
    sqrt(4) = 2, but x^2 = 4 -> x = +/- sqrt(4) = +/- 2.
    If what you said in your post was correct, the +/- wouldn't be needed in the quadratic formula as the square root would already include it.
    Ah I see. Took me a while to notice what the hell you guys were on about. That's not what I was intending in my explanation.
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    (Original post by miaofcourse)
    How do you factorise 2x^2 - 3xI understand you factorise by taking out a common factor, will the factor be X making it x^2 - 2x?? Or am I just being silly, I feel as if it is a simple question but I am going through a mind block! And I'm guessing it's not as simple as x(2x-3)??
    since it's in the form of ax^2 + bx and not ax^2 +bx + c (i.e. no c term) simply take out the common factor 'x', so you are correct.
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    (Original post by RDKGames)
    Ah I see. Took me a while to notice what the hell you guys were on about. That's not what I was intending in my explanation.
    For further information see http://www.thestudentroom.co.uk/show...5#post64637655
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    (Original post by BasharAssad)
    since it's in the form of ax^2 + bx and not ax^2 +bx + c (i.e. no c term) simply take out the common factor 'x', so you are correct.
    His original query was already answered some time ago, so there's no need for you to post a reply to it at this point.
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    (Original post by HapaxOromenon3)
    His original query was already answered some time ago, so there's no need for you to post a reply to it at this point.
    that's fine.
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    (Original post by RDKGames)
    To me it seems we have literally written the same thing. Maybe I just haven't explained it "by the book"
    No, again: \sqrt{4} \neq \pm 2 as you seem to think. That is plainly wrong.
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    (Original post by RDKGames)
    1.) When you square root something, the answer is always \pm of whatever you get because if you take the positive and square it, obviously you will get the positive, but if you square the negative, then you will ALSO get the positive due to the 2 negatives multiplying by each other. Hence there are 2 solutions for square roots.

    Example: \sqrt4=\pm2 because 2^2=4 and (-2)^2

    2.) Your answer is exactly the same. On the denominator, from the two fractions you can make one fraction under the same denominator by the following rule: \frac{1}{a}+\frac{1}{b}=\frac{b}  {ab}+\frac{a}{ba}=\frac{a+b}{ab}

    Once you have one fraction as the denominator, you can multiply the top and bottom of the overall fraction by the denominator of the smaller fraction if that makes sense: \frac{1}{\frac{a+b}{ab}}=\frac{a  b}{a+b} and this should help you arrive at the answer they want.

    3.) Again, exactly the same answer. Multiply the top and bottom by -1.
    Thanks soo much for you reply, it really is appreciated!! That kind of makes sense now, I just got a few more questions on these:

    1) Firstly, am I right by saying that I'm supposed to add the '+/-' sign when I square root the 'b^2' in order to remove the '^2' and get it onto the other side?

    2) Secondly, I'm lost on this question, when you say '
    Once you have one fraction as the denominator, you can multiply the top and bottom of the overall fraction by the denominator of the smaller fraction if that makes sense: \frac{1}{\frac{a+b}{ab}}=\frac{a  b}{a+b} and this should help you arrive at the answer they want.'; so far I have, 1 / R2 + R1 / R1R2 - is it possible for you to continue this working out following that part I never understood, by any chance?

    3) For this question, why would you multiply the top and the bottom by '-1'? Also, if they did this, then shouldn't there be a '-' sign on the bottom part as well as the top part?
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    (Original post by HapaxOromenon3)
    For question 1, when you have an equation of the form x^2 = some number, x could be the positive or negative square root. For example, x^2 = 4 -> x = +/- sqrt 4 = +/- 2, since squaring a negative number gives the same result as squaring the positive (remember minus times minus makes plus!).

    For question 2, Multiply the numerator and denominator of your answer by R1*R2 to see that it is equivalent to the required result. Remember that if you multiply/divide both the numerator and denominator of a fraction by the same thing, the value of the fraction stays the same.

    For question 3, just multiply the numerator ad denominator of your result by -1 to again see that it is equivalent.
    Thanks for your help too
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    (Original post by Ak786454)
    Thanks soo much for you reply, it really is appreciated!! That kind of makes sense now, I just got a few more questions on these:

    1) Firstly, am I right by saying that I'm supposed to add the '+/-' sign when I square root the 'b^2' in order to remove the '^2' and get it onto the other side?

    2) Secondly, I'm lost on this question, when you say '
    Once you have one fraction as the denominator, you can multiply the top and bottom of the overall fraction by the denominator of the smaller fraction if that makes sense: \frac{1}{\frac{a+b}{ab}}=\frac{a  b}{a+b} and this should help you arrive at the answer they want.'; so far I have, 1 / R2 + R1 / R1R2 - is it possible for you to continue this working out following that part I never understood, by any chance?

    3) For this question, why would you multiply the top and the bottom by '-1'? Also, if they did this, then shouldn't there be a '-' sign on the bottom part as well as the top part?
    1) Yes. b^2=a \rightarrow b=\pm\sqrt{a} (where a is whatever you want) but I'm unsure what you mean by "get it onto the other side"? If you square root b, then you square root the other side too, you don't move anything across.

    2) - We have: \frac{1}{\frac{R_1+R_2}{R_1R_2}}
    - Multiply top and bottom by R_1R_2 which gives: \frac{1\cdot R_1R_2}{\frac{R_1+R_2}{R_1R_2} \cdot R_1R_2}
    - Things cancel on the denominator and we are left with: \frac{R_1R_2}{R_1+R_2}

    3) Technically, you can just leave your answer as you found it firstly, but the answer in your book multiplied the fraction's numerator and denominator by -1 because it get rids of the negative on then numerator thus making it look simpler in a way.

    You found it to be: \frac{-fv}{f-v}
    Multiplying top and bottom by -1 doesn't change the value because: \frac{-fv}{f-v} \cdot 1 = \frac{-fv}{f-v} \cdot \frac{-1}{-1} = \frac{(-1)(-fv)}{(-1)(f-v)} = \frac{fv}{v-f}

    The tricks for 2) and 3) are literally the same, I hope you can see how this is helpful in achieving the answers in your book. If you multiply the numerator and denominator by the same factor, the quantity is unchanged because that very factor would cancel out anyway, however rather than cancelling it out, you can multiply across with other terms on the numerator and denominator.
 
 
 
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