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Edexcel IAL Mathematics C34- 17th January, 2017 Watch

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    (Original post by Hamzah249)
    I didn't have time to calculate exact time with pm, but time was 22.5 hours
    EDIT: that question was confusing, what was your lambda?
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    (Original post by Hamzah249)
    EDIT: that question was confusing, what was your lambda?
    Whaat? I got 6.4 hours I can't remember what Ii got for lambda tho
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    (Original post by wenogk)
    Whaat? I got 6.4 hours I can't remember what Ii got for lambda tho
    8/15?
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    (Original post by Hamzah249)
    8/15?
    I didn't get that Do you remember the question so we can re-do it?
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    (Original post by wenogk)
    I didn't get that Do you remember the question so we can re-do it?
    Yes.

    First part stated at time 4 pm there was no ice.
    At time 6pm there was 1.5mm of ice (x). Rate of increase of the thickness of ice was constant, meaning dx/dt = k.

    Find t in terms of x.
    I got t=4x/3

    part b was dx/dt = Lambda/(2x+1)

    i) find t in terms of x and lambda
    ii) calculate lambda
    iii)calculate t when x=3.

    As far as I can recall this was the whole question
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    Were there any new ideas introduced? Or was it the same as all the previous years?
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    (Original post by Hamzah249)
    Yes.

    First part stated at time 4 pm there was no ice.
    At time 6pm there was 1.5mm of ice (x). Rate of increase of the thickness of ice was constant, meaning dx/dt = k.

    Find t in terms of x.
    I got t=4x/3

    part b was dx/dt = Lambda/(2x+1)

    i) find t in terms of x and lambda
    ii) calculate lambda
    iii)calculate t when x=3.

    As far as I can recall this was the whole question
    Wow you have a very good memory. I got 15/8 as lambda...

    integration is:
    x^2 + x = (lambda)*t +C
    C is zero after subbing in t=0 and x=0 (at time 4pm)
    then, lambda= (x^2 + x)/t and subbing in t=2 and x=1.5 (at time 6pm) we get lambda as 15/8
    then, t=(x^2 + x)/(15/8)
    subbing in x=3, t=(9+3)/(15/8)=6.4 hours
    which is equivalent to 6 hours and 24 minutes hence the time would be 10:24pm.... I guess this is correct
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    (Original post by Fadel)
    Were there any new ideas introduced? Or was it the same as all the previous years?
    None at all, infact you know how every paper has this 1 tricky question you tend to lose significant marks on? This one had no tricky question pretty basic paper over all imo.
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    I'm going to do the Exam in June.

    Any tips for me?
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    (Original post by wenogk)
    Wow you have a very good memory. I got 15/8 as lambda...

    integration is:
    x^2 + x = (lambda)*t +C
    C is zero after subbing in t=0 and x=0 (at time 4pm)
    then, lambda= (x^2 + x)/t and subbing in t=2 and x=1.5 (at time 6pm) we get lambda as 15/8
    then, t=(x^2 + x)/(15/8)
    subbing in x=3, t=(9+3)/(15/8)=6.4 hours
    which is equivalent to 6 hours and 24 minutes hence the time would be 10:24pm.... I guess this is correct
    Yeah I guess you are right, I couldn't figure out this question till the very end because part b specifically stated a "Second model" has equation etc etc so I tried to work on it with only what was given in part b.
    Tried to rush through the question and I guess I got wrong lambda :/
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    (Original post by wenogk)
    Wow you have a very good memory. I got 15/8 as lambda...

    integration is:
    x^2 + x = (lambda)*t +C
    C is zero after subbing in t=0 and x=0 (at time 4pm)
    then, lambda= (x^2 + x)/t and subbing in t=2 and x=1.5 (at time 6pm) we get lambda as 15/8
    then, t=(x^2 + x)/(15/8)
    subbing in x=3, t=(9+3)/(15/8)=6.4 hours
    which is equivalent to 6 hours and 24 minutes hence the time would be 10:24pm.... I guess this is correct
    What was your area of parallelogram? I got 36(underroot2)
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    (Original post by Hamzah249)
    Yeah I guess you are right, I couldn't figure out this question till the very end because part b specifically stated a "Second model" has equation etc etc so I tried to work on it with only what was given in part b.
    Tried to rush through the question and I guess I got wrong lambda :/
    Ok ok dw im sure you did the other questions well, I actually didnt see the 14th question so I was doing the 13th question slowly and casually and I just flipped the page and when I saw it lmao :bricks: I managed to do all the parts in time except for the last two mark one(my brain wasn't functioning).
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    (Original post by Hamzah249)
    What was your area of parallelogram? I got 36(underroot2)
    I can't remember I didn't know what to do so I found the shortest perpendicular distance from point A to the line CD and multiplied by CD legnth
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    (Original post by wenogk)
    Ok ok dw im sure you did the other questions well, I actually didnt see the 14th question so I was doing the 13th question slowly and casually and I just flipped the page and when I saw it lmao :bricks: I managed to do all the parts in time except for the last two mark one(my brain wasn't functioning).
    That parts we missed were actually only worth 3-2 marks so I think it's fine there. But there's a possibility grade boundaries will be really high.
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    (Original post by thebrahmabull)
    I had it in the bag, only after I got out of the exam hall with a smiling face thinking I am getting an a* that I realised I didn't see any vectors question, which was weird. And then a stranger told me vectors came, it was the last one. No. 14. No.14????? There was a no. 14???? Fml
    Holy crap that sucks. Every C4 or C34 paper has a vector question.
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    (Original post by wenogk)
    I can't remember I didn't know what to do so I found the shortest perpendicular distance from point A to the line CD and multiplied by CD legnth
    What was your time in logarithm question. N=300/3+(17e^-0.2t)

    I got 4.32
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    (Original post by thebrahmabull)
    I had it in the bag, only after I got out of the exam hall with a smiling face thinking I am getting an a* that I realised I didn't see any vectors question, which was weird. And then a stranger told me vectors came, it was the last one. No. 14. No.14????? There was a no. 14???? Fml
    That's a nightmare.
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    Does anyone remember how much marks that time question you were discussing had?
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    (Original post by wenogk)
    Wow you have a very good memory. I got 15/8 as lambda...

    integration is:
    x^2 + x = (lambda)*t +C
    C is zero after subbing in t=0 and x=0 (at time 4pm)
    then, lambda= (x^2 + x)/t and subbing in t=2 and x=1.5 (at time 6pm) we get lambda as 15/8
    then, t=(x^2 + x)/(15/8)
    subbing in x=3, t=(9+3)/(15/8)=6.4 hours
    which is equivalent to 6 hours and 24 minutes hence the time would be 10:24pm.... I guess this is correct
    I got the same answer, 6.4 hours. Though my method was a bit different, I substituted in t=2 and x=1.5 instead of x=0 when t=0. Can't remember my lambda value.
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    (Original post by Fasih178)
    I got the same answer, 6.4 hours. Though my method was a bit different, I substituted in t=2 and x=1.5 instead of x=0 when t=0. Can't remember my lambda value.
    If you did definite integral I suppose you'll get full marks since the question was like 2 marks.
 
 
 
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