AQA MS03 w TSM EX JUN07

AQA MD01 w TSM EX JUN07

AQA MD01 w TSM EX JUN07
... the new routes This script clearly shows the new routes giving the new figures of 69 and 62, which meant that in the body of the script full marks were obtained Too many candidates will ‘work ... ‘work in their head’ and write down the best answer without any justification Mark Scheme MD01 Question 4a Student Response Commentary Candidates must know the difference between all the algorithms ... candidate knew that the exponential function had to be dealt with Many candidates were unsure as to how to proceed and used logs without realising the implications This solution showed a lack of...
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AQA MFP3 w TSM EX JUN07

AQA MFP3 w TSM EX JUN07
... the wrong answer 1 x + x In the exemplar the candidate clearly recognises 16 the need to use a law of logarithms and completes the solution within a few lines Part (c) was generally answered ... Most candidates used their answers to part (c)(i) as the limits but other values were used, not always with any justification In the exemplar, the candidate shows good examination technique by first ... incorrectly to ‘y = sin x + 3’ all marks would have been awarded (the examiners would have applied ISW for work after ‘y sec x = tan x + 3’) Mark Scheme MFP3 Question Student Response Commentary...
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AQA MM1A w TSM EX JUN07

AQA MM1A w TSM EX JUN07
... working to the required accuracy and hence ends up with a mark of zero It is worth noting that candidates are only penalised once for not working to the required accuracy, so this candidate would ... the weight of the balloon and no account is taken of the fact that it is accelerating In part (d), the candidate states the weight that had been calculated in part (b) However as the answer is ... common error Solutions of this type were awarded method mark Mark Scheme MM1A Question Student Response Commentary In part (c) of this question, a common error was to calculate an angle based on...
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AQA MM04 w TSM EX JUN07

AQA MM04 w TSM EX JUN07
... in part b) was to lose a π symbol resulting in the wrong answer Candidates showed good knowledge of suspension problems and even if errors were made in part b) full recovery marks were available ... 45 as 354 Explanations in a)ii) were good No follow through marks were available here as the question could be answered with reference to the diagram Part b) proved very discriminating with many ... “resolving the whole system” to get the answer as 500 N – failing to understand the implication of a reaction force at point C It was expected that candidates would use Newton’s third law at point...
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AQA MM05 w TSM EX JUN07

AQA MM05 w TSM EX JUN07
... Response Commentary A number of candidates ‘invented’ the answer given to part (a) This example shows a candidate who uses the extensions to be x in both strings, rather than the correct 2a ... speed which the question required The formula used, vmax = aω needed a to be the maximum displacement for the pendulum (where a = Al) rather than the maximum angular displacement which was A ... MM05 Question Student Response Commentary This example shows a candidate spending considerable time and effort instead of thinking and then using a more suitable approach as outlined in the Examiners...
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AQA MPC2 w TSM EX JUN07

AQA MPC2 w TSM EX JUN07
... question was answered well by the above average candidates but some left their answer as f(4x) which was not awarded the mark In the exemplar the candidate gives the other common wrong answer which ... MPC2 Question Student Response Commentry Part (a) was generally answered very well The only common error was in (iii) where the wrong answer ‘ x ’ was given by a minority ... Part (a) was generally answered well but some candidates failed to gain any credit due to poor examination technique In the exemplar the candidate shows good examination technique by writing ‘...
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AQA MPC3 w TSM EX JUN07

AQA MPC3 w TSM EX JUN07
... trigonometry questions if they give extra answers that are OUTSIDE of the given range However for extra answers within the range will be penalised Mark Scheme MPC3 Question 4b(ii) Student Response ... paper candidates were required to show that a given answer was true This was so that candidates could proceed with later parts of the question, BUT wrongs not make a right! Candidates were able to ... paper candidates were required to show that a given answer was true This was so that candidates could proceed with later parts of the question, BUT enough steps need to be shown as to justify...
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AQA MD01 w TSM EX JUN08

AQA MD01 w TSM EX JUN08
... diagram This’ solution’ shows a number of arrows on the diagram with no clear order shown The candidate appears to start at vertex1 but it is then unclear how the path follows on The candidate only ... alphabet! On the next line they have chosen to work with the first sublist only – again acceptable Next line working with the second subset is ok apart from their earlier mistake However they have ... candidate knowing something about odd vertices but not knowing exactly what to They have found AB, AC and AD without realising that pairs of vertices are required Again in their explanation they...
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AQA MD01 w TSM EX JUN09

AQA MD01 w TSM EX JUN09
... Although there were many fully correct responses to this question, there was a significant number who failed to write down the correct number of comparisons The number of swaps was well done, as ... correctly This is work that we would expect a student in Year 10 to be able to well It is essential that students practise drawing graphs accurately Although the line from (0, 60) to (40, 0) was an incorrect ... Commentary Candidates were given a piece of bookwork at the start of this question to help with the network given in part (b) This candidate correctly stated that there were edges in a minimum...
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AQA MD02 w TSM EX JUN08

AQA MD02 w TSM EX JUN08
... Scheme MD02 Question Student Response Commentary (a)(i) It is a good idea to explain what p represents before writing down expressions A better statement might have been that “Roseanne plays R1 with ... for Roseanne is explained in words MD02 Many candidates did not write such a statement and lost a mark (ii) Instead of using either of the two expressions used previously to show that the value ... , but what the candidate writes here, although badly worded, is understood The expected values when Collette chooses each of the columns are calculated correctly The diagram is a good example...
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AQA MD02 w TSM EX JUN09

AQA MD02 w TSM EX JUN09
... these were not required Many left these in their solution and this was not penalised The minimum of the column maxima was indicated with an arrow and further explanation showed why C1 was Colin’s ... the Simplex Method where fractions were used and the row operations were performed accurately Extra information was given regarding the pivot being used for the second iteration, which was not ... drew a wrong conclusion about the pivot Although this candidate does not use the word pivot, it is clear that the entry has been identified from the third row The row operations were clearly explained...
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AQA MFP1 w TSM EX JUN08

AQA MFP1 w TSM EX JUN08
... M15 6EX Dr Michael Cresswell, Director General Question Student Response MFP1 Commentary Most candidates answered this question well Common errors occurred in part (d), where, as in this example, ... part (b) In this x , but then after raising the x2 power by one divides by the old power rather than the new power Mark Scheme MFP1 Question MFP1 Student Response Commentary Many candidates appreciated ... equation would then become y (x+ 2) = ax (x+ 2) + b, or, Y = aX + b This script shows a typical candidate who struggled with the required algebraic multiplication by x + Mark Scheme MFP1 Question...
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AQA MFP1 w TSM EX JUN09

AQA MFP1 w TSM EX JUN09
... candidate was not sufficient to find (   ) The other common error, also shown here, was finding the product of the new roots which was 4   or 16  ; many used its value as 4( ) MFP1 Mark ... candidate shows to find when z  z  was real instead of letting the imaginary part, which was x  , equal zero, candidates looked for a far more complicated solution MFP1 Mark Scheme Question MFP1 ... Student Response Commentary In this example the candidate has just written down the equation Y = x log b + log a instead of showing the use of the two log laws and the intermediary result Y =...
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AQA MFP2 w TSM EX JUN09

AQA MFP2 w TSM EX JUN09
... particular, (as part (a) was completely correctly done by virtually all candidates) sufficient rows were written down by the candidate to show the cancellation Sometimes rows were written as 1/2(2-1) ... together with the method of showing how it was to be done Mark Scheme MFP2 Question Student Response MFP2 Commentary The candidate starts off well with ds/dx = √(1+(dy/dx)²) Many candidates started with ... candidates wrote down the answer without sufficient intermediate working In this case, the candidate went into considerable detail when evaluating (coskΘ+isinkΘ)(cosΘ+isinΘ) even to the extent of...
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AQA MFP3 w TSM EX JUN08

AQA MFP3 w TSM EX JUN08
... display in this case, without showing the working, no marks could have been awarded for method The candidate gave a wrong value for k Although no method was shown, the value given was the same as ... MFP3 Question Student Response Commentary Although this question was generally answered very well by candidates, the exemplar illustrates partial poor examination technique ... correct values for the four unknowns a, b, c and d A significant number of candidates, like the one in the exemplar, wasted time by d2 y finding an expression for which was not required in the solution...
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