Teacher Support Materials 2008 Maths FSMQ Paper Reference 6988/2 Copyright 2008 AQA and its licensors. All rights reserved. Permission to reproduce all copyrighted material has been applied for. In some cases, efforts to contact copyright holders have been unsuccessful and AQA will be happy to rectify any omissions if notified. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester 5 6EX. Dr Michael Cresswell, Director General.
Question 1 6988/2
Student Response 6988/2
6988/2 Commentary In part (a), nearly all candidates successfully plotted the data onto the distance/time graph. The line of best fit was accurately drawn by most candidates in part (b). In part (c), candidates attempted to find the gradient taking a suitable vertical and horizontal reading from their line of best fit but often overlooked the fact that the line s gradient was negative. Less than half the candidates used their gradient to obtain the formula for d in terms of t using their knowledge of the general equation for a straight line. Only the more able candidates realised that the solution to part (d) could be found by substituting d = 0 into their equation in part (c). Many candidates attempted to extend their line off the graph to find the number of days when the lava reached the village. A few made use of the gradient to find the answer but this often lacked the appropriate degree of accuracy. In part (e), very few candidates appreciated the link between the gradient of the line and the speed of the lava and made a fresh attempt to calculate the speed. A common error was to simply take two values from the table and divide a distance value by the corresponding time. The candidate in this illustration plotted the data onto the graph accurately in part (a) and drew a line of best fit in an appropriate position in part (b). In part (c) the candidate selected suitable measurements from the graph and the values used can be clearly seen on the graph. The candidate, like many others, calculated a correct value for the gradient but overlooked the fact that the gradient of the line was negative. The gadient-intercept method to find the equation of the straight line was appreciated with the candidate correctly reading the value 5.1 from their graph. In part (d) it was necessary for the candidate to realise that d = 0 had to be substituted into the equation from aprt (c) but d = 5 was used in error. The answer to part (e) was incomplete but it was obvious that the candidate had not seen the link between the gradient and the speed of the lava. Mark scheme (a) All points correctly plotted B2 B1 at least 4 correct (b) Line of best fit through their plots B1 (c) Gradient ( their vertical/ their horizontal) B2ft d = their 0.1t + their 5 B1ft B1 for positive gradient (d) Substitute d = 0 into their equation Or use their gradient 50 ( their 5/ their gradient) A1ft (e) Attempt vertical/horizontal their gradient A1ft TOTAL 10
Question 2
6988/2
Student response
6988/2
Commentary In part (a) candidates were often able to substitute the values of v and h into the given formula but then either did not know how to input these values into their calculator or failed to express a correct value into standard form. The common error in part (b)(i) was for candidates to omit the brackets when multiplying u by (h +10.3). This resulted in them incorrectly making the first step of the algebraic manipulation, stating uh + 10.3 = 10.3v. In parts (b)(ii) and (c) the majority of candidates found the manipulation of algebra linked with standard form values very difficult to come to terms with. Only the more able candidates achieved success in these parts. The candidate in this illustration correctly substituted the values for h and v into the given formula in part (a) and went onto work out the values of 10.3v and h + 10.3 accurately but then failed to obtain the correct value of u. In part (b) each step of the rearrangement was clearly shown with the correct answer given. Full marks for this part would have been gained without the final factorisation being completed. In part (d) the candidate again substituted values into the rearranged formula correctly but lost the accuracy mark as the division by 10.3 was overlooked in the final section of the calculation. The candidate showed a good understanding of a difficult rearrangement of formula in part (e) and this time also went on to complete the calculation correctly. Mark Scheme (a)(i) (b)(i) (b)(ii) (c) (10.3 2.75 10 2 ) 25 10.3 Or 8.02(4 ) 10-3 (cm³) A2 A1 if not in standard form 10.3v = u (h +10.3) u (h +10.3) 10.3 A1 OE 8.5 10 4 (30 10.3) 10.3 0.34255 seen (answer for numerator) 3.3(257...) 10 3 or 0.0033 (257...) ( cm3 ) A1 0.002 (h+10.3) = 10.3 0.006 0.002h + 0.0206 = 0.0618 0.002h = 0.0412 20.6 A1 Accept 21 TOTAL 10 0.28325 35.3
6988/2 Question 3
6988/2 Response
6988/2 Commentary Part (a) was answered well by most candidates but many failed to state the correct units for the answer and stating cm³ for the surface area was a common error. Candidates found difficulty putting their thoughts onto paper in part (b). Some good explanations were seen which expressed the link between 2πx ² and the areas of the two circular ends of the cylinder but the link between the curved surface area and 2πxy proved more demanding. Many candidates explained that from the formulae given on the data sheet, r = x and h = y could be substituted, resulting in the required conclusion. Some partial factorisation was seen in part (c) but it was disappointing to see so many candidates not understanding what was meant by the term factorise and gave expressions which bore little resemblance to the expression given in the question. In part (d) most candidates gave the correct term for the area of the circular base and many of these went onto successfully complete the expression for the total surface area in terms of π, x and y.very few candidates used linked the answers in parts (c) and (d) to the answer in part (e). Lengthy numerical calculations were often made to compare ratio of the two surface areas. The candidate in this illustration used the correct formula for the curved surface area of a cylinder in part (a) and then went on to calculate the correct answer and stated the units of the solution. In part (b) the candidate proceeded from using the general formula for the total surface area of a closed cylinder to showing how this formula could be modified to a formula for S in terms of x and y, correctly illustrating that r = x and h = y. There was full factorisation of the expression in part (c) and in part (d) the candidate showed how the total surface area could be found by linking the general formula to the expression that was required in terms of x and y. The candidate clearly saw the link between the answers in parts (b) and (d), correctly stating the conclusion in part (e) Mark Scheme 2 π 4 6 150.7(96 ) A1 cm 2 B1 πx 2 πx 2 B1 2 π x y B1 (c) 2πx( x y ) B2 (d) πx 2 πxy or π x y πx 2 πxy A1 2 times B1 TOTAL 11 (a) (b) (e) allow 151 or better B1 for partial factorisation OE
Question 4
6988/2 Student Response
Commentary It was pleasing to see many candidates correctly calculating the values for n and k correctly in parts (a) and (b) respectively. In part (c) there was only approximately one third of the candidates selecting the correct graph to show the relationship between n and x which possibly meant that a degree of guesswork was used! In this illustration it was pleasing to see the candidate showing the correct metods in parts (a) and (b) even though only one mark was available for a correct response. The correct graph, showing the relationship between n and x, was given in part (c). Mark Scheme (a) 25 B1 (b) 500 B1 (c) B B1 TOTAL 3
6988/2 Question 5
Student Response
6988/2
Commentary The correct distance between the two bus stops was found by many candidates in part (a). The vertical scale was sometimes read incorrectly with the vertical height being given as 17 metres per second rather than 16 metres per second. The link between area under the graph and distance travelled was appreciated by most candidates. In part (b), the common mistake was to calculate half the distance found in part (a) and to divide this by the maximum velocity; candidates failed to realise that all of this distance was not covered at the same speed. Very few candidates realised that it was necessary to first find the distance travelled at the maximum velocity by calculating 360 ½( 20 x 16) before dividing this answer by 16. Those who obtained an answer of 12.5 seconds did not always remember to add the time taken to travel the first 160 metres (20 seconds) to obtain the total time. The illustration shows in part (a) that the candidate appreciated that the area beneath the graph represented the distance travelled by the bus between the two bus stops and broke the trapezium into two triangles and a rectangle which enabled the correct answer to be calculated. The incorrect answer seen in part (b) was the common error made by candidates where in this example the candidate overlooked the fact that there was not a constant speed for half the distance between the bus stops. Mark Scheme (a) (b) 1 2 60 30 16 720(m) A1 360 12 20 16 200 16 12.5 A1 32.5 (sec) A1 TOTAL 6 Or 20 16 30 16 12 10 16 1 2
6988/2 Question 6
6988/2 Student Response
6988/2
Commentary Many correct expressions for the length and width of the shaded section were stated in part (a). A few candidates gave the terms as 20 - x² and 8 - x², rather than 20 2x and 8 2x. In part (b)(i) the product of the values found in part (a) were often equated to the area 124cm² but the area of the small white square was overlooked. This resulted in the quadratic equation containing the value 4x² rather than 3x². Candidates who wrote the correct opening equation sometimes found difficulty correctly multiplying the two brackets together. In part (b)(ii) most candidates attempted to solve the quadratic equation using the general formula method. Unfortunately many gave the value of -b as -56 rather than 56 which resulted in two negative answers for x. Only a very small number of candidates used factorisation to calculate the values of x. In part (b)(iii), some candidates who had the correct answers in part (b)(ii) failed to realise that the answer 18 was not appropriate in this question. Candidates who successfully found the values of x in part (b)(ii) often failed to attempt part (b)(iii). In this illustration it shows one of the few totally correct solutions to this question. In part (a) the expressions for the length and width of the shaded section were correctly stated. In part (b)(i) the candidate was able to show clearly that the the shaded section could be represented by the the area of the shaded rectangle less the area of the small square and this was equated with the area of 124 cm². The brackets were multiplied out correctly, like terms were collected and the final rearrangement resulted in the correct quadratic equation being obtained. As stated above, in part (b)(ii) most candidates used the formula method to solve the quadratic equation with mixed success. This candidate showed a good understanding of solving the quadratic equation by factorisation and correctly found the two values of x. In part (b)(iii) it was pleasing to see the candidate realising that only one of the solutions in part (b)(ii) was relevant to the diagram in this question. The correct area of the small square was then found. Mark Scheme (a) (b)(i) (b)(ii) 20 2x B1 8 2x B1 (20 2 x)(8 2 x) x 2 ( 124) 160 40 x 16 x 4 x 2 x 2 ( 124) Or 3 x 2 56 x 160 124 A1 AG, OE (3 x 2)( x 18) 0 or x (c) 2 3 4 9 2 3 or x 18 A1A1 23 or 0.44(44.) ( cm 2 ) A1 TOTAL 10 20 8 40 x 16 x 4 x x 2 56 2704 6 Condone 0.6 or better OE 2