Working Out Your Grade

Transcription

1 Working Out Your Grade Please note: these files are matched to the most recent version of our book. Don t worry you can still use the files with older versions of the book, but the answer references will be a bit different. If you re having any trouble with this, please give us a call on 000. Do a complete eam (i.e. all three papers). Use the answers and mark scheme to mark each eam paper. Use the tables below to record your marks. Paper Q Mark Q Mark 0 0 Total /0 Paper Paper Q Mark Q Mark 0 0 Total /0 Add together your marks for the three papers to give a total mark out of 0. Total Mark Paper Total + Paper Total + Paper Total Total Mark 0 Q Mark Q Mark 0 0 Total /0 Overall Percentage % (one last bit of maths for you to do) Look up your overall percentage in this table to see what grade you got. Overall Percentage Grade Important! The grade boundaries above are given as a guide only. Eam boards tinker with their boundaries each year, so any grade you get on these practice papers is no guarantee of getting that grade in the real eam but it should give you a pretty good idea. CGP 0

2 Paper A 0% of B 0. B B 0% of C 0. C A 0. (0. C) 0. C % of C [ mark] a) m represents the gradient and c represents the y-intercept The only graph with a positive gradient and a negative y-intercept is graph B [ mark] b) y is inversely proportional to has the equation y k which is shown by graph C [ mark] So... [ marks available mark for correct method, mark for digits in answer, mark for putting decimal point in correct position in answer] 0 (0 ) ( ) [ mark] 0 mm ( ) cm. cm [ mark] Round all numbers to significant figure [ mark] [ mark] 0. 0 [ mark] E F Elements of A are,,, Elements of B are,,,, ( is not in the universal set ) A B 0 [ marks available mark for identifying the elements of sets A and B, mark for identifying the elements in the intersection, mark for a completely correct Venn diagram] 0 a) a + b 0 a + b 0 [ mark] a + b 0 () + b 0 a + b b 0 a [ mark] b b [ mark] b) a b e o e o e o e o e o e o [ mark] e o [ mark] [ marks available in total as above] A B C A C B D H G Work out the fraction of the square made up by each shaded region: Region A region B Region C region D Overall fraction shaded ` j + ` j + + [ marks available mark for identifying the fraction of the square regions A and B make up, mark for identifying the fraction of the square regions C and D make up, mark for the correct answer] a) th term + ( ) so Fran is correct [ marks available marks for indicating that Fran is correct and finding that the th term is, otherwise mark for correct method to find the th term] b) A term in the sequence is the sum of the two previous terms. The net two terms are + [ mark] and + 0 [ mark] [ marks available in total as above] [ marks available mark for drawing an arc centre A with radius cm, mark for drawing the perpendicular bisector of the line BC, mark for indicating the correct region] q r The highest common factor of q and r is [ marks available marks for an answer of, otherwise mark for finding the prime factorisation of q (or for 0) and mark for finding the prime factorisation of r (or for )] ( 0 ) ( 0 ) ( ) 0 [ mark] 0. 0 [ mark] [ marks available in total as above] Convert area to m : 0 cm ( ) m [ mark] Pressure Force Area N/ m [ mark] [ marks available in total as above] CGP 0

3 Let be the value of the painting in January 0. In January 0 the painting was worth. In January 0 the painting was worth 0. (. ) 0. So the percentage decrease in value between January 0 and January 0 was ( 0.) 00 %. So the newspaper was incorrect. [ marks available marks for correct answer with fully correct eplanation, otherwise mark for saying the newspaper was incorrect with some eplanation, or for attempting to find. 0.] Shape A: Area p p 0 Shape B: Area p p 0 p p so the area of A is twice the area of B. [ marks available mark for the area of shape A, mark for the correct method to find the area of shape B, mark for the area of shape B and showing that the area of shape A is twice the area of shape B] Negative powers correspond to reciprocals and fractional powers correspond to roots. ` j ` j ` j [ mark] [ mark] ` j [ mark] [ mark] [ marks available in total as above] a) The cumulative frequency diagram starts at 0 minutes so that is the quickest possible winning time in 0. Difference: 0 minutes [ marks available mark for finding the quickest possible winning time in 0, mark for the correct answer] b) To find the median, read across at 0 and then read down: Median for 0. minutes. [ mark] The teams were faster on average in 0 than in 0 as the median was lower. [ mark] [ marks available in total as above] c) To find the upper quartile read across and down at, to find the lower quartile read across and down at : Upper quartile and lower quartile Interquartile range minutes [ mark] The times of the teams were less spread out in 0 than in 0 as the interquartile range was smaller, so times were more consistent in 0. [ mark] [ marks available in total as above] a) Read across and down at C. Temperature, T ( C) CGP Time, m (minutes) To reach C it takes minutes [ mark] b) Draw a tangent at m 0 (see graph) [ mark] Temperature, T ( C) Fall 0 C Across minutes Time, m (minutes) Gradient of tangent 0 [ mark]. C/minute So 0 minutes after the tea is made it is decreasing at a rate of. C/minute. [ mark for an answer in the range.-.] 0 Rationalise the denominator of the first term: [ mark] And simplify the second term: [ mark] So + + [ mark] Angle DBA [ mark] (alternate segment theorem) Angle OAD 0 [ mark] (a tangent and a radius are perpendicular) Obtuse angle COA angle CBA ( + ) [ mark] (angle at centre is twice angle at circumference) Angle CDA 0 0 [ mark] (opposite angles in a cyclic quadrilateral add up to 0 ) Angle DCO 0 0 [ mark] (angles in a quadrilateral add up to 0 ) [ marks available in total as above] There may be alternative ways of getting to the same answer. ( s + t ) a) t t (s + t ) t s + t [ mark] s t t t t t s [ mark] or s t [ marks available in total as above] b) s kt [ mark] s 0 and t so 0 k So k 0. [ mark] When t, s. 0 m [ mark] Therefore from part a): 0 [ mark] 0 +. [ mark] [ marks available in total as above] a) (n + )(n )(n + ) (n n + n )(n + ) [ mark] (n )(n + ) [ mark] n + n n [ mark] b) n(n + )(n + ) n(n + n + n + ) n(n + n + ) n + n + n [ mark] So n(n + )(n + ) (n + )(n )(n + ) n + n + n (n + n n ) n + (n + ) [ mark] n is an integer so (n + ) must be a multiple of. [ mark] An alternative method would be to notice that (n + ) appears as a factor of both parts and so can be taken out as a factor of the whole epression, i.e. (n + )[n(n + ) (n )(n + )]. The epression inside the square brackets can then be shown to simplify to give. The turning point of a graph written in the form y ( a) + b is at the point (a, b). The turning point is at (, ) so the curve must have the equation y ( ) + [ mark] Substituting y and k gives (k ) + [ mark] (k ) ± k [ mark] ± k So k or k [ mark] [ marks available in total as above] The equation could be solved by factorising or using the quadratic formula, but using the completed square form is simplest here.

4 Paper The upper bound is cm [ mark] Use Pythagoras theorem: a and so a [ mark] The answer could have been found by eliminating the wrong answers. E.g. [ marks available marks for drawing an isosceles trapezium, otherwise mark for drawing a shape that has two of the listed properties] Type the calculation directly into a calculator ensuring the whole of the denominator is square rooted [ mark]...0 [ mark] ( s.f.) [ marks available in total as above] The 0 at the end of the answer is important, otherwise the answer would only have significant figures and that s not enough. Probability of getting a white or a black tie [ mark] The ratio of black to white ties is :, so the probability of getting a white tie [ mark] So the probability of getting a white or a red tie , so she is correct [ mark] Ollie s epression: ( + ) ( + )( + ) [ mark] + + [ mark] Amie s epression: ( + )( + ) [ mark] So the two epressions are equivalent. a) kw [ mark] b) See diagram in part d). [ mark available for circling the point shown] c) Strong positive correlation [ mark] d) Ignore the outlier when drawing a line of best fit. Maimum speed (km/h) Maimum power (kw) Maimum speed 0 km/h (allow ± ). [ marks available mark for drawing a line of best fit (ignoring the outlier), mark for accurately reading from your graph the speed corresponding to a power of 0 kw] e) 0 kw lies outside of the range of data plotted on the scatter graph. [ mark] 0 Eterior angle of a regular nine-sided shape 0 [ mark] So angle ABC 0 0 [ mark] Angle ADC [ mark] Alternatively you could find the size of each interior angle (0 ) of the regular nine-sided shape and use this to find angle ABC. a) f() () [ mark] b) a, so a 0, a 0 [ mark] c) fg() ( + c) + c [ mark] fg(), so + c c, c [ mark] [ marks available in total as above] 0 litres of orangeade costs (.0) + (.0) 0.0 So litre of orangeade costs 0.0. litres of orangeade cost..0 [ marks available mark for using the ratio to find the cost of litres of orangeade, mark for dividing by, mark for multiplying by and mark for the correct final answer. ] There are several different methods you could use here. Any correct method with full working shown and a correct final answer would get marks. Volume of cylinder p..... cm [ mark] Mass of cylinder volume density g [ mark] Mass of cube.0... g so, volume of cube mass density cm [ mark] Side length [ mark]... cm. cm (to s.f.) [ mark] [ marks available in total as above] The single transformation is a rotation 0 clockwise, about the origin. [ marks available mark for rotation, mark for giving the angle of rotation together with the direction, mark for the centre] Scale factor for heights 0. [ mark] so scale factor for volumes..0 [ mark] Capacity of larger plant pot ml so the plant pot will hold litre ( 000 ml) of compost. [ mark] Angle ABC must be a right angle because side AC of the triangle is a diameter (angle in a semicircle).. cos [ mark] AC. So AC.0... cm [ mark] cos Circumference p.0... [ mark]... cm. cm ( s.f.) [ mark] [ marks available in total as above] a) Total takings [ mark] The possible scores for two spins of the spinner are Second Spin First Spin + CGP 0

5 So there are ways to get a or more. [ mark] The probability of getting a total score of or more is. [ mark] 00 0 winners epected [ mark] Prize money 0 0 So profit [ mark] [ marks available in total as above] There are other ways that you could have found the probability of scoring a total of or more, including systematic listing of the possible scores or a tree diagram. b) The spinner is equally likely to land in each of the sections. [ mark for any valid assumption linked to either the spinner being fair or the two spins being independent of one another] Let r 0. o Then 00r. o and 000r. o So 00r. o. o So r 00 [ marks available marks for fully correct proof, otherwise mark for finding 00r and 000r, with an attempt at subtraction] a) Account : The multiplier for an increase of.% is.0. Balance after years to the nearest penny Account : Balance after years to the nearest penny So Bayonie should invest in Account. [ marks available mark for correct method for working out the balance for Account, mark for correct method for working out the balance for Account, mark for finding correct final balances for both accounts, mark for a correct conclusion following on from calculated balances] Alternatively, you could compare.0.0 with to see that the overall percentage change is greater for Account. b) The multiplier for an increase of % is.0. So for two years investment the multiplier is.0.00 which corresponds to a.0% increase. [ mark] So 0.0% [ mark] % So 00% [ mark] So she invested 000. The question could be answered by forming and solving an equation: (.0) + 0. Factorise all the epressions. + ( + ) [ mark] ( ) ( + )( ) [ mark] ( + )( ) [ mark] a + b So ( ) + becomes a + b ( )( ) ( + )( ) ( + + ) Cancel common factors: a + b ( )( ) ( + )( ) ( + + ) a + b ( + ) a + b ( + ) a + b + a and b [ mark] [ marks available in total as above] CGP 0 a) There is a sign change in the value of + between the values 0 and 0., so a solution to + 0 lies in the interval 0 < < 0.. [ marks available mark for finding the value of the function at both 0 and 0., mark for an eplanation referring to the sign change] b) [ mark for correctly showing how to rearrange equation] c) and both round 0. to d.p. so the solution is 0. to d.p. [ marks available mark for carrying out the iteration correctly, mark for stopping when the values are equal to d.p., mark for the correct value of ] 0 a) y cos has been translated unit down. The curve is y cos [ mark] b) y cos has been translated 0 to the right. The curve is y cos( 0) [ mark] c) y cos has been reflected in the -ais. The curve is y cos [ mark] The curve could also have equation y cos( + 0 ). a) AC AB + BC b a [ mark] b) MN MA + AB + BN MA OA OM a a a [ mark] BN BD ^BC + CD h (a + b) b a [ mark] So MN a + b + (b a) b so AB and MN are parallel because MN b is a multiple of AB b. [ mark] b and b have a common factor of b so the two vectors must each be a scalar multiple of the other. + y 0 is the equation of a circle with centre (0, 0). A tangent is perpendicular to the radius at the point of contact. 0 Gradient of radius from (0, 0) to (, ) is [ mark] 0 The product of the gradients of perpendicular lines is, so gradient of tangent [ mark] and so the equation of the tangent is y + c [ mark] The tangent passes through (, ): + c so c 0 The equation of the tangent is y + 0 [ mark] [ marks available in total as above] Paper Reverse the number machine to find the input. ( + ), so the input is. [ mark] Probability of getting a number greater than. Probability that they both get a number greater than [ mark] 0... so 0. is closest. [ mark] When Natalie squares her number the final digit is, so her number must end in or. [ mark] It must be either,, or, but her number is not prime so it must be. [ mark] [ marks available in total as above]

6 The median is, so one of the unknown cards must be another. [ mark] The mean is so the sum of the cards is. The last card must be [ mark] The range is [ mark] Plan view Front elevation [ marks available mark for each correct drawing] Work back in Anna s sequence to find her first term: [ mark] So the first term of Carl s sequence is [ mark] and the first five terms of Carl s sequence are, 0,,,. His th term is. [ mark] Passes though the point (, ) Does not pass though the point (, ) Gradient equal to y y + Gradient not equal to y y y + [ marks available marks if all equations are in the correct places, otherwise mark if or lines are in the correct place] Bo B contains c + chocolates Bo C contains (c + ) c + chocolates [ mark for both epressions correct] Total number of chocolates in Boes A, B and C c + c + + c + c + [ mark] So c + 0 [ mark] c c, so the number of chocolates in Bo A is [ mark], the number in Bo B is + and the number in Bo C is. [ mark for boes B and C correct] [ marks available in total as above] 0 Area of one trapezium ( + ) [ mark]. cm [ mark] [ marks available in total as above] a) Dave s speed is 0 0 km/h [ mark] Olivia s speed is 0 + km/h So it takes her 0. hours [ mark] hours minutes [ mark] b) E.g. If Olivia drove a different distance at the same average speed, her journey time would be different [ mark]. Percentage of mortgage customers that are satisfied: 00 % [ mark] 0 Percentage of savings customers that are satisfied: 00. % [ mark] 0 So a greater percentage of mortgage customers are satisfied than savings customers. [ mark] Alternatively, you could compare decimals or consider the proportions of each type of customer that are not satisfied. a) (a ) a a [ mark] b) a 0 [ mark] c) So a b a b ( )a + b + a b [ marks available marks for the correct answer, otherwise mark if out of, a, b are correct] parts in the ratio. Angles in a triangle add to 0, so 0 parts 0 and part. So, the three angles in the triangle are ( ), ( ) and ( ) 0. One angle is 0, so the triangle is a right-angled triangle. [ marks available mark for finding the size of one part of the ratio, mark for finding the size of at least one angle in the triangle and mark for showing that one angle is 0.] a) y + y y 0 [ marks available mark for drawing each line correctly, mark for shading the correct region] b) There are pairs of integer coordinates within the region, so there are pairs of integer solutions. [ marks available mark for an answer of, mark for a correct eplanation or for listing the coordinate pairs] The integer solutions are (, ), (, ), (, ), (, ), (, ), (, ), (, ) and (, ). a) Angle AEB angle BDC 0. Angle CBD + angle ABE 0 Angle EAB + angle ABE 0 So angle CBD angle EAB. Two angles in the triangles match up, so the third angles must also match up, so triangles AEB and BDC are similar. [ marks available mark for showing that angles in the triangles match up, mark for a correct eplanation] b) AE and BD AE : BD : : [ marks available mark for finding the lengths of the two sides, mark for the correct answer] c) DC [ mark] From b) you know that corresponding sides of the two triangles are in the ratio :. : BE : CD BE : so BE 0 [ mark] The coordinates of E are (, 0) (, ). [ mark] a) The class widths are not all the same, so the manager should have plotted the frequency densities on the vertical ais. [ mark for an eplanation implying that the class widths are not all equal or mentioning that frequency densities should have been used] b) Calculate frequency densities using frequency class width Age, A years Frequency Frequency Density < A 0 0 < A < A < A < A < A < A CGP 0

7 Frequency Density Age, A years [ marks available mark for finding all frequency densities correctly, mark for bars correctly drawn, mark for vertical ais scaled and labelled] c) The ages of the members have a large range. There are a lot of people with ages much lower than the mean and a lot of people with ages much higher than the mean, so the mean is not a very typical age of the members. [ mark for a comment like these or that refers to the mean age being in a group which has a low frequency density] y a) 0 R [ marks available marks if R is in the correct position as shown above, otherwise marks if vertices of R are found correctly or mark if an image of the correct size is drawn but positioned elsewhere on the grid] Made in Europe Q Made outside Europe Total Antique 0 Not antique 0 Antique Total Made in Europe b) Probability 0 [ marks available marks for a correct answer, otherwise mark for either in the numerator or 0 in the denominator] 0 The upper bound for the radius is cm. The upper bound for the height is cm. [ mark for both upper bounds] Maimum volume of cone π cm [ mark] The lower bound for the rate that water leaves is: litres per minute cm per minute [ mark] So upper bound for time minutes [ mark] So Marion is not correct because it could take up to.0... minutes to empty. [ mark] [ marks available in total as above] Distance travelled in first 0 seconds m [ mark] Distance between 0 and 0 seconds (0 0) 00 m [ mark] Distance travelled in final 0 seconds 0. ( + ) 0 0 m [ mark] Total distance m [ mark] [ marks available in total as above] Let angle DAB. Start by working out the size of angle using the cosine rule. AB + AD BD cos AB AD + cos [ mark] 0... cos (0...) [ mark] BC can then be found using the cosine rule in triangle ABC. BC + ( cos ) [ mark] + cos BC cm ( d.p.) [ mark] [ marks available in total as above] There are alternative methods you could use to get the correct answer as long as you show your working and get the right answer you ll get full marks. Let Hannah s number be n, then Tim s number is n + Form the equation: + [ mark] n n + n + Therefore, + n n( n + ) n( n + ) n + n( n + ) (n + ) n(n + ) [ mark] n + n + n n n 0 [ mark] (n + )(n ) 0 [ mark] So n or n Hannah s number is negative so she was thinking of. [ mark] [ marks available in total as above] The quadratic formula could be used to solve the equation. [ marks available mark for putting both 0 and correctly into the table, mark for all remaining entries in table found correctly, and mark for completing the Venn diagram using and 0] 0 CGP 0