5w 3. 1MA0 Higher Tier Practice Paper 2H (Set D) Question Working Answer Mark Notes 1 (a) 5w 8 = 3(4w + 2) 5w 8 = 12w = 12w 5w 14 = 7w

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1 (a) 5w 8 = (4w + ) 5w 8 = w = w 5w 4 = 7w M for attempting to multiply both sides by as a first step (this can be implied by equations of the form 5w 8 = w +? or 5w 8 =?w + 6 i.e. the LHS must be correct M for isolating terms in w and the number terms correctly from aw + b = cw + d A cao M for (b) (x + 7)(x 7) B cao 5w 8 = 4w + M for isolating terms in w and the number terms correctly A cao (c) 4 4 x y B for x y or x y or x y (B for any two terms correct in a product eg. x 4 y n ) 4.5 4

2 cos x = 9. 6 M for cos x = 9. 6 or cos x = 0.66(6...) or cos x = x = cos 9. 6 = M for cos 9. 6 or cos 0.66(6...) or cos 0.67 A for Correct use of Pythagoras and then trigonometry, no marks until '7.55' '7.55' M for sin x = 9. 6 '7.55' or sin x = 9. 6 or tan x = 6. 4 sin '7.55' or cos x = '7.55' '7.55' M for sin 9. 6 or tan 6. 4 '7.55' sin 90 or sin '7.55' or cos A for SC B for (using rad) or (using grad)

3 5 50 = 0.5 h = 0 min = 0.46 h = 5 min or 0 (min) or 0.5(h) or 5 60 or M for 5 50 or or 5 (min) or 0.4(6)(h) or 0.4 (h) M(dep) or 0 5 A cao M for 60 5 (=.4) and 60.4 or 50 5 (= ) and 60 M(dep) 0 5 A cao

4 4 (a) M for x x x or 5 x or vol of cube = x or show vol cuboid = 0x A correct completion leading to x 0x 00 (b) x = 9 x = x = x = 4 4 x = 5 75 x = 6 56 x = x = 5. 8.(65) x = (608) x = (877) x = (464) x = 5.5.(75) x = (66) x = (9) x = () x = (79) x = (075) x = (90656) x = (09) B for a trial 5 x 6 evaluated correctly (B for any two trials evaluated correctly for positive values of x) B for a different trial 5. < x < 5.4 evaluated correctly B (dep on at least one previous B) for 5.4 Accept trials correct to the nearest whole number (rounded or truncated) if the value of x is to d.p., but correct to d.p. (rounded or truncated) if the value of x is to or more d.p. NB. Allow 00 for a trial of x =

5 5 (a) ( y8)( y) M (y ± 8)(y ± ) or y(y ) 8(y ) or y(y 8) (y 8) A cao (b)(i) t 5t = (t)( t) (t)( t) M (t + )(t + ) oe or t(t + ) + (t + ) or t(t + ) + (t + ) A (t)( t) (ii) This is always a product of two whole numbers each of which is greater than Correct explanation B ft from (i) for a convincing explanation referring to factors found in (i) 6 o x sin 60 x sin 60(= ) 7.7 x x M sin 60 = or sin 60 sin 90 oe sin 60 M (x = ) sin 60 or (x = ) sin 90 A x M cos(90 60) = M (x = ) cos(90 60) A

6 7 BD + = 6 oe M for BD + = 6 oe or 6 or seen BD= M for or (= ) (= ) '0.58' ' 0.58' '0.58' M for sin 40 = CD or cos 50 = CD sin 40 = CD '0.58' ' 0.58' '0.58' M for (CD =) sin 40 or cos50 CD = sin 40 A for M for BD + = 6 oe or 6 or seen M for or (=0.58..) ' 0.58' M for (BC =) 0.58 tan 50 or tan40 (=.6 ) M for '.6' ' ' A for (= ) = = ) 8.49 M (=5.5) '5.5' 00 M (dep) 64.8 oe A M 64.8 oe ( = 9.5(...)) M (dep) A

7 9 y > 5 7 M for clear intention to add to both sides (of inequality or y > 7 y equation) or clear intention to divide all three terms by or y > 7 or y < 7 or y = 7 7 y A or y > or y >. NB. final answer must be an inequality 7 (SC B for oe seen if M0 scored) 0 (a)(i) Explanation : Each member of the population has an equal chance of selection Each member of the population has an equal chance of selection B for explanation (ii) Description : Eg. number each student and use random select on a calculator Valid method B for an acceptable description (b) = M for '40' 00 oe or 0.96 A cao 7

8 Volume = M (=450) M (dep on st M) '450' 6.6 Mass = A cao SC: If no marks awarded then award B for an answer of 5940 Translation by 4 B Translation B 4 NB. Award no marks for a combination of transformations 8

9 For example UK Cheaper in US USA 4 M for.4.79 (= ) or.4.47 (=.88) $ per US gal ($)6.90(84) [$.5] M for or.79 '.88' per litre [.4] ( )0.56(5...) A for 6.90(84) per US gal ( )4.69(96) ( ).4(8...) C (dep on M) for $ 6.90(84) or $'6.9' and reaching a $ per litre ($).8(8) ($)0.8(...) conclusion consistent with their calculation Cost in per US gal of UK fuel=.4.79 = Cost in $ per US gal of UK fuel = $ = $ Cost in of US gal of US fuel = $.5.47 =.4 Cost in per litre of US fuel =.4.79 = 0. 56(5.. Cost in UK in per US gal =.4.79 (= ) Cost in USA in per US gal =.5.47 (=.48) Cost in UK is $ per litre =.4.47 (=.88) Cost in USA in $ per litre =.5.79 (=0.8...) M for.5.47 (=.48..) or.5.79 (=0.8) M for.4.79 or '0.8'.47 A for 0. 56(5 ) C (dep on M) for 0. 56(5...) or ' 0.57' and reaching a conclusion consistent with their calculation M.4.79 (= ) M.5.47 (=.48..) A 4.69(96) and.4(8...) C (dep on M) for '4.69(96)' or '4.70 AND.4(8...) and reaching a conclusion consistent with their calculation M for.4.47 (=.88) M for.5.79 (=0.8...) A for.8(8) and 0.8(...) C (dep on M) for $'.8(8)' and $'0.8(...)' and reaching a conclusion consistent with their calculation NB: Throughout values can be rounded or truncated to or more decimal places. In order to award the communication mark, correct currency must be shown with the calculated value(s) but these can still be rounded or truncated to one or more decimal places as they are being used for comparison. 9

10 4 AC AC 8.7 sin 49 sin64 M for sin 49 sin 64 oe 8.7 AC sin sin 49 sin64 M for (AC =) sin 64 (= 7.05 ) A for 7.(05 ) sin M for sin( ) ( ) A for BC 8.7 M for sin( ) sin sin'67' M for (BC =) sin 64 A for 8.9(0...) M for sin 49 A for oe (X is point such that AX is perpendicular to BC) M for AX = 8.7 sin 49 (= ) or XB = 8.7 cos 49 (= ) M for XB = 8.7 cos 49 (= ) and CX = tan 64 oe (=.0 ) A for 8.9(0...) or 5.7(07...) and.(0...) M for ½ ( ) oe A for

11 B for 9 or 9 or 9 (could be seen in working or on a tree diagram) or or or or or 5 4 M for M for A for 80 oe or 0.58(4...) B for 9 or 9 or or or M for M for A for 80 oe or 0.58(4...)

12 6 (a) M 7 6 X A 4 oe X X M Fully correct sample space with the correct X cases identified X X X A 4 oe (b) 5 6 M for identifying all possibilities of (,) and (,) and (,) X X X X X X X M A M A at least one of 7 6 (, ) or or 7 6 (, ) or oe or (, or ) (, ) Fully correct sample space with the correct cases identified 6 4 oe

13 7 B fully correct histogram with horizontal axis correctly scaled (B for 4 correct blocks or 5 correct blocks with incorrect or no scale ) (B for correct blocks of different widths or any correct blocks) SC : B for key, eg. cm = (trees) or correct values shown for (freq class interval) for at least frequencies (.5, 7, 4.5,,.5) Height h m Freq FD 0 < h 7.5 < h < h < h < h a =, b = 4, c = x = = 6 = = or = , M for (condone incorrect signs for 4 and ) M for 6 or A for one answer in the range.7 to.7 and one answer in the range 0.87 to 0.874

14 9 LQ = UQ = 45 4 M for 45 or or 4.5 or 9.5 or 7.75th or 8th or.5th or 4th (all of above may be seen in working space or indicated on S&L) or clear attempt to find UQ and LQ from a list of values or in stem and leaf diagram A cao 0 ( x) ( x) 6 6 x x 6 6 = 5x 9 M Use of common denominator of 6 (or any other multiple of 6) and at least one 6 numerator correct ( x) ( x) or Eg. 6 6 ( x) ( x) M 6 6 oe A cao 6 metres: 8 08 km. 6: : 8 0 : 5 00 : 5 00 M (indep) correct method to convert to consistent units M 8 '80 ' '6' (units may not be consistent) or 5 00 oe or 5 07 oe A : 5 00 or :

15 9 = M 0 (9 ) (=64) or BC = (M 9 (= 6) may be seen on diagram) BC = 8 M (indep) 9 + 'BC' where BC is a numerical value AC = M (dep on previous M) 8 '64' A.0.04 (a) b a B for b a or a + b (b) OP OA AP AP = 4 (b a) OP = a + 4 (b a) OP OB BP BP = 4 (a b) OP = b + 4 (a b) 4 (a + b) B for 4 (b a) ( OP ) OA AB OA AP or ( OP ) 4 M for or a ± 4 (b a) A for 4 (a + b) or 4 a + 4 b B for 4 (a b) ( OP ) OB BA OB BP or ( OP ) 4 M for or b ± 4 (a b) A for 4 (a + b) or 4 a + 4 b 5

16 4 4n + n + (4n n + ) = 4n + n + 9 4n + n 9 = 4n =8 n Proof M for out of 4 terms correct in expansion of either (n + ) or (n ) or ((n + ) (n ))((n + ) + (n )) A for 4n from correct expansion of both brackets A (dep on A) for 4n is a multiple of 8 or 4n = 8 n or 4n 8 = n 5 o A x x sin 0 x A o ( A ) xxsin0 shown M A x 0.5 x A A Ax 0.5 or C for completion with all steps shown o Height = x sin 0 = x x x x A M height = xsin 0 (= x) x A A Ax 0.5 or C for completion with all steps shown x Height = x sin 0 = x x A x x M for height = x sin 0 (= ) x A A Ax 0.5 or C for completion with all steps shown 6

17 6 (a) 6 4 M correct method to start to find BD or BO using triangle OBC or Let O be the centre of the triangle BCD (oe) base. Eg. OB + OC = 0 or BO = 50 or OB + OC = 0 ; OB 00 = 50 AO = AB OB = 50 BO = 50 (=7.07..) or BO = or 0 50 Vol = = BD or BD = 00 or BD= 00 (=4...) M (dep) correct method to find height of pyramid using triangle AOB Eg. AO = 0 ' 50 ' or AO = 50 or AO = 50 (=7.07..) 0 ' 50 ' M (indep) (but not A ) (b) Angle ABO = 45o Angle DAB = In ΔBAD, cos A = 0 0 ' 00 ' 0 0 = 0 90 M Angle DAB = A M cos BAD = A ' 00 ' 0 0 7

18 7 (a) 0.676, x M allow substitution of ±7 for c = 4 M 4 A answers in ranges and 5.7 to M (x + 4 ) oe 7 9 M for method leading to 6 4 A answers in ranges and 5.7 to 5.8 (b) y Put x and use part (a) Or 7y 9y 0 y ( 9) 4 7 ( ).48, 0.9 M A y x x or y (ft) answers in range and 0.9 to 0.94 M fully correct method which leads to 7y 9y = 0 or 7y + 9y + = 0 with correct method to solve (condone sign errors in substitution) A (ft) answers in range and 0.9 to

19 New Qn Question Number Paper Date Skill tested Maximum score Mean Score Mean Percentage Percentage scoring full marks a Q4b H 06 Solve linear equations in one unknown, with integer or fractional coefficients b Q4c H 06 Factorise quadratic expressions using the difference of two squares c Q4d H 06 Use instances of index laws, including use of fractional, zero and negative powers, and powers raised to a power Q6 H 06 Use the trigonometric ratios to solve -D and -D problems Q05 H Understand and use compound measures, including speed and density a Qa H Calculate volumes of right prisms and shapes made from cubes and cuboids b Qb H Use systematic trial and improvement to find approximate solutions of equations where there is no simple analytical method of solving them a Q4b H Factorise quadratic expressions b Q4c H Factorise quadratic expressions Q7 H Use the trigonometric ratios to solve -D and -D problems Q8 H 06 Understand, recall and use Pythagoras' theorem in -D, then in -D problems Q6 H Use percentages in real life situations Q08c H Solve simple linear inequalities in one variable, and represent the solution set on a number line a Qa H 06 Understand sample and population b Qb H 06 Understand sample and population Q H Understand and use compound measures, including speed and density Q0b H Describe and transform -D shapes using single translations Q0 H Calculate an unknown quantity from quantities that vary in direct or inverse proportion Q4 H 06 Use the sine and cosine rules to solve -D and -D problems Q5 H 06 Understand selection with or without replacement a Qa H Understand conditional probabilities b Qb H Understand conditional probabilities Q4 H Produce histograms from class intervals with unequal width Q H 06 Solve simple quadratic equations by using the quadratic formula

20 9 Q09b H Calculate median, mean, range, quartiles and interquartile range, mode and modal class - (SP.h) Q0 H Simplify rational expressions by cancelling, adding, subtracting, and multiplying Q9 H Write ratios in their simplest form Q5 H Understand, recall and use Pythagoras theorem in -D, then in -D problems a Q6a H 06 Calculate the resultant of two vectors b Q6b H 06 Solve geometrical problems in -D using vector methods Q H 06 Use algebraic manipulation to solve problems Q5 H Calculate the area of a triangle given the length of two sides and the included angle a Qa H Find the surface area and volumes of compound solids constructed from cubes, cuboids, cones, pyramids, spheres, 4 0. hemispheres, cylinders 0.6 6b Qb H Solve problems involving more complex shapes and solids, including segments of circles and frustums of cones a Qa H Solve simple quadratic equations by using the quadratic formula a Qb H Solve simple quadratic equations by using the quadratic formula