4 M1 for a correct method to find of 40; eg (= 16) A1 39. M1 for 1

Size: px

Start display at page:

Download "4 M1 for a correct method to find of 40; eg (= 16) A1 39. M1 for 1"

Alexis Harris
5 years ago
Views:

1 1 4 M1 f a crect method to find of 40; eg. 40 (= 16) f a crect method to find 8 of 40; eg (= ) M1 f a crect method to find of 40 and 8 of 40 M1 (dep on M1) f 80 "16" "" (= 39) = A1 OR oe M1 f 1 (= 3 ) and 1 8 (= 3 8 ) M1 f a crect method to find 3 of 40; eg (= 4) f a crect method to find 3 8 of 40; eg (= 1) M1 (dep on M1) f "4" + "1" (= 39) A oe w = P + 3 M1 f a clear intention to multiply both sides by add to both sides as a first step A1 f w = P + 3 oe 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 1

2 3 0 = 0. h = 30 min 60 = 0.416h = min 3 60 M1 f 0 30 (min) 0.(h) (min) 0.41(6)(h) 60 M1(dep) (6) 30 A1 cao OR M1 f 60 (=.4) and (= ) and 60 M1(dep) f 30 A1 cao 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0

3 4 1x + 9y = 1 x = 4 M1 f crect process to eliminate either x y 1x 16y = 104 y = 1 y = 4x + 3 = 7 OR 16x + 1y = 8 y = (condone one arithmetic err) A1 f either x = y = M1 (dep on 1 st M1) f crect substitution of their found value (indep) f crect process to eliminate the other variable (condone one arithmetic err) A1 cao f both x = and y = 9x 1y = 78 x = 0 SC: B1 f x = y = if M0 sced x = 4 + 3y = 7 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 3

4 (a) M1 f f f M1 (dep) f A1 f 4867.(0) cao (If crect answer seen then igne any extra years) Alternative method M f A1 f 4867.(0) cao [SC: 367.(0) seen B] (b) n M1 f an attempt to evaluate n f at least one value of n (not equal to 1) n (n ) (= ) and 400 A1 f cao n 1.07 evaluated, n 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 4

5 6 10 cos M1 f OR BC = cos 40 BC = 46.79(1...) 10 M1 f sin 70 = BC sin(180 70) sin(180 70) 10 M1 f BC = sin 70 A1 f 6.84(0 ) OR M1 f M1 f cos(180 70) M1 f ' 46.79(1...)' A1 f 6.84(0 ) M1 f perpendicular from A to BC, may be implied by crect wking M1 f 10 cos70 10 sin0 crect attempt to use sin cos M1 f 10 cos70 A1 f 6.84(0...) 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0

6 7 (i) 1 3 M1 crect algebraic expression f any relevant area ( x 6 3 x 4) ( x 1) (x + 6)(x 1) (x 1)(3x 4) 1 ( x 1)(3 x 4 ( x 6)) 1 (4 M1 f crect equation with at least one pair of brackets x x ) 119 expanded crectly shown A1 f completion to given equation (ii) (x ±1)(x ± 8) (= 0) 1 ( 1) x (x +1)(x 8) (=0) x and 8 given as solutions 8 A1 dep 3 M1 Start to solve quadratic condone one sign err in substitution if quadratic fmula used; allow in place of ( 1) M1 ft from an increct 3 term quadratic equation ft method from an increct 3 term quadratic equation Award all 3 marks if first M1 awarded and 8 alone given as final answer 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 6

7 8 (a) x 1 1 B1 cao (b) 4, 3, B f all 3 values and no extras (igne repeats) (B1 f crect values and no extras all 3 crect values and ) (c) y < 4 M1 f clear intention to add onto each side of an inequality ( equation) clear intention to divide all terms by as a first step (y =) 4 A1 cao 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 7

8 9 eg M1 (probabilities from selecting,, ) allow eg " " M1 (probabilities from selecting 1,, 3) allow M1 (probabilities f all combinations of 1,, 3) allow " " M1 complete crect method A1 oe eg., 0.98, MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 8

9 10 0 =,, 60 M1 f identifying the prime facts f of the 3 numbers 140 =,,, 7 0,140,40 (can be implied by a fact tree, repeated division Venn diagram) 40 =,, 3,, 7 F a complete Venn diagram f x and 140 with 0 in the intersection x = y = 40 At least the 1 st 3 multiples of 0 140x = 40 0 oe A1 (Allow 3 ) M1 f (= 0.8) ( = 8) M1 f oe oe A1 cao 1 D M1 f in the middle and 1 from 4(D L T ) 4 4 L 6 (L T D') 6(D T L') 6 M1 f any 4 crect entries T 3 A1 f all crect including outside the circles inside the rectangle /9 B1 ft from increct diagram 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 9

10 13 (a) 1 1 B1 (b) y = (x 6) / y = x 6 y + 6 = x x + 6 M1 f a crect flowchart including inverse A1 1MA1 practice paper 3H (Set 6) mark scheme: Version

11 14 B1 Recognition of angle LRM as required angle either drawn on diagram from wking PQ( ML) 0sin30 o (=10) MR 1 0 = 44 = 4 34 =3.3..) LR = 1 ( RQ) = 1 (10 3) M f a crect method to calculate PQ(ML) & MR MR & LR PQ(ML) & LR (NB: LR requires use of RQ = cos ) Or M1 f a crect method to calculate one of the sides PQ MR LR 10 sin MRL 4 34 ML MR 111 cos MRL 4 34 LR MR 10 ML tan MRL 111 LR.4 A1.38. M1 (Dep on M) Use of a crect trig ratio to find angle MRL 1MA1 practice paper 3H (Set 6) mark scheme: Version

12 1 1 M1 f 160r = M1 f (r=) 90 oe 160 M1 (dep M) f 160 A1 f 1.3 A1 f oe M1 f 160 r 90 = M1 f n n 100 = 90 M1 (dep M) f A1 f 1.3 A1 f 1 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 1

13 16 AC = cos M1 f AC = cos13 M1 f crect der of operations AC = A1 f AC = M1 f Area of ABC = ½ ½ (= 7.8..) + ½ sin13 Area of ADC = ½ sin13 M1 f ½ ½ sin13 A1 f an answer in the range (= ) 1MA1 practice paper 3H (Set 6) mark scheme: Version

14 17 x ( a t ) 4 M1 Crect rearrangement f crect expression f x x t a ( a t ) oe x t x x t a t 4 x 16a 4 oe oe x y a a oe x a a M1 Crect expressions f t t² f at² at in terms of x and a M1 f crect substitution of t and t² into expression f y 4 x y 16a 3 x a A1 Fully crect answer in required fm 1MA1 practice paper 3H (Set 6) mark scheme: Version

15 18 Area (1 < l < 6) = (0.1 1) (0.3 1) + (0.38 1) + (0. 1) + (0.16 1) = M1 f attempt to use frequency density width e.g M1 f (0.1 ) + (0.3 1) + (0.38 1) + (0. 1) + (0.16 ) 1.78 seen M1 f Total Area= (0.1 ) + (0.3 1) + (0.38 1) + (0. 1) + (0.16 ) = 1.78 Proption = ((0.1 1) + (0.3 1) + (0.38 1) + (0. 1) + (0.16 1))/ 1.78 A1 f answer which rounds to % equivalent vulgar fraction 7 89 OR M1 f attempt to use area e.g. sight of any one of 4.8, 6.4, 7.6, (cm²) oe M1 f (cm ) oe seen M1 f ( ) 3.6 oe A1 f answer which rounds to %% 7 89 equivalent vulgar fraction 1MA1 practice paper 3H (Set 6) mark scheme: Version 1.0 1

16 19 congruency proved 3 M1 f crect statement with crect reason explains why DE = FB M1 f a second crect statement with crect reason C1 f complete proof justifying congruency, eg SAS AAS Eg DAE = BCF (opposite angles of parallelogram are equal) AE = FC (E and F are midpoints of lines of equal length) AD = BC (opposite sides of parallelogram are equal) AED CFB (SAS) 1 C1 f relevant statement using congruency Eg DE and FB are cresponding sides of congruent triangles 1MA1 practice paper 3H (Set 6) mark scheme: Version

17 National perfmance data from Results Plus Original source of questions Mean sce of students achieving grade: Qn Spec Paper Session Max Qn Topic YYMM sce ALL A* A B C D E 1 MM H 1411 Q07 Fractions MM F 106 Q0 Rearranging equations MA0 F 111 Q3 Compound measures MM H 111 Q18 Simultaneous equations H 0906 Q19 Compound interest MM H 1111 Q17 Trigonometry MA0 3H 1606 Q17 Solving quadratic equations MM H 1406 Q10 Solve inequalities MA0 3H 1606 Q1 Probability MA0 4H 1606 Q10 LCM and HCF MM H 1306 Q1 Reverse percentages MA0 4H 1606 Q1 Venn diagrams MA0 H 130 Q17 Functions MA0(R) 4H 1606 Q17 Trigonometry AM H 1311 Q Proptional change MM H 1106 Q4 Sine and cosine rule MA0(R) 4H 1606 Q1 Rearranging equations AM1 1H 1111 Q Histograms and grouped data MM H 1311 Q6 Geometric proof MA1 practice paper 3H (Set 6) mark scheme: Version

OR Angle QCD = 54 Angle ACP = = 50 x =

OR Angle QCD = 54 Angle ACP = = 50 x = MA Practice Tests Set : Paper H (Regular) mark scheme Version.0 Angle BAC = 76 Angle BAP = 80 90 = 6 x = 76 6 Angle QCD = Angle ACP = 80 76 = 0 x = 80 90 0 0 o B f Angle BAC = 76 (could be just on the