AS Level / Year 1 Edexcel Maths / Paper 2

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Primrose Lawson
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1 AS Level / Year 1 Edexcel Maths / Paper 2 March 2018 Mocks 2018 crashmaths Limited

2 1 (a) For every (extra) 1 mile per hour Callum travels in his car, his car s average fuel consumption decreases by 0.46 miles per gallon Suitable interpretation containing all underlined elements oe (b) 50.1 (miles per gallon) Cao (c) (Likely to be) unreliable, since 85 is outside the data range (used to calculate the regression line) / it is extrapolation AO2.2 Unreliable + reason AO3.5b 3

3 2 (a) Beijing Cao AO1.2 (b) ( tr means a recorded value of) rainfall less than 0.05 mm Cao AO1.2 (c) (c/i) Mark (c/i) and (c/ii) together and see notes before marking idea that : large data set has many more than 8 data points, so using 8 points is easier to process / quicker to process / requires less analysis / etc. See notes AO2.4 (c/ii) idea that : large data set has many more than 8 data points, so using 8 points may not very representative / may lead to inaccurate/unreliable conclusions / etc. See notes AO2.4 (d) 1 st : any two of median, lower quartile and upper quartile correct 2 nd : completely correct box plot [NB: LQ = 0.2, median = 1.0, UQ = 5.5] (e) Suitable comparison in context, i.e. Heathrow had less rainfall on average than City X as the median is lower Heathrow had less variation in the amount of rainfall it received than City X, as the IQR/range is smaller A suitable comparison in the amount of rainfall in context. Must be a comparison + statistical reason. Must refer to median or IQR/range AO2.2 7

4 Question 2 Notes (c) For both marks, candidates need to convey the idea that the large data set has much more than 8 data points. However, this can be done in (c/i) OR (c/ii) OR both. Mark (c/i) and (c/ii) as one question out of 2 marks with: 2 marks ( ): idea that LDS has more than 8 data points in (c/i) or (c/ii) or both + one advantage + one disadvantage 1 mark ( B0): one advantage or one disadvantage OR one advantage + one disadvantage + does not convey idea that LDS has more than 8 data points 0 marks (B0 B0): no advantage or no disadvantage

5 3 (a) Speed is a continuous variable Justification (b) Width = 1 cm Correct width = 255 h h = 20 cm Correct height (c/i) Mean = ( ) + ( ) ( ) + ( ) = 70 mph (to 2 dp) = Convincing proof AO2.1 (c/ii) (d) Standard deviation = ( ) 2 = idea that it will decrease the standard deviation, as the data will be more concentrated about the mean Method to find standard deviation Correct standard deviation, awrt 7.9. Answer = 7.6 is SC1 (see notes) A1 Decrease + reason AO2.2 7 Question 3 Notes (c/ii): A1 candidates will get different answers depending on the accuracy of the mean they use. Accept all answers (resulting from correct workings) that round to 7.9. Special case 1 (SC1): correct use of mean = 70 in calculation leading to standard deviation = 7.6 is A1.

6 4 (a) Any two of: discrete number of trials 2 outcomes: success and failure fixed probability of success all trials are independent Any two conditions AO1.2 (b) H 0 : p = 0.25 H 1 : p 0.25 Correct hypotheses AO2.5 (c) [ X B(32,0.25) ] Sight of one of P(X 3) = , P(X 4) = P(X 12) = , P(X 13) = Critical region is { X 3}, { X 13} A1 A1 [3] (d) 14 is in the CR, so Jeremy s suspicion is supported / the deck of card s is not regular AO3.2a (e) Unlikely to be valid as (binomial distribution model used in) (d) requires the card selections to be independent AO3.5a 7

7 5 (a) (From a Venn diagram, it s clear that) P(A and B) = P(A) + P(B) P(A or B) Convincing proof AO2.1 = 8P(A and B) P(A and B) = 0.63 P(A and B) = 0.09 (b/i) P(A) = 0.36 P(not A) = = 0.64 Cao (b/ii) P(A and not B) = = 0.27 Cao (b/iii) P(A or not B) = = 0.73 Cao (c) P(A) P(B) = = Finds P(A)P(B) AO2.1 P(A and B) = , so the events A and B are not statistically independent Cso. Conclusion necessary AO2.1 A1 6

8 6 (a) v 2 = u 2 + 2a(h max ) 0 2 = (9.8)(h max ) h max = 222 2(9.8) = Uses correct equation(s) with correct magnitudes. Condone sign inconsistencies Correct max height above P. Awrt 25 m A1 So the max height of ball above the reservoir is 35 m Answer of 35 m or 34.7 m. Award ft for answer of 10 + their max height above P A1ft [3] (b) v 2 = ( 9.8)( 10) = 680 v = (c) So the velocity of ball as it reaches reservoir is 26 m/s a = = m/s 2 So deceleration of the ball is 8.7 m/s 2 Uses correct equation(s) with correct magnitudes. Condone sign inconsistencies Answer of 26 m/s or 26.1 m/s Correct deceleration given ft their (b). ft Answer of 8.7 m/s 2 or 8.69 m/s 2 A1 (d) 0.5g R = 0.5( ) Uses N2L with their deceleration R = N So magnitude of resistive forces on ball is 9.2 N Answer of 9.2 N or 9.25 N. A1 8

9 7 (a) a = 4 Cao AO2.2 Acceleration is 2bt 4b = 4 b = 1, but b > 0, so b = 1 Obtains an expression for the acceleration Correct value of b AO2.2 A1 [3] (b) Correct shape ft their value of b, with discontinuity at 5 Intersects velocity axis at 4 and time axis at 2 (or ft their a and b ) (c) d total = 0 2 (4 t 2 )dt 2 5 (4 t 2 )dt Correct partitioning of the integral, seen or implied at any stage AO2.2 4 t 2 dt = 4t t 3 { +c} 3 Integrates their a bt 2 with some values for a and b d total = 4(2) (5) 3 4(2) = 97 See notes 3 m Cao d A1 [4] 9

10 Question 7 Notes (c) 2 nd substitutes limits in the integral. The limits should either be: 0 and 5 OR limits arising from a correct partitioning of the integral based on their a bt 2. If they have a partitioned integral, they only need to evaluate the limits on one of the integrals for this mark.

11 8 (a) Resultant force is (4y + x)i + (e y 6)j Attempts to find resultant force on P (need not be simplified) AO3.1a At P, acceleration is 0, so net force is 0 4b + a = 0, e b 6 = 0 Forms two correct equations. Condone if still in terms of x and y AO3.1a d a = 4 ln6, b = ln6 Correct values of a and b A1 A1 4

12 9 (a) A :T 0.4(2g) = 2a Correct equation of motion of A AO3.4 B : 4.5g (T + ΔT ) = 4.5a 4.5g T 3 = 4.8a Correct equation of motion of B AO g + 4.5g 3 = 6.8a a = So acceleration in string is 4.9 m/s 2 Attempts to eliminate T to find a Correct acceleration = 4.9 m/s 2 or 4.89 m/s 2 (b) T = N Finds tension using their equations of motion and their acceleration F pulley = T 2 + (T + ΔT ) 2 Correct expression for F pulley. Need not see any substitution, the expression alone is OK A1 [4] ft F pulley = ( ) 2 + ( ( )) 2 = so resultant force is 28 N Correct resultant force = 28 N or 28.3 N A1 [3] (c) Any two suitable improvements. See notes Correct gradient of normal to C at A. Seen or implied AO3.5c AO3.5c 9

13 (c) Any two suitable improvements. For example: Question 9 Notes e.g. include a variable resistance instead of taking it to be constant e.g. include dimensions of the pulley in the model so the string is not necessarily parallel to it e.g. include a more accurate value of g in the model e.g. consider a string that not inextensible so acceleration is not the same on either side Ignore improvements that are linked to the distance travelled by the masses etc. as the model was not used to calculate these quantities.

14 Marks breakdown by AO AO Number of marks % AO AO AO

MATHEMATICS AS/P2/M18 AS PAPER 2

MATHEMATICS AS/P2/M18 AS PAPER 2 Surname Other Names Candidate Signature Centre Number Candidate Number Examiner Comments Total Marks MATHEMATICS AS PAPER 2 March Mock Exam (Edexcel Version) CM Time allowed: 1 hour and 15 minutes Instructions