Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 5 · Variant 3

1 The 26 members of the local sports club include Mr and Mrs Khan and their son Abad. The club is holding a party to celebrate Abad’s birthday, but there is only room for 20 people to attend. In how many ways can the 20 people be chosen from the 26 members of the club, given that Mr and Mrs Khan and Abad must be included? [2] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 M1 100947 A1 CAO 2

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2 Lakeview and Riverside are two schools. The pupils at both schools took part in a competition to see how far they could throw a ball. The distances thrown, to the nearest metre, by 11 pupils from each school are shown in the following table. Lakeview 10 14 19 22 26 27 28 30 32 33 41 Riverside 23 36 21 18 37 25 18 20 24 30 25 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with Lakeview on the left-hand side. [4] (b) Find the interquartile range of the distances thrown by the 11 pupils at Lakeview school. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Lakeview Riverside 9 4 0 1 8 8 8 7 6 2 2 0 1 3 4 5 5 3 2 0 3 0 6 7 1 4 Key: 6|2|3 means 26m for Lakeview and 23m for Riverside B1 Correct Lakeview labelled on left, leaves in order from right to left and lined up vertically, no commas. B1 Correct Riverside labelled on same diagram, leaves in order and lined up vertically, no commas. B1 Correct key for their diagram, need both teams identified and ‘m’ stated at least once here or in leaf headings or title. SC If 2 separate diagrams drawn: SC B1 if both keys meet these criteria. 4 2(b) UQ = 32, LQ = 19 M1 (30 ⩽ UQ ⩽ 33) – (14 ⩽ LQ ⩽ 22) IQR = 32 – 19 = 13 A1 WWW 2

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3 The times taken, in minutes, by 360 employees at a large company to travel from home to work are summarised in the following table. Time, t minutes 0 ≤t < 5 5 ≤t < 10 10 ≤t < 20 20 ≤t < 30 30 ≤t < 50 Frequency 23 102 135 76 24 (a) Draw a histogram to represent this information. 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(b) Calculate an estimate of the mean time taken by an employee to travel to work. 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Mark scheme: 3(a) Cw: 5 5 10 10 20 M1 At least 4 frequency densities calculated (f/cw), accept unsimplified and class widths 1 ± of true values. May be implied by graph. Fd: 4.6 20.4 13.5 7.6 1.2 A1 All heights correct on graph NOT FT B1 Bar ends at 0, 5, 10, 20, 30, 50 clear intention not to draw at 4.5 or 5.5 etc. B1 Axes labelled: Frequency density (fd), time (t) and mins (or appropriate title). Linear scales between 0 and 20.4 or above on vertical axis, and 0 and 50 or above on the horizontal axis. (Axes may be reversed.) 4 3(b) 2.5 23 7.5 102 15 135 25 76 40 24 360 × + × + × + × + × M1 Uses at least 4 midpoint attempts (e.g. 2.5 ± 0.5) in correct formula, accept unsimplified expression, denominator either correct or their Σfrequencies . 5707.5 41 15.9, 15 360 48   =     A1 Evaluated. 2

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4 Raj wants to improve his fitness, so every day he goes for a run. The times, in minutes, of his runs have a normal distribution with mean 41.2 and standard deviation 3.6. (a) Find the probability that on a randomly chosen day Raj runs for more than 43.2 minutes. 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(b) Find an estimate for the number of days in a year (365 days) on which Raj runs for less than 43.2 minutes. 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(c) On 95% of days, Raj runs for more than t minutes. Find the value of t. 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Mark scheme: 4(a) ( ) 43.2 41.2 43.2 ( 0.5556) 3.6 P X P Z P Z −   > = > = >     M1 Use of ±Standardisation formula once, allow continuity correction, not σ2, √ σ. 1 – ( ) Φ 0.5556 1 0.7108 = − M1 Appropriate area Φ, from final process, must be probability. 0.289 A1 AWRT 3 4(b) Probability = 1 – their (a) = 1 – 0.2892 = 0.7108 B1FT 1 – their (a) or correct. 0.7108 × 365 = 259.4 259, 260 B1FT FT their 4SF (or better) probability, final answer must be positive integer. 2 4(c) 1.645 z = ± B1 CAO, critical z value. 41.2 1.645 3.6 t − = − M1 Use of ±standardisation formula with µ, σ equated to a z-value, no continuity correction, allow σ2, √ σ. 35.3 t = A1 3

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5 A security code consists of 2 letters followed by a 4-digit number. The letters are chosen from {A, B, C, D, E} and the digits are chosen from {1, 2, 3, 4, 5, 6, 7}. No letter or digit may appear more than once. An example of a code is BE3216. (a) How many different codes can be formed? 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(b) Find the number of different codes that include the letter A or the digit 5 or both. 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A security code is formed at random. (c) Find the probability that the code is DE followed by a number between 4500 and 5000. 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Mark scheme: 5(a) M1 16 800 A1 2 Question Answer Marks Guidance 5(b) Method 1 [Identify scenarios] With A and no 5: 8 × 6P4 or (1 × 4 × 6 ×5 × 4 × 3) ×2 or 4C1 × 2! × 6P4 = 2880 With 5 and no A: 4P2 × 4 × 6P3 or (4 × 3 × 1 × 6 × 5 × 4) × 4 or 4P2 × 6C3 × 4! = 5760 With A and 5: 8× 4 × 6P3 or (4 × 1 × 1 × 6 × 5 × 4) × 8 or 4C1 × 2! × 6C3 × 4! = 3840 M1 One number of ways correct, accept unsimplified. M1 Add 2 or 3 identified correct scenarios only, accept unsimplified. [Total =] 12 480 A1 CAO Method 2 [total number of codes – number of codes with no A or 5] No A or 5 : (4 × 3 ) ×( 6 × 5 × 4 × 3) = 4320 M1 4P2 × 6P4 or 4C2 × 6C4 seen, accept unsimplified. Required number = their (a) – their 4320 M1 Their 5(a) (or correct) – their (No A or 5) value. 12 480 A1 Method 3 [subtracting double counting] With A 4P1 × 7P4 × 2 or 4C1 × 2 × 7C4 × 4! = 6720 With 5 5P2 × 6P3 × 4 or 5C2 × 2 × 6C3 × 4! = 9600 With A and 5 = 4P1 × 6P3 × 8or 4C1 × 2! × 6C3 × 4! × 8 = 3840 M1 One outcome correct, accept unsimplified. Required number = 6720 + 9600 – 3840 M1 Adding ‘with a’ to ‘with 5’ and subtracting ‘A and 5’. 12 480 A1 CAO 3 Question Answer Marks Guidance 5(c) Method 1 – number of successful codes divided by total (1 ×) 3 × 5P2 M1 3 × 5Pn, n = 2, 3. Condone 3 × 5C2, no + or –. Probability = 3 5 2 1 6 800 their P their × M1 Probability = 60 1 6 800 their their . 1 280 , 0.00357 A1 Method 2 – product of probabilities of each part of code 1 1 1 3 5 4 5 4 7 6 5 4   × × × × ×     or 1 1 3 5 2 5 4 7 4 P P × × × M1 1 1 5 4 k × × where 0 1 k < < for considering letters. M1 t 1 3 7 6 × × or 3 5 2 7 4 P t P × × where 0 1 t < < . 1 280 A1 CAO 3

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7 Box A contains 6 red balls and 4 blue balls. Box B contains x red balls and 9 blue balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B. (a) Complete the tree diagram below, giving the remaining four probabilities in terms of x. [3] Box A Box B Red Red 6 10 Blue Red 4 10 Blue Blue 4 (b) Show that the probability that both balls chosen are blue is [2] x + 10. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that the probability that both balls chosen are blue is 6.1 (c) Find the probability, correct to 3 significant figures, that the ball chosen from box A is red given that the ball chosen from box B is red. 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Mark scheme: 7(a) Probabilities: 1 10 x x + + , 9 10 x + , 10 x x + , 10 10 x + B1 One probability correct in correct position. B1 Another probability correct in correct position. B1 Other two probabilities correct in correct positions. 3 7(b) 4 10 10 10 their x × + M1 Method consistent with their tree diagram. 4 10 x + A1 AG 2 Question Answer Marks Guidance 7(c) 4 1 10 6 x = + 10 24, 14 x x + = = B1 Find value of x. Can be implied by correct probabilities in calculation. P(ARed|BRed) = P(ARed ∩ BRed) ÷ P(BRed) 6 1 10 10 6 1 4 10 10 10 10 x their x x x their their x x + × + + × + × + + = 6 15 10 24 6 15 4 14 10 24 10 24 × × + × = 3 8 73 120 B1 FT 6 1 10 10 x their x + × + as numerator or denominator of fraction. M1 6 1 4 10 10 10 10 x x their their x x + × + × + + seen anywhere. A1 FT Seen as denominator of fraction. 45 73 , 0.616[4…] A1 If B0 M0: SC B1 for 3 8 73 120 or 0.375 0.6083 SC B1 45 73 or 0.616. 5

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