Cambridge A Level Mathematics 9709 — 2012 May/June Paper 6 · Variant 2
9709/62/M/J/12 · 6 questions · 50 marks · ≈56 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper4 pages
Mark scheme6 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · The ages, x years, of 150 cars are summarised by Σx = 645 and Σ x2 = 8287.5
1 The ages, x years, of 150 cars are summarised by Σx = 645 and Σ x2 = 8287.5. Find Σ(x −x)2, where x denotes the mean of x. [4]
Mark scheme: 1 x = 3.4 B1 4.3 or 645/150 or 18.49 seen M1 Subst in correct formula to find sd or var or 82875. 2 sd = − 3.4 = 36.76 = 6.063 expand Σ ( x − x 2) correctly and substitute 150 Σ ( x − x 2) = 150× .6 063 2 M1 Mult by 150 = 5514 (5510) A1 [4] Answer rounding to 5510
Q2 · The random variable X has the probability distribution shown in the table
2 The random variable X has the probability distribution shown in the table. x 2 4 6 P(X = x) 0.5 0.4 0.1 Two independent values of X are chosen at random. The random variable Y takes the value 0 if the two values of X are the same. Otherwise the value of Y is the larger value of X minus the smaller value of X. (i) Draw up the probability distribution table for Y. [4] (ii) Find the expected value of Y. [1]
Mark scheme: 2 (i) B1 0, 2, 4 only seen for Y no probs needed. y 0 2 4 Accept other vals if P(value) = 0 seen in P(Y = y) 0.42 0.48 0.1 table, allow 0002244 with probs M1 Summing two or more 2-factor probs (can be implied) A1 One correct prob A1 [4] Correct table or list (ii) 0.96 + 0.4 = 1.36 B1ft [1] Ft their table for Y or X Σp = 1
Q3 · In Restaurant Bijoux 13% of customers rated the food as ‘poor’, 22% of customers rated…
3 In Restaurant Bijoux 13% of customers rated the food as ‘poor’, 22% of customers rated the food as ‘satisfactory’ and 65% rated it as ‘good’. A random sample of 12 customers who went for a meal at Restaurant Bijoux was taken. (i) Find the probability that more than 2 and fewer than 12 of them rated the food as ‘good’. [3] On a separate occasion, a random sample of n customers who went for a meal at the restaurant was taken. (ii) Find the smallest value of n for which the probability that at least 1 person will rate the food as ‘poor’ is greater than 0.95. [3]
Mark scheme: 3 (i) P(2 < X < 12) = 1 – P(0, 1, 2, 12) M1 Using binomial with 12Csomething and powers summing to 12, Σp = 1 = 1 – (0.35)12 – (0.65)(0.35)1112C1 – A1 Correct unsimplified answer (0.65)2(0.35)1012C2 – (0.65)12 = 1 – 0.0065359 = 0.993 A1 [3] Accept 0.994 from correct working only (ii) 1 – (0.87)n > 0.95 M1 Equality or inequality in (0.87 or 0.78 or 0.35), power n or n – 1, 0.95 or 0.05 0.05 > (0.87)n M1 Attempt to solve an equation with a power in (can be implied) n = 22 A1 [3] Correct answer
Q4 · The back-to-back stem-and-leaf diagram shows the values taken by two variables A and B
4 The back-to-back stem-and-leaf diagram shows the values taken by two variables A and B. A B (3) 3 1 0 15 1 3 3 5 (4) (2) 4 1 16 2 2 3 4 4 5 7 7 7 8 (10) (3) 8 3 3 17 0 1 3 3 3 4 6 6 7 9 9 (11) (12) 9 8 8 6 5 5 4 3 2 1 1 0 18 2 4 7 (3) (8) 9 9 8 8 6 5 4 2 19 1 5 (2) (5) 9 8 7 1 0 20 4 (1) Key: 4 16 7 means A = 0.164 and B = 0.167. (i) Find the median and the interquartile range for variable A. [3] (ii) You are given that, for variable B, the median is 0.171, the upper quartile is 0.179 and the lower quartile is 0.164. Draw box-and-whisker plots for A and B in a single diagram on graph paper. [3]
Mark scheme: 4 (i) A: median = 0.186, B1 IQ range = 0.198 – 0.179 M1 Subt LQ from their UQ = 0.019 A1ft [3] Correct IQ range ft dp in wrong place (ii) A B1ft 2 correct boxes ft (i) OK if superimposed B B1 2 pairs correct whiskers lines up to box not inside 0.15 0.16 0.17 0.18 0.19 0.20 0.21 B1 [3] Correct uniform scale from at least 0.15 to 0.21 seen. No scale no marks (ii) unless perfect A and B with all 10 values shown GCE AS/A LEVEL – May/June 2012 9709 62
Q5 · An English examination consists of 8 questions in Part A and 3 questions in Part B
5 An English examination consists of 8 questions in Part A and 3 questions in Part B. Candidates must choose 6 questions. The order in which questions are chosen does not matter. Find the number of ways in which the 6 questions can be chosen in each of the following cases. (i) There are no restrictions on which questions can be chosen. [1] (ii) Candidates must choose at least 4 questions from Part A. [3] (iii) Candidates must either choose both question 1 and question 2 in Part A, or choose neither of these questions. [3]
Mark scheme: 5 (i) 11C6 = 462 B1 OR A3 B3 or A4 B2 or A5 B1 or A6 = 8C3 + 8C4 × 3C2 + 8C5 × 3C1 + 8C6 B1 = 56 + 210 + 168 + 28 = 462 [1] (ii) 8C4 × 3C2 + 8C5 × 3C1 + 8C6 M1 ∑ 2 or more two-factor terms, P or C any numbers = 210 + 168 + 28 B1 Any correct option unsimplified = 406 A1 [3] Correct answer (iii) 9C4 + 9C6 = 126 + 84 M1 Summing 9Cx + 9Cy can be mult by 2 no other terms B1 126 or 84 seen or unsimplified 9C4, 9C6 = 210 A1 Correct answer OR 1,2 in A tog with : A1B3 + A2B2 + A3B1 + M1 ∑ 5 or more 2-factor 6Px or 6Cx with 3Cx or A4B0 + 1,2 out of A : A3B3 + A4B2 + 3Px only (can be mult by 2) A5B1 + A6B0 = 6C1 + 6C2 × 3C2 + 6C3 × 3C1 + 6C4 + 6C3 × B1 3 or more correct unsimplified options 3C3 + 6C4 × 3C2 + 6C5 × 3C1 + 6C6 = 6 + 45 + 60 + 15 + 20 + 45 + 18 + 1 = 210 A1 Correct answer OR 462 – 9C5 – 9C5 M1 subt two 9Cx options from their (i) B1 9C5 seen oe if using this method = 210 A1 [3] Correct answer
Q6 · A box of biscuits contains 30 biscuits, some of which are wrapped in gold foil and some…
6 A box of biscuits contains 30 biscuits, some of which are wrapped in gold foil and some of which are unwrapped. Some of the biscuits are chocolate-covered. 12 biscuits are wrapped in gold foil, and of these biscuits, 7 are chocolate-covered. There are 17 chocolate-covered biscuits in total. (i) Copy and complete the table below to show the number of biscuits in each category. [2] Wrapped in gold foil Unwrapped Total Chocolate-covered Not chocolate-covered Total 30 A biscuit is selected at random from the box. (ii) Find the probability that the biscuit is wrapped in gold foil. [1] The biscuit is returned to the box. An unwrapped biscuit is then selected at random from the box. (iii) Find the probability that the biscuit is chocolate-covered. [1] The biscuit is returned to the box. A biscuit is then selected at random from the box. (iv) Find the probability that the biscuit is unwrapped, given that it is chocolate-covered. [1] The biscuit is returned to the box. Nasir then takes 4 biscuits without replacement from the box. (v) Find the probability that he takes exactly 2 wrapped biscuits. [4]
Mark scheme: 6 (i) wrapped unwrapped total choc 7 10 17 B1 One correct row or column numbers not choc 5 8 13 total 12 18 30 B1 [2] All correct including labels (ii) 12/30 (0.4) B1ft [1] Ft their table (iii) 10/18 (5/9) (0.556) B1ft [1] Ft their table (iv) 10/17 (0.588) B1ft [1] Ft their table (v) P(2 wrapped) = 12/30 × 11/29 × 18/28 × 17/27 × 4C2 M1 Mult by 4C2 M1 12 × 11 × 18 × 17 seen in num M1 30 × 29 × 28 × 27 seen in denom = 0.368 (374/1015) A1 Correct answer OR (12C2 × 18C2)/30C4 M1 12C2 seen mult or alone in num (not added) M1 18C2 seen mult or alone in num (not added) M1 30C4 seen in denom = 0.368 A1 [4] Correct answer GCE AS/A LEVEL – May/June 2012 9709 62 42 − 411. M1 Standardising no cc no sq rt no sq 7 (i) P(> 42) = P z > 4.3 = P(z > 0.2647) = 1 – 0.6045 = 0.3955 A1 Correct prob rounding to 0.395 or 0.396 Prob = (0.3955)(0.6045)23C1 M1 Binomial 3Cx powers summing to 3, any p, Σp = 1 = 0.433 or 0.434 A1 [4] Rounding to correct answer
What was in this paper
The subtopics covered by these 6 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
What you needed in this session
Cambridge’s own grade thresholds for 2012 May/June, Paper 6 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.