UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2010 question paper

for the guidance of teachers

9709 MATHEMATICS
9709/63 Paper 6, maximum raw mark 50

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.

bestexamhelp.com
Mark schemes must be read in conjunction with the question papers and the report on the
examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
 Page 2 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 63

Mark Scheme Notes

Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are
not lost for numerical errors, algebraic slips or errors in units. However, it is not usually
sufficient for a candidate just to indicate an intention of using some method or just to
quote a formula; the formula or idea must be applied to the specific problem in hand,
e.g. by substituting the relevant quantities into the formula. Correct application of a
formula without the formula being quoted obviously earns the M mark and in some
cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated method mark is earned (or
implied).

B Mark for a correct result or statement independent of method marks.

• When a part of a question has two or more "method" steps, the M marks are generally
independent unless the scheme specifically says otherwise; and similarly when there are
several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a
particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme.
When two or more steps are run together by the candidate, the earlier marks are implied and
full credit is given.

• The symbol √ implies that the A or B mark indicated is allowed for work correctly following
on from previously incorrect results. Otherwise, A or B marks are given for correct work only.
A and B marks are not given for fortuitously "correct" answers or results obtained from
incorrect working.

• Note: B2 or A2 means that the candidate can earn 2 or 0.

B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether
a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless
otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working
following a correct form of answer is ignored.

• Wrong or missing units in an answer should not lead to the loss of a mark unless the
scheme specifically indicates otherwise.

• For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,
or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated
above, an A or B mark is not given if a correct numerical answer arises fortuitously from
incorrect working. For Mechanics questions, allow A or B marks for correct answers which
arise from taking g equal to 9.8 or 9.81 instead of 10.

© UCLES 2010
 Page 3 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 63

The following abbreviations may be used in a mark scheme or used on the scripts:

AEF Any Equivalent Form (of answer is equally acceptable)

AG Answer Given on the question paper (so extra checking is needed to ensure that
the detailed working leading to the result is valid)

BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely
clear)

CAO Correct Answer Only (emphasising that no "follow through" from a previous error
is allowed)

CWO Correct Working Only – often written by a ‘fortuitous' answer

ISW Ignore Subsequent Working

MR Misread

PA Premature Approximation (resulting in basically correct work that is insufficiently

accurate)

SOS See Other Solution (the candidate makes a better attempt at the same question)

SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a
case where some standard marking practice is to be varied in the light of a
particular circumstance)

Penalties

MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or

part question are genuinely misread and the object and difficulty of the question
remain unaltered. In this case all A and B marks then become "follow through √"
marks. MR is not applied when the candidate misreads his own figures – this is
regarded as an error in accuracy. An MR–2 penalty may be applied in particular
cases if agreed at the coordination meeting.

PA –1 This is deducted from A or B marks in the case of premature approximation. The

PA –1 penalty is usually discussed at the meeting.

© UCLES 2010
 Page 4 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 63

1 Normal B1
mean 60 kg, variance 90 kg2 B1 Any sensible values (mean 40–80 kg,
variance 16–225 kg2), could give s.d.
4–15 kg
[2]

2 (i)
x 1 2 3 4 5 M1 1, 2, 3, 4, 5 seen, together with some
Prob k 2k 3k 4k 5k probabilities involving k but not x
M1 summing probs involving k to 1
15k = 1
k = 1/15 (0.0667) A1 correct answer
[3]

(ii) E(X) M1 using Σpx no dividing

= k + 4k + 9k + 16k + 25k
= 55k = 11/3 (3.67) A1ft correct answer, ft on 55k, 0 < k < 1
[2]

3 (i) Y Y = young, M = middle-aged, O = old

0.7
Ph 0.25 M M1 Correct shape with Ph, NPh first
0.68 0.05
O
Y
0.32 0.26
NPh 0.10 M
0.64
O A1 All probabilities and correct
[2]

0.68 × 0.25
(ii) P(Ph│M) = B1 For correct numerator using cond prob
0.68 × 0.25 + 0.32 × 0.1 formula with numerator < denominator
M1 For attempt at P(35 – 60 years old),
involving the sum of two 2-factor probs,
seen anywhere
= 0.842 (170/202) A1 Correct answer
[3]

4 (i) x = 60 + 245/70 M1 245/70 seen

= 63.5 A1 Correct answer
[2]

(ii) Σ(x – 50) = Σx – Σ50 M1 Any valid method, involving 70

= 245 + 70 × 60 – 70 × 50
= 945 A1 Correct answer
[2]

(iii) coded mean = 945/70 = 13.5

2
Σ( x − 50) 2  945  2
−  = 10.6 M1 Using variance formula with coded mean
70  70 
Σ(x – 50)2 = 20623 (20600) A1 Correct answer
[2]

© UCLES 2010
 Page 5 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 63

5 (i)
2 to 4 4 to 6 6 to 7 7 to 8 8 to 10 10 to 16 M1 Using fd to evaluate freqs
20 44 34 30 30 36 A1 Any four correct
A1 All correct
[3]

(ii) mid-points 3, 5, 6.5, 7.5, 9, 13 M1 5 or 6 correct mid-points

E(X) = (3 × 20 + 5 × 44 + 6.5 × 34 + 7.5 × 30 +
9 × 30 + 13 × 36) / 194 = 1464/194

= 7.55 A1ft Correct answer, ft on 6 correct mid-

points and the frequencies in their table
[2]

(iii) p = 60/194 (0.309) B1ft 60/194 seen, ft on (their 30 + their 30) /

their total
P(1) = 2 × (60/194)(134/193) M1 multiplying a probability by 2
= 8040/18721 (0.429) A1 Correct answer
[3]
14 14
6 (i) P12 M1 P12 seen oe
= 4.36 × 1010 A1 Correct answer
[2]

(ii) business people 3! = 6 B1 3! oe seen, not in denominator

students 5! = 120 B1 5! oe seen, not in denominator
married couples 3P2 × 2 × 2 = 24 B1 24 oe seen, not in denominator
total ways = 17280 B1 correct final answer
[4]

(iii) Mrs Brown 3 B1 any 2 of 3, 10, 5 oe seen, not in

Mrs Lin 10 denominator
Student 5
Prob = 3 × 10 × 5 × 11P9 / (i) B1 11
P9 seen multiplied
M1 dividing by their (i)

= 0.0687 A1 correct answer

[4]
OR1 3/14 × 10/13 × 5/12 = 150/2184 (0.0687) B1 any 2 of numerators 3, 10, 5 oe seen
B1 denominators 14, 13, 12 of 3 fractions
M1 multiplying 3 separate fractions
A1 correct answer

OR2 1 − 3/14 = 11/14 B1 1 − 3/14 seen

1 − 11/14 × 5/13 = 127/182 B1 1 − 11/14 × 5/13 seen
8/14(4/13 × 12/12 + 9/13 × 7/12) + M1 attempt to find P(Mrs Lin not behind a
3/14(3/13 × 12/12 + 10/13 × 7/12) student and Mrs Brown not in front row),
= 1206/2184 involving 8/14 × prob + 3/14 × prob
1 − (1524 + 1716 − 1206)/2184 = 150/2184 A1 correct answer

© UCLES 2010
 Page 6 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 63

7 (i) z = 0.807 B1 0.807 seen

10 − 8.2
0.807 = M1 standardising, must have σ, no sq rt, no
σ cc and a z-value
s = 2.23 A1 correct answer
[3]

1
(ii) P(> 1 min from mean) = P(mod z > ) M1 standardising, their sd, no cc and adding
2.23 two areas
= P ( z > 0.4484) M1 using 1 – Φ(z)
= (1 – 0.6729) × 2
= 0.654 A1 correct answer
[3]

(iii) P(> 2 longer) = 1 – P(0, 1, 2 longer) M1 binomial term 6Cxpx(1 − p)6 − x

= 1 – {(0.79)6 + 6C1(0.21)(0.79)5 + A1 correct unsimplified answer

6
C2(0.21)2(0.79)4}

= 0.112 A1 correct answer

[3]

(iv) µ = 35 × 0.5 = 17.5 B1 17.5 and 8.75 or 8.75 seen

σ2 = 35 × 0.5 × 0.5 = 8.75

 15.5 − 17.5 
P(X < 16) = Φ  M1 standardising, with or without cc, must
 8.75  have sd in denom
M1 continuity correction 15.5 or 16.5 only,
seen
= 1 – Φ(0.676) M1 using 1 – Φ(z)
= 1 – 0.7505
= 0.2495 (0.249 or 0.250) A1 correct answer
[5]
OR 35C00.500.535 + 35C10.510.534 + 35C20.520.533 +... M1 binomial term 35Cx0.5x0.535 − x
= 8582372584/235 = 0.250 A1 at least 2 correct terms (x Þ 0) seen
M1 summing 16 or 17 terms
A1 correct expression
A1 correct answer

© UCLES 2010