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
9709/62 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.

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 62

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
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GCE A LEVEL – October/November 2010 9709 62

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 62

1 4p + 5p2 + 1.5p + 2.5p + 1.5p = 1 M1 Summing 5 probs to = 1 can be implied

10p2 + 19p – 2 = 0

p = 0.1 or –2 A1 For 0.1 seen with or without –2

p = 0.1 A1 Choosing 0.1 must be by rejecting –2

[3]

2 (i) Σ(x – 50) = 824 – 16 × 50 = 24 B1 Correct answer

2
Σ( x − 50) 2  Σ( x − 50)  2
−  = 6.5 M1 Consistent substituting in the correct
16  16  coded variance formula OR valid method
for Σx2 then expanding Σ(x – 50)2, 3 terms
at least 2 correct
Σ(x – 50)2 = 712 A1 Correct answer
[3]

(ii) new mean = 896/17 (= 52.7) B1 Correct answer

2
712 + 22 2  24 + (72 − 50) 
new var = −  M1 Using the correct coded variance formula
17  17  with n = 17 and new coded mean2 OR
their (Σx2 + 722)/17 – their new mean2
new sd = 7.94 A1 Rounding to correct answer, accept 7.95
or 7.98 or 7.91
[3]

© UCLES 2010
 Page 5 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 62

2 4 8
3 P(E and 12) = × = (2/45) M1 2/5 or 3/5 mult by dice-related
5 36 180 probability seen anywhere
2 4
A1 × seen oe
5 36
3 1 8 11
P(12) = × + = (0.0611) M1 Summing two 2-factor probs involving
5 36 180 180 2/5 and 3/5
A1ft 3/5 × 1/36 + their P(E and 12), ft their
P(E 12)
P( E and 12)
P( E 12) = M1dep Subst in condit prob formula, must have
P(12) a fraction
8
= (0.727) A1 Correct answer
11 [6]
OR list
Even: 2 and (4,3) or (3,4) or (2,6) or (6,2) M1 List attempt evens
4 and ditto
Gives 8 options A1 8 options

Odd: 1 and (6,6) or 3 and (6,6) or 5 and (6,6) M1 List attempt odds
Gives 3 options A1 3 options

Prob(E│12) = 8/11 M1 (Their even)/(their total)

A1 Correct answer

4 (i)
sugar flour B1 Correct stem must be integers.
(stem and leaves can be in reverse order)
194 159
195 3 B1 Correct leaves flour must be single and
81 196 24 ordered
7 197 7
943 198 B1 Correct leaves sugar must be single and
4 199 ordered
8 200
741 201

key
1 196 2

means 1.961 kg for sugar and 1.962 kg for flour B1ft Correct key needs all this, ft if single
leaves and 1.96 etc in stem
[4]

(ii) med = 1.989 kg B1 correct median

IQ range = 2.011 – 1.977 M1 subt their LQ from their UQ, UQ > med,
LQ < med

= 0.034 kg A1 Correct answer

[3]

© UCLES 2010
 Page 6 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 62

367 − 320
5 (i) Zotoc: z = = 2.176 M1 Standardising either car’s fuel, no cc, no
21.6 sq, no √
367 − 350
Ganmor: z = = 2.267
7.5

P(Zotoc) = 0.985 A1 Correct answer

P(Ganmor) = 0.988 A1 Correct answer

[3]

(ii) z = 0.23 B1 ± 0.23 seen

x − 320
0.23 = M1 Standardising either car, no cc, no sq rt,
21.6
no sq

x = 324.968 M1ind 320 + d – 320 i.e. just d on num

d = 4.97 A1 Correct answer, –4.97 gets A0

[4]

6 (i) constant/given prob, independent trials, B1 One option correct

fixed/given no. of trials, only two outcomes B1 Three options correct
[2]

(ii) P(8, 9, 0, 1) = M1 One term seen involving (0.3)x(0.7)9 – x(9Cx)

9
C8(0.3)8(0.7) + (0.3)9 + (0.7)9 + 9C1(0.3)(0.7)8 A1 Correct unsimplified expression

= 0.196 A1 Correct answer

[3]

(iii) mean = 90 × 0.3 = 27

var = 18.9 B1 Expressions for 27 and 18.9 (4.347) seen
 35.5 − 27 
P(X > 35) = 1 – Φ   M1 Standardising one expression, must have
 18.9  sq rt in denom, cc not necessary

= 1 – Φ(1.955) = 0.0253 M1 Continuity correction applied at least

 26.5 − 27  once
P(X < 27) = Φ  = 1 – Φ(0.115) M1 (1 – Φ1) + (1 – Φ2) accept (0.0329 + 0.5)
 18.9  if no cc
= 0.4542
Total prob = 0.480 accept 0.48 A1 Rounding to correct answer
[5]

© UCLES 2010
 Page 7 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – October/November 2010 9709 62

7 (i) 4M 2W or 5M 1W M1 At least 1 of 10C4 × 9C2 and 10C5 × 9C1

seen
chosen in 10C4 × 9C2 + 10C5 × 9C1 A1 Correct unsimplified expression
= 9828 A1 Correct answer
[3]
9
(ii) C3 × 8C1 + 9C4 = 798 M1 One of 9C3 × 8C1 and 9C4 × (8C0) seen

Prob = 798/9828 = 0.0812 A1 Correct answer

[2]

(iii) Albert + not T... 9C3 × 8C2 + 9C4 × 8C1 M1 One of 9C3 × 8C2 or 9C4 × 8C1 or
9
= 3360 C5 × (8C0) seen
Tracey + not A... 9C4 × 8C1 + 9C5
= 1134 A1 Unsimplified 3360 or 1134 seen

Number of ways = 4494 A1 Correct final answer

[3]

(iv) 6! – 4! × 5 × 2 or 6! – 5! × 2 (= 480)
OR 4! × 5 × 4 or 4! × 5P2 (= 480) B1 6! – 4! × 5 × 2 or 6! – 5! × 2
or 4! × 5 × 4 or 4! × 5P2
prob = 480/6! = 2/3 (0.667) M1 dividing by 6!
A1 correct answer
[3]
OR using probabilities…as above

OR Women together 5!/4! (= 5)

Women not together = 15 – 5 = 10 B1 5 or 10 seen
total ways MMMMWW = 6!/4!2! = 15 M1 Dividing by 15
prob = 2/3 A1 Correct answer

© UCLES 2010