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FP1 Complex Number Answers

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June 2013R Q6

4725 Mark Scheme June 2005

) (i) p = 2

(ii) a = 44

1

1

1

1

1 A1ft

1

Or use substitution u = x2 quadratic equation of correct

a nge and square

Obtain

(b

M A M A M

3

2

2

1

Write down a orm or re rraf

x2 + 4x + 16 = 0 Use sum or product of roots to obtain 6p =12

r 6p3 = 48 Obtain p = 2 Attempt to find

O

Σαβ numerically or in terms f p or substitute their 2, 4 or 6 in equation

Obtain 11p2

o

9.

⎜ 0 1 ⎠⎟

, e.g. (0,1) transforms to (3,1)

(iii) M =

(iv)

Mk =

1B1

1B1

1

B1

1

1

A1

A1

2

12

ach column correct

ne example or sensible explanation

t CD ) btain given answer

xplicit check for n = 1or n = 2

Induction hypothesis that result is true for Mk Attempt to multiply MMk or vice versa Element 3(2k+1 –1) derived correctly

ll other elements correct

xplicit statement of induction conclusion

(i) ⎛ 2 0 ⎞ ⎝

(ii) Shear

⎝

⎛⎜

20 31 ⎠

⎞⎟

⎝

⎛⎜

2k

0 3(2

k – 1)1 ⎠

⎞⎟ .

⎝

⎛⎜ 2

k + 1

0 3(2

k + 1 – 1)1 ⎠

⎞⎟ .

B

B

M1 A

M

M A1

2

2

6

E

O

Attempt to find DC (noO

E

A E

54

PMT

(Q9, June 2005)

1

4725 Mark Scheme January 2006

24

Mark Total 1.

(i) 2 + 16i –i –8i2 10 +15i (ii)

15(10 + 15i) or 2 + 3i

M1 A1 M1 A1 A1ft

2 3 5

Attempt to multiply correctly Obtain correct answer Multiply numerator & denominator by conjugate Obtain denominator 5 Their part (i) or 10 + 15i derived again / 5

2.

12 = 16 × 1 × 2 × 3

16 n(n + 1)(2n + 1) + (n + 1)2

16 (n + 1)(n + 2){2(n + 1) + 1}

B1 M1 DM1 A1 A1

5 5

Show result true for n = 1or 2 Add next term to given sum formula, any letter OK Attempt to factorise or expand and simplify Correct expression obtained Specific statement of induction conclusion, with no errors seen

3.

(i)

2⎣

⎡⎢

21

13 ⎦

⎤⎥ – 1

⎣

⎡⎢

11

13 ⎦

⎤⎥ + 3

⎣

⎡⎢

11

21 ⎦

⎤⎥

2 x 5 – 1 x 2 +3 x -1 5 (ii)

M1 A1 A1 B1ft

3 1 4

Show correct expansion process, allow sign slips Obtain correct (unsimplified) expression Obtain correct answer State that M is non-singular as det M non-zero, ft their determinant

4.

u2 + 4u + 4 u3 + 6u2 + 12u + 8

u = 3 5

x = 2 + 3 5

B1 M1 A1 A1ft A1ft

5 5

u + 2 squared and cubed correctly Substitute these and attempt to simplify Obtain u3 – 5 = 0 or equivalent Correct solution to their equation Obtain 2 + their answer [ Decimals score 0/2 of final A marks]

PMT

(Q2, Jan 2006)

2


28

7. (i)

A2 = ⎝

⎛⎜

40 01 ⎠

⎞⎟ A

3 = ⎝

⎛⎜

80 01 ⎠

⎞⎟

(ii) An = ⎝

⎛⎜

2n

0 01 ⎠

⎞⎟

(iii)

M1 A1 A1 B1 B1 M1 A1 A1

3 1 4 8

Attempt at matrix multiplication Correct A2 Correct A3 Sensible conjecture made State that conjecture is true for n = 1 or 2 Attempt to multiply An and A or vice versa Obtain correct matrix Statement of induction conclusion

8. (i)

a⎣

⎡⎢

a2

01 ⎦

⎤⎥ – 4

⎣

⎡⎢

11

01 ⎦

⎤⎥ + 2

⎣

⎡⎢

11

a2 ⎦

⎤⎥

a2 – 2a

(ii)

a = 0 or a = 2

(iii) (a)

(b)

M1 A1 A1 M1 A1A1ft B1 B1 B1 B1

3 3 4 10

Correct expansion process shown Obtain correct unsimplified expression Obtain correct answer Solve their det M = 0 Obtain correct answers Solution, as inverse matrix exists or M non-singular or detM 0≠ Solutions, eqn. 1 is multiple of eqn 3

PMT

(Q7, June 2006)

3

PhysicsAndMathsTutor.com

4725 Mark Scheme Jan 2007

21

(ii)

±−1 i 3 (iii)

M1 A1 A1 B1

B1

B1

3

3

7

Attempt to solve quadratic equation or substitute x + iy and equate real and imaginary parts Obtain answers as complex numbers Obtain correct answers, simplified Correct root on x axis, co-ords. shown Other roots in 2nd and 3rd quadrants Correct lengths and angles or co-ordinates or complex numbers shown

6. (i)

un+1 – un = 2n +4

(ii)

B1

M1

A1

B1

M1

M1

A1

A1

3

5

8

Correct expression for un+1

Attempt to expand and simplify

Obtain given answer correctly

State u1 = 4 ( or u2 = 10 )and is divisible by 2 State induction hypothesis true for

un

Attempt to use result in (ii)

Correct conclusion reached for un+1

Clear,explicit statement of induction conclusion

7. (i) α + β = – 5 αβ = 10

(ii) α2 + β2 = 5

(iii)

01212 =+− xx

B1 B1

M1

A1

B1

M1

A1

B1ft

2

2

4

8

State correct values

Use (α + β)2 – 2αβ

Obtain given answer correctly, using value of -5 Product of roots = 1

Attempt to find sum of roots

Obtain 105 or equivalent

Write down required quadratic

equation, or any multiple.

PMT

(Q6, Jan 2007)

4


20

1 EITHER

a = 2

,32=b OR

a = 2 32=b

M1 A1 M1 A1 M1 M1 A1 A1

4

4

Use trig to find an expression for a (or b) Obtain correct answer Attempt to find other value Obtain correct answer a.e.f. (Allow 3.46 ) State 2 equations for a and b Attempt to solve these equations Obtain correct answers a.e.f. SR ± scores A1 only

2 (13 = ) 1

4 × 12 × 22

14 n2(n + 1)2 + (n + 1)3

14 (n + 1)2(n + 2)2

B1 M1 M1(indep) A1 A1

5

5

Show result true for n = 1 Add next term to given sum formula Attempt to factorise and simplify Correct expression obtained convincingly Specific statement of induction conclusion

3 3Σr2 – 3Σr + Σ 1

3Σr2 = 1

2 n(n + 1)(2n + 1)

3Σr = 3

2 n(n + 1)

Σ 1 = n

n3

M1 A1 A1 A1 M1 A1

6

6

Consider the sum of three separate terms Correct formula stated Correct formula stated Correct term seen Attempt to simplify Obtain given answer correctly

4

(i) 21

⎝

⎛⎜

5– 3

– 11 ⎠

⎞⎟

(ii)

21

⎝

⎛⎜

223

0– 5 ⎠

⎞⎟

B1 B1 M1 M1(indep) A1ft A1ft

2

4 6

Transpose leading diagonal and negate other diagonal or solve sim. eqns. to get 1st column Divide by the determinant or solve 2nd pair to get 2nd column Attempt to use B-1A-1 or find B Attempt at matrix multiplication One element correct, a.e.f, All elements correct, a.e.f. NB ft consistent with their (i)

PMT

(Q2, June 2007)

5


17

6 (i)

(ii) 232 + i

B1 B1 B1 B1 B1 B1 M1 A1

5

3 8

Horizontal straight line in 2 quadrants Through (0, 2) Straight line Through O with positive slope In 1st quadrant only State or obtain algebraically that y = 2 Use suitable trigonometry Obtain correct answer a.e.f. decimals OK must be a complex number

7 (i) a = -6

(ii) A-1 = ⎟⎟⎠

⎞⎜⎜⎝

⎛ −+ aa 2

316

1

6

26

4 , +−

+ == aa

a yx

M1 A1 B1 B1ft M1 A1ft A1ft

2

5 7

Use det A = 0 Obtain correct answer Both diagonals correct Divide by det A Premultiply column by A-1, no other method Obtain correct answers from their A-1

8 (i) u2 = 4, u3 = 9, u4 = 16 (ii) un = n2 (iii)

M1 A1 B1 B1 M1 A1 A1

2

1

4 7

Obtain next terms All terms correct Sensible conjecture made State that conjecture is true for n = 1 or 2 Find un+1 in terms of n Obtain ( n + 1 )2 Statement of Induction conclusion

9 (i) 3223 33 βαββαα +++ (ii) Either 7,5 ==+ αββα 2033 =+ βα x2 – 20x + 343 = 0 Or

075 31

32

=+− uu 0343203 =+− uu

M1 A1 B1 B1 M1 A1

M1 A1ft M1 A1 M2 A2

2

6

8

Correct binomial expansion seen Obtain given answer with no errors seen State or use correct values Find numeric value for 33 βα + Obtain correct answer Use new sum and product correctly in quadratic expression Obtain correct equation

Substitute 31

ux = Obtain correct answer Complete method for removing fractional powers Obtain correct answer

2

PMT

(Q8, Jan 2008)

6



4725 Further Pure Mathematics 1

1 (i) B1 Two elements correct 1 15 1⎛ ⎞⎜ ⎟−⎝ ⎠

B1 All four elements correct 2

(ii) EITHER B1 Both diagonals correct

2 115 43

−⎛ ⎞⎜ ⎟−⎝ ⎠

B1 Divide by determinant

2 OR B1 Solve sim. eqns. 1st column correct B1 2nd column correct 2 (i) 5 B1 Correct modulus 0.927 or 53.1o B1 Correct argument, any equivalent form 2 (ii)(a) B1 Circle centre A (3, 4) B1 Through O, allow if centre is (4, 3) 2 (b) B1 Half line with +ve slope B1 Starting at (3, 0) B1 Parallel to OA, (implied by correct arg shown) 3

A(3, 4)

O

3 (i) ( 1)!

rr +

M1 Common denominator of (r + 1)! or r!(r + 1)!

A1 Obtain given answer correctly 2

(ii) 11( 1)n

−+ !

M1 Express terms as differences using (i)

A1 At least 1st two and last term correct M1 Show pairs cancelling A1 Correct answer a.e.f. 4 4 B1 Establish result is true, for n = 1 ( or 2 or 3 ) M1 Attempt to multiply A and An, or vice versa M1 Correct process for matrix multiplication A1 Obtain 3n + 1, 0 and 1 A1 Obtain ½(3n + 1 – 1) A1 Statement of Induction conclusion, only

if 5 marks earned, but may be in body of working

6

17

PMT

7

(Q4, June 2008)

7

(i) nnnn 613613 11 +++ +− (ii)

B1 M1 A1 B1 B1 B1 B1

3 4 7

Correct expression seen Attempt to factorise both terms in (i) Obtain correct expression Check that result is true for n =1 ( or 2) Recognise that (i) is divisible by 7 Deduce that un+1 is divisible by 7 Clear statement of Induction conclusion

8 (i) (ii) 2,6 kk ==+ αββα k)24(=− βα (iii) ∑ = k6'α 1)('' −−−= βααββα 1)24('' 2 −−= kkβα 01)24(6 22 =−−+− kkkxx

M1 A1 B1 B1 M1 A1 B1ft M1 A1ft B1ft

2 4 4

10

Expand at least 1 of the brackets Derive given answer correctly State or use correct values Find value of βα − using (i) Obtain given value correctly ( allow if –6k used ) Sum of new roots stated or used Express new product in terms of old roots Obtain correct value for new product Write down correct quadratic equation

9 (i) (ii) 12

112

1311 +− −−+ nn

(iii) 3

4

M1 A1 M1 M1 A1 A1 M1 A1 B1ft

2 6 1 9

Use correct denominator Obtain given answer correctly Express terms as differences using (i) Do this for at least 1st 3 terms First 3 terms all correct Last 3 terms all correct ( in terms or n or r) Show pairs cancelling Obtain correct answer, a.e.f.( in terms of n) Given answer deduced correctly, ft their (ii)

PMT

(Q7, Jan 2009)

8

9. (i)

111

2111

211 aa

a

aa 22 2 (ii) a = 0 or 1 (iii) (a) (b)

M1 A1 A1 M1 A1ft A1ft B1 B1 B1 B1

3

3

4 10

Correct expansion process shown Obtain correct unsimplified expression Obtain correct answer Equate their det to 0 Obtain correct answers, ft solving a quadratic Equations consistent, but non unique solutions Correct equations seen & inconsistent, no solutions

10. i) u2 = 7 u3 =19 (ii) un = 2(3n – 1 )+ 1 (iii) un+1 = 3(2(3n – 1)+1) – 2 un +1 = 2(3n) + 1

M1 A1 A1 M1 A1 B1ft M1 A1 A1 B1

3

2

5 10

Attempt to find next 2 terms Obtain correct answers Show given result correctly Expression involving a power of 3 Obtain correct answer Verify result true when n = 1 or n = 2 Expression for un +1 using recurrence relation Correct unsimplified answer Correct answer in correct form Statement of induction conclusion

PMT

(Q10, June 2009)

9



19

−−

∆3

5475

1 aa

8 (i) M1 Attempt to equate real and imaginary parts of (x + iy)2 & 5 – 12i 522 =− yx and 6−=xy A1 Obtain both results, a.e.f M1 Obtain quadratic in x2 or y2 M1 Solve to obtain 3)(±=x or 2)(±=y ± (3 – 2i) A1 5 Obtain correct answers as complex nos -------------------------------------------------------------------------------------------------------------------------------------- (ii) B1ft Circle with centre at their square root B1 Circle passing through origin B1ft 2nd circle centre correct relative to 1st B1 4 Circle passing through origin 9 9 (i) M1 Show correct expansion process for 3 × 3 or multiply adjoint by A M1 Correct evaluation of any 2 × 2 at any stage det A = 66 −=∆ a A1 Obtain correct answer

A-1 = −−

−−−+−

∆63321214113

1 aaa

M1 Show correct process for adjoint entries

A1 Obtain at least 4 correct entries in adjoint B1 Divide by their determinant A1 7 Obtain completely correct answer ----------------------------------------------------------------------------------------------------------------------------------------- (ii) M1 Attempt product of form A-1C or eliminate to get 2 equations and solve A1A1A1 Obtain correct answer ft all 3 4 S.C. if det now omitted, allow max A2 ft 11 10 (i) B1 Correct M2 seen

M2 = 1041 M3 =

1061 M1 Convincing attempt at matrix

multiplication for M3 A1 3 Obtain correct answer -----------------------------------------------------------------------------------------------------------------------------------------

(ii) Mn = 1021 n B1ft 1 State correct form, consistent with (i)

--------------------------------------------------------------------------------------------------------------------------------------

PMT


20

10 (iii) M1 Correct attempt to multiply M & Mk or v.v. A1 Obtain element 2( k + 1 ) A1 Clear statement of induction step, from correct working B1 4 Clear statement of induction conclusion, following their working ---------------------------------------------------------------------------------------------------------------------------------------- (iv) B1 Shear DB1 x-axis invariant DB1 3 e.g. ( 1, 1 ) → ( 21, 1 ) or equivalent using scale factor or angles 11

PMT

(Q10, Jan 2010)

10

1 B1 M1 M1 A1 A1 5

5

Establish result true for n = 1or n = 2 Add next term to given sum formula Attempt to factorise or expand and simplify to correct expression Correct expression obtained Specific statement of induction conclusion

2 (i) M1

A1 2 Obtain a single value Obtain correct answer as a matrix

( )7−

(ii) BA =

M1 A1 B1 B1ft 4

6

−−

123205

−−−

2011207

Obtain a 2 2 matrix All elements correct 4C seen or implied by correct answer Obtain correct answer, ft for a slip in BA

×

3 Either

nnnnnn ++−++ )1(2)12)(1(32

)12)(12(31 +− nnn Or

==

−n

r

n

r

rr1

22

1

2 4

)12)(1(4)14)(12(2 61

61 ++×−++× nnnnnn

)12)(12(31 +− nnn

M1 M1

A1 M1 A1 A1 6 M1

M1 A1 M1 A1 A1

6

Express as a sum of 3 terms Use standard sum results Correct unsimplified answer Attempt to factorise Obtain at least factor of n and a quadratic Obtain correct answer a.e.f. Express as difference of 2 series

Use standard result Correct unsimplified answer Attempt to factorise Obtain at least factor of n Obtain correct answer

2r

PMT

(Q1, June 2010)


1

1 (i) ) B1B1 2 Each element correct SC (7,9) scores B1 ----------------------------------------------------------------------------------------------------------------------------------------------- (ii) B1* Obtain correct value depB1 2 Clearly given as a matrix -----------------------------------------------------------------------------------------------------------------------------------------------

(iii) M1 Obtain 22 × matrix

A1 Obtain 2 correct elements A1 3 Obtain other 2 correct elements 7

( 97

( )18

−−

26412

2. (i) – 12 +13i B1B1 2 Real and imaginary parts correct ------------------------------------------------------------------------------------------------------------------------------------------------ (ii) B1 z* seen M1 Multiply by w*

3714

3727 − i A1 Obtain correct real part or numerator

A1 4 Obtain correct imaginary part or denom. Sufficient working must be shown 6 3 B1* Establish result true for n = 1 or 2 M1* Use given result in recurrence relation in a relevant way A1* Obtain 2n + 1 correctly depA1 4 Specific statement of induction conclusion 4

4 Either B1 Correct value for stated or used M1 Express as sum of two series

r

)1(2

)1(4

22 +++ nbnnna A1 Obtain correct unsimplified answer

M1 Compare coefficients or substitute values for n a = 4 b = – 4 A1 A1 6 Obtain correct answers Or M1 Use 2 values for n a + b = 0 4a + b = 12 A1 A1 Obtain correct equations M1 Solve simultaneous equations a = 4 b = – 4 A1 A1 Obtain correct answers 6 5 B1 (A-1)-1 = A seen or implied M1 Use product inverse correctly A2 A1cao 3 Obtain correct answer 3

PMT

(Q3, Jan 2011)

12

11

1 (i) B1 3B seen or implied

B1 2 elements correct B1 3 Other 2 elements correct, a.e.f., including brackets

01244 a

(ii) M1 Sensible attempt at matrix multiplication

for AB or BA A1 2 Obtain correct answer 5

14344 aa

2 B1 Establish result true for n = 1 or 2 M1* Add next term to given sum formula DM1 Combine with correct denominator A1 Obtain correct expression convincingly A1 5 Specific statement of induction conclusion, provided 1st 4 marks earned 5

3 B1 Obtain correct det M1 Equate their det to 0 A1 3 Obtain correct answers 3

162k

4k

4 M1 Express as sum of two series nnnn 2)14)(12(23 2

161 A1 A1 Each term correct a.e.f.

M1 Attempt to factorise A2 6 Completely correct answer, ( A1 if one factor not found ) 6

)34(2 2 nn

2a 5 (i) B1 Correct modulus

3,60arg a , 1.05 B1 2 Correct argument (ii) B1 Circle B1 Centre )3,1(

B1 Through origin, centre )3,1( and another y intercept B1 Vertical line B1* Through a or their centre, with +ve gradient DB1 Correct half line 6 8

PMT

(Q2, June 2011)

13



7

Question Answer Marks Guidance 7 (i) M1 Attempt at matrix multiplication A1 Obtain M2 correctly A1 Obtain given answer correctly [3]

7 (ii) 3 0

3 1 1

n

n

B1 B1

3 elements correct 4th element correct

[2]

7 (iii) B1 Show that their result is true for n = 1 or 2 M1 Attempt to find Mk .M or vice versa 1

1

3 03 1 1

k

k A1 Obtain correct answer

B1 Complete statement of induction conclusion Must have 1st 3 marks [4]

8 (i) M1 Combine with a common denominator A1 Obtain given answer correctly [2] 8 (ii) M1 Express terms using (i) A1 At least 1st two and last two correct

1n

n M1 Show terms cancelling

A1 Obtain correct answer, in terms of n [4]

PMT

(Q7, Jan 2012)

14Question Answer Marks Guidance

EITHER 3 M1 Use sum of root and conjugate a = –8 A1 Obtain correct answer M1 Use product of root and conjugate b = 25 A1 Obtain correct answer [4] OR M1 Substitute or conjugate into equation 4 + 3i M1 Equate real and imaginary parts a = –8 A1 Obtain correct answer b = 25 A1 Obtain correct answer 4 M1 Express as sum of 3 series M1 Use standard series results, at least 1 correct A1 Two terms correct 312 2( 1)(2 1) ( 1) 2n n n n n n A1 Third term correct

M1 Obtain factor of n 2( 1)n n A2 Obtain correct answer c.a.o.

Allow A1 for

)22(212n

[7] 5 B1 Verify result true when n = 1 M1* Add next term in series DepM1 Attempt to obtain correctly A1 Show sufficient working to justify correct

expression

B1 Clear statements of Induction processes, but 1st 4 marks must all be earned.

[5]

PMT

(Q5, June 2012)

15


8

Question Answer Marks Guidance 9 (i) M1 Attempt at complete expansion

A1 Obtain correct unsimplified answer A1 Obtain given answer correctly [3] 9

(ii)

Either

4 3, ,p

B1

State (anywhere) correct values for , ,

M1 Express given expression as a single fraction A1 Obtain correct expression using (i) M1 Use their values for sum of roots etc. in their expression 16 6

9p

A1 Obtain correct answer

Or

3 2 29 (6 16) (8 ) 1 0u p u p u 16 6

9p

[5]

B1 M1 A1 M1 A1

Use substitution 1 / u Rearrange appropriately and square out Obtain correct co-efficients of u3 and u2 Use (+/-)b/a from their cubic Obtain correct answer

10 (i) 23

, 25

, 27

B1 B1 B1

B1 x 3, Obtain 3 correct values Justify given answer

[3] 10 (ii) 2

2 1n

M1 Fraction, in terms of n, with correct numerator or denominator A1 Obtain correct answer a.e.f.

[2] 10 (iii) 2

2 1 1( )n

B1ft Verify result true when n = 1, for their (ii), or n = 2, 3 or 4 M1 Expression for un + 1 using recurrence relation in terms of n using their (ii)

A1 Correct unsimplified answer A1 Correct answer in correct form B1

[5] Specific statement of induction conclusion, previous 4 marks must be earned, n=1 must be verified

PMT

(Q10, Jan 2013)

16


5

Question Answer Marks Guidance

1 M1 Use correct trig expression √3 A1 Obtain correct answer M1 Correct expression for modulus 2√3 A1FT Obtain correct answer aef 3 – 3i B1FT Correct conjugate seen or implied –√3 i B1FT Correct answer [6] 2 (i) 7 23 B1B1 Each element correct, missing brackets B1 only [2] 2 (ii) 6 15

4 10

M1 A1 A1

Obtain 2 × 2 matrix Obtain 2 correct elements Obtain other 2 correct elements

det CB = 0 A1FT Obtain their det CB, must be a 2 × 2 matrix singular A1FT Correct conclusion from their det CB [5] 3 2 2 11 and 6 5x y xy M1 Attempt to equate real and imaginary parts of (x + iy)2 and 11 + 12√5 A1 Obtain both results cao M1* Obtain a quadratic in x2 or y2 ±( 2√5 + 3i) DM1 Solve a 3 term quadratic to obtain a value for x or y A1

A1 Obtain 1 correct answer as complex number Obtain only the other correct answer

[6] 4 B1 Establish result true for n = 1 or n = 2 M1 Multiply M and Mk, either order 12 2 2 2k or 1 12 2 2k k A1 Obtain correct element A1 Obtain other 3 correct elements A1 Obtain 22 2k convincingly B1 Specific statement of induction conclusion, provided 5/5 earned so far and verified for n = 1 [6]

PMT

(Q4, June 2013)

17


FP1 Series Answers

D1 Algorithms – Sorting PhysicsAndMathsTutor.com

1. (a) E.g: 650 431 245 643 455 710 234 162 452 134 M1

650 643 710 455 431 245 234 162 452 134 A1

650 710 643 455 431 245 452 234 162 134 A1 ft

710 650 643 455 431 452 245 234 162 134 A1 ft

710 650 643 455 452 431 245 234 162 134 A1

5

(b) Bin 1 710 + 245 Bin 3 643 + 162 + 134 Bin 5 431 M1A1

Bin 2 650 + 234 Bin 4 455 + 452 A1A1(ft) 4

(c) 10004116 = 4.116 5 bins needed optimal M1A1(ft) 2

[11]

2. (a) 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 2.6 4.0 3.2 4.0 2.6 0.6 2.5 0.5 0.4 0.3 1.0 3.2 0.4 M1 4.0 4.0 3.2 2.6 0.6 2.5 0.5 1.0 0.4 0.3 4.0 0.5 A1 4.0 4.0 3.2 2.6 0.6 2.5 1.0 0.5 0.4 0.3 2.5 A1ft 4.0 4.0 3.2 2.6 2.5 0.6 1.0 0.5 0.4 0.3 1.0 A1ft 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3 A1 cso 5

Notes

1M1 Pivot, p, chosen. List sorted, >p, p.

1

2A1ft 2nd pass correct and chosen next two pivots correctly for sublists >1

3A1ft 3rd pass correct and next pivot for sublist >1 chosen correctly.

4A1 cso.

Misread in part (a)

• If they have misread a number at the start of part (a), so genuinely miscopied and got for example 0.1 instead of 1.0 then mark the whole question as a misread – removing the last two A or B marks earned. This gives a maximum total of 9.

• If they misread their own numbers

Edexcel Internal Review 6


during the course of part (a) then count it as an error in part (a) but mark parts (b) and (c) as a misread. So they would lose marks in (a) for the error and then the last two A or B marks earned in (b) and (c) – giving a maximum of 8 or maybe 7 marks depending on how many marks they lose in (a).

The most popular misread is the one listed above – where 1.0 has changed to 0.1 giving

4.0 4.0 3.2 2.6 2.5 0.6 0.5 0.4 0.3 0.1 at the end of (a) for this one (b) and (c) are:

(b) Length 1: 4

Length 2: 4

Length 3: 3.2 0.6 left column & 1.0 in place M1

Length 4: 2.6 1.0 0.4 0.6 & 0.5 A1

Length 5: 2.5 0.5 0.3 0.4 A1

All correct (c.s.o) A1 4

Note

Length 1: 4

Length 2: 4

Length 3: 3.2 0.6 0.1

Length 4: 2.6 0.5 0.4 0.3

Length 5: 2.5

(c) 19.1/4 = 4.775 so 5 lengths needed, accept total is 19.1m, or refer to 0.9 ‘spare . B1

Yes, the answer to (b) does use the minimum number of bins. DB1 2

Note

18.2/4 = 4.55 so 5 bins, or total is 18.2 or 1.8 ‘spare’

Yes answer in (b) uses the minimum number of bins.

Alternate

Choosing middle left 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 0.5) 0.6 4.0 2.5 3.2 2.6 4.0 1.0 0.5 0.4 0.3 (pivots 3.2, 0.4) 4.0 4.0 3.2 0.6 2.5 2.6 1.0 0.5 0.4 0.3 (pivots 4.0, 2.5) 4.0 4.0 3.2 2.6 2.5 0.6 1.0 0.5 0.4 0.3 (pivot 0.6) 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3



Choosing first 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 0.6) 4.0 2.5 3.2 2.6 4.0 1.0 0.6 0.5 0.4 0.3 (pivots 4.0, 0.5) 4.0 2.5 3.2 2.6 4.0 1.0 0.6 0.5 0.4 0.3 (pivots 2.5, 0.4) 4.0 3.2 2.6 4.0 2.5 1.0 0.6 0.5 0.4 0.3 (pivot 3.2) 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3

OR (alternate choosing first) 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 0.6) 4.0 2.5 3.2 2.6 4.0 1.0 0.6 0.5 0.4 0.3 (pivots 4.0, 0.5) 4.0 4.0 2.5 3.2 2.6 1.0 0.6 0.5 0.4 0.3 (pivots 2.5, 0.4) 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3 (pivots 3.2) 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3 4.0 4.0 3.2 2.6 2.5 1.0 0.6 0.5 0.4 0.3

Sorting into ASCENDING order (full marks if then reversed, otherwise MISREAD)

Middle left 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 0.5) 0.4 0.3 0.5 0.6 4.0 2.5 3.2 2.6 4.0 1.0 (pivots 0.4, 3.2) 0.3 0.4 0.5 0.6 2.5 2.6 1.0 3.2 4.0 4.0 (pivots 2.5, 4.0) 0.3 0.4 0.5 0.6 1.0 2.5 2.6 3.2 4.0 4.0 (pivot 0.6) 0.3 0.4 0.5 0.6 1.0 2.5 2.6 3.2 4.0 4.0

Middle right 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 2.6) 0.6 2.5 0.5 0.4 0.3 1.0 2.6 4.0 3.2 4.0 (pivots 0.4, 3.2) 0.3 0.4 0.6 2.5 0.5 1.0 2.6 3.2 4.0 4.0 (pivots 0.5, 4.0) 0.3 0.4 0.5 0.6 2.5 1.0 2.6 3.2 4.0 4.0 (pivot 2.5) 0.3 0.4 0.5 0.6 1.0 2.5 2.6 3.2 4.0 4.0 (pivot 1.0)

First (1) 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 0.6) 0.5 0.4 0.3 0.6 4.0 2.5 3.2 2.6 4.0 1.0 (pivots 0.5, 4.0) 0.4 0.3 0.5 0.6 2.5 3.2 2.6 1.0 4.0 4.0 (pivots 0.4, 2.5) 0.3 0.4 0.5 0.6 1.0 2.5 3.2 2.6 4.0 4.0 (pivot 3.2) 0.3 0.4 0.5 0.6 1.0 2.5 2.6 3.2 4.0 4.0

First (2) 0.6 4.0 2.5 3.2 0.5 2.6 0.4 0.3 4.0 1.0 (pivot 0.6) 0.5 0.4 0.3 0.6 4.0 2.5 3.2 2.6 4.0 1.0 (pivots 0.5, 4.0) 0.4 0.3 0.5 0.6 2.5 3.2 2.6 1.0 4.0 4.0 (pivots 0.4, 2.5) 0.3 0.4 0.5 0.6 1.0 2.5 3.2 2.6 4.0 4.0 (pivot 3.2) 0.3 0.4 0.5 0.6 1.0 2.5 2.6 3.2 4.0 4.0

[11]



3. (a) M J E K H B L P N D B M1 1A1

B M J E K H L P N D H

B E D H M J K L P N D L 2A1ft

B D E H J K L M P N (E) K P

B D E H J K L M N P (J) N 3A1ft

B D E H J K L M N P (M)

Sort completed 4A1 5

Note

1M1: quick sort, pivots, p, identified, two sublists one

p.

If choosing one pivot only per iteration, M1 only.

1A1: first pass correct, next pivot(s) chosen consistently.

2A1ft: second pass correct, next pivot(s) chosen consistently

3A1ft: third pass correct, next pivot(s) chosen consistently

4A1: cso List re-written or end statement made or each element been chosen as a pivot.

(b) 62101

=

+ Katie reject left M1

92107

=

+ Natsuko reject right 1A1

82

87=

+ Miri reject right 2A1ft

7 = Louis name found 3A1 4

Note

1M1: binary search, choosing pivot rejecting half list.

If using unordered list then M0.

If choosing J M1 ony

1A1: first two passes correct, condone ‘sticky’pivots here, bod.

2A1ft: third pass correct, pivots rejected.

3A1: cso, including success statement.



Special case

If just one letter out of order, award maximum of M1A1A0A0

[9]

4. e.g. 52 48 50 45 64 47 53 M1

52 50 48 54 47 53 45

52 50 54 48 53 47 45 A1

52 54 50 53 48 47 45

64 52 53 50 48 47 45 A1ft

64 53 52 50 48 47 45 A1 4

No further changes – list sorted

M1 Bubble sort – 1st pass complete – end term 64 or 45, consistent L → R or R → L shuffle, Quick etc gets M0

A1 First 2 passes correct} condone shrinking list

A1ft Next 2 passes correct (if L → R next pass} condone shrinking list

A1 Final pass and final statement/rewritten list cso – must see whole list [4]

Notes

Bubble R → L

52 48 50 45 64 47 53 M1

64 52 48 50 45 53 47

64 53 52 48 50 45 47 A1

64 53 52 50 48 47 45 A1

No further changes – list sorted A1

Misreads – sorting into ascending order

(note – if candidates reverse list full credit is gained.

L → R (ascending – misread) MR

52 48 50 45 64 47 53 M1

48 50 45 52 47 53 64

48 45 50 47 52 53 64 A1

45 48 47 50 52 53 64

45 47 48 50 52 53 64 A1

No further changes – list sorted A1 (4-2 for misread)



R → L

52 48 50 45 64 47 53 M1

45 52 48 50 47 64 53

45 47 52 48 50 53 64 A1

45 47 48 52 50 53 64

45 47 48 50 52 53 64 A1

No further changes – list sorted A1 (4-2 for misread)

5. e.g.

747474747474

286363686868

63546863

63

546854

54

5454

54

492849

49

37492837

37

683737282828

54546368

M1A1

A1ft

A1

4937

(28)

sort complete

4

∴Ali, Sophie, Eun–Jung, {Katie + Marciana}, Peter, Rory, Bobby A1 M1 Pivot clear list > P >. Bubble sort etc. M0 A1 1st pass correct, next pivots correctly selected consistently A1ft 2nd + 3rd passes correct, pivots for next pan selected consistently each time. Penalise fragmented list here (or lists rewritten or all chosen as pivots) A1 c.s.o. + stop statement (o.e.). Penalise non-sig no. errors here. Penalise “sloppyness” here A1 c.a.o. accept c.a. even if MR

[5]

Alternative correct answers

(i)

74747474

28636868

636863

5454

54285454

495449

37492837

68373728

546368

M1A1

A1ft

4937 (54)



(ii)

74747474

28636868

635463

54685454

5454

492849

37492837

68373728

546374

M1A1

A1ft

4928 (54)

(iii)

74747474

28636868

636863

5454

542854

49542849

3749492837

6837373728

546374

M1A1

A1ft

544928 (68)

(37)

1st in list

(iv)

7474

74

2828636868

68

63635463

63

5454545454

54

545449545454

54

4949374949494949

3737683737373737

686828

28

742863

(68) 545449

(37)

M1

A1ft

A1

Ali, Sophie, Eun-Jung, Katie + Marciana, Peter, Rory, Bobby

MISREADS (–2 for MR)

(a)

74282828

28543737

634949

54375454

5454

497463

37637468

68686874

544937

M1A1

A1ft

6368 (54)

MR

(b)

74282828

28493737

633749

5454

547454

49637463

3754637468

6868686874

54493768

M1A1

A1ft

5463

(28)

MR



(c)

7428282828

2854493737

63493749

543754

5454

497463

37637468

68686874

54544928

M1A1

A1ft

6374

MR

(d)

74282828

28493737

633749

5454

54745454

496363

37547468

68686874

544928

M1A1

A1ft

6374 (54)

MR

If candidates reverse list then restore full marks.

names or numbers

Bobby, Rory, Peter, Katie + Marciana, Eun-Jung, Sophie, Ali

6. (a) E.g. M1 A1 A1ft A1ft A1 5 650650650710710

431643710650650

245710643643643

643455455455455

455431431431452

710245245452431

234234452245245

162162234234234

452452162162162

134134134134134

(b) Bin 1 710 + 245 Bin 3 643 + 162 + 134 Bin 5 431 M1 A1 Bin 2 650 + 234 Bin 4 455 + 452 A1ft A1 4

(c) eg.

10004116 = 4.116 ∴ 5 bins needed ∴ optimal M1 A1ft 2

[11]

7. (a) eg. 45 37 18 46 56 79 90 81 51 or 37 18 45 56 79 46 90 81 51 or 45 37 46 18 51 56 79 90 81 M1A1 2

(b) 56 45 79 46 37 90 81 51 18 or 90 45 56 37 79 46 18 81 51 M1A1 2



(c)

+

2111 = 6 value 44 discard top M1

+

2117 = 9 value 71 discard top A1

+

21110 = 11 value 94 discard bottom A1

list reduces to 10th value. This is 73 so 73 has been located as the 10th value A1 4

[8]

8. (a) The list is not in alphabetical order B1 1

(b) Use of Bubble Sort or Quick Sort M1 e.g.

Bubble sort

GBBBBBB

NGCCCCC

MNGEEEE

YMNGGGG

LYMNLLL

BLYMNMM

CCLYMNN

EEELYPP

SPPPPYS

PSSSSSY

1 pass2 pass3 pass4 pass5 pass6 pass

st

nd

rd

th

th

th

No more changes

Quick sort

GBBBBBB

NGGCCCC

MNCGEEE

YMEEGGG

LYLLLLL

BLNNNMM

CCMMMNN

EEYSPPP

SSSPSSS

PPPYYYY

1 pass2 pass3 pass4 pass5 pass6 pass

st

nd

rd

th

th

th

No sublists > 2 and no more changes

A1 A1ft No more changes No sublists > 2 + no more changes A1cso 4



(c) 1 2 3 4 5 6 7 8 9 10 B C E G L M N P S Y

2

]110[ + = 6 Manchester discard first half of list and pivot M1 A1

2]107[ + = 9 Southampton discard last half of list and pivot

2]87[ + = 8 Plymouth discard last half of list and pivot A1ft

Final term 7 Newcastle ∴ word found at 7 A1cso 4 [9]

9. (a) e.g.

RBBBB

PHGGG

BGHHH

YKKKK

TRRMM

KPPPP

MYMRR

HTTTT

WMYWW

GWWYY

M1 A1 A1 ft A1 ft A1 ft

5

(b)

+

2110 = 6 Palmer; reject Palmer → Young M1 A1

+

215 = 3 Halliwell; reject Boase → Halliwell A1

+

254 = 5 Morris; reject Morris

List reduces to Kenney – name found, search complete A1 4 [9]

10. (a) (i) le ft to right or right to le ft

25 22 30 18 29 21 27 21 25 22 30 18 29 21 27 21 M1 25 30 22 18 29 21 27 21 25 22 30 18 29 27 28 21

25 30 22 29 18 21 27 21 25 22 30 29 18 27 21 21

25 30 22 29 21 18 27 21 25 30 22 29 18 27 21 21

25 30 22 29 21 27 18 21 30 25 22 29 18 27 21 21 A1 (pass) (ii) 25 30 22 29 21 27 21 18 30 29 25 22 27 18 21 21

30 25 29 22 27 21 21 18 30 29 27 25 22 21 18 21

30 29 25 27 22 21 21 18 30 29 27 25 22 21 21 18

30 29 27 25 22 21 21 18 30 29 27 25 22 21 21 18



30 29 27 25 22 21 21 18

(b) (i) rod 1 30 18 2 29 21 3 27 22 M1 (to the 22) 4 25 21 A1 2

(ii) 193 ÷ 50 = 3.86, ∴ 4 rods needed, so minimum M1 A1 2 [9]



1. No Report available for this question.

2. Many candidates scored at least 8 marks here. In part (a) a minority produced an ascending list and failed to reverse it. Some candidates did not choose their pivots consistently, swapping between middle right and middle left pivots. The decimals here caused some problems and even though the original list was printed in the answer booklet, a surprising number of candidates initially lost one item or changed one, most commonly 1.0 became 0.1. Some candidates found only one pivot per row, with some not explicitly choosing pivots when sublists of length 2 happened to be in order – most frequently the two 4.0s and the 1.0, 0.6 at the end. Good presentation, with a list spread evenly, in columns, across the page, helps here. (Vertical listing is rarely successful). Part (b) was generally well done, the two most popular errors being to put 0.6 in bin 5 or 0.4 in bin 5. A significant number who had sorted the numbers into increasing order in part (a) proceeded to use a “first fit increasing” method here. In part (c) most candidates calculated the lower bound correctly. Other candidates correctly stated that since the five largest items were over half a bin in size they could not share a bin, so at least 5 bins would be needed. A few simply stated ‘yes’ without justification, gaining no credit.

3. This was generally well done. A disappointingly large number of candidates only chose one pivot per iteration, rather than choosing one pivot per sublist, and some candidates used lengthy methods of presentation that isolated each sublist in turn, making it difficult to see if they were choosing more than one pivot per iteration. The examiners would advise candidates to refrain from showing this unnecessary detail and simply indicate the pivots selected at each iteration. Some candidates did not select a pivot where the sublist was of order two, with the two items being in the correct order, and some did not consistently pick ‘middle left’ or ‘middle right’ when the sublist was of even order. Candidates are reminded that when the items are being transferred to the next line, the order of the items should be preserved, so if item Y is to the left of item X in the current line, neither of them being a pivot, then Y should be to the left of X in the next line. The best candidates allowed each item to become a pivot before declaring the sort complete. Some candidates did not check that their final list was in alphabetical order. In part (b) some candidates tried to apply the algorithm to the original unsorted list given at the start of (a) and others did not discard the pivot at each stage, but generally the binary search was very well done. A few candidates selected J as the first pivot, the specification makes it clear that candidates must take the ‘middle right’ where necessary.

4. This proved a good starter question for the candidates, with the vast majority scoring full marks. Only a few candidates sorted the list into ascending order, and very few incorrect methods were seen, but a disappointing number of candidates did not seem to be aware that a bubble sort should be performed consistently in one direction. Amongst those candidates using the correct method, more marks were lost by those misreading their own writing and changing one number into another than those lost making errors in applying the algorithm. Some candidates omitted a ‘stop’ statement. Candidates were asked to give the state of the list after each pass, but many showed each exchange and some each comparison, which wasted time, many of these candidates needed to use additional sheets to show all of this working and many got into time difficulties later on in the paper.



5. The vast majority of candidates showed that they understood the concept of quick sort, with very few bubble sorts seen. Most candidates chose to start with one of the 54’s as a pivot and a number of candidates were unsure what to do with the second 54. Some chose 2 pivots initially, or created an incorrect order where the two 54s were next to each other. However, most candidates dealt well with this situation. Other common errors were: not identifying a pivot towards the end of the quick sort, where two numbers were already in the correct order, fragmenting the list rather than selecting pivots concurrently and the regularly seen re-ordering of the sub-lists. Many candidates did not produce a list of students in order.

6. Some very good answers were seen to part (a), but many candidates produced disappointing attempts. Poor presentation and lack of concentration accounted for most errors in part (a); there was inconsistent choice of pivots, numbers that disappeared from the list, numbers that mutated into other numbers and, of course, numbers being reordered in the list. A large minority sorted the list into ascending order. A number of candidates are only selecting one pivot per pass, which rather defeats the object of a quick sort. Only a very few Bubble sorts were seen. Candidate would help themselves hugely by not fixing the position of the pivots until the line after they are selected, this would avoid the need to try to cram numbers into the ever-decreasing space formed by their previously chosen pivots. Candidates could then use the whole width of the line each time. Part (b) was usually well done. Some used the first fit algorithm and many put 134 into bin 5 rather than bin 3. Part (c) was often well attempted with the majority of candidates giving a clear, arithmetical argument.

7. This was generally well done. Many candidates completed the quick sort, wasting time. Some candidates did not understand the difference between an exchange and a pass in a bubble sort. Most candidates carried out the search well, but many did not give the location of the value. A large number are still assuming that the item is in the list, making statements such as ‘down to one item so found’. A surprisingly large minority of candidates used the mean of the end numbers in the remaining list to create a ‘pivot’ which is unacceptable.

8. This question was often well answered. Most candidates correctly competed part (a), although a very few stated that the list should be in ascending rather than alphabetical order. Most could correctly name and use a suitable sorting algorithm in part (b), although some did not make their stopping statement clear and a few used a shuttle sort (not in this specification) stating that it was a bubble sort. A surprisingly large minority confused the order of the alphabet with S and P (and then M and N) most frequently transposed. Part (c) was usually well done but candidates must make their pivots – and the order in which they select their pivots, clear. Candidates must remember to discard their pivots and note that the specification instructs them to ‘round up’. Once again the stopping/found statement was sometimes missing, and some candidates assumed the presence of N, stating that once they had got down to 1 term only, that term must be N.



9. Many candidates were able to gain full marks on this question. The most common errors in part (a) were in re-ordering the letters in the sub-lists and choosing the pivots inconsistently. A surprising number of candidates seemed unsure of the alphabet. Part (b) was well done by the majority of candidates. A surprising number tried to use an unsorted list for their search, gaining no marks and others omitted to discard the pivot. The commonest error was in failing to select Morris after correctly selecting Palmer then Halliwell. A few candidates did not make the order in which they selected the pivots clear making it impossible to give credit.

10. Most candidates were able to complete the bubble sort correctly, although a number of shuttle

sorts were seen from a few candidates. A number of candidates did not complete a final pass, (or stated that they had performed a final pass and found no further exchanges). The majority were able to complete the bin packing but a number were unable to show that they had used a minimum number of bins, once again the lower bound would have helped here.


D1 Algorithms - Bin packing PhysicsAndMathsTutor.com

1. (a) e.g. total weight is 239, lower bound is 239 3.9860

= so 4 bins. M1 A1 2

Note

1M1: Any correct statement, must involve calculation 1A1: cao (accept 4 for both marks)

(b) Bin 1 : 41 Bin 4 : 36 M1 A1 Bin 2 : 28 + 31 Bin 5 : 32 A1 3 Bin 3 : 42 Bin 6 : 29

Note

1M1: Bins 1 and 2 correct and at least 6 values put in bins 1A1: Bins 1, 2, 3 and 4 correct. 2A1: All correct

Misread in (b) First Fit Decreasing

Bin 1: 42 Bin 2: 41 Bin 3: 36 Bin 4: 32 28 Bin 5: 31 29 (Remove up to two A marks if earned – so M1 max in (b) if first 4

bins correct.)

(c) Full Bins : 28 + 32 31 + 29 M1 A1 2 The other 3 items (42, 41, 36) require 3 separate bins

Note

1M1: Attempt to find two full bins and allocate at least 6 values 1A1: cao

(d) There are 5 items over 30. No two of these 5 can be paired B2, 1, 0 2 in a bin, so at least 5 bins will be required.

Note

1B1: Correct argument may be imprecise or muddled (bod gets B1) 2B1: A good, clear, correct argument. (They have answered

the question ‘why?’) [9]

2. (a) needed4so38.360230 = M1 A1 2

Note

1M1: Their 230 divided by 60, some evidence of correct method 3.8 enough.

1A1: cso 4.



(b) Bin 1: 32 17 9 M1 A1

Bin 2: 45 12 A1

Bin 3: 23 28 A1 4

Bin 4: 38 16

Bin 5: 10

Note

1M1: Use of first fit. Probably 32, 45 and 17 correctly placed.

1A1: 32, 45, 17, 23, 38 and 28 placed correctly

2A1: 32, 45, 17, 23, 38, 28, 16, 9 placed correctly.

3A1: cao

Special case for (b) misread using first fit decreasing.

Give M1A1 (max)

Bin 1: 45 12

Bin 2: 38 17

Bin 3: 32 28

Bin 4: 23 16 10 9

M1 for placing 45, 38, 32, 28 and 23 correctly

A1 for cao.

(c) e.g. Bin 1: 32 28

Bin 2: 38 12 10 M1 A1

Bin 3: 45 9

Bin 4: 23 17 16 A1 3

Note

1M1: Use of full bin – at least one full bin found and 5 numbers placed.

1A1: 2 full bins found

Eg [32+28 and 38+12+10] [23+28+9 and 16+12+32]

[32+28 and 23+16+12+9] [38+12+10 and 23+28+9]

2A1: A 4 bin solution found. [9]



3. (a) 100502 = 5.02 so 6 tapes. Ml

Al 2

1M1 (502 ±40) ÷ 100 (maybe implicit)

1A1: cao 6 tapes

(b) Bin l: 29, 52 Bin 5: 47,38 Ml Bin 2: 73 Bin 6: 61 Bin 3: 87 Bin 7: 41 Al Bin 4: 74 Al 3

1M1 Bin 1 correct and at least 8 values put in bins

1A1: Condone one error, (e.g. extra, omission, ‘balanced’ swap).

2A1: All correct

(c) Binl: 87 Bin 4: 61,38 M1 Bin 2: 74 Bin 5: 52,47 A1 Bin 3: 73 Bin6: 41,29 Al 3

1M1 Bin 1 correct and at least 8 values put in bins

1A1: Condone one error, (e.g. extra, omission, ‘balanced’ swap).

2A1: All conect [8]

4. (a) E.g. M1 A1 A1ft A1ft A1 5 650650650710710

431643710650650

245710643643643

643455455455455

455431431431452

710245245452431

234234452245245

162162234234234

452452162162162

134134134134134

(b) Bin 1 710 + 245 Bin 3 643 + 162 + 134 Bin 5 431 M1 A1 Bin 2 650 + 234 Bin 4 455 + 452 A1ft A1 4

(c) eg.

10004116 = 4.116 ∴ 5 bins needed ∴ optimal M1 A1ft 2

[11]



5. (a) Bin 1 – 75+20 Bin 2 – 70 +20 M1 Bin 3 – 60+40 A1 Bin 4 – 50+35 A1 Bin 5 – 20

5 Planks needed: cost £15 A1 Wastage = 5+10+0+15+80 = 110cm A1 5

(b) Bin 1 (1.5m) – 75 + 70 Bin 1 (1m) –75 + 20 M1 Bin 2 (1.5m) – 60 + 50 + 40 or Bin 2 (1.5m) –70 + 60 + 20 Bin 3 (1m) – 35 + 20 + 20 + 20 Bin 3 (1.5m) –50 + 40 + 35 + 20 A1

Cost £11 A1 1.5m lengths better value than 1m lengths to use as many as possible A1 4

[9]

6. (a) (i) le ft to right or right to le ft

25 22 30 18 29 21 27 21 25 22 30 18 29 21 27 21 M1 25 30 22 18 29 21 27 21 25 22 30 18 29 27 28 21

25 30 22 29 18 21 27 21 25 22 30 29 18 27 21 21

25 30 22 29 21 18 27 21 25 30 22 29 18 27 21 21

25 30 22 29 21 27 18 21 30 25 22 29 18 27 21 21 A1 (pass) (ii) 25 30 22 29 21 27 21 18 30 29 25 22 27 18 21 21

30 25 29 22 27 21 21 18 30 29 27 25 22 21 18 21

30 29 25 27 22 21 21 18 30 29 27 25 22 21 21 18

30 29 27 25 22 21 21 18 30 29 27 25 22 21 21 18

30 29 27 25 22 21 21 18

(b) (i) rod 1 30 18 2 29 21 3 27 22 M1 (to the 22) 4 25 21 A1 2

(ii) 193 ÷ 50 = 3.86, ∴ 4 rods needed, so minimum M1 A1 2 [9]



1. This was a good source of marks for well-prepared candidates with over 75% gaining at least 7 marks, though the final part challenged all but the most able.

Part (a) was almost always completed correctly.

Part (b) was often well done, although some used first fit decreasing and 29 was sometimes either omitted from the list or changed to 39.

Most candidates were able to complete part (c) correctly but a few only listed one full bin.

In part (d) many incorrectly stated that the full bin solution was optimal, others made some vague reference to the statues being too heavy, relatively few attempted a valid numerically based answer and managed to express it clearly.

2. This was often a good source of marks for candidates. Most were able to calculate the lower bound getting 3.8 and therefore 4 bins. Most candidates applied first fit correctly, although some candidates did not offer each item to each bin in turn, starting with bin 1 each time, so often the 9 was the first item to be misplaced. Some candidates wasted time by replacing each letter by a number, or explaining in lengthy detail the steps they took to place each item. Not all candidates realised that they should fill as many bins as possible when using full bin, most found one full bin, but only the best found two.

3. This proved an accessible first question and was well answered by many candidates. Some candidates probably spent too long on this question, drawing out very neat and accurate bar graphs in (b) and (c), where numbers in bins were perfectly acceptable. Most candidates calculated the lower bound correctly in part (a) although some attempted a full bin solution and others divided by 9, some having calculated 5.02 rounded down to 5. Apart from the usual omissions of data, part (b) was usually well answered, the most common slips being to swap the 38 and the 41, or to use the 52 to start off the second bin. First–Fit increasing was disappointingly often seen in part (c), but those who used the correct algorithm were usually successful with the only common error being misplacing the 38.

4. Some very good answers were seen to part (a), but many candidates produced disappointing attempts. Poor presentation and lack of concentration accounted for most errors in part (a); there was inconsistent choice of pivots, numbers that disappeared from the list, numbers that mutated into other numbers and, of course, numbers being reordered in the list. A large minority sorted the list into ascending order. A number of candidates are only selecting one pivot per pass, which rather defeats the object of a quick sort. Only a very few Bubble sorts were seen. Candidate would help themselves hugely by not fixing the position of the pivots until the line after they are selected, this would avoid the need to try to cram numbers into the ever-decreasing space formed by their previously chosen pivots. Candidates could then use the whole width of the line each time. Part (b) was usually well done. Some used the first fit algorithm and many put 134 into bin 5 rather than bin 3. Part (c) was often well attempted with the majority of candidates giving a clear, arithmetical argument.



5. Many candidates scored full marks in part (a); however some candidates used the values in ascending order scoring zero. There was no need to use a formal sort to put the list in order. Many candidates gave the cost of the wasted wood rather than the cost of the five planks. In part (b) many candidates found an optimal solution, but many then went on to consider just the amount of wood wasted, rather than showing that the cost depended upon the value for money of each plank and therefore maximising the number of 1.5m lengths used.

6. Most candidates were able to complete the bubble sort correctly, although a number of shuttle

sorts were seen from a few candidates. A number of candidates did not complete a final pass, (or stated that they had performed a final pass and found no further exchanges). The majority were able to complete the bin packing but a number were unable to show that they had used a minimum number of bins, once again the lower bound would have helped here.


D1 Linear programming – Graphical PhysicsAndMathsTutor.com

1. (a) To show a strict inequality B1 1

(b) There must be fewer than 10 children B1

There must be between 2 and 10 adults inclusive B2, 1, 0 3

(c) 2x + 3y ≥ 24 B1

x ≤ 2y B1 2

(d)

B1ft (2x +3y = 24)

B1ft (x = 2y) B1ft (shading) B1 4



(e) Minimum 0 Children 8 Adults –8 Passengers M1A1

Maximum 9 Children 10 Adults –19 Passengers B1 B1 4 [14]

2. (a) To indicate the strict inequality B1 1

Note

1B1: CAO

(b) 3x = 2y and 5x + 4y = 80 added to the diagram. B1, B1 R correctly labelled. B1 3

Note

1B1: 3x = 2y passing through 1 small square of (0, 0) and (12, 18), but must reach x = 15

2B1: 5x + 4y = 80 passing through 1 small square of (0, 20) and (16, 0) (extended if necessary) but must reach y = 6

3B1: R CAO (condoning slight line inaccuracy as above.)

(c) [Minimise C =] 500x + 800y B1, B1 2

Note

1B1: Accept expression and swapped coefficients. Accept 5x + 8y for 1 mark 2B1: CAO (expression still ok here)



(d) Point testing or Profit line M1 A1 Seeking integer solutions M1 (11, 7) at a cost of £ 11 100. B1, B1 5

Note

1M1: Profit line [gradient accept reciprocal, minimum length line passes through (0, 2.5) (4, 0)] OR testing 2 points in their FR near two different vertices.

1A1: Correct profit line OR 2 points correctly tested in correct FR (my points)

e.g

(7 113 , 10 11

10 ) = 12 363 117 or (7, 11) = 12 300

(8, 10) = 12 000

(8, 11) = 12 800

(11 51 , 6) 10 400 or (11, 6) = 10 300

(15, 6) 12 300 or (15, 7) = 13 100

(15,22 21 ) = 25 500 or (15, 22) = 25 100

(11, 7) = 11100

2M1: Seeking integer solution in correct FR (so therefore no y = 6 points) 1B1: (11, 7) CAO 2B1: £11 100 CAO

[11]

3. (a) x + 2y ≤ 12 (150x + 300y ≤ 1800) M1A1 2

Note

1M1 – correct terms, accept = here, accept swapped coefficients.

1A1 – cao does not need to be simplified.

(b) 0.9x + 1.2y ≤ 9 M1

→ 3x + 4y ≤ 30 (*) A1 cso 2

Note

1M1 – correct terms, must deal with cm/m correctly, accept = here.

1A1 – cso answer given.



(c) (You need to buy) at least 2 large cupboards. B1 1

Note

1B1 – cao ‘at least’ and ‘2’ and ‘large’.

(d) Capacity C and 140%C

So total is Cx Cy100140

+ M1

Simplify to 7y + 5x (*) A1cso 2

Note

1M1 – ‘1.4’ or ‘5 × 40%’maybe ‘5+2’ seen, they must be seen to engage with 140% in some way.

1A1 – cso answer given.

(e)

Graph:

y ≥ 2 B1

0.9x + 1.2y ≤ 12 (3x + 4y ≤ 30) B1

x + 2y ≤ 12 (150x + 300y ≤ 1800) B1

Lines labelled & drawn with a ruler B1

Shading, Region identified B1, B1 6



Note

Lines should be within 1 small square of correct point at axes.

1B1 – correctly drawing y = 2.

2B1 – correctly drawing 3x + 4y = 30 [0.9x + 1.2y = 12]

3B1 – correctly drawing x + 2y = 12 [150x + 300y = 1800], ft only if swapped coefficients in (a) (6,0) (2,8).

These next 3 marks are only available for candidates who have drawn at least 2 lines, including at least one ‘diagonal’ line with negative gradient.

4B1 – Ruler used. At least 2 lines labelled including one ‘diagonal’ line.

5B1 – Shading, or R correct, b.o.d. on their lines.

6B1 – all lines and R correct.

(f) Consider points and value of 5x + 7y: M1A1

Or draw a clear profit line

(7,2) → 49 or (7 ⅓,2) 50 ⅔, or (7.3, 2) → 50.5

(6,3) → 51

(0,6) → 42 A1

(0,2) → 14

Best option is to buy 6 standard cupboards and 3 large cupboards. A1 4

Note

1M1 At least 2 points tested or objective line drawn with correct m or 1/m, minimum intercepts 3.5 and 2.5.

1A1 – 2 points correctly tested or objective line correct.

2A1 – 3 points correctly tested or objective line correct and distinct/labelled.

3A1 – 6 standard and 3 large, accept (6,3) if very clearly selected in some way.

[17]



4. (a) 7x + 5y ≤ 350 M1 A1 2

Note

1M1: Coefficients correct (condone swapped x and y coefficients) need 350 and any inequality

1A1: cso.

(b) y ≤ 20 e.g. make at most 20 small baskets B1

Y ≤ 4x e.g. the number of small (y) baskets is at most 4 times the number of large baskets (x). B1 2

{E.g if y = 40, x =10, 11, 12 etc. or if x = 10, y = 40, 39, 38}

Note

1B1: cao

2B1: cao, test their statement, need both = and < aspects.



(c)

Draw three lines correctly B3 2 1 0

Label R B1 4

Note

1B1: One line drawn correctly

2B1: Two lines drawn correctly

3B1: Three lines drawn correctly. Check (10, 40) (0, 0) and axes

4B1: R correct, but allow if one line is slightly out (1 small square).

(d) (P=) 2x + 3y B1 1

Note

1B1: cao accept an expression.



(e) Profit line or point testing. M1 A1

x = 35.7 y = 20 precise point found. B1

Need integers so optimal point in R is (35, 20); Profit (£)130 B1;B1 5

Note

1M1: Attempt at profit line or attempt to test at least two vertices in their feasible region.

1A1: Correct profit line or correct testing of at least three vertices.

Point testing: (0,0) P= 0; (5,20) P = 70; (50,0) P = 100

7

92073131P20,

725020,

7535 ==

=

also (35, 20) P = 130. Accept (36,20) P = 132 for M but not A.

Objective line: Accept gradient of 1/m for M mark or line close to correct gradient.

1B1: cao – accept x co-ordinates which round to 35.7

2B1: cao

3B1: cao [14]



5. (a)

Graph B1B1 B1 (lines) B1 (shading) B1 (R found) B1 (labels) 6

(b) Point testing or Profit line method M1 A1 Minimum point (0, 80); Value of 80 B1 A1 Maximum point (24, 96); Value of 168 B1 A1 6

[12]



6. (a) y ≥ 2x B2,1,0 2

(b) x + 2y = 160 correctly drawn y = 60 correctly drawn and distinctive (strict inequality) shading correct B4,3,2,1,0 4

(c) R correct B1ft 1

(d) Profit line added or Point testing seen M1 correctly done. A1 70 boxes identified B1 3

(e) (P =) 1.2x + 1.4y B1 1

(f) Profit line added or Point testing seen M1 correctly done A1ftA1 (32, 64) identified B1 4

(g) £128.00 B1 1 [16]

7. (a) To show a strict inequality B1 1

(b) There must be fewer than 10 children B1 There must be between 2 and 10 adults, inclusive. B2,1,0 3

(c) 2x + 3y ≥ 24 B1 x ≤ 2y B1 2



(d)

12

10

8

6

4

2

2 4 6 8 10 12 14 Child

Adult y

x

x = 10

y = 10

y = 2

x y = 2

2 + 3 = 24x y

R

2x + 3y = 24 B1ft x = 2y B1ft shading B1ft B1(R) 4

(e) minimum 0 children 8 adults – 8 passenger M1A1 maximum 9 children 10 adults – 19 passengers B1B1 4

[14]

8. (a) Maximise, (P =) 15x + 15y B1, B1

Subject to 3x + 10y ≤ 3000 B3, 2, 1, 0 5 3x + 2y ≤ 1200 B3, 2, 1, 0 x ≥ 2y x, y ≥ 0



(b)

200 400 600 800 1000

3 + 2 = 1200x y

x y = 2

3 + 10 = 3000x y

y

x

600

500

400

300

200

100

0

Profitline

FeasibleRegion

B6, 5, 4, 3, 2, 1, 0 6

(c) Profit line or vertex testing, (300, 150). Profit = £67.50 M1 A1ft A1ft 3

(d) Production of stickers should be increased since this would be more the intersection point further from the origin. B2, 1ft, 0 2

(e) e.g. The constraint line, would be for outside the feasible region B2, 1, 0 2 – so they would not effect it.

[18]

9. (a) Maximum (P=) 0.4x + 0.2y B1 accept 40x + 20y

Subject to x ≤ 6.5 y ≤ 8 x + y ≤ 12 B5,4,3,2,1,0 6 y ≤ 4 – x y ≥ 0



(b) Point testing or Profit line (6.5, 5.5) ⇒ 65ω type x and 55ω type y M1A1A1 3

(c) P = 0.4(65ω) + 0.22(55co) M1 = £3.7 co A1 2

[11]

10. (a) Maximise P = 30x + 40y (or P = 0.3x + 0.4y) B1

subject to x + y ≥ 200 B1 x + y ≤ 500 B1

x ≥ 10020 (x + y) ⇒ 4x ≥ y M1 A1

x ≤ 10040 (x + y) ⇒ 3x ≥ 2y A1 6



(b)

B5,4,3,2,1,0 5 (x + y = 200, x + y = 500) B1 ft (y = 4x) B1 ft (2y = 3x) B1 ft (labels) B1 ft FR B1 (NB: Graph looks OK onscreen at 75% magnification but may print out misaligned)



(c) Point testing or profit line M1 Intersection of y = 4x and x + y = 500 A1 (100, 400) Profit = £190 (units must be clear) A1 3

[14]

11. (a)

15

10

5

(b)

(c)

y 3x 15 + = y 2x =

2y = 7

x y 12 + =

5 10 x

y

R

0

B5, 4, 3, 2, 1, 0 5



(b) Either point testing or profit line M1

,2134)

213,

21(8 B,

6125)

213,

65(3A →→

Accept C (4,8) → 48 and D (3,6) → 36

Profit line gradient –52 A1

Identifies A (365 , 3

21 ) cost 25

61 A1, A1 4

(c) Either point testing or profit line M1

A (365 , 3

21 ) → not integer so try (4, 4) → 20 Profit line

B (821 , 3

21 ) → not integer so try (8,4) → 32

→ try (7,5) → 31 gradient –23

Accept C (4,8) → 28 and D (3,6) → 21 A1

Identifies (8,4) profit 32. A1 A1 4 [13]

12. (a) (P =) 300x + 500y B1

(b) Finishing 3.5x + 4y ≤ 56 ⇒ 7x + 8y ≤ 112 (or equivalent) B1 Packing 2x + 4y ≤ 40 ⇒ x + 2y ≤ 20 (or equivalent) B1 3



(c)

14

12

10

8

6

4

2

4 8 12 16 20x

y

16.5

11

x y + 2 = 20

3.5 + 4 = 56x y

2 + 3= 33

x y

F. R.Profit line

B4, 3, 2, 1, 0 4

(d) e.g.: Point testing: test corner points in feasible region find profit at each and select point yielding maximum Profit line: draw profit lines select point on profit line furthest from the origin B2,1,0 2

(e) Using a correct, complete method M1 make 6 Oxford and 7 York Profit = £5300 A1 ft A1 ft 3

(f) The line 3.5x + 4y = 49 passes through (6, 7) so reduce finishing by 7 hours M1 A1 ft A1 3

[15]



1. No Report available for this question.

2. This question gave rise to a good spread of marks. Most candidates completed part (a) correctly although some very lengthy responses were seen. 5x + 4y = 80 was drawn correctly more often than 3x = 2y in part (b), with many candidates drawing the latter with a negative gradient. Pleasingly most candidates used a ruler to draw their lines, a great improvement on previous years. The feasible region was often incorrectly identified and labels were often absent.

Most were able to complete part (c) correctly.

Those who used the objective line method in (d) usually gained more marks than those who used the point testing method. Some of those using the latter method seemed confused by the y-axis scale and only considered vertices with even values of y, many tested points by reading from the graph rather than solving simultaneous equations.

A large number of solutions had y = 6 despite answering part (a) correctly. Some found the maximum solution. Many did not make their method clear.

3. Most candidates were able to score at least 12 out of 17 marks on this question. Parts (a), (b) and (c) were usually correct, with only a very few making slips with the inequality in (a) or muddling ‘small’ with ‘large’ in part (c). The units in part (b) caused difficulty for some candidates, but most changed all lengths into cm and proceeded correctly. Many candidates struggled with part (d). When the answer is printed on the paper candidates must ensure that their reasoning is both clear and convincing, disappointingly, many candidates were not able to derive the given result and in particular many ‘derivations’ attempted to start with 1.4y = x. There were many fully correct graphs, helped by widespread use of rulers, a big improvement from past papers. Three correct lines almost always invariably led to the correct region. As always, some lost a mark because they did not label their lines and/or R. In (f) both the vertex testing and profit line methods were often successful. As always it is vital that the method is clearly seen, some lost all 4 marks in (f) because they merely described the use of a profit line but failed to draw it. Others drew a very short profit line – candidates should use sufficiently large values for the axes intercepts to ensure an accurate gradient. Those using the vertex method should be reminded that all vertices should be tested, a number of candidates only tested one or two vertices.



4. There were some very good, and very poor, solutions seen to this question. Almost all candidates were able to write down the correct inequality in part (a) with only a very few getting the wrong coefficients or replacing the inequality with an equals sign. Part (b) proved challenging for many candidates. Candidates struggled in particular to interpret y ≤ 4x. The usual error was to confuse ‘small’ with ‘large’ but many failed to refer to, or reversed, the inequality. The most able described the inequality in terms of percentages; where this was seen it was almost always correct. Most candidates drew 5x + 7y = 350 and y = 20 correctly. Most candidates used a ruler and most plotted the axes interceptions accurately. Unsurprisingly y = 4x caused the most difficulty, often replaced by x = 4y. Most candidates used shading sensibly although some shaded so scruffily that they obscured their line. Most candidates labelled R correctly; most candidates did not label their lines. Most candidates were able to write down the correct objective function. Part (e) was often poorly done with many candidates failing to make their method clear; if using the objective line method candidates MUST draw an objective line, and of a sensible length, so that its accuracy can be checked; if using point testing then the points and their values must be stated. As always those who use the objective line method are more successful than those who use point testing. When point testing, all vertices in the feasible region must be tested. Many candidates assumed that the point (36, 20) was a vertex; it was pleasing to see a small number of scripts where this was tested and found to be outside the feasible region. Others found the precise point but then did not seek integer solutions to complete their answer.

5. While a number of candidates scored the full 12 marks on this question, many could only achieve a handful of marks.

The first two lines were often drawn correctly however y = 4x was frequently incorrect. The most common error was to draw y = (1/4) x, not drawing the line long enough, or not noticing the differing scales on the axes. The shading on two of the lines was often correct but only the better candidates were able to get the shading correct on all three lines, so many gave the FR as the central triangle. Many candidates did not label their lines. There continues to be evidence that candidates are not going in to the examination properly equipped with (30cm) rulers.

In part (b) candidates who gained full marks did so most frequently and easily by using the objective line method. If errors were seen in this method, it was often due to a reciprocal gradient, or an inability to read the scale on the graph accurately. A significant number of candidates continue to state they are using this method without showing any evidence of this, these gain no credit. If the objective line method is being used, the examiners need to see an accurately drawn objective line (of decent length). Those who chose to use the point testing method frequently lost marks through the inaccuracy of their extreme points, or by not testing all of their extreme points. There was a tendency to try to read the coordinates from the LP-graphs, and unsurprisingly the point ( )92,95 6215 was therefore rarely seen. Candidates should be reminded that if they choose to use the vertex method they must show their complete working, and will be expected to use simultaneous equations to find the exact coordinates of any vertices. A few candidates stated that they were using this method but then tested points other than the vertices of their FR, another frequently seen error was to find the coordinates of the vertices but then not use them to find any F values.



6. This question proved challenging for many. Some candidates used very lengthy methods to determine the equation of the straight line in part (a), and many got the 2 on the wrong side. Many were able to determine the correct inequality but others reversed it. The lines in part (b) were sometimes very disappointing, many failed to label their lines and/or make y = 60 distinctive, but most disappointing of all was the lack of ruler use and inaccurate plotting of axis intercepts. Part (c) was often well done with examiners following through on candidate’s lines were possible. Many candidates identified an optimal point in part (d) but some failed to state that 70 was the total number of boxes, the method used was not always apparent to the examiners and candidates need to be reminded that they should always make their method clear on this ‘methods’ paper.

Most of the candidates who did display their method used point-testing. Almost all the candidates were able to find an objective function in part (e). Once again many candidates did not make their method clear in part (f). If point-testing candidates should list the points they are testing and show the value of the objective at each point. If using a profit line they should clearly draw a profit line - of good length and label it. Point testing proved a more popular method in this question, but few tested all five vertex points. Poor notation was often seen with a number of candidates writing 32x + 64y instead of x = 32, y = 64. Those who located the optimal point in part (e) were usually successful in calculating the profit in part (f).

7. Parts (a) and (b) were often well–answered, the commonest error was imprecise use of language, especially use of the phrase ‘between 2 and 10’, without the word ‘inclusive’. Although there were the usual sign reversals and some strict inequalities seen, part (c) was often well–answered with the large majority of candidates successfully finding the first inequality. The second inequality proved more difficult, with the 2 often on the wrong side. Most candidates were able to draw 2x + 3y = 24, but there was confusion (both ways) between x = 2y and y = 2x. Some candidates made shading errors and even the best rarely followed the instructions to label the feasible region. Candidates MUST show their working when seeking maximum and minimum points. There was often no evidence of any method being used. If candidates use the point testing method, they should state the points they are testing and the result of the test, if they are using the ‘profit line’ method, they must draw in a clearly labelled profit line – and this must be long enough for examiners to confirm the accuracy of the gradient. Some candidates tested their point using the cost equation rather than simply totalling the number of passengers. Many candidates simply stated the maximum and minimum totals rather than the number of adults and



8. Most candidates were able to make some progress with part (a), most correctly stated the objective function but often the objective was omitted. The non-negativity constraints were often omitted and many had difficulty in finding the x ≥ 2y inequality. The examiners were all disappointed by the standard of the graph work seen in part (b). Lines were often imprecisely drawn or omitted, x = 2y (if found in part (a)) was often incorrectly drawn. Labels, scales and/or shading were often omitted and the feasible region was not always indicated. Not all candidates used sharp pencils and rulers. If candidates are going to use the profit line method in part (c) they must draw in, and label, a profit line which should long enough to enable examiners to check the gradient. If candidates are going to use the point testing method they must state and test every vertex point in the feasible region, not just the most likely point. Many candidates did not state the profit, and of those that did, some did not state units. Those who drew a correct graph generally answered part (d) well. Part (e) was often well-answered but there were many irrelevant comments seen.

9. Many omitted the instruction to maximise the objective. Most candidates were able to write down at least 3 constraints correctly, although few obtained them all. y=0, was often omitted and there were often errors in writing down x + y =12 and y = 4x. The solution of those candidates using point testing in part (b) was often spoilt by using incorrect coordinates. The y coordinate of the point with x coordinate 6.5 was frequently misstated. Those using the objective line method must draw the line clearly on the diagram, and in such a way that the gradient can be see to be correct. The clearest way in this case is to draw the line from e. g. (2, 0) to (0, 4). A number of candidates did not pick up that the values of x and y represented 1000’s of batteries and this together with problems with pounds and pence caused much confusion in part (c), although many completely correct answers were seen.

10. This question was often poorly done. Poor algebra was often seen in part (a). Most candidates were able to state the objective function but did not state that this was to be maximised. The x + y inequalities were better handled. Only the better candidates were able to correctly handle the 20% and 40% inequalities. A number tried to combine the four inequalities into two constraints. In part (b) the lines x + y = 200 and x + y = 500 were plotted correctly but the other lines very poorly plotted. Many candidates drew vertical or horizontal lines. Candidates should use rulers and sharp pencils to draw lines. Labels were frequently omitted both of the lines and the feasible region. Most who used the ‘profit line’ method got the correct answer although a few drew lines with the reciprocal gradient. Some candidates using the ‘vertex method’ considered that they only need check two points and not all four.



11. In part (a) the lines x + y = 12 and y + 3x = 15 were well drawn by most candidates. The lines 2y = 7 and y = 2x and the identification of the feasible region were less well done. Many did not make any distinction between way they drew the line y = 2x and the other lines, not picking up on the strict inequality. Others did not use rulers to draw their lines and others did not label their lines. In parts (b) and (c) candidates could use either the point testing method or the profit line method: generally those using the latter were more successful.

The presentation of the solution was sometimes very difficult to follow. Those using the profit line method must draw, and label, a profit line so that its gradient can be checked and the method seen. Those using point testing must make the points they are testing and the value of the function at these points clear. The fractions caused problems for many candidates. Part (c) was less well done with many candidates ignoring the integer constraint.

12. Parts (a) and (b) were well done by the vast majority of candidates. Most candidates were able

to draw the lines correctly in part (c) but many did not label their lines or the axis. Some very poor choices of scale were often seen, so that sometimes the whole feasible region was not shown, or was too small to be useful. Most candidates were able to score some credit in part (d) but few were able to give complete, clear explanations. A surprisingly large number of candidates failed to show any evidence of working in part (e), no point testing or drawing of a profit line giving no marks. Part (f) was often omitted, but well done by those who did attempt it, with candidates either using their graphs or testing in all three inequalities.


D2 Linear programming - Simplex algorithm PhysicsAndMathsTutor.com

1. (a) b.v x y z r s Value Raw ops

z 21 0 1

41 0 20 R1 ÷ 4 M1 A1

s 0 4 0 21

− 1 120 R2 – 2R1 M1 A1 ft

P 8 –8 0 5 0 400 R3 + 20R1 A1 ft 5

(b) P + 8x – 8y + 5r = 400 B1ft 1

(c) Not optimal since there is a negative number in the profit row B1ft 1 [7]

2. (a) 2 6 0P x y z− − − = B1 1

Note

1B1: cao

(b)

b.v x y z r s t Value

r 0 1 2 1 0 0 24

s 2 1 4 0 1 0 28

t –1 12 3 0 0 1 22

P –1 –2 –6 0 0 0 0

b.v. x y z r s t Value Row Ops.

r –1 12 0 1 –

12 0 10 R1 – 2R2 M1 A1

z 12

14 1 0

14 0 7 R2 ÷ 4 M1 A1ft

t – 52 –14 0 0 –

34 1 1 R3 – 3R2 A1

P 2 – 12 0 0 32 0 42 R4 + 6R2 5

b.v. x y z r s t Value Row Ops.

y –2 1 0 2 –1 0 20 R1 ÷ 12 M1 A1ft

z 1 0 1 – 12 12 0 2 R2 –

14 R1 M1 A1

t –3 0 0 12 –1 1 6 R3 +14 R1

P 1 0 0 1 1 0 52 R4 + 12 R1 4


D2 Linear programming - Simplex algorithm PhysicsAndMathsTutor.com

Notes

1M1: correct pivot located, attempt to divide row 1A1: pivot row correct including change of b.v. 2M1: (ft) Correct row operations used at least once or stated correctly. 1A1ft: Looking at non zero-and-one columns, one column ft correct 2A1: cao. 3M1: (ft) Correct pivot identified – negative pivot gets M0 M0 1A1: ft pivot row correct including change of bv – but don’t penalise

b.v. twice. 4M1: (ft) Correct row operations used at least once or stated correctly. 1A1: cao

Misread Alternative 1

Increasing x first, b.v. x y z r s t Value row ops

r 0 1 2 1 0 0 24 R1 no change

x 1 21 2 0 12 0 14 R2 ÷ 2

t 0 1 5 1 12 1 36 R3 + R2

P 0 23− –4 0 2

1 0 14 R4 + R2

then y next, b.v. x y z r s t Value row ops

y 0 1 2 1 0 0 24 R1 ÷ 1

x 1 0 1 21− 12 0 2 R2 – 12

1 R

t 0 0 3 –1 12 1 12 R3 – R1

P 0 0 –1 23 2

1 1 50 R4 + 123 R

then z. b.v. x y z r s t Value row ops

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