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MARKING SCHEME

TRIAL STPM TERM 3 2013

MATHEMATICS M (950/3)

SECTION A (45 Marks)

No. SCHEME MARKS1

(a) M1A1A1(b) n > 69.2

n = 70

M1

M1

A1

6 marks

2 (a)

M1

M1

M1A1

(b) maximum profit, B1

(c) , where p = price per unit

When p = 20, = RM120

M1

M1

A1

8 marks

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3 Initial tableau

Basic x y Solution 2 3 1 0 8 1 6 0 1 10p -1 -3 0 0 0

Second tableau

Basic x y Solution 0 1 3y 1 0

p 0 0 5Final tableau

Basic x y Solutionx

1 0

2

y 0 1 p

0 0 6

x = 2 , y = , p = 6

B1B1

M1A1

M1A1

A1

7 marks

4 (a)

= RM1280

M1

A1

(b)(i) = 154.92

(ii)

= 15.5 (iii)

= 40

M1

A1

M1

A1

M1A1

8 marks

or = 15.48

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5 (a)Reduced form of the matrix : Marys expected gain :

p : The probability that Mary chooses strategy A.1-p: The probability that Mary chooses strategy B.

The optimal strategy for Mary : James expected gain :q : The probability that James chooses strategy Q.

1- q: The probability that James chooses strategy R.

The optimal strategy for James :

B1

M1

A1

M1

A1

(b)The value of the game = 2( The game is in favour of Mary and she is expected to win on average

RM for each game.

B1

B1

7 Marks

6(a) Critical activities : A, D, F, H

The minimum completion time is 11 days.

(b) Activity Total Float

B 4

C 6

E 4

G 4

B1B1

B1

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(c)

(d)

Project can be completed with 2 workers.

A

B

C

D

E

F

G

H

2 B C E G

1 A D FH

D1Float

D1

critical

activities

D1noncritical

activities

D1 D1*other

correctdiagramsare

acceptable

B1

9 marks

0 1 2 3 4 5

Days

6 7 8 9 10 11

Activity

0 1 2 3 4 5

Days

6 7 8 9 10 11

No.

ofworkers

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SECTION B (15 Marks)

No. SCHEME MARKS

7 (a)

= 48.99

M1

A1

(b)

= 18.37 Maximum inventory level,

= 30.62

M1

A1

A1

(c)

= 26.13 days

M1

A1

(d)

= 16.33 days

B1

M1

A1

(e) = (1.875 6)18.37= -7.12

M1

A1

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(f)

*Accept other correct graphs of Q*, S

*.

D1:correct

shape

D1:Q*,S*,T*

D1:All

correct

15 marks

8 (a) Plan A:

=RM1,816.05

Plan B: =RM 2,492.84

M1A1

A1

M1A1

A1

(b) Plan A: Total repayment = 360 x 1816.05 = RM653,778Plan B: Total repayment = 180 x 2492.84 = RM448,711.20Difference = RM653,778RM448,711.20 = RM205,066.80

M1A1M1A1

A1

(c) Difference of repayment will be the interest saved .Therefore, theinterest saved is RM205,066.80

B1B1

(d) M1A115 marks

26.13

31

-19

Inventory level

Time