tunku ampuan durah 2013 m3(a)

8/13/2019 TUNKU AMPUAN DURAH 2013 M3(A)

1/7

SMK TUNKU AMPUAN DURAH, SEREMBAN

TRIAL STPM 2013 - TERM 3

MATHEMATICS M (950/3)

MARK SCHEME


2/7

Zadoc2013

SECTION A

1. (a) Givenx= number of textbook.

Total cost, C(x) = Fixed cost + Variable cost

= Fixed cost + (Cost per unit)x

= 80 000 + 3x B1

(b) Total revenue, R(x) = (Price per unit)(x)

R(x) = px B1

For break-even, C(x) = R(x)

80 000 + 3(4000)=p(4000) M1

p = RM23 A1 [4]

2. (a)Loan amount ,P = 230 00010%(230 000) = 207 000 B1

Interest rate per period, 0.075 0.0062512

rim

Number of payments, n = 12 30 = 360 B1

Monthly payment 1 1

n

iR P

i

360

207000 0.006251447.37

1 1 0.00625R RM

M1A1

(b) To settle the four payments 1 1

ni

A Ri

4

1 0.00625 11447.37

0.00625A

= RM 5 843.98 M1A1

(c ) number of payments = 36012(10) = 240

1 1 ni

A Ri

240

1 1 0.006251447.37

0.00625A

= RM 179 665.12 M1A1

[8]


3/7

Zadoc2013

3. n = 300, D = 12000 , c = RM30, C0= RM25, Ch= 0.20 x 30 = RM6 , Cb=RM7.50

(a) Q* 02

h b

h b

DC C C

C C

2 12000 25 6 7.5424.26 4246 7.5

units M1A1

(b) S* * h

h b

CQ

C C

6424 188.44

6 7.50

M1A1

d12000

40300

B1

number of days to wait*S

d

188.444.7

40 M1A1 [7]

4.

(a) Max{min}=3, Min{max} = 3 B1

thus, value of the game, v = 3 > 0 B1

Not a fair game. (Player X has the advantage.) B1

b) Player Yplays Y1, Player Xplays X2 B1, B1 [5]

5.

Key: Network & keyD2

ESTM1A1

LSTM1A1

(b) Critical Path= B-C-E-F-H & Min.completion time = 52 wks B1B1

(c) Yes, since the project completion time is 52 weeks= 1 year B2 [10]

PLAYER Y RowminY1 Y2 Y3

PLAYER X

X1 -4 -1 3 -4

X2 3 4 7 3

X3 2 1 5 1

Column max 3 4 7

A 6

0 2

B 8

0 0

C 15

8 8

D 4

23 25

E 6

23 23

F 15

29 29

G 12

29 32

H 8

44 44

START

0 0

END

52 52

Act Dur

EST LST


4/7

Zadoc2013

6. Letx= number of product A

y= number of product B

LP model : Maximise profitz=x+ 1.5y subject to B1

x+y1500 B1

y0.5x B1x 0 , y 0 B1

200

1400

1200

800

600

400

y

x0

|

200|

400

|

600

|

800

|

1000|

1200|

1400

y= 0.5x

y+x = 1500

x+1.5y = 300

(1000,500)

1000 Lines correctD1

Search lineD1

RegionD1

From the graph, max pt (1000,500)

Manufacturer produces 1000 product A and 500 product B A1, A1

to obtain a maximum profit = 1000 + 1.5(500) M1

= RM 1750 A1 [11]


5/7

Zadoc2013

SECTION B

7. (a) 2 1800q p

2

2

4 1800

0.25 1800

q p

q p

2

1800 0.25p q A1

Revenue, R q pq

21800 0.25q q

31800 0.25R q q q M1A1

Profit = RevenueCost

P(q) =R(q)C(q)

= 1800q0.25q3(630q0.1q3) M1

= 1170q0.15q3 A1

(b) P(q) = 0

11700.45q2 = 0 M1

q2= 2600

q= 50.99 51 units

'' 0.9P q q

Whenx= 51, '' 51 0.9 51 0P M1Thus, profit is maximum whenx= 51 A1

Maximum profit = 1170(51)0.15(51)3

= RM 39 772.35 A1

(c )(i) At equilibrium point, Demand = Supply

2 1800 2 9 200p p M11800 9 200

10 2000

200

p p

p

p

Whenp= 200, q= 2 1800 200 80 Thus, the equilibrium price = RM200

equilibrium quantity = 80 units A1(both)

(ii)

D1, D1

If charges more than the equilibrium price, there will be a surplus of supply B1

while if charges less, there will be a shortage of supply. B1 [15]

0

q

p1800

84.85 -

22.22 200

80 -

Demand curve

Supply curve


6/7

Zadoc2013

8.

Bob RowMin

I II III IV

Ali

I 2 3 2 5 2

II -3 1 3 -4 -4

III 4 1 6 0 0

IV 3 -2 4 -5 -5

Colmn max 4 3 6 5

(a) Max{min}= 2 , Min{max} = 3 M1

Play-safe strategy: Ali plays I, Bob plays II A1

Since max{min} min{max} , no saddle point.

Thus, game is unstable. B1

(b)

2 3 2 5

3 1 4 3

4 1 6 0

3 2 4 5

2 3 2 5

4 1 6 0

3 2 4 5

2 3 5

4 1 0

3 2 5

2 3 5

4 1 0

M1

Bob

I II IV

Alip I 2 3 5

1p III 4 1 0

BobsStrategy Expected pay-off of Ali

I y1=2p + 4(1p) = 42p

II y2= 3p + (1p) = 2p + 1

IV y3= 5p

M1

p

Expected pay-off

1 -

6 -

0 1

y3

y2

4 -

y1

Expected gain;

42p= 2p+ 1

4p= 3

3

4p

Graph D1

M1

A1


7/7

Zadoc2013

To find Bobs Optimal Strategy: delete column IV

1 3 2 1

1 3 4 2

c bq

a b c d

q

2 3 1 4 1

1

2

q q q q

q

Optimal Strategy: Ali3 1

, 0, , 04 4

Bob1 1

, , 0, 02 2

Value of the game = 3 1

2 1 24 2

M1A1

Ali has the advantage B1

[15]

Bob

I II

q 1-q

AliI 2 3

III 4 1

M1

A1

A1

A1

OR

tunku ampuan durah 2013 m3(a)

Documents