Prelim Examination 2011 / 2012

MATHEMATICS Standard Grade - Credit Level Paper I Time allowed - 55 minutes

Read Carefully 1. 2. 3.

Answer as many questions as you can. Full credit will be given only where the solution contains appropriate working. You may not use a calculator

 Pegasys 2011

FORMULAE LIST

The roots of ax  bx  c  0 2

are

x 

Sine rule:

a b c   sin A sin B sin C

Cosine rule:

a 2  b 2  c 2  2bc cos A

Area of a triangle:

Area 

Standard Deviation:

 Pegasys 2011

1 2

s

b

b 

2

 4 ac



2a

cos A 

or

b2  c2  a 2 2bc

ab sin C



 ( x  x )2 n 1



( x) 2 n n 1

 x2 

KU RE

1.

Evaluate

26  4  18  09  3 .

2.

Evaluate

6 23 of

3.

A function is given as f ( x)  x 3  20 .

 54

2



 34 .

3

Hence find x when f ( x)  44 .

4.

2

Solve algebraically the equation 5x 

5.

2x  1  45 . 3

3

In the diagram below triangles ABC and ADE are mathematically similar. BC  12 cm, DE  9 cm and AE  21 cm. C E

12 cm

9 cm

21cm

B D A Find the length of CE.

 Pegasys 2011

3

KU RE 6.

Halley's comet travels in a wide loop around our solar system. At its closest point to the earth it is travelling at an average speed of 1  4  10 5 miles per hour. At this speed how far, in miles, will it travel in a week? Give your answer in scientific notation correct to 2 significant figures.

7.

4

A formula is given as V  2u  3t 2 . Change the subject of the formula to t.

8.

3

A survey is undertaken to investigate the total delivery time, in days, from order to arrival of customised computer systems, ordered from two different web based companies. The survey was conducted over a single financial year. The following data was obtained.

Median

Lower Quartile

Upper Quartile

42

28

22

32

38

25

18

34

Company

Minimum

Maximum

(no. of days)

(no. of days)

MEGASYS Ltd

12

RAM-BO

14

(a)

Draw an appropriate statistical diagram to illustrate these two sets of data.

(b)

Calculate the semi-interquartile range for Megasys Ltd.

(c)

Which company showed a better consistency of delivery? Give a reason for your answer.

 Pegasys 2011

3 2 1

KU RE 9.

A line passes through the points ( 0 , 4 p 2 ) and ( 2t  4 p , t 2 ) as shown. y ( 2t  4 p , t 2 ) 4 p2

o

(a)

x

Show clearly that the gradient (m) of this line can be written as m 

(b)

10.

t  2p . 2

4

Hence find the equation of this line when t  5 and p  2 .

3

Merlin is making up one of his favourite potions. It has three powdered ingredients called misill, canthor and ruari. For the potion the ratio of misill to canthor to ruari must be 2 : 3 : 12 . (a)

(b)

If he wishes to use 21 grammes of canthor in the potion, how many grammes of ruari will he need?

2

He decides, in the end, to make up a potion using exactly 66 grammes of ruari. What will be the total weight, in grammes, of the completed potion?

End of Question Paper

 Pegasys 2011

3

Prelim Examination 2011 / 2012

MATHEMATICS Standard Grade - Credit Level Paper II Time allowed - 80 minutes

Read Carefully 1. 2. 3.

Answer as many questions as you can. Full credit will be given only where the solution contains appropriate working. You may use a calculator

 Pegasys 2011

FORMULAE LIST

The roots of ax  bx  c  0 2

are

x 

Sine rule:

a b c   sin A sin B sin C

Cosine rule:

a 2  b 2  c 2  2bc cos A

Area of a triangle:

Area 

Standard Deviation:

 Pegasys 2011

1 2

s

b

b 

2

 4 ac



2a

cos A 

or

b2  c2  a 2 2bc

ab sin C



( x  x ) n 1

2

( x) 2 x  n n 1 2



KU RE 1.

A patient in a hospital is injected with 200 mg of a drug. (a)

It is known that for each hour after the injection the number of milligrams of the drug left in the body is 15% less than at the beginning of that hour. How many milligrams of the drug are left in the patient’s body at the end of 3 hours?

(b)

3

The patient is given a second drug. It is known that, for this second drug, at the end of each hour the number of milligrams of the drug left in the body is 12% less than at the beginning of that hour. At the end of one hour the patient had 123 2 mg of the second drug left in his body. Calculate the size of the initial dose, of this second drug, given to the patient.

2.

Solve algebraically the equation 4 tan x   3  0 ,

3.

2

for 0  x  360 .

3

A sketch of an antique writing box is shown below. It is a prism with its end face made up of a rectangle and a right angled triangle.

35cm

20cm

6cm 50cm

w

(a) (b)

Calculate the width (w) of the box giving your answer correct to the nearest whole number. Hence calculate the volume of the writing box in cubic centimetres.

 Pegasys 2011

3 4

KU RE 4.

Susan has a contract for her mobile phone. She pays a fixed sum of £28 per month. She has unlimited texts and 3 hours free call time. Once her free call time has been used up all further calls costs her 18 pence per minute. (a) (b) (c)

What will Susan's total phone bill be in a month where her total call time is 230 minutes?

2

Write down a formula for the total cost, £C, for Susan's phone in a month where her total call time is t minutes and t  180 .

2

One month Susan's bill was £50.50. What was her total call time, in minutes, for this month?

5.

3

In the diagram ABCD represents a steel framework with BCD being a triangular steel plate. Angle ADB is a right angle. AB  14  6 cm, BAC  70  , BDC  44  and DCB  80 

C 80o D

44o

70o A 14  6 cm

(a)

Find the length of DB.

(b)

Calculate the area of triangle BCD.

 Pegasys 2011

B

3 5

KU RE 6.

Solve the equation 3 x ( x  2)  1 .

Give your answers correct to 1 decimal place.

7.

5

A clock has a pendulum swinging below it. When the clock is ticking the pendulum travels along an arc of a circle, centre O. The length of the connection cord OQ is 15 centimetres. The length of the arc followed by the pendulum as it swings from P to Q is 18cm.

O 15 cm

P

Q

Find the size of angle POQ, the angle through which the pendulum swings from P to Q.

8.

5

The lift, L, produced by the wings of an aircraft varies directly as the total surface area of the wings, A, and as the square of the aircraft's speed, s. (a)

Write down a relationship connecting L, A and s.

(b)

If the aircraft's speed is doubled what effect does this have on the lift?

 Pegasys 2011

1 2

KU RE 9.

Two snowboarders are timed between two points down a short part of a run. The total distance each of them travel, between these points, is also recorded.

Boarder A travelled a distance of 9x metres in x  2 seconds. Boarder B travelled a distance of 7 x  2 metres in x  1 seconds. (a) (b)

Write down an expression, in terms of x, for the average speed of Boarder A and a similar expression for the average speed of Boarder B.

2

Given now that both the boarders had the same average speed, show clearly that the following equation can be formed. 2x 2  7x  4  0 .

4

(c)

Hence solve the equation to find x, where x  0 .

2

(d)

What was the average speed of the two boarders, in metres per second, for their run?

1

End of Question Paper

 Pegasys 2011

Credit Level Paper 1 ~ 2011/2012 Qu

Marking Scheme

Give one mark for each ● 20·37 2 marks KU

Illustrations for awarding mark

1

ans :

2

1 carries out division 2 completes calculation ans : 1/3

3

1 evaluates bracket 2 knows how to complete calculation 3 completes calculation ans : x = 4 2 marks RE

x3 – 20 = 44; x3 = 64 x=4

4

1 1 equates and begins to solve 2 answer 2 ans : x = 8 3 marks KU

15x + 2x – 1 = 135 17x – 1 = 135; 17x = 136 x=8

5

1 multiplies to remove fraction 1 2 simplifies and starts to solve 2 3 solves 3 ans : 7 cm 3 marks RE

4/3 [12/9] AC = 21 × 4/3 = 28 cm EC = 28 – 21 = 7 cm

6

1 finds linear scale factor 2 solves for x 3 calculates CE ans : 2·4 × 107 miles

7

1 19·09 ÷ 3 = 6·03 2 26·5 – 6·03 = 20·37 3 marks KU 1 2 3

1 2 3

1/20 20/3 × 1/20 1/3

4 marks KU

1 knows to multiply by 24 and 7 2 starts to evaluates 3 rounds to 2 sig. figs. 4 gives answer in Scientific notation v  2u 3 marks KU ans: t  3

1 1·4 × 105 ×24 × 7 2 23 520 000 3 24 000 000 4 2·4 × 107 miles

1

1

2

subtracts 2u from both sides



divides both sides by 3

3

takes square root of both sides

 Pegasys 2011

2 3

3t 2  v  2u v  2u t2  3 v  2u t 3

Qu 8a

ans :

Give one mark for each ● diagram drawn 3 marks RE

1 2 boxplots drawn 2 correct whiskers for each 3 correct box for each b

ans: 5 1 2

1 2 3

knows how to find SIQR answer

ans:

9a

1 states more consistent with reason ans: proof 4 marks RE

10a

b

Megasys with reason

1 2

(32 – 22) ÷ 2 5

1

Megasys; smaller SIQR

1

m

1 mark RE

1

knows how to find gradient

2

subs values in formula



3

factorises numerator

3

4

factorises denominator and simplifies

4

ans:

see end of marking scheme

2 marks KU

c

b

Illustrations for awarding mark

2y = x + 32 [or equi.]

y 2  y1 x 2  x1

t 2  4 p2 m 2t  4 p  0 (t  2 p )(t  2 p ) m ....... (t  2 p )(t  2 p ) t  2 p m  2(t  2 p) 2

2

3 marks RE

1 subs values to find gradient 2 subs values to find point 3 subs into equation of straight line ans: 84 grammes 2 marks KU

1 2 3

m=½ (18, 25) [or (0, 16)] y – 25 = ½ (x – 18) [or y = ½ x + 16]

1 2

1 2

times 7 12 × 7 = 84 grammes

1 2 3

times 5·5

finds multiplication factor answer

ans: 93· 5 grammes 1 2 3

3 marks RE

finds multiplication factor calculates amount of each ingredient answer

misill = 5·5 × 2 = 11; canthor = 5·5 × 3 = 16·5

11 + 16·5 + 66 = 93·5 grammes Megasys Ltd Ram - Bo

10

Total:  Pegasys 2011

KU 19

20

RE 19

30

40

Credit Level Paper 2 ~ 2011/2012 Qu 1a

ans : 1 2 3

Marking Scheme

Give one mark for each ● 3 marks KU 122·825 mg

correct multiplier knows how to decrease for 3 hours answer

b

ans:

140 mg

2

1 method 2 answer ans : 143·1o, 323·1o

3a

1 evaluates tan xo 2 finds one value for x 3 finds second value for x ans : 32 cm

1 …… × 0·85 2 200 × 0·853 3 122·825 mg

2 marks RE 1 88% = 123·2 ÷ 88 × 100 2 140 mg 3 marks KU 1 tan xo = –0·75 [– ¾] 2 143·1o 3 323·1o 3 marks RE

1

14

1

 assembles facts in RAT 2 uses Pythagoras 3 answer properly rounded 20 800 cm3

Illustrations for awarding mark

 2 3

35

35² – 14² = 1 029 √1 029 = 32 cm to nearest cm

b

ans:

1 evidence of 2 areas being added 2 ½ × 32 × 14 = 224 cm2 3 6 × 32 = 192 cm2; 416 cm2 4 416 × 50 = 20 800 cm3

4a

1 knows to find cross sectional area 2 finds area of triangle 3 finds area of rectangle and total area 4 multiplies total to answer ans : £37 2 marks KU 1 2

1 2

50 × 0·18 + 28 £37

1 2

t – 180 28 + 0·18(t – 180)

1 2 3

28 + 0·18(t – 180) = 50·5 0·18(t – 180) = 22·5; (t – 180) = 125 t = 305 minutes

b

4 marks KU

method answer

ans :

28 + 0·18(t – 180)

2 marks RE

1 expression for number of minutes 2 expression for cost c

ans:

305 minutes

1 equates formula to 50·5 2 starts to solve 3 answer

 Pegasys 2011

3 marks RE

Qu 5a

ans: 1 2 3

b

6

Give one mark for each ● 13·7 cm 3 marks KU 1 2 3

evidence and choice of correct ratio sin 70o = DB/14·6 DB = 14·6 × sin 70o = 13·7 cm

1 realises the need use sine rule 2 subs values

1

3 4

3 4 5

evidence 13  7 CB 13  7 DC or   sin 80 sin 44 sin 80 sin 56 CB = 9·7 cm or DC = 11·5 cm A = ½ × 13·7 × 9·7 × sin 56o or A = ½ × 13·7 × 11·5 × sin 44o 55 cm2

1 2

3x2 + 6x – 1 = 0 evidence

ans

knows to use trig subs values answer 55 cm²

5 marks RE

2

finds CB or DC knows how to find area

5 answer ans : 0·2; –2·2 1 2

5 marks KU

multiplies brackets and equates to 0 knows to use quadratic formula

7

3 subs values correctly 4 calculates b2 – 4ac 5 both answer correctly rounded ans : 57·3o 5 marks RE

8a

1 uses correct diameter 2 calculates circumference 3 sets up equal ratios 4 starts to solve 5 answer ans : L = kAs2 1

b

Illustrations for awarding mark

1 2

strategy answer

 Pegasys 2011



4 5

 6  6 2  4  3  1 6 √48 0·2; –2·2

1 d = 36 cm [may be in formula] 2 C = 3·14 × 36 = 113·04 3 18/113·04 = angle/360 4 angle = (18 × 360) / 113·04 5 57·3o 1 mark KU

finds relationship

ans : multiplied by 4

3

1

L = kAs2

1 2

subs values to find answer multiplied by 4

2 marks RE

Give one mark for each ● 9x 7x  2 ; 2 marks KU x  2 x 1

Qu 9a

ans : 1

 2 c

1

writes expression for boarder A writes expression for boarder B

ans :

proof

2

4 marks RE

1 equates expressions 2 3 4 c

x=4

ans : 1

6 m/s

 Pegasys 2011

1 2

(2x + 1)(x – 4) = 0 x = –½ or 4

1 mark KU

subs values into either expression

For PI & PII

2 3 4 2 marks KU

1 knows to factorise 2 solves and discards d

9x x2 7x  2 x 1 9x 7x  2 = x2 x 1 9x(x + 1) = (x +2)(7x + 2) 9x2 + 9x = 7x2 + 16x + 4 2x2 – 7x – 4 = 0

1

cross multiplies multiplies brackets simplifies to answer

ans :

Illustrations for awarding mark

1 9(4) / 4 + 2 or (7(4) + 2) / 4 + 1 = 6 m/s

Total :

KU

26

RE

26

Totals :

KU

45

RE

45

Standard Grade Credit Paper 1 - Marks Grid Question Number

2011/2012

KU

RE

/19

/19

1 2 3 4 5 6 7 8(a) 8(b) 8(c) 9(a) 9(b) 10(a) 10(b) Total

 Pegasys 2011

Standard Grade Credit Paper 2 - Marks Grid Question Number

2011/2012

KU

RE

/26

/26

1(a) 1(b) 2 3(a) 3(b) 4(a) 4(b) 4(c) 5(a) 5(b) 6 7 8(a) 8(b) 9(a) 9(b) 9(c) 9(d) Total

 Pegasys 2011