Edexcel Mathematics Higher Tier, May 2009 (1380/3H) · PDF fileEdexcel Mathematics Higher Tier, May 2009 (1380/3H) (Paper 3, non-calculator) 0775 950 1629 Page 3

Edexcel Mathematics Higher Tier, May 2009

(1380/3H) (Paper 3, non-calculator)

www.chattertontuition.co.uk 0775 950 1629 Page 1

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Question 1 a)

Walk Car Other Total

Boy 15 25 note 1 14 54

Girl 22 note 2 8 note 4 16 46 note 3

Total 37 33 note 5 30 note 6 100

Note 1: 54 – (15 + 14) = 54 – 29 = 25

Note 2: 37 – 15 = 22

Note 3: 100 – 54 = 46

Note 4: 46 – (22 + 16) = 8

Note 5: 25 + 8 = 33

Note 6: 14 + 16 = 30

Check that total of bottom row makes 100 (37 + 33 + 30 = 100�)

b) There were 100 children in total and 37 of these walked to school so the probability that this child

walked to school is ��

��

Question 2 a) group all the terms

2x + 8y

b) 2 compasses would cost 2 x c pence = 2c

4 rulers would cost 4 x r pence = 4r

total cost would be 2c + 4r




Question 3 a)

x -2 -1 0 1 2 3

Y -11 -7 note 1 -3 1 note 2 5 note 3 9

Note 1: (4 x -1) – 3 = -4 – 3 = -7

Note 2: (4 x 1) – 3 = 4 – 3 = 1

Note 3: (4 x 2) – 3 = 8 – 3 = 5

b)

Question 4 a) substituting P with 50

50 = 4k -10

rearranging with k on the left hand side

4k – 10 = 50

add 10 to both sides

4k = 60

divide both sides by 4

k = 15

b) substituting n with 2 and d with 5

y = (4 x 2) – (3 x 5)

y = 8 – 15 = -7




Question 5 a)

The point 0 is the origin which is the coordinate (0,0)

b) Shape P has moved 3 units to the right and 1 unit down. The is known as a vector translation

where the vector is given by � ��

Question 6 a) the shape is a rectangle so the opposite sides are equal in length to each other so we have

4x + 1 = 2x + 12

b) subtracting 2x from both sides (so that we only have x terms on one side of the equation)

2x + 1 = 12

subtract 1 from both sides of the equation

2x = 11


x = 5.5 cm

c) now we know that x = 5.5cm, we know the length of each side of the rectangle.

The shorter side is length 5.5cm, the longer side is (4 x 5.5) + 1 = 22 + 1 = 23 cm

Perimeter (length all the way around the rectangle) is

5.5 + 23 + 5.5 + 23 = 57cm




Question 7 a) 3.22 x 4.8 is roughly 3 x 5 = 15 so our exact answer will be 15.456 (using the same digits as given)

Also if we know that 322 x 48 = 15456 then we need to move the decimal point 2 places to get 3.22

and then a further one place to get 4.8. At the moment the decimal point is after the 6 (15456.) so it

will now need to be after the 5 (15.456)

b) 0.322 x 0.48 is roughly 0.3 x 0.5 = 0.15 so our exact answer will be 0.15456

Also we need to move the decimal point 3 places to get 0.322 (from 322) and a further 2 places to

get 0.48 (from 48) so we are moving it 5 places in total to 0.15456 (from 15456)

c) 15456 ÷ 4.8 is roughly 15000 ÷ 5 = 3000 so our exact answer would be 3220

Also If we first rearrange our original given equation so that

15456 ÷ 48 = 322 and see how this compares to our new equation 15456 ÷ 4.8

15456 has been unchanged but we are now dividing by 4.8 instead of 48 so our answer will be 10

times bigger than before. Answer is 3220

Question 8 a) divide both sides by 2

x2 = 36

square root both sides

x = �6

b)

so 72 = 2 x 2 x 2 x 3 x 3 = 23 x 32

72

2 36

6 6

2 3 2 3




Question 9 a)

b)

Question 10

1 litre = 1000 mililitres

multiply by 40

40 litres = 40,000 mililitres

Water leaks out at the rate of 125 mililitres per second. We need to divide 40,000 by 125.

If we first put this calculation as a fraction then we may be able to cancel it down: �,��

�� =

,��

= �,��

� = 320 seconds

Question 11 a) 62.5 (this is the smallest it can be and still be rounded to 63 (to the nearest centimetre))

b) 63.49� (this is the largest it can be and still be rounded to 63 (to the nearest centimetre)) (63.5

would have also been acceptable in this type of question)




Question 12 The first condition leads to us drawing a circle of radius 4cm with B as the centre

The second condition leads to us drawing the angle bisector of angle A (which gives the locus of the

points that are equidistant from AB and AC)

Both must be drawn using a compass and ruler, and the examiners will be looking for correct

construction arcs.

Question 13 a)Which of the following types of magazines have you read in the past month (place a tick in the

relevant box(es))?

None

Education

Fashion

Sport

Other

b)

How many magazine’s have you read in the past month (place a tick in only one box)?

0

1 – 3

4 – 6

7 – 9

More than 9




Question 14 round each number to just one significant figure

� � ��

�.��

in order to get rid of the decimal on the bottom multiply the top and the bottom by 100

��

� =

��

= 700 x 40 = 28,000

Question 15 a) 6.4 x 104 (as we have effectively moved the decimal point 4 places)

b)1.56 x 10-5

(we have made 156 smaller by 100 so to compensate we need to make -7 bigger by

100)

Question 16 a) Find something that goes into both terms. This is 2x.

2x(2x – 3y)

b) Find two numbers that multiply to give -6 and add to give +5. The two numbers are +6 and -1.

(x + 6)(x – 1)




Question 17 a)

b) to find the median we find half of the total frequency. Half of 120 is 60 so draw a line across from

60 on the vertical axis until it meets the curve. Then drop the line down to meet the x axis. This

meets the x axis at 240. The line is shown above in red dashes.

240 is an estimate of the median

c)The women had a lower median than the men (205 compared to 240) so the women spent less

money during their summer holidays than the men did.

Question 18 a) A tangent will always make a right angle where it meets the radius. Hence triangle OAD is a right

angled triangle with angle OAD = 90⁰. Angles in a triangle add up to 180⁰ so

angle AOD = 180 – (36 + 90) = 180 – 126 = 54⁰

b) i) angle ABC = half of angle AOC = ½ of 54⁰ = 27⁰

ii) “angle at centre is twice the angle at the circumference”




Question 19 a) we simply look to see where the two lines meet. This will give us the solutions to the

simultaneous equations:

x =2, y = 3

b) Any line which is parallel to y = ½x + 2 will have the same gradient as this line. The gradient is ½.

We now need to find the y intercept. This is straightforward since they have given us the co-

ordinate (0,4) which is the point where the line meets the y axis. The y intercept is 4.

Our new line is y = ½x + 4

Question 20 a) subtract t from both sides (so that we have t on just one side of the equation and the side where it

will be positive)

2t + 1 � 12

subtract 1 from both sides

2t � 11


t � 5.5

b) t has to be less than 5.5 but also has to be a whole number. The largest it can be is 5.

Question 21 M � L

3

we can replace � with = as long as we now have a constant k

M = k x L3

substituting in the values of l and m we are given,

160 = k x 23

160 = k x 8

k x 8 = 160


k = 20

we have M = 20 x L3

When L = 3

M = 20 x 33

M = 20 x 27 = 540




Question 22 First we must calculate the frequency density for each class of data.

Frequency density = frequency ÷ class width

Insert two new columns

Length (x) minutes frequency Class width Frequency Density

0 � x � 5 4 5 0.8

5 � x � 15 10 10 1

15 � x � 30 24 15 1.6

30 � x � 40 20 10 2

40 � x � 45 6 5 1.2

We can see that the highest frequency density is 2 so we can now decide on the scale (4cm to every

1 of frequency density) (we use most of the space given but also have a scale that is easy to plot)




Question 23 a)

b)He will win both games if he wins the first game and wins the second game

0.5 x 0.5 = 0.25

Question 24 a) to prove that any two triangles are congruent we must prove one of the following:

all three sides are the same length (SSS)

two sides and the angle in between are the same (SAS)

two angles and one side are the same (AAS) or

Right angle, hypotenuse and one other side (RHS) this is the one I will use

side AB = side AC (as triangle is equilateral) hypotenuse

side AD = side AD (as same line) other side

we know the triangles are right angled (given in question) right angle

These three points prove that the triangles ADC and ADB are congruent

b) As ADC and ADB are congruent so side BD = side DC = ½ of BC

Now BC = AB (as triangle ABC is an equilateral triangle)

So BD = ½BC = ½AB

0.2

0.2

0.2

0.2

0.3

0.3

0.3

0.5

0.5

0.5

0.5

1st

game 2nd game

win

win

win

win

draw

draw

draw

draw

lose

lose

lose

0.3

lose




Question 25 a)

u = 2½ = ��

v = 3⅓ = ��

��

= 1 ÷ u = 1 ÷ �� = 1 x �

� =

��

�� = 1 ÷ v = 1 ÷

��

= 1 x ��

= �

��

��

+ �� =

�� +

��

=

�� +

��

= �

��

��

= �

��

multiply both sides by f

1 = ��

multiply both sides by 10

10 = 7f


��

= f

f = 1��

b)subtract �� from both sides of the equation

��

� �� -

��

multiply both sides of the equation by u

1 = ��

- ��

multiply both sides by f

f = u - ��

multiply both sides by v




fv = uv – uf

factorise the left hand side to get u on its own

fv = u(v – f)

divide both sides by (v – f) ��

�� = u

Question 26 a) we start with y = f(x), the curve has been translated (moved) 4 places in the positive x direction so

the new curve is y = f(x - 4) note that when the transformation effects the x it effects it in the

opposite way to what you would think

if the original equation was y = x2 then the new equation would be y = (x – 4)2

b) the curve has had two transformations. It has been stretched by a scale factor of ½ parallel to the

x axis (squashed), and stretched by a scale factor of 3 parallel to the y axis

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Edexcel Mathematics Higher Tier, May 2009 (1380/3H) · PDF fileEdexcel Mathematics Higher Tier, May 2009 (1380/3H) (Paper 3, non-calculator) 0775 950 1629 Page 3