Name
Class
Worked Solutions
Pre Public Exam
Paper 3H
June 2016
Higher Tier
Edexcel Style
Calculator
Time
1 Hour 30 mins
Marks Available 80
Commissioned by The PiXL Club Ltd.
Question
Mark
Maximum
mark
1
4
2
5
3
6
4
4
5
5
6
3
7
6
8
1
9
6
10
4
11
3
12
6
13
3
14
6
15
3
16
5
17
5
18
5
Total
80
Question 1.
The diagram shows a trapezium PQRS and two identical semicircles.
Q
P
R
S
18cm
The centre of each circle is on PQ.
Work out the area of the shaded region.
Give your answer correct to 3 significant figures.
Area of trapezium =
𝟏𝟔
𝟐
(14+18) = 8 × 32 = 256 cm2
Radius of each semi-circle = 14 ÷ 4 = 3.5
Area of two semi-circles 𝑨 = 𝝅 × 𝟑. 𝟓𝟐 = 38.48451001
Area of shaded region = 256- 38.48451001 = 217.51549 = 218 3sf
(Total 4 marks)
Question 2.
Sanjay is going on holiday to America.
The exchange rate is £1 = $1.45075
Sanjay changes £675 to dollars.
(a) Work out how many dollars he should get.
Give your answer correct to the nearest dollar.
675 × 1.45075 = 979.25625 = $979 to nearest dollar
979 dollars
(2)
Sanjay sees a pair of trainers in America.
The trainers cost $135.
Sanjay does not have a calculator.
He uses £2 = $3 to work out the approximate cost of the trainers in pounds.
(b) Use £2 = $3 to show that the approximate cost of the trainers is £90.
$135 ÷ 3 = 45
45 × 2 = 90
(2)
(c) Is using £2 = $3 instead of using £1 = $ 1.45075 a sensible way for Sanjay to work out the cost of
the trainers in pounds?
You must give a reason for your answer.
Yes it is.
It makes it much easier to calculate the cost of something to the nearest pound without using a
calculator. Using $1.5 instead of $1.45075 makes sense.
(1)
(Total 5 marks)
Question 3.
Here are the first five terms of a Fibonacci sequence.
1
3
4
7
11
The rule for the sequence is:
‘The next term is the sum of the previous two terms.’
(a) Find the 8th term of this sequence.
6th term: 7+11 = 18
7th term: 11 + 18 = 29
8th term: 29 + 18= 47
47
(1)
The first three terms of a different sequence which follows the same rule are:
m
2n
m + 2n
(b) Show that the 6th term of this sequence is 3m +10n
4th term: m + 2n + 2n = m + 4n
5th term: m + 4n + m + 2n = 2m + 6n
6th term: 2m + 6n + m + 4n = 3m + 10n
(2)
Given that the 3rd term is 5 and the 6th term is 23,
(c) Find the value of m and the value of n
3m +10n = 23
3m +10n = 23
m + 2n = 5
×3
3m + 6n = 15
-
4n = 8
n=2
m + 2n = 5
m + 4 = 5 so m = 1
m=1
n=2
(3)
(Total 6 marks)
Question 4.
= 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
A = 𝑥: 𝑥 is a factor of 12
1, 2, 3, 4, 6
B = 𝑥: 𝒙 is a multiple of 3 3, 6, 9
(a) Complete the Venn diagram to show the elements of each set.
A
5
2
7
108
4
B
1
3
6
9
(3)
(b) A is changed to A = 𝑥: 𝑥 is a factor of 15 .
James says that the number of elements in A U B stays the same.
Is James right?
Give a reason why.
NO, it doesn’t stay the same, since 6 is not a factor of 15
(1)
(Total 4 marks)
Question 5.
Julie is thinking of having a water meter.
These are the two ways she can pay for the water she uses.
Watermeter
Nowatermeter
Afixedchargeof£104.82
Plus
£1.95foreverycubicmetreofwaterused
Achargeof£283peryear
1cubicmetre=1000litres
Julie uses on average, 140 litres of water each day.
Use the information above to determine whether or not Julie should have a water meter.
With the water meter cost would be:
140 litres = 0.140 cubic metres
Cost of 0.140 cubic metre per day: 0.140 × £1.95 = £0.273
Cost over 365 days: 365 × £0.273 = £99.645
Total cost £104.82 + £99.645 = £204.465 which is approx £205
Since £205 is less than £283- it is cheaper for Julie to have a water meter.
(Total 5 marks)
Question 6.
!
!
Debbie cuts a length of ribbon 2 ! metres long into ! metre pieces.
How many whole pieces can she cut and what length of ribbon is left over?
Give your answer as a mixed number.
𝟏𝟏
𝟒
𝐱
𝟑
𝟐
=
𝟑𝟑
𝟖
𝟏
=4𝟖
𝟏
4𝟖
(Total 3 marks)
Question 7.
Sarah works for a company that delivers parcels.
One day the company delivered 80 parcels.
The table shows information about the weights, in kg, of these parcels.
(a) Complete the cumulative frequency table.
19
36
51
63
73
80
(1)
(b) On the grid draw a cumulative graph to represent this data.
(2)
Sarah says,
"75 % of the parcels weigh less than 3.4 kg."
(c) Is Sarah correct?
You must show how you get your answer.
UQ = 3.6 > 3.4 so no she is not correct
(3)
(Total 6 marks)
Question 8.
Write 8.9 x 10-5 as an ordinary number
0.000089
(Total 1 mark)
Question 9.
(a) Factorise x2 + 9x + 20
(x + 4) (x + 5)
(2)
(b) Solve 7x – 4 < 2x + 15
5x < 19
x<
𝟏𝟗
𝟓
x<
𝟏𝟗
𝟓
(2)
(c) p is an integer with -4 ≤ 2p < 9
Write down all the values of p
-2 ≤ p < 4.5
-2, -1, 0, 1, 2, 3, 4
2, -1, 0, 1, 2, 3, 4
(2)
(Total 6 marks)
Question 10.
The function f is such that
f(x) = 3x – 2
(a) Find the value of
(i)
f(0)
(3 x0) – 2 = -2
-2
(ii)
f(-2)
(3 x -2) – 2 = -8
-8
(2)
(b) Find the value of x for which f(x) = 16
16 = 3x – 2
18 = 3x
x=6
6
(2)
(Total 4 marks)
Question 11.
Solve x2 + 3x – 6 = 0
Give your answers correct to 3 significant figures.
x=
x=
!𝟑± ( !𝟑 𝟐 !𝟒 𝟏 !𝟔 )
𝟐(𝟏)
!𝟑±√𝟑𝟑
𝟐
x= -4.372
or x= 1.372
x= -4.372
x= 1.372
(Total 3 marks)
Question 12.
Terry asked 500 people which fruit they liked from apples, oranges and bananas.
294 people liked oranges
231 people liked only oranges
34 liked oranges and bananas only but not apples
28 liked oranges and apples only but not bananas
127 liked bananas only.
10 people liked bananas and apples but not oranges
(a) Work out the probability that a person selected at random likes only apples.
oranges
apples
69
1
231
28
65
5
34
10
127
bananas
𝟔𝟗
𝟓𝟎𝟎
(4)
(b) Given that a person chosen at random likes bananas, work out the probability that this person also
likes just one other fruit.
𝟒𝟒
𝟏𝟕𝟐
(2)
(Total 6 marks)
Question 13.
Diagram not drawn accurately.
P
Q
C
S
R
PCR and QCS are diameters of a circle. C is the centre of the circle.
Prove that triangles PQR and QRS are congruent.
RHS
SCQ = PCR both diameters of circle (hypotenuse)
PQR = QRS = 90° angles in a semicircle are right angled
QR shared side
or
SCQ = PCR both diameters of circle
PQR = QRS = 90° angles in a semicircle are right angled
RPQ = QSR angles in same segment are equal.
So PRQ = SQR (180° = in a triangle).
Therefore PQR and QCS are congruent ASA (Angle Side Angle)
(Total 3 marks)
Question 14.
(a) Show that the equation x2 – 7x – 3 = 0 has a solution between x =7 and x = 8
F(7) = -3, F(8) = 5
(7) – 7(7) -3 = -3
(8) – 7(8) -3 = 5
Change of sign therefore at least one root in interval 7 < x < 8
2
2
(b) Show that the equation x2 – 7x – 3 = 0 can be rearranged to give x = 7 +
2
x = 7x + 3 (÷ x)
(2)
!
!
𝟑
x=7+𝒙
(1)
(c) Starting with
x0 = 7 use the iterative formula xn+1 = 7 +
solution of x2 – 7x – 3 = 0 to one decimal place.
x1 = 7.42857....
x1 = 7 +
x2 = 7.40384....
x2 = 7 +
!
!!
twice, to find an estimate for the
𝟑
𝟕
𝟑
𝒙𝟏
7.4
x= 7.4
(3)
(Total 6 marks)
Question 15.
Joanna has five cards labelled 3, 4, 5, 6 and 7
She draws two cards at random and makes a 2-digit number with the first card drawn used to make the tens
digit.
(a) How many possible outcomes are there?
5x4
20
(2)
(b) What is the probability of making 57?
P(57) = 1/20
𝟏
𝟐𝟎
(1)
(Total 3 marks)
Question 16.
Diagram not to scale
E
A
F
B
D
H
3cm
C
G
10cm
The volume of this box is 150cm2
A pencil is placed in the box and wedged between points E and C.
Find the size of the angle the pencil makes with the plane CDHG.
Give your answer to 2 significant figures.
150÷ (3x10) = 5 = EH
CH2 = 102 + 32= 109
𝟓
Tan ECH = √𝟏𝟎𝟗 ECH = 26°
or
150÷ (3x10) = 5 = FG
CH2 = 102 + 32= 109
CE2 = 109 + 52 = 134
CE = √134
Sin ECH =
= (sin-1
𝟓
𝟏𝟑𝟒
𝟓
𝟏𝟑𝟒
)
26°
26°
(Total 5 marks)
Question 17.
A point P (3,5) lies on the circumference of the circle x2 + y2 = 34.
(a) Find the gradient of the line joining P to the origin (0,0)
𝟓!𝟎
𝟑!𝟎
𝟓
𝟑
(1)
(b) Find the equation of the tangent to the circle at A in the form y = mx + c
Gradient of tangent = (-3)/5
y = (-3)/5x + c
5 = (-3)/5 (3) + c
c = 34/5
𝟑
𝟑𝟒
y=-𝟓X+ 𝟓 (4)
(Total 5 marks)
Question 18.
12b
X
Y
9a
Z
O
(a) Find the vector XZ.
-9a+12b (1)
When OZ is extended, W is a point such that OW is twice OZ.
!
M is the point with position vector OM = ! OX
N is the point on YW such that YN : NW = 1 : 2
(b) Prove that MN is parallel to XY and state the relationship between them.
OM = 6a
𝟏
YN = 𝟑(-9a+12b)=-3a+4b
MN = 3a +12b-3a+4b
MN = 16b
𝟒
MN parallel to XY as MN = 𝟑(XY
(4)
(Total 5 marks)
TOTAL FOR PAPER IS 80 MARKS