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Transcript
Mathematics STEP
Dr Chris Warner
College Teaching Officer
Director of Studies in Mathematics
Educational Consultant
Robinson College
Why STEP?
•
•
•
•
•
Reveals greater depth of understanding
Measures ability to think “outside the box”
Tests lateral thinking
Tests ability to apply knowledge
A level is not (we believe) a good (enough)
preparation for mathematics in top
universities
Robinson College
STEP resources
• Cambridge Assessment’s Admissions Testing Service
web site:
− http://www.admissionstestingservice.org/for-testtakers/step/about-step/
•
OpenBook publishers web site:
− Advanced Problems in Mathematics: Preparing for University
http://www.openbookpublishers.com/product/342/advanced
-problems-in-mathematics--preparing-for-university
‒ purchasable book and downloadable PDF by Stephen Siklos
‒ hints and full solutions
‒ a good starting point for preparation
Robinson College
STEP resources
• Mathematics Department web site:
‒
NRICH: STEP preparation archive
https://nrich.maths.org/10622
‒
Meikleriggs Mathematics http://meikleriggs.org.uk/
‒ complete solutions to STEP papers (latest 2010)
‒
STEP support programme: Online course to help prepare for
STEP https://maths.org/step/
Robinson College
STEP administration
• Administered by Cambridge Assessment
‒
‒
‒
‒
STEP = “Sixth Term Examination Paper”
well-established Mathematics university
admissions test
managed by a specialist team whose remit is
assessments relating specifically to university
entrance
dedicated website:
http://www.admissionstestingservice.org/for-testtakers/step/about-step/
Robinson College
STEP administration
• Administered by Cambridge Assessment
‒
website provides information on
‒ method of entry
‒ Dates
‒ Syllabus
‒ downloadable past papers
Robinson College
STEP specification
• Test summary
‒
‒
‒
‒
‒
‒
‒
three papers
candidates typically sit 2 papers
pen and paper test
3 hours
six questions from 13
no calculators
taken in late June
Robinson College
STEP papers and offers
• Three STEP mathematics papers I, II, and III
‒
thirteen questions:
‒ 8 pure
‒ 3 mechanics
‒ 2 statistics and probability
‒ assessed on answers to at most 6 questions
‒
five grades (from highest to lowest) S, 1, 2, 3,
and U
Robinson College
STEP papers and offers
• Cambridge offer usually includes grades in
two papers
‒ not taking the full Further Mathematics A level or
equivalent: Papers I and II;
‒ offer is usually 1,1; sometimes 1,2
‒ taking Further Mathematics A-level or equivalent:
Papers II and III
‒ offer is usually1,1
Robinson College
STEP papers and offers
• Paper I
‒
‒
‒
‒
‒
syllabus based on single subject A level pure mathematics
content slightly more than common core
mechanics and probability and statistics sections equivalent
to two or three A-level modules
intended for candidates not taking Further Mathematics
questions easier than paper II
• Paper II
‒
‒
same content as Paper I
harder questions
Robinson College
STEP papers and offers
• Paper III
‒
‒
syllabus based on further maths A level
questions same level of difficulty as paper II
Robinson College
Preparing for STEP
• Can candidates prepare without help?
–
we make adjustments
– arguably these may not go far enough
– students need to be able to cope when they start
– some “training” is essential
– the Cambridge Mathematics Department runs an online STEP
support programme
– more likely to relax on STEP offers in August for candidates
who have no input from their school
Robinson College
Preparing for STEP
• How teachers can help
–
STEP is aimed at the top few students so use as
differentiated material
– it can be hard to do this
– … but it is not impossible, particularly for further
mathematics groups
–
Remind students that training has a huge effect
– as teachers would expect
– so start early and practice regularly and frequently
– perseverence is good preparation for University and life!
Robinson College
Preparing for STEP
• How teachers can help
–
reassure students that STEP is indeed “very hard”
(as quite a few teachers have told me)
– questions are hard compared to A level
– there is a commonality of style
–
encourage students to try the online STEP support
programme
– reminder: this is aimed to “close the gap” for candidates
who do not receive help in school; who are in schools with
few applicants for Cambridge; who trigger social deprivation
“flags”
Robinson College
A STEP paper I question
A positive integer with 2𝑛 digits (the first of which must not be
0) is called a balanced number if the sum of the first 𝑛 digits
equals the sum of the last 𝑛 digits. For example 1634 is a 4-digit
balanced number, but 123401 is not a balanced number.
(i) Show that seventy 4-digit balanced numbers can be made
using the digits 0, 1, 2, 3, and 4.
1
(ii)Show that 𝑘 𝑘 + 1 4𝑘 + 5 4-digit balanced numbers can
6
be made using the digits 0 to 𝑘.
You may use the identity
𝑛
2
𝑟=0 𝑟
1
6
= 𝑛 𝑛 + 1 2𝑛 + 1 .
Robinson College
What skills are being tested?
• Being systematic and methodical
• Spotting patterns
• Creating formulas to count the number of
possible sums and grouping the terms
appropriately
• Using the formula for sum of an arithmetic
sequence (in formula booklet)
• Using given formula for sum of first 𝑛 square
numbers
Robinson College
One possible solution
Consider combinations of first two digits and present in a
table:
00
01
02
03
04
10
11
12
13
14
20
21
22
23
24
30
31
32
33
34
40
41
42
43
44
Robinson College
One possible solution
The first half of the four digit number cannot start with a 0
whereas the second half can start with a 0. The number of
possibilities for each sum are therefore:
Sum: 1
First
1
half
Second 2
half
2
2
3
3
4
4
5
4
6
3
7
2
8
1
3
4
5
4
3
2
1
Robinson College
One possible solution
Thus the number of 4-digit balanced numbers is
1×2 + 2×3 + 3×4 + 4×5 +
4 × 4 + 3 × 3 + 2 × 2 + 1 × 1 = 70
There are two qualitatively different types of term. The
first four take the form of a sum of 𝑟(𝑟 + 1) for 𝑟 = 1 to 4
and the second four take the form of a sum of 𝑟 2 for 𝑟 = 1
to 4.
Robinson College
One possible solution
For four digit balanced numbers made using the digits 0 to
𝑘, the sum of terms generalises straightforwardly to:
𝑘
𝑟=1 𝑟
𝑟+1 +
𝑘
2
𝑟
𝑟=1
=
𝑘
2
(2𝑟
𝑟=1
+ 𝑟)
1
1
= 2 × 𝑘 𝑘 + 1 2𝑘 + 1 + 𝑘(𝑘 + 1)
6
2
1
1
= 𝑘 𝑘 + 1 4𝑘 + 2 + 3 = 𝑘 𝑘 + 1 4𝑘 + 5
6
6
Robinson College