Download Guide to Leaving Certificate Mathematics Ordinary Level Paper 1

Guide to Leaving Certificate Mathematics
Ordinary Level
Dr. Aoife Jones
Paper 1
For the Leaving Cert 2013, Paper 1 is divided into three sections. Section A is entitled
Concepts and Skills and contains four questions, all of which must be completed.
These are designed to test the student’s understanding of basic mathematical concepts.
In Paper 1, this corresponds to an understanding of the basics of algebra and number
systems. Questions that appear here tend to be on topics that do not lend themselves
to the larger-scale problems that occur later on in the exam. Topics that can appear in
this section include, but are not limited to:
(i) prime numbers and factors
(ii) inequalities and numberlines
(iii) indices and scientific notation
(iv) profit and loss, compound interest, income tax
(v) quadratic equations and simultaneous equations
(vi) and also topics that may appear in Section B.
[See Example 1.]
Section B is entitled Contexts and Applications and contains two questions. Again,
no choice is given and both questions must be answered. These questions allow for more
extensive testing of a student’s understanding of particular subjects. It is important to
note that the questions asked here may span over a number of topics that were traditionally asked in separate questions. As a result, students must be vigilant when approaching
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these questions; they must now keep to mind their entire armoury of mathematical approaches when attempting any one question. Topics that can appear in this section
include, but are not limited to:
(i) complex numbers
(ii) patterns
(iii) series and sequences
On this latter topic, students are expected to be able to identify patterns that are described in words or in diagrams, and to express the pattern in algebraic form. This topic
is new to the syllabus with the introduction of Project Maths and aims to enable students
to better understand the role that maths plays in natural world [see Example 2.]
Section C is entitled Functions and Calculus (old syllabus) and contains three
questions, of which students must answer two. This topic remains unchanged from
previous years for the Leaving Cert in 2013 only, and so the questions closely follow those
asked in the past exam papers. Students should look at Questions 6, 7 and 8 from Paper
1 of the old exam papers, as the content has not changed and the questions still follow
the same outline.
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Example 1 - Paper 1 Section A Concepts and Skills
2011 (Project Maths Trial Schools) Paper 1 Q1
(a) Explain what a prime number is.
Definitions are being given a new emphasis with the introduction Project Maths. It is no
longer sufficient to know whether a number is prime, but we must understand why.
Solution A prime number is any number whose only factors are 1 and itself.
(b) Express 2652 as a product of prime numbers.
All numbers can be written as a product of prime numbers. It is just a matter of breaking
the factors down until they are all prime.
Solution
2652
2652
2652
2652
=
=
=
=
2 × 1326
2 × 2 × 663
2 × 2 × 3 × 221
2 × 2 × 3 × 13 × 17
(c) The number 261 −1 is a prime number. Using your calculator, or otherwise, express its
value, correct to two significant figures, in the form a×10n , where 1 ≤ a < 10 and n ∈ N.
Most calculators will do this for you.
Solution 261 − 1 = 2.305843 × 1018 = 2.3 × 1018 , correct to two significant figures.
(d) Use your answer to part (c) to state how many digits there are in the exact value of
261 − 1.
Tip: If having difficulties with hard questions, we can try to think of a simple example
and solve that. For example, 2.3 × 101 is 23, which has two digits. 2.3 × 102 = 230,
which has 3 digits. It is always one more than the power of the ten.
Solution 261 − 1 = 2.3 × 1018 has 18 + 1 = 19 digits.
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Example 2 - Paper 1 Section B Contexts and Applications
2011 (Project Maths Trial Schools) Sample Paper 1 Q5
Sı́le is investigating the number of square grey tiles needed to make patterns in a
sequence. The first three patterns are shown below, and the sequence continues in the
same way. In each pattern, the tiles form a square and its two diagonals. There are no
tiles in the white areas in the patterns there are only the grey tiles.
Figure 1: (taken from the Leaving Cert paper, available from the State Examinations
Commission website)
(a) In the table below, write the number of tiles needed for each of the first five patterns.
Patterns
1 2 3
No. of tiles 21 33
4 5
The number of tiles for the 3rd pattern can simply be counted. We should recognise that
the first pattern is of width 5, the next of width 7 and the next of width 9. Therefore
the next two will be of width 11 and 13. For the 4th and 5th patterns, a good approach
is to actually draw out the tiles:
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Figure 2:
4th pattern
5th pattern
Solution
Patterns
1 2 3 4 5
No. of tiles 21 33 45 57 69
(b) Find, in terms of n, a formula that gives the number of tiles needed to make the nth
pattern.
We note that the number of tiles goes up by 12 each time. This means it is an arithmetic
sequence, where the common difference is d = 12. We then use the formula for the
nth term of an arithmetic sequence Tn . For this we must know a, the first term of the
sequence.
Solution
Tn =
=
=
=
a + (n − 1)d
21 + (n − 1)12
21 + 12n − 12
9 + 12n
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(c) Using your formula, or otherwise, find the number of tiles in the tenth pattern.
When a question offers the choice of solving a problem by “using a previous solution“
or “otherwise”, then the use of the previous solution is always preferable. The exception
is if we have been unable to find the previous solution, in which case this phrasing is
explaining that the problem can still be solved using an alternative method. In this case,
that could involve either adding the 12 tiles to the number of tiles required for each
pattern until reaching the tenth pattern. Alternatively, we could try to draw the pattern
of width 23 tiles.
Tn
⇒ T10
T10
T10
=
=
=
=
9 + 12n
9 + 12(10)
9 + 120
129
(d) Sı́le has 399 tiles. What is the biggest pattern in the sequence that she can make?
This question clearly requires a whole number as the solution. Also, it is not possible
to simply round the answer using the standard rounding-off convention. If we find it is
possible to make the 3.6th pattern, for example, then clearly she can only make the third
pattern and not the fourth. For this reason, the number must be rounded down.
Solution We use Tn = 9 + 12n and we are told that Tn ≤ 399. Therefore:
Tn
399
399 − 9
390
390
12
n
=
≥
≥
≥
9 + 12n
9 + 12n
12n
12n
≥ n
≤ 32.5
So Sı́le can make the 32nd pattern (and not the 33rd ).
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(e) Find, in terms of n, a formula for the total number of tiles in the first n patterns.
When the terms of a sequence are added, this is known as a series. In this case, we have
an arithmetic sequence where d = 12 and a = 21 as before, but we use the formula for
the sum of n terms of an arithmetic sequence:
Sn =
=
=
=
=
=
n
[2a + (n − 1)d]
2
n
[2(21) + (n − 1)(12)]
2
n
[42 + 12n − 12]
2
n
[30 + 12n]
2
30n 12n2
+
2
2
15n + 6n2
(f) Sı́le starts at the beginning of the sequence and makes as many of the patterns as
she can. She does not break up the earlier patterns to make the new ones. For example,
after making the first two patterns, she has used up 54 tiles, (21 + 33). How many
patterns can she make in total with her 399 tiles?
This question simply requires using the formula we found in the previous section but with
Sn = 399.
Sn
⇒ 399
0
0
0
2n + 19 = 0
2n = −19
19
n=−
2
n
=
=
=
=
=
15n + 6n2
15n + 6n2
6n2 + 15n − 399
2n2 + 5n − 133 , dividing both sides by 3
(2n + 19)(n − 7)
or
or
n−7=0
n=7
or
n=7
=
7 , (as we disregard negative answers here)
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So Sı́le can make the 1st , 2nd , 3rd , 4th , 5th , 6th and 7th patterns, which is 7 patterns in
total.
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Paper 2
For the Leaving Cert 2013, Paper 2 is divided into just two sections. The entire of Paper
2 has had a Project Maths re-lift, and there are no questions that follow the previous
syllabus. As in Paper 1, Section A is entitled Concepts and Skills. It contains six
questions, all of which must be completed. There is one choice given, as students are
allowed the option of answering Question 6A or 6B. These six questions are designed to
test the student’s understanding of the basic mathematical concepts for examination in
Paper 2. For the first five questions, topics that can appear include, but are not limited
to:
(i) coordinate geometry of the line and circle
(ii) enlargements
(iii) area and volume
(iv) permutations
(v) and also topics that may appear in Section B.
[See Example 3.]
Question 6 currently offers a choice of two topics, and will continue to do so for the
duration of the three year roll-out of the new syllabus (2012, 2013, 2014). After this
time, there will be no choice given. Question 6A is based on synthetic geometry, which
consists of being able to use certain terms related to logic and deductive reasoning
(theorem, proof, axiom, corollary, converse, implies), as well as being able to carry out a
number of geometric constructions (angle of 60◦ without using a protractor or set square,
tangent to a given circle at a given point on it, parallelogram given the length of the
sides and the measure of the angles, circumcentre and circumcircle of a given triangle
using only straight-edge and compass, incentre and incircle of a given triangle using only
straight-edge and compass, centroid of a triangle). Question 6B consists of answering a
problem-solving question based on the geometry theorems studied for the Junior Certificate.
In Paper 2, Section B is entitled Contexts and Applications and contains two
questions, both of which must be answered. Again, these questions allow for more
extensive testing of a student’s understanding of particular subjects, and each question
is likely to span a number of topics. In particular, students should be aware of the links
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between probability and statistics and the strong possibility of these being tested together
in one question. Topics that can appear in this section include, but are not limited to:
(i) probability
(ii) statistics
(iii) trigonometry
[See 2011 (Project Maths Trial Schools) Paper 2 Q7 as an example.]
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Example 3
2011 (Project Maths Trial Schools) Paper 2 Q3
A plastic toy is in the shape of a hemisphere. When it falls on the ground, there are
two possible outcomes: it can land with the flat side facing down or with the flat side
facing up. Two groups of students are trying to find the probability that it will land with
the flat side down.
(a) Explain why, even though there are two outcomes, the answer is not necessarily equal
to 12 .
For questions like this that are asked in words, it is suitable for us to give an answer in
words.
Solution The answer would only be equal to 12 if both outcomes are equally likely, which
we cannot know for definite. The two sides of the toy are shaped differently which may
affect the outcome.
(b) The students estimate the probability by experiment. Group A drops the toy 100
times. From this, they estimate that it lands flat side down with probability 0.76. Group
B drops the toy 500 times. From this, they estimate that it lands flat side down with
probability 0.812.
(i) Which groups estimate is likely to be better, and why?
Students are expected to know that probabilities estimated from experiments improve in
accuracy as the number of trials increases.
Solution Group B’s estimate will be better as they have carried out more trials, leading
to better accuracy.
(ii) How many times did the toy land flat side down for Group B?
If students get confused about what to do with the numbers, it is good practice to imagine
what we would do if the numbers were easier. For example, if the toy was dropped 6
times and it was found to have a probability of 0.5 of landing flat side down, we would
know it landed flat side down 3 times. This is because 0.5 is equal to 12 , and a half of 6
is 3. Or, in other words, 0.5 × 6 = 3.
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Solution
0.812 × 500 = 406 times
(iii) Using the data from the two groups, what is the best estimate of the probability that
the toy lands flat side down?
Etimates are improved by using the most number of trials. Therefore the best estimate
that can be made here is when we include the results from both experiments.
Solution Group B found the toy landed flat side down 0.812 × 500 = 406 times. Group
A found the toy landed flat side down 0.76 × 100 = 76 times. In total, the toy landed
flat side down 406 + 76 = 482 times out of 600 times.
Best estimate of probability of landing flat side down =
12
482
.
600