Download Solutions

The National Bank
Junior Mathematics Competition
Solutions Booklet
Department of Mathematics and Statistics
University of Otago
Visit us at www.maths.otago.ac.nz/nbjmc
Year 9 Prize Winners
First
Second
Third
Yi Lin Zheng
Byung Cheol Cho
Eric Jou
Diocesan School for Girls
Auckland Grammar School
Mission Heights Junior College
Top 30 prizes:
Frank Zhou, Pakuranga College
Jun Lee, The Correspondence School
Ian Seong, Burnside High School
Xiaohan Chen, Kristin School
Joon Park, Macleans College
Luke Naylor, Palmerston North Boys’ High School
Logan Glasson, Burnside High School
Natalia Chen, St Andrew’s College
Jaynesh Narayan, Burnside High School
Jenny Cao, St Cuthbert’s College
Yung (Justin) Koh, ACG Strathallan College
Mikael Koo, Burnside High School
Sun Jae Lee, Auckland Grammar School
Sheng Cao, St Paul’s Collegiate School
Arthur Joe Yanagisawa, Auckland Grammar School
Yoonjae Kim, Christchurch Girls’ High School
Andrew Wang, Macleans College
Young Lu, St Cuthbert’s College
Chris Wong, Burnside High School
Sabrina Yeh, St Cuthbert’s College
Angela Chu, Villa Maria College
Hannah Suh, Westlake Girls’ High School
Wei Kit Yoon, Avondale College
Kevin Yu, Lynfield College
Jennifer Zhan, Macleans College
Robin Laven, St Peter’s College (Palmerston North)
John McLachlan, The Correspondence School
Year 10 Prize Winners
First
Second
Third
Brian Ng
Richard Chou
Raymond Lee
Hutt Valley High School
Macleans College
Macleans College
Top 30 prizes:
Francesco Kook, Kristin School
Arun Chockalingam Shanmuganathan, Auckland Grammar
Andy Chen, Macleans College
School
Sally Park, St Kentigern College
Hin Loh, Macleans College
Rachel Boswell, New Plymouth Girls’ High School
Franklin He, Rutherford College
Basil Connor, Wellington College
Callum Johns, Spotswood College
Phillip Zhang, Macleans College
Kent Jianrong Liang, Auckland Grammar School
Pak-Hang Henry Yuen, Auckland Grammar School
Kieran James Sim, Auckland Grammar School
Timothy Gray, Macleans College
An-Ran Chen, Macleans College
Maya Wilde, Wellington East Girls’ College
Merrila Babu John, St Andrew’s College
Anita Miranda, Baradene College
Howell Zicong Fu, Auckland Grammar School
Nalin Choudhary, Lynfield College
Segar Manoharan, Hamilton Boys’ High School
Tomy Jeon, Palmerston North Boys’ High School
Wendy He, Lynfield College
Robert Shin, Macleans College
Lea Kapelevich, Epsom Girls’ Grammar School
Nicholas On, Wellington College
Joey Chen, Rangitoto College
Year 11 Prize Winners
First
Second
Third
Michael Wang
Andrew Carrell
Warren Wang
Macleans College
Christ’s College
Macleans College
Top 30 prizes:
Eun Hoi Koo, Auckland Grammar School
Mirendra Aruldasan, Botany Downs Secondary College
Ha-Young Shin, Christchurch Boys’ High School
Soobin An, Westlake Girls’ High School
Oliver Dunn, Christ’s College
Thomas Prebble, King’s College
Arkar Ian Thein, Auckland Grammar School
Luke Xing-Yan Tian, Auckland Grammar School
Scott Wong, King’s College
Danielle Johnstone, Middleton Grange School
Jay Chan, Wellington College
Philip Yuan-Ho Chien, Auckland Grammar School
Jin Soo Kim, Auckland Grammar School
Paul Yi Hsinn Liao, Auckland Grammar School
Patrick Dawson, Logan Park High School
Juntao Chen, Macleans College
You-Chei (Aileen) Sung, Rangitoto College
Vicki Cho, Burnside High School
Thomas Elliott Adams, Auckland Grammar School
Alexander Guang Ding, Auckland Grammar School
Thomas Campbell Thorpe Riley, Auckland Grammar School
Kaishuo Yang, Auckland Grammar School
Boyd Siripornpitak, Christ’s College
Bianca Yow, Corran School
Ashish Pandey, King’s College
Boxuan Fan, Lynfield College
Chang Zhai, St Paul’s Collegiate School
This year, many of the questions involved topics which students
might encounter later on. Many of the questions had easy
starters, but they got harder. It was possible for a student to get
100%. The winner from Year 11 did just that.
y
A
Question One (Year Nine and below only)
(a)
Draw a number line from -5 to 5 IN YOUR ANSWER
BOOKLET. Number all the integers.
The number line drawn had to clearly have zero labeled,
B
and have the negative numbers in the correct order. Very
well answered. Missing out zero was done by very few.
C
(b)
The diagram shows two number lines at 90° to each other.
The points A (1, 5) and B (3, 1) are shown. Copy the
x
diagram into your Answer Booklet and add and label with
their letters the points C (-1, 1) and D (-3, -5).
This question was meant to be encouraging. As long as
NOT TO SCALE
each point was in the correct quadrant it earned full
marks. Surprisingly, many students got C wrong (bottom
D
right quadrant or actually on one of the axes) and some
got D wrong. Most couldn't do the rest of the question.
(c)
Use the Theorem of Pythagoras
2
2
2
h =a + b
where h is the hypotenuse (the longest side) of an appropriate right angled triangle to show that the distance from
A to B is 2√5. (Answers found with a ruler will receive no credit.)
Markers were told not to worry about setting out in this question. Unsighted on most papers.
2
2
2
h
=
a + b
2
2
2
h
=
2 +4
2
h
=
20
h
=
2√5
Many students used their calculators to find the decimal value, but this earned nothing unless they either reached
20 or showed that the decimal value was equivalent to a number of decimal places to 2√5. A comment about
calculators is needed here. You just about cannot do the competition without one.
(d)
Find the area of the quadrilateral ABCD found in this question.
Note that the quadrilateral should have been ABDC, but this didn't seem to affect very many pupils.
There are several ways to break the area up into triangles. Probably the easiest is
Area
=
area ABC + area BCD
=
½44+½46
=
20
The markers were instructed not to worry about units. Of course follow-on was possible here, but part (b) was
regarded as so easy, so none was given.
(e)
The line AB has slope of 2 while the line CM, where M is the midpoint of AB, has slope of -½. In a few words
tell us what this says about the two lines.
They are perpendicular to each other. Note that the signs of the two slopes are both wrong – this did not seem to
affect the answer however for any student. It was barely noticed by the students, although teachers commented
about it in emails.
Question Two (all students)
This question was a little too hard as a starter for the older students. One regret of the organisers was that we didn't swap
questions 2 and 3, as question 3 proved to be easier. Students from schools who had previously encountered matrices
found the going a lot easier, however.
A 2 by 2 matrix (plural matrices) is an array of numbers. Among other applications, they can be used for sending codes in
times of war. To decode the message, you have to know the inverse matrix. If
[
1 d −c
det −b a
]
[ ]
a c
b d
is the matrix, then
is the inverse.
The det is a number found by calculating the value of ad – bc.
(a)
[ ]
2 4
2 3
Find the det of the matrix
det
=
23–24
=
-2
Generally well answered, although 6 – 8 = 2 was too common.
Some candidates 'flipped' the value, so that their determinant was –½. In most cases this meant that credit was not given.
(b)
In a few words or by a formula, explain when the inverse matrix doesn't exist.
It doesn't exist if the determinant (det) equals 0. Or equivalent. Answered correctly reasonably often.
(c)
Show that the inverse matrix of
[ ] [
2 4
2 3
is
−1.5 2
1
−1
]
assuming that you can multiply right through by
the det. Show all necessary working.
det
=
-2
Inv
=
Inv
=
[
[
3 −4
−2 2
−1.5 2
1
−1
1/-2
]
The product of two matrices can be found by
]
(No marks for this as it was given in the question.)
[ ][ ] [
a c
b d
e g
f h
=
aecf
bedf
agch
bgdh
]
where a, b, c, d, e, f, g, h
are all real numbers.
(d)
If A = 1, B = 2, . . . O = 17, . . . etc, what 2 by 2 matrix results when you use the matrix to code the message
'FOOD'? (You will need to find the matrix representing the word FOOD then multiply the two matrices together
with the matrix for FOOD going second).
Note that O = 15, and not O = 17. This mistake was spotted by many candidates, and it is regretted. There were
four possible answers which might gain full marks, two each for the case O = 15, as matrix multiplication is, in
general, not commutative (order matters), and two if the candidate used O = 17, as in the question. Only one of
these (O = 17, FOOD matrix second) is shown below; the process is basically the same for the other cases.
[ ][ ]
[
]
[ ]
2 4 6
2 3 17
1268
=
1251
80 50
=
63 46
17
4
3416
3412
(The second matrix was worth half marks if there was nothing else.)
It wasn't uncommon to see a correct answer, although arithmetic mistakes were unfortunately common.
(e)
What 2 by 2 matrix results when a matrix is multiplied by its inverse?
[ ]
1 0
0 1
Probably the best way to get this answer (it is not a proof) was to multiply the two matrices
given in the question together. Some of those who had encountered matrices before described it
as the identity matrix, although the actual matrix was needed for any credit here.
Question Three (all students)
This question was the easiest in the competition. Full marks were common.
A prime number can be defined to be a whole number larger than 1, with only two factors (numbers which divide exactly
into it), 1 and itself. For example, the number 43 is prime, because only 1 and 43 itself divide into 43 exactly.
(a)
Write down the first twelve prime numbers in numerical order (smallest to largest).
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. Generally well done, although it was quite common for 2 to be missing,
and 39 (= 3 13) was common at the end. Some candidates didn't seem to be able to add to 12, giving more than
12 primes, or too few. Another common mistake was to have 1, which the question defines not to be prime.
(b)
The number 2009 is not prime, but can be written as the product of more than one prime number. Show how to
reach this answer.
2009 = 7  7  41
287 7 was common, from people who thought that 287 is a prime number.
(c)
The thirteenth prime number is 41. This number led the famous mathematician Euler, for a short while, to believe
that a formula, n2 – n + 41, where n is any whole number, existed for which the result is prime. Substitute the
numbers n = 2 and n = 10 to find two more prime numbers.
n = 2 gives 43.
n = 10 gives 131.
Well answered, generally, although some did 22 – 10 + 41 = 35 or the like.
(d)
Explain in a few words how the formula doesn't work if n = 41 itself.
If n =41 then the number 41 itself must be a factor; hence the result cannot be prime.
Well answered. Most people saw that adding 41 'undoes' subtracting 41, leaving only 41 2 as the result, although
to be considered for higher honours a pupil generally needed to demonstrate the concept of 'factor'.
(e)
The formula also does not work if n = 42. Show working, without simply using your calculator (although you
may use it to check your result – this will not be worth any marks) to show that substituting n = 42 into the
formula gives a result which is not prime.
If n = 42 then we have
n2 - n + 41
=
422 - 42 + 41
=
422 - 1
=
43  41 (difference of squares)
which cannot be prime.
There were no marks for just finding the answer unless the students showed that it isn't prime, for example, by
calculator. This was the part of the question which was more poorly answered. Many candidates just thought they
had to give the value. Arithmetic mistakes often led to a multiple of 3 or 5 also.
Question Four (all students)
It can be very difficult to prove the general result, so we only asked for a conjecture at the end of the question.
Bobo the Clown is standing directly in front of a pool, and he is carrying a number of cards with him in a bag. They are
either marked with the word 'FORWARD' or the word 'BACK'. If he draws out at random the word 'FORWARD' he must
step into the pool and so get wet. If the word he draws says 'BACK', he steps backward one pace and so doesn't get wet,
although he still might do so later, depending on the order with which he draws the cards.
(a)
If there are two words in the bag, one marked with 'FORWARD' and the other marked with 'BACK', and he
draws one of them at random, what is the probability that he has to step forward and so get wet?
0.5, or equivalent. Many candidates wrote things like '1 : 1' or 'one in two' as their answer. Expressing
probabilities as decimals or fractions is not universally known. Well answered just the same. For many
candidates, this was the only correct answer in the question, and was probably the easiest part of the paper.
(b)
Assume that Bobo acts on the instructions as soon as he draws the card out. For example, he draws a card which
states 'FORWARD', and he must immediately step forward into the pool, in which case he would fall in and no
further cards would be drawn. In other words, the cards are drawn at random and they are written on a board.
How many different sequences could possibly be seen on the board when either all cards are drawn or the clown
is in the pool, if there are six cards in the bag, three identical ones marked with the word 'FORWARD' and the
others identically marked with 'BACK'?
Well answered, although there were 'easy' marks given for showing a suitable method, for example,a probability
tree (not shown), even if the final answer wasn't correct.. A list does just as well. In full, the following are
'possible'. For the purposes of this question, and the next, any paths with more 'Forwards' than 'Backs', or at
least 'Forwards' coming before 'Back', terminate with Bobo falling in the pool. The following paths are 'possible',
if we let 'F' stand for 'Forward', and B for 'Back':
FFFBBB (an example of a path where Bobo falls in, so it is really equivalent to just F)
FFBFBB
FFBBFB
FFBBBF
FBFBBF
FBBFFB
FBBFBF
FBBBFF
FBFBFB
FBFFBB
FBBFFB
FBBFBF
FBBBFF
BFFFBB
BFFBFB
BFFBBF
BFBFBF
BBFFBF
BBFBFF
BBBFFF
Of these 20 listed paths, nine (answer) are valid (could be written on the board).
A common error here was to list just four paths, three of which result in Bobo falling in. This is probably due to
the probability given in the next part. Only partial credit could be obtained in such a case.
(c)
(d)
If there are six cards altogether, three marked 'FORWARD' and three marked 'BACK', and Bobo acts
immediately on the card which he has just drawn, show that the probability that he will fall in is 0.75 (or
equivalent).
Of the 20 paths listed above, 15 result in his falling in. So the answer is 15/20, or 0.75.
Alternatively, via a probability tree the probabilities of the four cases where Bobo falls into the pool that were
written on the board can be calculated. The probabilities (0.5, 0.15, 0.05, 0.05) sum to 0.75.
Some students felt that since there were only nine valid paths, of which four resulted in Bobo falling in, the
probability was 4/9. The error here is that the valid paths are not equally frequent; some are more frequent than
others.
If Bobo has only four cards, two with 'FORWARD' and two with 'BACK', then the probability of his falling into
the pool is 2/3. Use this, and your answers to parts (a) and (c), to make a conjecture as to what the probability of
his falling in when there are 2n cards, where n is a natural number. You do not have to prove your conjecture.
This question aimed to see if students could generalise. Any answer with a variable (unless it was just 2n) was
given some credit. However the best answer was n/(n + 1) or equivalent. It is reasonably hard to prove this, so a
proof was not expected.
Question Five (all students)
An equilateral triangle of side length 1 unit is completely covered by rows of circles as in the diagram, which is not drawn
to scale. The diagram below shows the case where there are two rows. In this question the overlapping circles have exactly
enough area to cover the triangle (with some spare). They pass through the corners of the triangle.
This question was felt to be too hard for the juniors, although some managed it. It required Trigonometry, Trig without
right angles, and Pythagoras (given in Question One). Answers here are in surd form. This was not essential. Many
students overcomplicated the question and struggled to explain exactly what they were doing. We felt justified in setting
this question, because quite frankly much of the rest of the competition was relatively simple, and this question is designed
to identify the top participants.
(a)
Calculate the area of the triangle.
To calculate the height, Pythagoras gives
h2 = 12 – 0.52
which leads to
h = √3/2
The area then is
A = ½bh
A = √3/4.
(b)
Show that if there is only one circle covering the triangle, then it must have radius of 1/√3 units.
Here the 'best' and most 'elegant' way (there are others) to answer the question is to use the Sine Rule, not
known (yet) by many. Most of the best candidates knew all about it.
1
r
=
sin 120 sin 30
1/ 2
r=
 3 /2
1
r=
3
(c)
(d)
If two rows of circles cover the triangle as in the diagram what is the radius of each circle?
In this question, credit was given for saying that the three circles each pass through the centre of the sides of the
large triangle, and intersect in the middle. So the radius is half that given above. A decent explanation of this
halving was essential for full credit here.
For the case in the diagram below, a 'rosette' is formed by the overlapping circles. Find the length of one branch
of the rosette.
Some students didn't know what a 'rosette' is, and said so. To these students, we apologise if the question wasn't
clear enough. Some students seemed to know at once what was being asked. The winner at Year 11 (amongst a
few others) assumed two possible answers, and gave them both.
Apologies also for the fact that there appears to be four triangles in the diagram – there should have been only
one.
Let the length be x.
tan 30°
=
x/½
x
=
½ (1/√3)
x
=
1/2√3
Alternatively (and more commonly), many
students noted that the length of the rosette
was the same as the radius of one of the small circles, so gave the same number as in 5(c). This was enough for
full credit in this question.
A Problem Solving Resource Book for Teachers and Parents
The resource book A Decade of Problems, which contained all the questions from The National Bank
Competition over the ten year period 1991 to 2000, has now sold out. However, it is still possible for you to get
Ten of the Best
Junior Mathematics Competition Resource Book
2001 – 2005 and 1986 – 1990
with full worked solutions
•The questions and solutions from 1986 to 1990 were previously published in booklet form back in
1990. These booklets are no longer readily available, so it was decided to republish these, along with
the questions and model solutions from 2001 to 2005, in a form suitable for photocopying and
distributing to students.
•In this booklet each question has been edited onto a single page so that teachers may photocopy
questions as desired. The questions may be handed out to classes (or individual students) as
exercises, or for homework.
•The solutions are placed immediately next to the questions and may also be copied if desired.
•This booklet will be a useful resource for all students, not just those intent on sitting the competition.
•Please order your copies using this order form.
•Payment is required with the order as we have tried to keep the costs to a minimum by not involving
the financial wizards with their invoices etc. A receipt will be sent with your books.
Ten of the Best
Junior Mathematics Competition Resource Book
2001 – 2005 and 1986 – 1990
with full worked solutions
Copies of the book are available from the address below. The cost is
2
$25 per book (including G.S.T. and postage) for a single copy
3
$20 per book for 3 copies or more.
Please send
OR
copies of Ten of the Best to
NAME:
ADDRESS:
AMOUNT ENCLOSED: $
($25 per book, or $20 per book for 3 or more copies)
Please make cheques payable to
University of Otago
and send orders to
Department of Mathematics and Statistics
University of Otago, P.O. Box 56,
Dunedin.