Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 = ½44+½46 = 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 = 23–24 = -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 = aecf bedf agch bgdh ] 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 1268 = 1251 80 50 = 63 46 17 4 3416 3412 (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.