The University of Melbourne / BHP Billiton School Mathematics Competition 2004 Junior Division

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Samuel Harmon
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1 The University of Melbourne / BHP Billiton School Mathematics Competition 2004 Junior Division (1) At a party each man shook hands with everyone except with his wife, and no handshakes took place between the women present. If 20 married couples attended the party, how many handshakes were there? Number the men from 1 to 20. Man 1 shakes hands with 38 people. Man 2 shakes hands with 37 people, which excludes himself, his wife and Man 1. Continue this process until Man 20, who shakes hands with 19 people. Therefore, the total number of handshakes is: which equals 570. This sum can be found easily as follows. Let S = Also S = , writing the numbers down in reverse order. Now add these two equations to give: 2S = 20 57, or S = = 570. (2) A shipping clerk wishes to determine the weights of five boxes. Each box weighs a different amount less than 100 kg. Unfortunately the only scales available measure weights in excess of 100 kg. The clerk therefore decides to weigh the boxes in pairs, so that each box is weighed with every other box. The weights for the pairs of boxes are (in kilograms) 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. From this information the clerk can determine the weight of each box. What are the weights of the boxes? 1

2 Let the weights (in kilograms) be a, b, c, d and e, where a < b < c < d < e. Then: a + b = 110 (1) a + c = 112 (2) d + e = 121 (3) c + e = 120 (4) Since each box is used in four weighings, 4(a + b + c + d + e) = = 1156 a + b + c + d + e = 289 (5) Adding equations (1) and (3) gives: a + b + d + e = 231 Then from (5), c = 58. a = 54, e = 62, b = 56 and d = 59. (3) In the following diagrams, ABC is an equilateral triangle. The points D, E and F are the mid-points, respectively, of the sides AB, AC and BC. The middle triangle DEF is removed from the triangle ABC. Call this Step 1. A A D E D E B F C B F C Step 2 consists of removing the middle triangles from the remaining triangles ADE, DBF and EFC, as shown. In a similar way, Step 3 consists of removing the middle triangles from

3 those triangles that still remain. If this process is continued, what proportion of the triangle ABC remains after Step 4? (Write your answer in the form a, where a and b are positive b integers). After Step 1, the proportion of triangle ABC remaining equals After Step 2, the proportion of triangle ABC remaining equals 1 1 (3 1 ) After Step 3, the proportion of triangle ABC remaining equals (9 1 ) After Step 4, the proportion of triangle ABC remaining equals ( = 256 = Alternatively, the proportion of triangle ABC remaining after step n equals (3/4) n. 256 ). (4) In the following multiplication problem A, B, C, D and E are positive integers which are all different. Determine their values. It is obvious that either A = 1 or A = 2. Now A B C D E 4 E D C B A 4E = A, A + 10, A + 20 or A + 30 (1) Therefore A is even, and thus A = 2. Now, 4A = 8 = E, E 1, E 2 or E 3, therefore E = 8, 9, 10 or 11. From (1), 4E = 2, 12, 22 or 32, so E = 3 or 8. Therefore, E = 8, and we have: Therefore, 2 B C D D C B 2

4 D = 4B, 4B + 1, 4B + 2 or 4B + 3 (2) Since D 9, and 4B 9, B = 0 or 1 (since A = 2). Since B > 0, B = 1. Therefore, from (2), D = 4, 5, 6 or 7. However, 4D + 3 = 11, 21 or 31, so D = 7. It now follows readily that C = 9. The solution is therefore: (5) Solve the following cross-number puzzle (each entry is one of the integers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9): The clues are: Across Down 1. An integer multiple of An integer multiple of A product n, where n is a positive integer. 5. A product of two or more consecutive prime numbers. 2. A value of 2 n, where n is a positive integer. 3. An integer multiple of 11.

5 No answer can begin with 0. (A prime number is one whose only factors are itself and 1). To get started, it is easy to check that 4 Across can only be = 120 or = 720. Similarly, 2 Down can only be 128, 256 or 512. Therefore 4 Across must be 720 and 2 Down must be So the entries are (6) The positive integers 1, 2, 3, 4, 5, 6, 7, 8 and 9 are arranged along the sides and the corners of a triangle, as illustrated: Find an arrangement so that the sums of the squares of the four positive integers along each side are equal. Let the sum of the squares along any side equal t, and let the numbers at the corners of the triangle be a, b, c and the other numbers be d, e, f, g, h, i as shown. Then: a 2 + d 2 + e 2 + b 2 = t b 2 + f 2 + g 2 + c 2 = t

6 a 2 + h 2 + i 2 + c 2 = t Adding these equations gives: (a 2 + b 2 + c 2 ) + ( ) = 3t a 2 + b 2 + c 2 = 3t 285 = 3(t 95). i.e. the sum of the squares at the corners must be a multiple of three. Only 2, 5 and 8 are eligible for the corners, so 3t = = 378, giving t = 126. This leads to the following solution: Side 1: 2, 4, 9, 5 Side 2: 5, 1, 6, 8 Side 3: 8, 3, 7, 2. (7) One hour after a train has departed Station A for Station B a fault occurs which requires the train to reduce its speed to 3 of its normal time-table speed. As a result the train reaches 5 Station B 2 hours late. If the fault had occurred 50 kilometers nearer to Station B the train would have arrived 30 minutes earlier. Find the distance between Stations A and B and the train s normal time-table speed. Let the normal time-table speed of the train be v k.p.h, let the distance between A and B be d kilometres, and let the normal time for the trip be t hours. Then the following equations can be derived from the information given: d = vt (1) 1 + d v 3 5 v = t + 2 (2) 50 + v v + d 50 v 3 5 v = t = t (3) Using (1), equation (2) gives t = 4, and substituting d = 4v and t = 4 in (3) gives v = 200 Therefore, d =

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 2017 Junior Preliminary Problems & Solutions

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 017 Junior Preliminary Problems & s 1. If x is a number larger than, which of the following expressions is the smallest? (A) /(x 1) (B) /x (C) /(x