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Top PDF 2016IMAS First Round Junior Division Problems

2016IMAS First Round Junior Division Problems

2016IMAS First Round Junior Division Problems

... Record your answers on the reverse side of the Answer Sheet (not on the question paper) by FULLY filling in the circles which correspond to your choices.. Your Answer Sheet will be read[r] ...

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2018IMAS First Round Junior Division Problems

2018IMAS First Round Junior Division Problems

... When your teacher gives the signal, begin working on the problems. I I nt n te er rn na at t io i on n al a l M Ma at t he h em ma at ti ic cs s A A ss s se es ss sm me en n ts t s f fo or r S Sc ch ho oo ol ls s ...

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2017IMAS First Round Junior Division Problems

2017IMAS First Round Junior Division Problems

... When your teacher gives the signal, begin working on the problems. I I nt n te er rn na at t io i on n al a l M Ma at t he h em ma at ti i cs c s A A ss s se es ss sm me en n ts t s f fo or r S Sc ch ho oo ol ls s ...

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2014IMAS First Round Junior Division Problems

2014IMAS First Round Junior Division Problems

... Record your answers on the reverse side of the Answer Sheet (not on the question paper) by FULLY filling in the circles which correspond to your choices.. Your Answer Sheet will be read[r] ...

9

2015IMAS First Round Junior Division Problems

2015IMAS First Round Junior Division Problems

... (A)20 (B)31 (C)42 (D)53 (E)64 19. In an election between four candidates, they are supported respectively by 11, 12, 13 and 14 of the first 50 voters. Six more votes are to be cast, each for one of the four ...

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2016IMAS Second Round Junior Division Problems

2016IMAS Second Round Junior Division Problems

... z Do not open the contest booklet until you are told to do so. z Be sure that your name and code are written on the space provided above. z Round 2 of IMAS is composed of three parts; the total score is 100 ...

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2016IMAS First Round Junior Division Full Solutions

2016IMAS First Round Junior Division Full Solutions

... 2 20162016 = 2016 (2016 × − = 1) 2015 2016 2017 × × = × × × × × × 2 3 5 7 13 31 2017 , so the prime factors of the positive integer which satisfy the conditions should be 11, 17, 19, ...

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2016IMAS First Round Upper Primary Division Problems

2016IMAS First Round Upper Primary Division Problems

... Record your answers on the reverse side of the Answer Sheet (not on the question paper) by FULLY filling in the circles which correspond to your choices.. Your Answer Sheet will be read[r] ...

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2016IMAS First Round Middle Primary Division Problems

2016IMAS First Round Middle Primary Division Problems

... 18. Hanna makes four straight cuts on a round cake. Into at most how many pieces can she cut it? (A)8 (B)10 (C)11 (D)12 (E)14 19. Eight children are sharing 61 balloons. Each gets at least one, and everyone ...

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2018IMAS Second Round Junior Division Problems

2018IMAS Second Round Junior Division Problems

... Do not open the contest booklet until you are told to do so. Be sure that your name and code are written on the space provided above. Round 2 of IMAS is composed of three parts; the total score is 100 ...

9

2017IMAS Second Round Junior Division Problems

2017IMAS Second Round Junior Division Problems

... z Do not open the contest booklet until you are told to do so. z Be sure that your name and code are written on the space provided above. z Round 2 of IMAS is composed of three parts; the total score is 100 ...

9

2014IMAS Second  Round Junior Division Problems

2014IMAS Second Round Junior Division Problems

... Do not open the contest booklet until you are told to do so. Be sure that your name and code are written on the space provided above. Round 2 of IMAS is composed of three parts; the total score is 100 ...

9

2015IMAS Second  Round Junior Division Problems

2015IMAS Second Round Junior Division Problems

... Do not open the contest booklet until you are told to do so. Be sure that your name and code are written on the space provided above. Round 2 of IMAS is composed of three parts; the total score is 100 ...

7

2018IMAS First Round Junior Division Full Solutions

2018IMAS First Round Junior Division Full Solutions

... If 2 and 8 are both to the right of 4, the sequence partially looks like 4ab2c8, where 1 must be a. If 4 is the first from the left, then the sequence looks like 41b2c8de, 3 and 6 must be at b, d, two choices for ...

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2017IMAS First Round Junior Division Full Solutions

2017IMAS First Round Junior Division Full Solutions

... (A)12 (B)18 (C)24 (D)30 (E)36 【Solution 1】 From the given information, one student will participate in two events and each of other two students participate on exactly one event. Let the student (the one will participate ...

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2014IMAS First Round Junior Division Full Solutions

2014IMAS First Round Junior Division Full Solutions

... the first row, it follows that the difference of the sum of three numbers in the second row and the sum of three numbers in the first row equal the sum of three numbers in the first row, that is; the ...

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2015IMAS First Round Junior Division Full Solutions

2015IMAS First Round Junior Division Full Solutions

... Subtracting this from the total surface area of the individual cubes, we have 14 6 42    42 . Answer: (C) 19. In an election between four candidates, they are supported respectively by 11, 12, 13 and 14 of the ...

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2016IMAS Second Round Junior Division Full Solutions

2016IMAS Second Round Junior Division Full Solutions

... the first 10 kg and below, the postage price is $6 per kg; for each successive kilogram after 10 kg, the postage price per kg is slightly lower than that of the first ...the first 10 kg than that of ...

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2018IMAS First Round Middle Primary Division Problems

2018IMAS First Round Middle Primary Division Problems

... Four students namely Annie, Benny, Charlie and Deany all paid the same amount of money to buy some number of notebooks together.. After distributing the notebooks, Annie, Ben[r] ...

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2018IMAS First Round Upper Primary Division Problems

2018IMAS First Round Upper Primary Division Problems

... Questions 21-25, 6 marks each 21. Mike constructs a sequence in the following way: the first two terms are 1 and 2. Starting from the third term, each term is the smallest possible integer that is not relatively ...

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