[PDF] Top 20 2014IMAS First Round Junior Division Problems
Has 1969 "2014IMAS First Round Junior Division Problems" found on our website. Below are the top 20 most common "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] ... See full document
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2016IMAS First Round Junior Division Problems
... 8. The chart below shows the sale figures of a certain merchandise in 2014 and 2015 by the season. How many more items were sold in 2015 than in 2014? (A)23 (B)48 (C)85 (D)90 (E)110 9. ABC is an equilateral ... See full document
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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 ... See full document
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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 ... See full document
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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 ... See full document
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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 ... See full document
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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 ... See full document
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2014IMAS First Round Middle Primary Division Problems
... 22. The area of a triangle is 50 cm 2 . Each side is divided into five equal parts, and some pairs of division points are joined as shown in the diagram. What is the total area, in cm 2 of the shaded regions? 23. ... See full document
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2014IMAS First Round Upper Primary Division Problems
... 22. The inside lane of a track has length 400 m and the outside lane has length less than 500 m. From the marked line, as shown in the diagram, Max and Lynn start running counterclockwise along the track at the same ... See full document
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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 ... See full document
9
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 ... See full document
9
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 ... See full document
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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 ... See full document
9
2016IMAS First Round Junior Division Full Solutions
... 3 2 5 2 2016 − 2016 = 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, …, 29, 37, …, 2011, 2027, …. ... See full document
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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 ... See full document
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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 ... See full document
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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 ... See full document
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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] ... See full document
9
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] ... See full document
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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 ... See full document
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