Top PDF 2018IMAS Second Round Upper Primary Division Full Solutions

2018IMAS Second Round Upper Primary Division Full Solutions

2018IMAS Second Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2018IMAS Second Round Upper Primary Division Problems

2018IMAS Second Round Upper Primary Division Problems

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2017IMAS Second Round Upper Primary Division Full Solutions

2017IMAS Second Round Upper Primary Division Full Solutions

If there are less than 4 rows with pieces more than 1, covering 4 rows with the most pieces will leave all remaining row each contains no more than one piece. The remaining pieces can be covered by 4 columns, contradiction. (5 marks) Put 13 pieces in the squares as the picture below will ensure that any 4 rows and 4 columns can not cover all pieces. (5 marks) Noticing that pieces in the upper left

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2014IMAS Second  Round Upper Primary Division Full Solutions

2014IMAS Second Round Upper Primary Division Full Solutions

Answer: 45 number sentences 10. An ant starts from the top left corner square of a 3 5  chessboard. It moves from a square to the adjacent square in the same row or column. After visiting every square exactly once, it ends up at the square in the middle row second column from the right. How many different paths can the ant follow?

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2015IMAS Second  Round Upper Primary Division Full Solutions

2015IMAS Second Round Upper Primary Division Full Solutions

Answer: 3600 dollars. 7. The diagram shows a path consisting of four semi-circular arcs. Each arc is of length 100 m and uses a different side of a square as its diameter. Initially, Jane is at A and Yves at B. They start walking counter-clockwise at the same time. Jane’s speed is 120 m per minute and Yves’s is 150 m per minute. Each pauses for 1 second whenever they are at the points A, B, C or D. How many seconds after

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2016IMAS Second Round Upper Primary Division Full Solutions

2016IMAS Second Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2018IMAS First Round Upper Primary Division Full Solutions

2018IMAS First Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2018IMAS Second Round Middle Primary Division Full Solutions

2018IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2017IMAS First Round Upper Primary Division Full Solutions

2017IMAS First Round Upper Primary Division Full Solutions

【Solution】 If there is no carry when the number is added by 1, sum of digits increases by 1, contradiction to that both sum of digits are divided by 4. If there is a carry, the last digit is 9. Since the sum of all digits changes by 8 when one carry happens and by 17 when two carries happens. It follows that only one carry should happen and the original number has the sum of all digits divided by 4. In order to take the number to be maximal, the first digit should also be 9. The second digit can be either 2 or 6. The largest one is 969.

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2016IMAS First Round Upper Primary Division Full Solutions

2016IMAS First Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2014IMAS First Round Upper Primary Division Full Solutions

2014IMAS First Round Upper Primary Division Full Solutions

Thus, Max must cut a minimum of 50 × 4 + 5 × 12 = 260 cm paper strips. Hence, we select option (A). Answer: (A) 14. The tap is leaking at the rate of one drop per second. The volume of each drop is 0.05 mL. At 9 pm, Wendy puts an empty measuring cup under the tap. Some time during the night, she finds the cup partially filled, as shown in the diagram. No water is lost from the cup. At what time is the water level in the measuring cup closest to the time in the following?

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2015IMAS First Round Upper Primary Division Full Solutions

2015IMAS First Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2017IMAS Second Round Middle Primary Division Full Solutions

2017IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2014IMAS Second  Round Middle Primary Division Full Solutions

2014IMAS Second Round Middle Primary Division Full Solutions

By observation, we discover that each of the number in the first set when divided by 3 will give a remainder of 1, each number in the second set when divided by 3 will give a remainder of 2 and each number in the third set when divided by 3 or when each of them is divided by 3 will give a remainder of 0. Hence, according to the meaning of the problem, we must form a three-digit number where the digit in each place is selected from each set and the sum of all the digits formed when divided by 3 will give a remainder of 0, that is; the arrangement of all the three-digit numbers that formulate is always divisible by 3. Hence, we just need to find all those three-digit numbers from the above that are divisible by 2; or the digit in the units place is an even number.
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2015IMAS Second  Round Middle Primary Division Full Solutions

2015IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2016IMAS Second Round Middle Primary Division Full Solutions

2016IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2017IMAS Second Round Upper Primary Division Problems

2017IMAS Second Round Upper Primary Division Problems

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2014IMAS Second  Round Upper Primary Division Problems

2014IMAS Second Round Upper Primary Division Problems

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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2015IMAS Second  Round Upper Primary Division Problems

2015IMAS Second Round Upper Primary Division Problems

After making your choice, fill in the appropriate letter in the space provided. Each correct answer is worth 4 marks. There is no penalty for an incorrect answer. Questions 6 to 13 are a short answer test. Only Arabic numerals are accepted; using other written text will not be honored or credited. Some questions have more than one answer, as such all answers are required to be written down in the space provided to obtain full marks. Each correct answer is worth 5 marks. There is no penalty for incorrect answers.

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2016IMAS Second Round Upper Primary Division Problems

2016IMAS Second Round Upper Primary Division Problems

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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