ARITMETICA E ALGEBRA PRACTICE WITH CLIL 1. Find the Least Common Multiple (LCM) and Greatest Common Divisor (GCD) of the following couples of numbers. a. 12; 15 d. 60; 90 b. 26; 85 e. 35; 56 c. 120; 135 f. 96; 128 2. Find the LCM and GCD of the following three numbers. a. 10; 20; 25 b. 120; 150; 180 c. 12; 30; 60 d. 27; 51; 81 e. 24; 72; 144 f. 25; 45; 65 3. Convert the following decimals to fractions. a. 0.75 b. 1.66 c. 0.625 d. 12.8 e. 1.2 13 f. 0.123 4. Calculate the sum, difference, product and quotient of fractions and edit the result to the simplest form of a fraction. 5 2 7 14 a. _ _ d. _ : _ 2 5 5 3 5 1 15 25 b. _ + _ e. _ : _ 6 4 16 24 11 6 3 3 c. _ _ f. _ _ 8 10 12 5 5. The Merchant s puzzle Of the Merchant the poet writes, «Forsooth he was a worthy man withal . He was thoughtful, full of schemes, and a good manipulator of figures. «His reasons spake he eke full solemnly. Sounding alaway the increase of his winning . One morning, when they were on the road, the Knight and the Squire, who were riding beside him, reminded the Merchant that he had not yet propounded the puzzle that he owed the company. He thereupon said, «Be it so? Here then is a riddle in numbers that I will set before this merry company when next we do make a halt. There be thirty of us in all riding over the common this morn. Truly we may ride one and one, in what they do call the single file, or two and two, or three and three, or five and five, or six and six, or ten and ten, or fifteen and fifteen, or all thirty in a row. In no other way may we ride so that there be no lack of equal numbers in the rows. Now, a party of pilgrims were able thus to ride in as 146 many as sixty-four different ways. Prithee tell me how many there must perforce have been in the company. The Merchant clearly required the smallest number of persons that could so ride in the sixty-four ways. [H.E. Dudeney, The Canterbury Puzzles, 2009, Dodo Press] Mathemagics! 1. Get any deck of cards. Shuffle the deck many times and ask a friend to take out no more than 10 cards from its top: he/she will keep the cards hidden in their pocket without saying its total number. 2. Now ask your friend to memorize the card which, in the original deck, is occupying the position equivalent to the total number of cards of the first draw. 3. Let him/her choose the first and last names of a celebrity. Now, ask to draw from the deck a card for each letter in the chosen name and to lay them down on the table. To let him/her understand this passage, ask the names chosen and show how to do it. Then, put the cards back on the top of the deck. 4. Place back on the original deck all the cards previously taken out and let your friend try alone step 3. 5. Now ask to show the card which is next to the last ones drawn after the celeb s name. 6. The card revealed will be the one your friend has memorized in step 2. Let us reveal the trick The success in this game depends solely and exclusively on its correct performance, carefully following all steps. It is based on the fact that the events are mutually exclusive; the only limitation is about the number of the first picking not exceeding 10 cards. Let us look at an example to analyze the mechanism. 1. Imagine your friend has taken out 5 cards, though you do not know (you will realize this is not an information you may need).