So the edge bordering the 84 and 70 faces has length 14, and the edge bordering the 84 and 30 edges has length 6. The only way to make 84 with one each of these common factors are 14 from the top line and 6 from the bottom. The common factors between 84 and 30 are 1, 2, 3 and 6.The common factors between 84 and 70 are 1, 2, 7 and 14.The 30 face could be 1 x 30, 2 x 15, 3 x 10 or 5 x 6. The 70 face could be 1 x 70, 2 x 35, 5 x 14 or 7 x 10. įirst we need to work out the possible side lengths, by seeing which two numbers multiply to be the area of each face. Let’s now draw the string divided into twentieths: So, we cut off 8/20, subtract 14m and are left with 5/20. In order to compare the fractions 2/5 and then 1/4, lets change them to the lowest common denominator, which is 20. In other words, we are left with 1/4 of the original length. Then 14m, and are left with a piece that is a third of the size of what was cut. It requires us to think more visually about the string: We cut 2/5 of it. Interestingly, the Singapore method of solution is different. Which rearranges to:7L/5 = 56, or L = 40. By substituting the first equation in the second we have 2L/5 + 14 = 9L/5 –42. Let L be the original length of the string, and R be what is remaining once you have cut the string twice. Oh Mary! This is how I would have solved it, using equations. What is the length of the remaining string? The ratio of the length of string remaining to the total length cut off is 1 : 3.
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