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1. On a highway, car A and car B travel in an opposite direction from each other from the same starting point. The speeds of cars A and B are 75km/h and 85km/h respectively. How long (in min) will it take for cars A and B to be at least 40km apart ?2. A store bought 250 eggs for $200 and found that 25 of them... 顯示更多 1. On a highway, car A and car B travel in an opposite direction from each other from the same starting point. The speeds of cars A and B are 75km/h and 85km/h respectively. How long (in min) will it take for cars A and B to be at least 40km apart ? 2. A store bought 250 eggs for $200 and found that 25 of them were cracked. The rest of the eggs will be sold at a percentage profit of not greater than 35%. Find the maximum selling price of each egg. 3. The ingredients of the nuts of Brand X,Brand Y and Breand Z in percentage are as follows : Brand X Y Z Walnut 30% 25% 55% Almond 30% 50% 30% Hazelnut 40% 25% 15% How many kg of nuts from Brand Z should be added to 0.5kg of nuts from Brand X and 1kg of nuts from Brand Y, such that there are not less than 20% of hazelnuts in the mixture ? 以上問題請詳解.

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1. On a highway, car A and car B travel in an opposite direction from each other from the same starting point. The speeds of cars A and B are 75km/h and 85km/h respectively. How long (in min) will it take for cars A and B to be at least 40km apart? A. Let the time taken for cars A and B to be at least 40km apart be y hour. 85y+75y ≦ 40 160y ≦ 40 y ≦ 40/160 y ≦ 1/4 Therefore, the time taken for cars A and B to be at least 40km apart is 60/4 = 15 mins 2. A store bought 250 eggs for $200 and found that 25 of them were cracked. The rest of the eggs will be sold at a percentage profit of not greater than 35%. Find the maximum selling price of each egg. A. Let the selling price of each egg be $y (250-25)y-200 /200 *100% ≦ 35% 225y-200 /200 ≦ 35% 225y-200 ≦ 0.35*200 225y-200 ≦ 70 225y ≦ 70+200 225y ≦ 270 y ≦ 270/225 y ≦ 1.2 Therefore, the maximum selling price of each egg is $1.2 3. The ingredients of the nuts of Brand X, Brand Y and Brand Z in percentage are as follows: Brand X Y Z Walnut 30% 25% 55% Almond 30% 50% 30% Hazelnut 40% 25% 15% How many kg of nuts from Brand Z should be added to 0.5kg of nuts from Brand X and 1kg of nuts from Brand Y, such that there are not less than 20% of hazelnuts in the mixture? A. Let z kg of nuts from Brand Z should be added to 0.5kg of nuts from Brand X and 1kg of nuts from Brand Y, such that there are not less than 20% of hazelnuts in the mixture. [(15%)z+0.5(40%)+1(25%)]/(z+0.5+1) ≧ 20% 0.15z+0.2+2.5 ≧ 0.2(1.5+z) 0.15z+2.7 ≧ 0.3+0.2z 0.05z ≧ 2.7-0.3 0.05z ≧ 2.4 z ≧ 48 Therefore, 48 kg of nuts from Brand Z should be added to 0.5kg of nuts from Brand X and 1kg of nuts from Brand Y, such that there are not less than 20% of hazelnuts in the mixture. 以上問題請詳解.

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