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What can cause problems in questions like this one is

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What can cause problems in questions like this one is not keeping clear which base the discount or markup should be applied to. A related difficulty crops up in working with foreign exchange rates, as Example M.7.2 shows. E XAMPLE M.7.2: If bread is $1.50/loaf, then a dollar is worth 1/1.50 = 2/3 of a loaf of bread. In exchange, the price of a unit of good A in terms of good B is the reciprocal of the price of a unit of good B in terms of good A. This relationship is just as true when it is two different currencies being exchanged as it is for two goods in barter exchange. Foreign exchange rates can be and are expressed in two ways. Suppose that the Canadian dollar is worth U.S.$0.80. Then we can also say that the U.S. dollar is worth 1/0.8 = $1.25 Canadian. In this case, the price of 1 Canadian dollar is 80 U.S. cents, or U.S.$0.80. When we say that the Canadian dollar has depreciated against the U.S. dollar, we mean that $1 Canadian buys or exchanges for fewer U.S. dol- lars. If the Canadian dollar now falls to U.S.$0.70 per Canadian dollar, it has depreciated by (0.7–0.8)/0.8 = – 1/8 = – 12.5%. If the Canadian dollar has depre- ciated by 12.5%, then it is now worth 1 – .125 = .875 of what it was worth initial- ly, in exchange with U.S. dollars. If the Canadian dollar has depreciated , then nec- essarily the U.S. dollar has appreciated against the Canadian dollar, in this case by (1/.875) – 1 = 8/7 – 1 = 1/7, or 14.29%. Our third example deals with weights and weighting. 75 kg 25 kg x metres F S 2 metres M7-2 MATH MODULE 7: PROPORTIONS, WEIGHTS, AND PERCENTAGES FIGURE M.7-1
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MATH MODULE 7: PROPORTIONS, WEIGHTS, AND PERCENTAGES M7-3 E XAMPLE M.7.3: Imagine a father (who weighs 75 kg) and his 25-kg son on a 4-metre-long see-saw, supported at the middle. The son is located with his centre of gravity at the right end of the see-saw, 2 metres from the middle. For the see-saw to be perfectly bal- anced, where must the father’s centre of gravity be? Answer : For a balance of forces, the following relation must hold: m F d F = m S d S , (M.7.1) where m F and m S are the father’s and son’s masses in kg, and d F and d S are their distances from the centre or fulcrum, in metres. Substituting known values, we have 75 d F = 25(2), and so d F = 25(2)/75 = 2/3 metres left of the middle. The “teeter-totter theorem” has many applications, some of which are not quite as obvious as the above one. The final example provides one such application.
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