2 24 pazoozas per dollar depreciation of the dollar

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lar by 20% makes the exchange rate = 2(1 + 0.2) = 2.4 pazoozas per dollar. Depreciation of the dollar by 20% then makes the exchange rate = 2.4 (1 – 0.2) = 1.92 pazoozas per dollar. The net effect is a depreciation of the dollar by (1.92 – 2)/2 = –0.04, or 4%. 6. Initially, the foreign exchange rate = 2 pazoozas per dollar. Depreciation of the dollar M ATH M ODULE S olutions to Exercises 7
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by 20% makes the exchange rate = 2(1 – 0.2) = 1.6 pazoozas per dollar. Appreciation of the dollar by 20% then makes the exchange rate = 1.6 (1 + 0.2) = 1.92 pazoozas per dollar. If the dollar has depreciated by 4% against the pazooza in this and the pre- ceding question, then the pazooza has appreciated against the dollar by 1/(1 – 0.04) –1 = 4.167%. 7. Let us denote K/L by k , K A /L A by k A , K B /L B by k B , L A /L by l A and L B /L by l B . Then we have k = l A k A + l B k B , and (since l A = 1 – l B ), k = (1 – l B ) k A + l B k B . [Check this relation to show that it is correct, by substituting in and cancelling.] (a) With K = 360 machine-days and L = 120 labour-days, we have k = 3 = (1 – l B ) k A + l B k B = (1 – l B )(2.4) + l B (4) = 2.4 + 1.6 l B . Therefore l B = (3 – 2.4)/1.6 = 3/8, and so L B = (3/8) L = (3/8)(120) = 45 labour-days , while L A = (5/8) L = (5/8)(120) = 75 labour-days . (b) Now we have k = 3 = (1 – l B ) k A + l B k B = (1 – l B )(8/3) + l B (14/3) = 8/3 + 2 l B . Therefore l B = (3 – 8/3)/2 = 1/6, and so L B = (1/6) L = (1/6)(120) = 20 labour-days , while L A = (5/6) L = (5/6)(120) = 100 labour-days . For K/L to increase in both sec- tors, the proportion of labour in the relatively capital-intensive sector (Blankets) must shrink . (c) If total capital increases to 432 machine-days, then aggregate K/L = k increases from 3 to (432/120) = 3.6. We now have k = 3.6 = (1 – l B ) k A + l B k B = (1 – l B )(2.4) + l B (4) = 2.4 + 1.6 l B . Therefore l B = (3.6 – 2.4)/1.6 = 3/4, and so L B = 3/4 L = 3/4(120) = 90 labour-days , while L A = 1/4 L = 1/4(120) = 30 labour-days . This result actually illustrates an important theorem (Rybczynski): if K , and with it the total
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