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　　GRE数学货币问题介绍

　　Currency problems are really quasi-weighted-average(准加权平均)problems, because each item (bill or coin ) in a problem is weighted according to its monetary value. Unlike weighted average problems, however, the “average value” of all the bills or coins is not at issue. In solving currency problems, remember the following:

　　1. You must formulate algebraic expressions involving both number of items (bills or coins) and value of items.

　　2. You should covert the value of all moneys to a common unit (that is, cents or dollars) before formulating an equation. If converting to cents, for example, you must multiply the number of nickels by 5, dimes by 10, and so forth.

　　GRE数学货币问题实例讲解

　　1. Mike has $ 2. 05 in dimes and quarters. If he has four fewer dimes than quarters, how much money does he have in dimes?

　　解：Letting x equal the number of dimes, x + 4 represents the number of quarters. The total value of the dimes ( in cents ) is 10x, and the total value of the quarters ( in cents) is 25 (x+4)，or 25x+ 100.

　　Given that Mike has $ 2. 05, the following equation emerges ：

　　10x+25x+100 = 205，所以x = 3

　　Mike has three dimes, so he has 30 cents in dimes.

　　2. 9. The charge for a telephone call made at 10:00 a. m. from City Y to City X is $ 0. 50 for the first minute and $ 0.34 for each additional minute. At these rates, what is the difference between the total cost of three 5-minute calls and the cost of one 15-minute call?

　　(A) $ 0. 00

　　(B) $ 0. 16

　　(C) $ 0. 32

　　(D) $ 0. 48

　　(E) $ 1. 00

　　上午10点时从Y城向X城打电话，第一分钟收费0. 5美元，此后每附加一分钟收0. 34 美元。以这个标准计费，打3个5分钟的电 话与打一个15分钟的电话之间的差价是多 少?

　　解：本题的正确答案是(C)。3个5分钟的电话费为：3X (0.5 + 0. 34 X 4) = 5. 58

　　15 分钟的电话费：0.5 + 0.34X14 = 5.26 差价为：5. 58 — 5. 26 = 0. 32 本题还有一个较为简便的算法，3个5分钟的 电话与一个15分钟的电话的时间.相同，它们 惟一的区别是3个5分钟的电话比一个15分 钟的电话多了两个“第一分钟”，考生只要找出 这两个“第一分钟”与两个“非第一分钟”的差 价就可以了，即为 2X (0.5—0. 34) =0. 32。

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