AMC8 1998

AMC8 1998 · Q25

AMC8 1998 · Q25. It mainly tests Word problems (algebra), Arithmetic misc.

Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and Toy to double the amount that each has. Jan then gives enough to Ami and Toy to double their amounts. Finally, Toy gives Ami and Jan enough to double their amounts. If Toy has \$36 when they begin and $36 when they end, what is the total amount that all three friends have?

三位慷慨的朋友，各有一些现金，按以下方式重新分配金钱：Ami 给 Jan 和 Toy 足够的钱，使她们各自的金额翻倍。Jan 然后给 Ami 和 Toy 足够的钱，使她们各自的金额翻倍。最后，Toy 给 Ami 和 Jan 足够的钱，使她们各自的金额翻倍。如果 Toy 开始和结束时都有 \$36，他们三人总共有多少钱？

(A) \$108 \$108

(B) \$180 \$180

(D) \$252 \$252

(E) \$288 \$288

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Since Toy begins with \$36 and her amount is doubled in the first two exchanges, her amounts are \$36, \$72, \$144, and \$36. This means that she gave away \$108, and this is exactly enough to double the amounts of Ami and Jan. So, the total must be $2(\$108)+\$36=\$252$.

答案（D）：由于 Toy 一开始有 \$36，并且在前两次交换中她的金额都翻倍，所以她的金额分别是 \$36、\$72、\$144 和 \$36。这意味着她送出了 \$108，而这恰好足以使 Ami 和 Jan 的金额都翻倍。因此，总额必为 $2(\$108)+\$36=\$252$。

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