2010 AMC 10A 第 3 题

先试着解答 2010 AMC 10A 第 3 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2010 AMC 10A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

3.

Tyrone 有 9797 颗弹珠,Eric 有 1111 颗弹珠。之后 Tyrone 给了 Eric 一些弹珠,使得最后 Tyrone 的弹珠数是 Eric 的两倍。Tyrone 给了 Eric 多少颗弹珠?
Tyrone had 9797 marbles and Eric had 1111 marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
33
1313
1818
2525
2929
答案:D
知识点:一次方程比与比例
小提示:
弹珠总数保持不变。
The total number of marbles stays fixed
大提示:
最后的数量比为 2:12:1。
The final amounts are in the ratio 2:12:1
解答:
设 Eric 最后有 xx 颗弹珠,则 Tyrone 最后有 2x2x 颗。
总弹珠数为 97+11=10897 + 11 = 108,所以 3x=108 3x = 108 x=36。x = 36\text{。}
因此 Tyrone 最后有 36⋅2=7236 \cdot 2 = 72 颗弹珠,他给出了 97−72=2597 - 72 = 25 颗。
所以正确答案是 D。
Let xx be the number of marbles that Eric ends up with. Then Tyrone ends up with 2x.2x.
The total number of marbles is 97+11=108,97 + 11 = 108, so 3x=108 3x = 108 x=36.x = 36.
Then, Tyrone ends up with 36⋅2=7236 \cdot 2 = 72 marbles. This means he has to give away 97−72=2597 - 72 = 25 marbles.
Thus, D is the correct answer.
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