【OG20-P288-401题】A group consisting of several families visited an _OG17新题

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【OG20-P288-401题】

A group consisting of several families visited an amusement park where the regular admission fees were ¥5,500 for each adult and ¥4,800 for each child.Because there were at least 10 people in the group, each paid an admission fee that was 1 0% less than the regular admission fee. How many children were in the group?

(1) The total of the admission fees paid for the adults in the group was ¥29,700.

(2) The total of the admission fees paid for the children in the group was ¥4,860 more than the total of the admission fees paid for the adults in the group.

A

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
分析A选项
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B

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
分析B选项
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C

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
分析C选项
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D

EACH statement ALONE is sufficient.
分析D选项
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E

Statements (1) and (2) TOGETHER are NOT sufficient.
分析E选项
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当前版本由 huan~ 更新于2018-09-29 14:51:20 感谢由 huan~ 对此题目的解答所做出的贡献。

人数A+C≥10→价格A：5500*90%=4950；C：4800*90%=4320
条件1：4950A=29700→A=6→C≥4，不充分
条件2：4320C-4950A=4860→48C-55A=54，不充分
1+2：A=6，C=8，充分

题目讨论（4条评论）

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你这只猪

方程联立求解吧

0 0 回复 2020-06-08 14:46:25
园园爱吃肉

看错题了，以为就10个人。。。

0 0 回复 2019-11-30 16:33:42
hannahyz

需要解决的问题是：48C-55A=54是否有唯一解（在A+C>10，A、C均为正整数的情况下） 48*C-54=55*A 化成：6（8*C-9）=5*11*A 因此8*C-9 需要被55整除，A 被6整除。8C-9和A可以按比例无限扩大，比如 8C-9=55，A=6 8C-9=550，A=60 因此该二元一次不定方程没有唯一解，不充分。

2 0 回复 2019-09-17 11:00:02
cbc021

注意理解题意：因为团体成员超过10人，成年人和孩子的门票费都是标价的90%。

0 0 回复 2018-02-21 13:31:24