2020小升初分班考 第6讲 逻辑推理二(含答案)

2020小升初分班考 第6讲 逻辑推理二(含答案)
2020小升初分班考 第6讲 逻辑推理二(含答案)

2020小升初专项训练班讲义

第六讲逻辑推理(二)数字游戏

◇专题知识简述◇

上一讲我们介绍了有关逻辑推理问题的简单例子,它并没有用到专门的数学原理,而是直接运用正确推理,解决逻辑问题的.这一讲我们将利用图表解决一些较为复杂的逻辑推理问题。

◇例题解析◇

例11 B一次数学考试,共六道判断题.考生认为正确的就画“√”,认为错误的就画“×”.记分的方法是:答对一题给2分;不答的给1分;答错的不给分.已知A、B、C、D、E、F、G七人的答案及前六个人的得分记录在表中,请在表中填出G的得分,并简单说明你的思路。

分析由于E得了9分,说明他只答错了一道题.先假定答错的是第1题,这样就有一个标准答案,并由此可分析其他人的得分.如出现矛盾,再假定E答错的是第2题,…,直到判断出E答错的题号为止.有了正确的答案,就可以写出G的得分。

(更正C为4,F为6分)

解:假设E的第1题答错,那么A至少错3道题,一题未答,最多得5分,与A得7分矛盾.所以E第1题答对。

假设E第2题答错,可知A最多得3分,矛盾.所以E第2题答对。

假设E第3题答错,则B最多得3分,矛盾.所以E第3题答对。

假设E第6题答错,则D最多得3分,矛盾.所以E第6题答对。

由于E得9分,因此E只答错一题,因此E第4题答错,于是A的第2、4两题对,3、6两题错.而A得7分,说明A的第5题是对的.由A、E 两人的答案,可得一标准答案如下表:

按此标准评分,与题中所给A、B、C、D、E、F得分相符合,所以E 的第4题确实答错了.上表的答案是正确的.故可知G得8分。

例12A 李英、赵林、王红三人参加全国小学生数学竞赛,他们是来自金城、沙市、水乡的选手,并分别获得一、二、三等奖.现在知道:

①李英不是金城的选手;

②赵林不是沙市的选手;

③金城的选手不是一等奖;

④沙市的选手得二等奖;

⑤赵林不是三等奖。

根据上述情况,王红是__的选手,他得的是__等奖。

解:为了便于分析,我们画表帮助思考.

根据条件①②,在相应的格中打上“×”。

由条件④得出:如果王红是沙市的选手,他得二等奖,那么由条件③可知:金城选手不是一等奖,只能是三等奖.又因为李英不是金城选手,只有赵林得三等奖.这与条件⑤矛盾.所以王红不是沙市选手,沙市选手应该是李英,他得二等奖.这样金城的选手只能是王红,他得三等奖。

例13B 李云和他哥哥参加一次集会,同时出席的还有其他两对兄弟.见面后有的人握手问候,没有人和自己的兄弟问候,也没有人和同一个人握两次手.事后李云发现除自己外每个人握手次数互不相同,问李云握了几次手?李云的哥哥握了几次手?

解:设除李云(用0表示)之外的五个人分别是A、B、C、D、E,他们握手的次数分别是0次、1次、2次、3次、4次,那么他们的握手情况可以用右图来表示,其中一条连线表示握过手一次,没有连线即表示没握过手。

(更正C的圈4应改为圈2)

从图中很容易看出:李云握手2次。

那么,谁是李云的哥哥呢?因为A是唯一没有和E握过手的人,所以A、E是一对兄弟.D只和A、B没握过手,而A已经是E的兄弟了,所以B、D也是一对兄弟.这样只剩下C是李云的哥哥,他握手的次数也为2次.

例14A 红、黄、蓝、白、紫五种颜色的珠子各一颗,分别用纸包着,在桌子上排成一行,有A、B、C、D、E五个人,猜各包珠子的颜色,每人只猜两包。

A猜:第二包是紫的,第三包是黄的;

B猜:第二包是蓝的,第四包是红的;

C猜:第一包是红的,第五包是白的;

D猜:第三包是蓝的,第四包是白的;

E猜:第二包是黄的,第五包是紫的。

猜完后,打开各纸包一看发现每人都只猜对了一包,并且每包只有一人猜对.请你判断他们各猜对了哪一包?

解:我们把题目中的条件列成一个表,就更清楚了。

根据已知条件,每一包都只有一人猜对,而第一包只有C猜,所以C 猜对了第一包,是红的;又根据每人只猜对了一种,所以C猜第五包是白的,猜错了;第五包只有C、E两人猜,所以E猜第五包是紫的,猜对了;那么E猜第二包是黄的,猜错了;紫颜色的珠子,只有A、E两人猜,那么A猜第二包是紫的,猜错了;第二包有A、B、E三人猜,其中A、E都猜错了,所以B猜第二包是蓝的,猜对了;那么B猜第四包是红的,猜错了;D猜第三包是蓝的,也猜错了;所以A猜对的是第三包,是黄的;D 猜对的是第四包,是白的。

总结以上推理判断,A猜对了第三包是黄的,B猜对了第二包是蓝的,C猜对了第一包是红的,D猜对了第四包是白的,E猜对了第五包是紫的。

注如果题中只给了一个条件:“每人都只猜对了一包”,你能判断他们都猜对了哪包吗?

例15 有A、B、C三个足球队,每两队都比赛一场,比赛结果是:A有一场踢平,共进球2个,失球8个;B两战两胜,共失球2个;C共进球4个,失球5个,请你写出每队比赛的比分。

分析解决本题首先要明白两点常识:

①一个队踢进一个球,对方就失去一个球,所以三个队的总进球数应等于总失球数;

②两个队踢平,显然该场球的进、失球的总数应相等。

根据已知条件,可以列成表格如下:

解:已知每两个队要赛一场,一共要赛三场球.B是两战两胜,显然一场胜A,另一场胜C;A踢平一场无疑是与C比赛的这场球。

由总进球数等于总失球数,则B队的进球数应为9个。

因为A与C两队进球总数是6个,那么除去A、C对B的那两场球赛中,踢进B队的那2球外,剩下的4个球便是A与C踢平那一场中双方各自踢进对方的进球数的和,因此A与C踢成2比2。

现在从C的进球数分析,由于C进球4个,除去与A两平外,另外进的两个球是对B比赛进的球数;再从C的失球数分析,因为C对A失两球,表中C共失了5个球,因此另外失的3个球就是对B失的球数.所以C对B是2比3。

再因为B进球共9个,除去对C进的3个球,那么对A就进了6个球,A对B没有进球,所以B对A是6比0。

例16 北京至福州列车里坐着6位旅客:A、B、C、D、E、F.分别来自北京、天津、上海、扬州、南京和杭州,已知

①A和北京人是医生;E和天津人是教师;C和上海人是工程师。

②A、B、F和扬州人参过军,而上海人从未参军。

③南京人比A岁数大;杭州人比B岁数大;F最年轻。

④B和北京人一起去扬州;C和南京人一起去广州。

试根据已知条件确定每位旅客的住址和职业。

分析由于职业可由住址确定,所以只需考虑确定旅客的住址。

解:下面我们利用表格进行推理.表格中记号“√”表示这个人是来自这个城市;记号“×”表示这个人不来自这个城市。

由①可知,A、C、E既不是北京人,也不是天津、上海人;由②可知,A、B、F不是上海人,也不是扬州人.于是得到D是上海人.那么他不是其他城市的人.如图(a)。

由③知,A和F不是南京人,那么A一定是杭州人.而其他旅客都不是杭州人.如下图(b)。

由④可知,B不是北京人,也不是南京人;C不是南京人,那么B是天津人,C是扬州人;故F是北京人,E是南京人.如下图(c)。

综合上述推理,我们得到:

A是医生,来自杭州;B是教师,来自天津;

C是工程师,来自扬州;D是工程师,来自上海;

E是教师,来自南京;F是医生,来自北京。

例17 甲、乙、丙三人分别在北京、天津、上海的中学教数学、物理、化学.已知

①甲不在北京;

②乙不在天津;

③在北京的人不教化学;

④在天津的人教数学;

⑤乙不教物理。

根据以上情况判断,甲、乙、丙三人分别在何处教何课程?

分析根据已知条件,我们把人、地区、科目这三类分别用点表示在三个集合内.规定:两者之间有关系用实线连接,没有关系用虚线连接.这样把问题转化为用图进行推理(如图(a)).据此,下面的结果是显然的:①如果某一点用虚线连接某一个集合的两个点,则这点与这一集合内的第三个点应连实线;②如果在以不同集合内的点为顶点的三角形中两条边是实线,则第三条边也应该是实线.这样,上述三角形中若一条边为虚线,另

一条边为实线,则第三条边一定为虚线.这两条结论是解题的依据.解题的关键是找到三个以实线为边的三角形。

解:根据题意,甲与北京、乙与天津、乙与物理、北京与化学之间连虚线;天津与数学之间连实线(如上图(b)).这样,根据上面的结论,乙与数学应连虚线,乙与化学应连实线。

从而天津与化学连虚线,上海与化学连实线,乙与上海连实线(如下页图(c)),即乙在上海教化学.由图(c)进一步可以看出,甲与上海应连虚线,甲与天津连实线.因而甲与数学连实线(如下页图(d)).由此得出:甲在天津教数学,而余下就是丙在北京教物理.

◇练习巩固◇

1.A、B、C、D四位同学参加60米赛跑的决赛.赛前,四位同学对比赛结

果各说了如下的一句话:

A说:“我会得第一名.”

B说:“A、C都不会取得第一名.”

C说:“A或B会得第一名.”

D说:“B会得第一名.”

结果有两位同学说对了.试问:谁会获得这次决赛的第一名?

2.A、B、C、D四人同住一间寝室,其中一人在修指甲,一人在洗头,

一人在画画,另一人在看书,已知:

①A不在修指甲,也不在看书;

②B不在画画,也不在修指甲;

③若A不在画画,则D不在修指甲;

④C既不在看书,也不在修指甲;

⑤D不在看书,也不在画画。

请问:他们各自在干什么?

3.张、王、李三人分别出生在北京、上海和武汉,他们分别是歌唱演

员、相声演员和舞蹈演员.已知:①小王不是歌唱演员,小李不是相声演

员;②歌唱演员不出生在上海;③相声演员出生在北京;④小李不出生在武汉.试分别确定他们的出生地和职业。

4.有甲、乙、丙、丁四人同住在一座四层的楼房里,他们之中有工程

师、工人、教师和医生.如果已知:

①甲比乙住的楼层高,比丙住的楼层低,丁住第四层;

②医生住在教师的楼上,在工人的楼下,工程师住最低层。

试问:甲、乙、丙、丁各住在这座楼的几层?各自的职业是什么?

◇练习答案◇

1.利用图表可得A是第一名。

2.方法1:由①②③④⑤知,既不是A、B在修指甲,也不是C在修

指甲,以及A、C.D不在看书,所以B在看书,修指甲的是D.但“D修指

甲”与③的有条件的结论矛盾.所以③的条件是不成立的.这就得到A在画画.由④知C在洗头。

方法2:可用图表法进行推理。

3.小李是上海人,舞蹈演员;小王是北京人,相声演员;

小张是武汉人,歌唱演员。

4.甲:教师,住二层;乙:工程师,住一层;

北京小升初分班考试英语试题及答案

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