What is the number of integers that are common to both set S and set T ?
(1) The number of integers in S is 7, and the number of integers in T is 6.
(2) U is the set of integers that are in S only or in T only or in both, and the number of integers in U is 10.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
U is the set of integers that are in S only or in T only or in both, and the number of integers in U is 10.。所以U是S, T的并集。根据公式A∪B =|A∪B| = |A|+|B| - |A∩B |(∩:重合的部分) ,6+7-10=3, 交集部分是3.
【引用】U is the set of integers that are in S only or in T only or in both, and the number of integers in U is 10.。所以U是S, T的并集。
韦恩图:
s+d=7, t+d=6, s+d+t=10,求d=?
用三个不同的方程式解一元三次方程,能解出来。将s和t都转换成d的表达式,归在三者相加等于10那个方程式里。
条件1只知道S和T各自的元素数量,完全不知道交集的情况;
条件2知道U=(S+T+交集)的元素数量,还是不知道交集的元素数量。
二者结合可以知道S\T\交集各自的元素数量。
c
为啥是C?
用韦恩图是最快理解此类题目的方式
公共的整数个数为7 + 6 - 10 = 3,其中 7 + 6 相当于公共的部分算了两遍
但怎么样推出U是P和S的并集呢?条件2是啥意思呀?我的理解是U中的任意一个数在P或者S中能找到。比如P=1,2,3,4,5,6,7然后S=8,9,10,11,12,13,这样的话U=4,5,6,7,8,9,10,11,12,13也满足条件2?但是P和S没有重合...
条件2就可以判断出。U中的整数分为三个部分,只在S中的整数,只在T中的整数,既在S中又在T中的整数
派大星为什么老是在发言里面做题(黑人问号???)
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