26、问某数为3的倍数?
除21后余数为奇数;
除27后余6;
【答案】B
【思路】由1得21K+奇数被3除,即为奇数是否被3整除,不一定。
由2得27K+6 定能被3整除,所以选B
睛天猪
【答案】B
【思路】由A,X=21N+2n+1=3/3),由于不能确定2n+1是3的倍数,故A不充分;由B,X=27N+6=3,B充分
27、某人得到5个帐单要支付,每个价格不同,她随便支付,问她最先支付最贵的,最后支付最便宜的机率?
【答案】P3,3/P5,5=1/20
【思路】分子把最贵的,最便宜的固定了顺序,就等于把中间3个全排列
睛天猪
【答案】P3,3/P5,5
【思路】将此题化为:求从1-5中,先取1,最后取5的概率。因为1、5的次序一定,故只需算中间三个数有多少种取法即可,即P3,3。用此结果除以随意取5个数的取法,即为所求。
28、9的20次方是十位数是:0,9的19次方问的十位数是?
【答案】8
【思路】9的N次方个位数是9,1的循环,因此9的19次方个位数必为9。设十位数为X而9的20次方十位数为0,那么9X+8一定为10的倍数,所以X只能是8
睛天猪
【答案】8 分页标题#e#
【思路】9的n次幂,其个位数有如下规律:n为奇数,个位为9;n为偶数,个位为1。因此有9的19次幂个位为9,设9的19次幂十位为X,则9的20次幂十位表示为9X结果的个位数+8=0,故X=8。
29、X的平方大于Y的平方?
X大于Y;
XY大于0;
【答案】E
【思路】1反例0》-1,与2》1 结论不一定
2 反例好举
睛天猪
【答案】E
【思路】举反例:-1-2-1平方-2平方;55平方3平方
上例的两组数同样适用。
30、一对夫妇生四个孩子,两男两女的概率?
【答案】3/8
【思路】4个孩子中选2个为男孩,剩下为女孩C^4=3/8
睛天猪
【答案】 CC/2^4
【思路】 生孩子为独立重复事件^-^,每次两种可能,生4次共有2^4种可能;由于不是生男就是生女,故分子只需考虑一种情况即可。从四个中任选两个,两个都是男/女的取法有C4,2C2,2,结果与前述2^4相除即得解。
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