GRE考试中,虽说考生在GRE数学部分有着很大的优势,但对于复习也不能轻视,要把这个优势发挥最大的作用,提升GRE成绩,下面是GRE数学模拟练习题。
1. 从1,2,3,4,5,6,7,8,9中选出三个数字组成一个三位数,这个三位数的digits中有两个相同,另一个digit与其它两个都不同,问共有多少个这样的三位数?
72
144
180
216
54
2. 符合 X^2+Y^2=100的整数解共有多少对?
3. A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box,what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
1/4
3/8
1/2
5/8
3/4
4. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parrallel to the x-axis. The x-and y-coordinates of P,Q,and Rare to be integers that satisfy the inequalities -4=5,6=16,how many different triangles with these properties could be constructed?
110
1100
9900
10000
12100
5. 要组成三位电话号码,第一位不能是0或1,三位数中相邻两位不能为同一数. ,求可以组成多少个这样的电话号码?
参考答案:
1.解:因为三个数中两个数相同,所以从9个数中取出两个的可能性为,三个数中那个不同的数和可能位置为3个,取出的两个数哪个做不同的数的可能性为两种
2.解:本题的意为一个半径为10的圆中的整数对有多少对。首先,是一种特殊情况,另四个坐标轴上各有10对解,共41对,对1至7之间的数字而言,任一对均满足条件,则对,对于8,满足条件的有1,2,3,4,5,6六个数,则:
624=48对解
对于9,满足条件的有1,2,3,4四个数,则:
424=32
所以总数为41+196+48+32=317对解。
3.解:本题因为奇数和偶数个数相同,和也只有两种方式,所以三个数的和为奇数的可能性为1/2。也可以考虑从这100个数中任取三个数的奇偶,可能性为:
奇奇奇,奇偶偶,奇奇偶和偶偶偶,其中奇奇奇和奇偶偶两种情况的和为奇数,所以也得到可能性为1/2
4.解:本题首先应考虑赵有三条形的构成,对于-4=5 10个数而言,须取两个点构成直角边,且有顺序问题,因为直角可在两边,则应为;对于6=16,也应取两个点构成另一直角边,也存在顺序问题
5.解:
第一位只能选2-9,有8个可能性
第二位只能选不是第一位的数,有9个可能性
第三位只能选不是第二位的数,有9个可能性
因此共有899=648种方法
以上是小编为大家整理的GRE数学模拟练习题,希望考生多加练习,为GRE提分。
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