11. If 1+x+x2+x3=60, then the average (arithmetic mean) of x, x2 , x3 and x4 is equal to which of the following?
A 12x
B 15x
C 20x
D 30x
E 60x
Ans: B
解析： (x+x2+x3+x4)/4=x(1+x+x2+x3)/4=60x/4=15x
解题要点：先根据概念，列出表达式，再化简利用上已知条件。
12. The relationship between the area A of a circle and its circumference C is given by the formula A=kC2 ,where k is a constant. What is the value of k?
A 1/4π
B 1/4
C 1/2π
D 2π
E 4π2
Ans: A
解析： A=πr2 C=2πr
A=kC2 πr2=k(2πr)2
k=πr2/4π2r2=1/4π
解题要点：半径r是用来表示圆的周长C和面积A的共同的指标，可利用它作桥梁，来求k值
13. The sequence of numbers , , …… …….is defined by an=1/n-1/n+2, for each integer n≥1. What is the sum of the first 20 terms of this sequence?
A (1+1/2)-1/20
B (1+1/2)-(1/21+1/22)
C 1-(1/20+1/22)
D 1-1/22
E 1/20-1/22
Ans: B
解析：把数列的前两项相加得(1-1/3)+(1/2-1/4)=(1+1/2)-(1/3+1/4)
把数列的前三项相加得(1-1/3)+(1/2-1/4)+(1/3-1/5)=(1+1/2)-(1/4+1/5) 把数列的前四项相加
(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)=(1+1/2)-(1/5+1/6)
以此类推，得到规律，数列的前k项的和的通式为(1+1/2)-(1/(k+1)+1/k+2)
因此，数列的前20项的和为
(1+1/2)-(1/21+1/22)
解题要点：涉及数列的题最重要的是通过对前面几项的计算，归纳出一个通项。
14. The table below shows the frequency distribution of the values of a variable Y. What is the mean of the distribution?
Give your answer to the nearest 0.01
Y Frequency
1/2 2
3/4 7
5/4 8
3/2 8
7/4 9
Ans: 1.29
解析：
解题要点：理解频数分布表格的含义
常考的两个量mean median
mean: x1,x2,……xn的平均值为(x1+x2……+xn)/n
median: 将一个序列按从小到大值进行排列，位于序列最中间的为其中位数。当序列为奇数项时，最中间的为中位数;当序列为偶数项时，中间两个数的平均数为中位数。
15. Let S be the set of all positive integers n such that n2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S?
Indicate all such integers
A 12
B 24
C 36
D 72
Ans: AC
解析： n的平方为24和108的倍数，则先找出24和108的最小公倍数
24=23*3 108=22*33
因此，它们的最小公倍数是23*33
又因为n的平方是24和108的倍数，指数应该为偶数，所以最小的
n2=(24)(34)
n=(22)(32)=36
S中n的值是36的正倍数即可，则S由36, 72,108,144,……组成
因为要求的是能整除S中所有数的一个数
所以选项中只有12，和36可以。
解题要点： 将合数化成几个质数相乘的形式
16. The ratio of 1/3 to 3/8 is equal to the ratio of
A 1 to 8
B 8 to 1
C 8 to 3
D 8 to 9
E 9 to 8
Ans: D
解析： 1/3:3/8=24*1/3:24*3/8=8:9
解题要点：两分式化比例，分母同乘公倍数。
17. A reading list for a humanities course consists of 10 books, of which 4 are biographies and the rest are novels. Each student is required to read a selection of 4 books from the list, including 2 or more biographies. How many selections of 4 books satisfy the requirement
A 9
B 115
C 130
D 144
E 195
Ans: B
解析： 10本书，4本biographies,6本novels
1. 选择4本biographies，1种选法
2. 选择3本biographies，有C34种选法，1本novel有C16种选法
共有C34*C16=24种选法
3. 选择2本biographies，有C24种选法，2本novel有C26种选法
共有C24*C26=90种选法
共有1+24+90=115种选法
18. In a graduating class of 236 students,142 took algebra and 121 took chemistry What is the greatest possible number of students that could have taken both algebra and chemistry?
( )students
Ans: 121
解析：
从图中可得
x≤121
x最大为121
解题要点：集合问题可以利用韦恩图
19. in the figure above ,if m // k and s=t+30 ,then t=
A 30
B 60
C 75
D 80
E 105
Ans: C
解析：因为m // k，同位角相等，s的补角也为t
因为s=t+30, s+t=180
所以t+30+t=180 t=75
解题要点：几何问题一大解决方法是将题中的已知信息添加到图上，再建立数量关系。
20. If 2x=3y=4z=20,then 12xyz=
A 16000
B 8000
C 4000
D 800
E 10
Ans: C
解析：方法一
2x=3y=4z=20 x=10 y=20/3 z=5
12xyz=12*10*20/3*5=4000
方法二
12xyz=(2x)(3y)(4z)/2=20*20*20/2=4000