题目

1. If k is an integer and 121 < k^2 < 225,then k can have at most how many values?

A. 3   B. 4   C. 5    D. 6    E. 7

答案：D.

解析：k可以是12,13,14,-12,-13,-14.

2. Runner A ran 4/5 kilometer and Runner B ran 800 meters.

Quantity A: The distance that A ran

Quantity B: The distance that B ran

A. Quantity A is greater.

B. Quantity B is greater.

C. The two quantities are equal.

D. The relationship cannot be determined from the information given.

答案：C

解析：Quantity A=(4/5)*1000=800=Quantity B。

3. x < y < z,

Quantity A: (x + y + z)/3

Quantity B: y

A. Quantity A is greater.

B. Quantity B is greater.

C. The two quantities are equal.

D. The relationship cannot be determined from the information given.

答案：D

解析：已知条件太少，AB大小都可以根据不同数据列出可能答案，因此无法判断准确唯一解，选D。

4. 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 requirements?

A. 90

B. 115

C. 130

D. 144

E. 195

答案：D

解析：10本书中，4本biographies，6本novel。現在biographies要选2本以上。

思路1：正面解法

所以可能情形有bio选2本(配nov选2)+bio选3本(配nov选1本)+bio选4本(配nov选0本)。

即(C4取2乘C6取2)+(C4取3乘C6取1)+(C4取4乘C6取0)=115

思路2：反向解法

用全部扣除bio选0本(配nov选4)再扣bio选1本(配nov选3)

利用C10取4-C4取0*C6取4-C4取1*C6取3 = 210 - 15 - 80 = 115

5. 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?

答案：121

解析：可能不少人会用142+121-236=27的思路来解题，但问题是least possible number，同时也没有表明只能选两门课，因此有可能还有其他課可选，因此正确答案是眼前两个交集的最大数，也就是142和121取最小数，121个学chemistry的都选了algebra。所以为121。