如何备考gmat数学考试部分呢?对于一般的考生来说，首先需要对gmat考试中数学部分的基本概念有所掌握，然后再去总结和归纳解题技巧和方法，下面就来看看gmat数学考试中的算术概念有哪些需要记忆。

本期热点导读：

1.GMAT数学考试中的公理有哪些

2.GMAT数学基本概念解读技巧

3.GMAT数学满分备考及应试中的细节

4.GMAT数学备考策略三步曲

5.GMAT数学考试中的应试技巧

正文：

　　1.平均数(AVERAGE OR ARITHMETIC MEAN)

　　2.中数(MEDIAN)

　　TO CALCULATE THE MEDIAN OF N NUMBERS，FIRSTSGROUPSTHE NUMBERS FROM LEAST TO GREATEST;IF N IS ODD，THE MEDIAN IS DEFINED AS THE MIDDLE NUMBER，WHILE IF N IS EVEN，THE MEDIAN IS DEFINED AS THE AVERAGE OF THE TWO MIDDLE NUMBERS. FOR THE DATA 6, 4, 7, 10, 4, THE NUMBERS, IN ORDER, ARE 4, 4, 6, 7, 10, AND THE MEDIAN IS 6, THE MIDDLE NUMBER. FOR THE NUMBERS 4, 6, 6, 8, 9, 12, THE MEDIAN IS (6+8 )/2 = 7. NOTE THAT THE MEAN OF THESE NUMBERS IS 7.5.

　　3.众数(MODE)：一组数中的众数是指出现频率最高的数。

　　例：THE MODE OF 7，9，6，7，2，1 IS 7。

　　4.值域(RANGE)：表明数的分布的量，其被定义为最大值减最小值的差。

　　例：THE RANGE OF–1，7，27，27，36 IS 36-(-1)= 37。

　　5.标准方差(STANDARD DEVIATION)：

　　ONE OF THE MOST COMMON MEASURES OF DISPERSION IS THE STANDARD DEVIATION. GENERALLY SPEAKING, THE GREATER THE DATA ARE SPREAD AWAY FROM THE MEAN, THE GREATER THE STANDARD DEVIATION. THE STANDARD DEVIATION OF N NUMBERS CAN BE CALCULATED AS FOLLOWS:

　　(1)FIND THE ARITHMETIC MEAN ;

　　(2)FIND THE DIFFERENCES BETWEEN THE MEAN AND EACH OF THE N NUMBERS ;

　　(3)SQUARE EACH OF THE DIFFERENCES ;

　　(4)FIND THE AVERAGE OF THE SQUARED DIFFERENCES ;

　　(5)TAKE THE NONNEGATIVE SQUARE ROOT OF THIS AVERAGE.

　　NOTICE THAT THE STANDARD DEVIATION DEPENDS ON EVERY DATA VALUE, ALTHOUGH IT DEPENDS MOST ON VALUES THAT ARE FARTHEST FROM THE MEAN. THIS IS WHY A DISTRIBUTION WITH DATA GROUPED CLOSELY AROUND THE MEAN WILL HAVE A SMALLER STANDARD DEVIATION THAN DATA SPREAD FAR FROM THE MEAN.

（未完见下一页）

本期热点导读：

1.GMAT数学考试中的公理有哪些

2.GMAT数学基本概念解读技巧

3.GMAT数学满分备考及应试中的细节

4.GMAT数学备考策略三步曲

5.GMAT数学考试中的应试技巧

　　6.排列与组合

　　THERE ARE SOME USEFUL METHODS FOR COUNTING OBJECTS AND SETS OF OBJECTS WITHOUT ACTUALLY LISTING THE ELEMENTS TO BE COUNTED. THE FOLLOWING PRINCIPLE OF MULTIPLICATION IS FUNDAMENTAL TO THESE METHODS.

　　IF A FIRST OBJECT MAY BE CHOSEN IN M WAYS AND A SECOND OBJECT MAY BE CHOSEN IN N WAYS, THEN THERE ARE MN WAYS OF CHOOSING BOTH OBJECTS.

　　AS AN EXAMPLE, SUPPOSE THE OBJECTS ARE ITEMS ON A MENU. IF A MEAL CONSISTS OF ONE ENTREE AND ONE DESSERT AND THERE ARE 5 ENTREES AND 3 DESSERTS ON THE MENU, THEN 5×3 = 15 DIFFERENT MEALS CAN BE ORDERED FROM THE MENU. AS ANOTHER EXAMPLE, EACH TIME A COIN IS FLIPPED, THERE ARE TWO POSSIBLE OUTCOMES, HEADS AND TAILS. IF AN EXPERIMENT CONSISTS OF 8 CONSECUTIVE COIN FLIPS, THE EXPERIMENT HAS 28 POSSIBLE OUTCOMES,SWHERESEACH OF THESE OUTCOMES IS A LIST OF HEADS AND TAILS IN SOME ORDER.

　　☆阶乘：FACTORIAL NOTATION

　　假如一个大于1的整数N，计算N的阶乘被表示为N!，被定义为从1至N所有整数的乘积，

　　例如：4! = 4×3×2×1= 24

　　注意：0! = 1! = 1

　　☆排列：PERMUTATIONS

　　THE FACTORIAL IS USEFUL FOR COUNTING THE NUMBER OF WAYS THAT A SET OF OBJECTS CAN BE ORDERED. IF A SET OF N OBJECTS IS TO BE ORDERED FROM 1ST TO NTH, THERE ARE N CHOICES FOR THE 1ST OBJECT, N-1 CHOICES FOR THE 2ND OBJECT, N-2 CHOICES FOR THE 3RD OBJECT, AND SO ON, UNTIL THERE IS ONLY 1 CHOICE FOR THE NTH OBJECT. THUS, BY THE MULTIPLICATION PRINCIPLE, THE NUMBER OF WAYS OF ORDERING THE N OBJECTS IS

　　N (N-1) (N-2)…( 3) (2) (1) = N!

　　FOR EXAMPLE, THE NUMBER OF WAYS OF ORDERING THE LETTERS A, B, AND C IS 3!, OR 6：ABC, ACB, BAC, BCA, CAB, AND CBA.

　　THESE ORDERINGS ARE CALLED THE PERMUTATIONS OF THE LETTERS A, B, AND C.也可以用P 33表示.

　　例如：1, 2, 3, 4, 5这5个数字构成不同的5位数的总数为5! = 120

　　☆组合：COMBINATION

　　A PERMUTATION CAN BE THOUGHT OF AS A SELECTION PROCESS IN WHICH OBJECTS ARE SELECTED ONE BY ONE IN A CERTAIN ORDER. IF THESGROUPSOF SELECTION IS NOT RELEVANT AND ONLY K OBJECTS ARE TO BE SELECTED FROM A LARGER SET OF N OBJECTS, A DIFFERENT COUNTING METHOD IS EMPLOYED.

　　SPECIALLY CONSIDER A SET OF N OBJECTS FROM WHICH A COMPLETE SELECTION OF K OBJECTS IS TO BE MADE WITHOUT REGARD TO ORDER,SWHERES0≤K≤N . THEN THE NUMBER OF POSSIBLE COMPLETE SELECTIONS OF K OBJECTS IS CALLED THE NUMBER OF COMBINATIONS OF N OBJECTS TAKEN K AT A TIME AND IS CKN.

　　从N个元素中任选K个元素的数目为:

　　CKN. = N!/ (N-K)! K!

　　例如：从5个不同元素中任选2个的组合为C25 = 5!/2! 3!= 10

　　排列组合的一些特性(PROPERTIES OF PERMUTATION AND COMBINATION)

　　☆加法原则：RULE OF ADDITION

　　做某件事有X种方法，每种方法中又有各种不同的解决方法。例如第一种方法中有Y1种方法，第二种方法有Y2种方法，等等，第X种方法中又有YX种不同的方法，每一种均可完成这件事，即它们之间的关系用“OR”表达，那么一般使用加法原则，即有：Y1+ Y2+。。。+ YX种方法。

　　☆乘法原则：RULE OF MULTIPLICATION

　　完成一件事有X个步骤，第一步有Y1种方法，第二步有Y2种方法，。。。，第X步有YX种方法，完成这件事一共有Y1· Y2·。。。·YX种方法。

　　以上只是GMAT考题中经常涉及到的数学—算术方面的问题，今后我们将陆续在新开辟的“网上课堂”中介绍代数、几何以及系统的习题、讲解，以帮助大家在GMAT数学考试中更好地发挥中国学生的优势，拿到让美国人瞠目结舌的成绩!

　　通过上面对gmat数学考试中的常见算术概念的介绍，相信对于很多计划参加gmat考试的人来说，可以参考上述的信息来准备和计划gmat数学考试了。

　　想要获得更多咨询服务点击进入 >>>>有问题?找免费的澳际专家咨询! 或联系QQ客服： ，也可以通过在线咨询处留言，把您最关心的问题告诉我们。

GMAT数学基础概念之算术概述gmat数学

　　如何备考gmat数学考试部分呢?对于一般的考生来说，首先需要对gmat考试中数学部分的基本概念有所掌握，然后再去总结和归纳解题技巧和方法，下面就来看看gmat数学考试中的算术概念有哪些需要记忆。

本期热点导读：

1.GMAT数学考试中的公理有哪些

2.GMAT数学基本概念解读技巧

3.GMAT数学满分备考及应试中的细节

4.GMAT数学备考策略三步曲

5.GMAT数学考试中的应试技巧

正文：

　　1.平均数(AVERAGE OR ARITHMETIC MEAN)

　　2.中数(MEDIAN)

　　TO CALCULATE THE MEDIAN OF N NUMBERS，FIRSTSGROUPSTHE NUMBERS FROM LEAST TO GREATEST;IF N IS ODD，THE MEDIAN IS DEFINED AS THE MIDDLE NUMBER，WHILE IF N IS EVEN，THE MEDIAN IS DEFINED AS THE AVERAGE OF THE TWO MIDDLE NUMBERS. FOR THE DATA 6, 4, 7, 10, 4, THE NUMBERS, IN ORDER, ARE 4, 4, 6, 7, 10, AND THE MEDIAN IS 6, THE MIDDLE NUMBER. FOR THE NUMBERS 4, 6, 6, 8, 9, 12, THE MEDIAN IS (6+8 )/2 = 7. NOTE THAT THE MEAN OF THESE NUMBERS IS 7.5.

　　3.众数(MODE)：一组数中的众数是指出现频率最高的数。

　　例：THE MODE OF 7，9，6，7，2，1 IS 7。

　　4.值域(RANGE)：表明数的分布的量，其被定义为最大值减最小值的差。

　　例：THE RANGE OF–1，7，27，27，36 IS 36-(-1)= 37。

　　5.标准方差(STANDARD DEVIATION)：

　　ONE OF THE MOST COMMON MEASURES OF DISPERSION IS THE STANDARD DEVIATION. GENERALLY SPEAKING, THE GREATER THE DATA ARE SPREAD AWAY FROM THE MEAN, THE GREATER THE STANDARD DEVIATION. THE STANDARD DEVIATION OF N NUMBERS CAN BE CALCULATED AS FOLLOWS:

　　(1)FIND THE ARITHMETIC MEAN ;

　　(2)FIND THE DIFFERENCES BETWEEN THE MEAN AND EACH OF THE N NUMBERS ;

　　(3)SQUARE EACH OF THE DIFFERENCES ;

　　(4)FIND THE AVERAGE OF THE SQUARED DIFFERENCES ;

　　(5)TAKE THE NONNEGATIVE SQUARE ROOT OF THIS AVERAGE.

　　NOTICE THAT THE STANDARD DEVIATION DEPENDS ON EVERY DATA VALUE, ALTHOUGH IT DEPENDS MOST ON VALUES THAT ARE FARTHEST FROM THE MEAN. THIS IS WHY A DISTRIBUTION WITH DATA GROUPED CLOSELY AROUND THE MEAN WILL HAVE A SMALLER STANDARD DEVIATION THAN DATA SPREAD FAR FROM THE MEAN.

（未完见下一页）

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