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SAT模拟考试题:SAT数学练习题(18)

2016-08-11 16:59:05来源:网络

  SAT模拟考试SAT数学练习题(18):本文是sat数学模拟题。新东方在线SAT频道为大家整理了SAT模拟考试题,供考生们参考,以下是详细内容。

  Question 1: If a§b = 2a - 3b for all real numbers a and b, what is 2(3§2)?

  Question 2: A regular polygon has 6 sides. What is the degree measure of the angle, within the polygon, between any two sides?

  Question 3: A cubic box 'A' has sides of length a and another cubic box 'B' has sides of length 3a. How many boxes A could fit into a single box B?

  Question 4: If a two-sided coin is flipped 4 times, what is the probability that at least one head will show up?

  Question 5: What is the value of (2x + 1 - 2x) / (2x - 2x - 1)?

  Question 6: Side AB of the ABCD square is 10, AE = ED and BF = FC. What is the ratio between the area of parallelogram BFDE and the area of triangle AEB?

SAT模拟考试题:SAT数学练习题(18)

  Question 7: If 15% of x is 45, what is 20% of x?

  Question 8: Given the | x - 3 | = 5 equation, what is the average of its solutions?

  Question 9: In the x, y plane, what is the area of the quadrilateral delimited by the x axis, the y axis, the x = 5 line and the y = -5 line?

  Question 10: If the average of x and y is 3, what is the average of x, y and 6?

  SAT考试Grid-in练习题3参考答案

  Question 1: If a§b = 2a - 3b for all real numbers a and b, what is 2(3§2)?

  Answer: 1

  Explanation: 2(3§2) = 2(2·3 - 3·2) = 20 = 1

  Question 2: A regular polygon has 6 sides. What is the degree measure of the angle, within the polygon, between any two sides?

  Answer: 120

  Explanation: The sum of all of the angles of a polygon is (n - 2)·180o, where n is the number of sides of the polygon. For a polygon with 6 sides, this is (6 - 2)·180o = 720o

  The polygon is regular, so all angles are congruent. The degree measure of the angle, within the polygon, between any two sides is 720o/6 = 120o.

  Question 3: A cubic box 'A' has sides of length a and another cubic box 'B' has sides of length 3·a. How many boxes 'A' could fit into a single box 'B'?

  Answer: 27

  Explanation: The volume of box 'A' is a3 while the volume of box 'B' is (3·a)3.

  The number of boxes A that could fit into a box B is the ratio between the volumes of the 2 boxes. This is (3·a)3 / a3 = 27

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本文关键字: SAT模拟考试 SAT数学

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