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SAT数学Problem Solving类型练习三

2019-10-03 13:11:20来源:网络

  为了帮助大家高效备考SAT,新东方在线SAT频道为大家带来*SAT数学Problem Solving类型练习三,希望对大家SAT备考有所帮助。更多精彩尽请关注新东方在线SAT频道!

  Question #1: In the x,y plane, which of the following statements are true?

  I. Line y + x = 5 is perpendicular to line y - x = 5.

  II. Lines y + x = 5 and y - x = 5 intersect each other on the y axis.

  III. Lines y + x = 5 and y - x = 5 intersect each other on the x axis.

  (a) I and III are both true.

  (b) I is the only true statement.

  (c) II is the only true statement.

  (d) I and II are both true.

  Answer: y + x = 5 can be written as y = -x + 5. The slope of this equation is m1 = -1.

  y - x = 5 can be written as y = x + 5. The slope of this equation is m2 = 1.

  m2 = -1/m1 so the 2 lines are perpendicular.

  We also need to find where the 2 lines intersect. If we add the 2 equations, 2·y = 10, y = 5.

  From the first equation, x = 5 - y = 5 - 5 = 0. In conclusion the lines intersect at (0, 5) and this point is on the y axis.

  In conclusion I and II statements are correct.

  Question #2: If a is an integer chosen randomly from the set {3, 5, 6, 9} and b is an integer chosen randomly from the set {2, 3, 4}, what is the probability that a/b is an integer?

  (a) .125

  (b) .250

  (c) .333

  (d) .5

  (e) .55

  Answer: We have 4 possible integers for a and 3 for b, so the number of possible combinations for a/b is 4 · 3 = 12.

  a/b is an integer only for 4 combinations:

  1. a = 3 and b = 3

  2. a = 6 and b = 2

  3. a = 6 and b = 3

  4. a = 9 and b = 3

  The probability that a/b is an integer is 4/12 = 1/3 = .333.

  Question #3: What is the value of integer a, if x = 2 is a solution of the equation √(a + x) = 2·x?

  (a) a = 10

  (b) a = 12

  (c) a = 14

  (d) a = 16

  (e) a = 18

  Answer: If we square the equation we get a + x = 4·x2

  By replacing x with 2, a + 2 = 4·22, so a + 2 = 16.

  In conclusion, a = 14.

  Question #4: What is the value of (3x + 1 - 3x) / (3x - 3x - 1)?

  (a) 6

  (b) 3x

  (c) 3x + 1

  (d) 3x - 1

  (e) 3

  Answer: The numerator of the fraction is: 3x + 1 - 3x = 3x·(3 - 1) = 2 · 3x

  The denominator of the fraction is: 3x - 3x - 1 = 3x - 1·(3 - 1) = 2 · 3x - 1

  We can write the fraction as (2 · 3x) / (2 · 3x - 1) = 3x / 3x - 1 = 3

  Question #5: Two diameters of a circle create an angle AOB of 45o between them. What is the length of arc AB if the radius of the circle is 10/??

  (a) 5/2

  (b) 3/2

  (c) 2

  (d) 4

  (e) 6

  Answer: The circumference of the circle is 2·?·r = 2·?·10/? = 20.

  The ratio between the length of arc AB and the circumference of the circle is equal between the ratio between the 45o angle and 360o.

  In conclusion, AB = 20 · 45o/360o = 20/8 = 5/2.

  Question #6: A bus travels from town A to town B for 2 hours at a speed of 60 miles/hour. The bus stops in town B for 2 hours and then travels from town B to town C for 1 hour, at a speed of 50 miles/hour. What is the average speed of the bus?

  (a) 30miles/hour

  (b) 31miles/hour

  (c) 32miles/hour

  (d) 34miles/hour

  (e) 40miles/hour

  Answer: The distance the bus travels from A to B is 60 miles/hour · 2 hours = 120 miles.

  Then, the bus travels from B to C: 50 miles/hour · 1 hour = 50 miles.

  The total distance traveled is 120 + 50 = 170 miles and the total time is 2 hours + 2 hours stop + 1 hour = 5 hours.

  In conclusion the average speed was 170 miles / 5 hours = 34 miles/hour.

  Question #7: If a·b + b·c + c·a = 0, what is (a + b)2 + (b + c)2 + (c + a)2?

  (a) a2 + b2 + c2

  (b) 2·(a2 + b2 + c2)

  (c) (a2 + b2 + c2)/2

  (d) a2 + a + b2 + b + c2 +c

  (e) (a + b + c)/2

  Answer: (a + b)2 + (b + c)2 + (c + a)2 = 2·(a2 + b2 + c2) + 2·(a·b + b·c + c·a)

  Since a·b + b·c + c·a = 0, the correct result is 2·(a2 + b2 + c2).

  Question #8:

  Column AColumn B

  x2 + 1x + 1

  (a) The quantity in Column A is greater then the quantity in Column B.

  (b) The quantity in Column B is greater then the quantity in Column A.

  (c) The two quantities are equal.

  (d) The relationship cannot be determined from the information given.

  Answer: We need to compare x2 + 1 with x + 1. This results in a comparison between x2 and x.

  For some x, x2 will be greater than x, e.g. for x = 2. For others, e.g. x = 1/2, x2 will be lower so the relationship cannot be determined from the information given.

  Question #9:

  2·m - n = 4

  m + 2·n = 12

  Column AColumn B

  (m + n)261

  (a) The quantity in Column A is greater then the quantity in Column B.

  (b) The quantity in Column B is greater then the quantity in Column A.

  (c) The two quantities are equal.

  (d) The relationship cannot be determined from the information given.

  Answer: From the first equation, n = 2·m - 4. Then, the first equation will be m + 2·(2·m - 4) = 12

  m + 4·m - 8 = 12 so 5·m = 20 and m = 4

  From the first equation, n = 2·m - 4 = 2·4 - 4 = 4

  Column A expression will be (m + n)2 = (4 + 4)2 = (8)2 = 64

  The quantity in Column A is greater than the quantity in Column B.

  Question #7: If a and b are positive integers and a·b = 200, which of the following can be the sum a + b?

  (a) 40

  (b) 46

  (c) 33

  (d) 55

  (e) 50

  Answer: 200 = 2·2·2·5·5.

  If a and b are positive integers, the 2 numbers and their sum can be:

  2 + 100 = 102

  4 + 50 = 54

  5 + 40 = 45

  8 + 25 = 33

  10 + 20 = 30

  (c) is the correct answer.

  以上就是关于“SAT数学Problem Solving类型练习三”的内容,更多精彩内容,请关注SAT频道!

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