SAT2数学Level 2试题1 含答案.

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• Question 1: What is the closest approximation of the solution of the equation 2^{x - 1} = 3^{x + 1}?

  (a) -4.42

  (b) -5.81

  (c) -3.22

  (d) 4.93

  (e) 3.33

  • log(2^{x - 1}) = log(3^{x + 1}) (x - 1)log2 = (x + 1)log3 x(log2 - log3) = log3 + log2 x = (log3 + log2)/(log2 - log3) x is aprox. = -4.418

• Question #2: What is the range of (x - y) if 3 < x < 4 and -2 < y< -1?

  (a) 4< x-y <5

  (b) 1< x-y <3

  (c) 1< x-y <5

  (d) 4< x-y <6

  (e) 3< x-y <6

  • Answer: We can determine the range of -y: 1 < -y < 2 We determine the range of x-y by adding the ranges of x and -y: Therore, 4< x-y <6

• Question #3: A bus travels the distance d from New York to Boston. t₁ hours after the bus lt New York, a car starts to travel the same distance d from New York to Boston. Both vehicles reach Boston at the same time. Find an expression for d as a function of t₁, the speed of the bus v₁ and the speed of the car v₂.

  (a) d = v₁t₁/(v₂ - v₁)

  (b) d = v₁v₂t₁/(v₂ - v₁)

  (c) d = v₁t₁/(v₂ + v₁)

  (d) d = v₁v₂t₁/(v₂ + v₁)

  (e) d = v₁v₂t₁

  • Answer: If t is the duration of the travel of the bus, d = v₁t d = v₂(t - t₁) Therore, v₁t = v₂(t - t₁) t = v₂t₁ / (v₂ - v₁) d = v₁v₂t₁ / (v₂ - v₁)

• Question #4:

  Find the value of x if: x + y + z = 5 x + y - z = 3 x - y = 2

  (a) -3

  (b) -1

  (c) 1

  (d) 3

  (e) 5

  • Answer: We notice that if we add equations 1 and 2 we can eliminate z: x + y + z + x + y - z = 5 + 3 2·(x + y) = 8 x + y = 4 At this point we have a system of 2 linear equations with 2 variables, x and y: x - y = 2 x + y = 4 If we add these 2 equations, we get 2·x = 6 and x = 3.

• Question #5: A camera has a price of 300 dollars. Its price is lowered 10% and then increased 10%. What is the final selling price of the camera?

  (a) $297

  (b) $303

  (c) $310

  (d) $330

  (e) $303

  • Answer: After it is lowered by 10%, the price is $300 - $30 = $270. A 10% increase of a $270 price is $27, so the final price is $270 + $27 = $297.

• Question #6: The equation 2x² - 2x - 60 = 0 has the following 2 solutions:

  (a) {-5, 5}

  (b) {5, -6}

  (c) {-5, -6}

  (d) {-5, 6}

  (e) {5, 6}

  • Answer: 2x² - 2x - 60 = 0 x² - x - 30 = 0 The sum of the solutions is 1 and the product is -30 so the solutions are {-5, 6}

• Question #7: The side of a cube is two times the radius of a sphere. What is the ratio of the volume of the cube to the volume of the sphere?

  (a) 6/¶

  (b) 3/¶

  (c) ¶/6

  (d) ¶/4

  (e) ¶

  • Answer: If l is the side of the cube, the volume of the cube is: V_cube = l³. The volume of the sphere is V_sphere = 4¶r³/3 = 4¶(l/2)³/3 = (¶/6)l³ V_cube/V_sphere = 6/¶

• Question #8: If tan(2x) = 2, tan(x) is equivalent to which of the following expressions?

  (a) sin(2x)/cos(2x)

  (b) 1 + tan(x)

  (c) 1 - tan(x)

  (d) (1 - tan(x))(1 + tan(x))

  (e) (1 - tan(x))²

  • Answer: tan(2x) = 2tan(x)/(1 - tan²(x)) 2 = 2tan(x)/(1 - tan²(x)) tan(x) = 1 - tan²(x) tan(x) = (1 - tan(x))(1 + tan(x))

• Question #9: What is the closest approximation of the value of angle a in the figure below, if AB = 7, BC = 11 and CA = 5?

  (a) 96.4^o

  (b) 100.8^o

  (c) 144.9^o

  (d) 132.2^o

  (e) 135.9^o

  • Answer: BC² = AB² + CA² - 2·AB·CA·cos(a) 11² = 5² + 7² - 2·5·7·cos(a) 121 = 25 + 49 - 70cos(a) cos(a) = -47/70 a = arccos(-47/70) = 132.2^o

• Question #10: For some positive real number ‘a’, the first 3 terms of a geometric progression are a - 1, a + 3 and 3a + 1. What is the numerical value of the fourth term?

  (a) 25

  (b) 36

  (c) 32

  (d) 100

  (e) 9

  • Answer: a + 3 = k(a - 1) 3a + 1 = k(a + 3) (a + 3)(a + 3) = (3a + 1)(a - 1) a² + 6a + 9 = 3a² - 2a - 1 2a² - 8a -10 = 0 a² - 4a -5 = 0 the solutions of this equation are 5 and -1. The only positive solution is 5, so the progression is 4, 8, 16. The fourth term will be 16·2 = 32

转载请注明来自澳际留学

• Question 1: What is the closest approximation of the solution of the equation 2^{x - 1} = 3^{x + 1}?

  (a) -4.42

  (b) -5.81

  (c) -3.22

  (d) 4.93

  (e) 3.33

  • log(2^{x - 1}) = log(3^{x + 1}) (x - 1)log2 = (x + 1)log3 x(log2 - log3) = log3 + log2 x = (log3 + log2)/(log2 - log3) x is aprox. = -4.418

• Question #2: What is the range of (x - y) if 3 < x < 4 and -2 < y< -1?

  (a) 4< x-y <5

  (b) 1< x-y <3

  (c) 1< x-y <5

  (d) 4< x-y <6

  (e) 3< x-y <6

  • Answer: We can determine the range of -y: 1 < -y < 2 We determine the range of x-y by adding the ranges of x and -y: Therore, 4< x-y <6

• Question #3: A bus travels the distance d from New York to Boston. t₁ hours after the bus lt New York, a car starts to travel the same distance d from New York to Boston. Both vehicles reach Boston at the same time. Find an expression for d as a function of t₁, the speed of the bus v₁ and the speed of the car v₂.

  (a) d = v₁t₁/(v₂ - v₁)

  (b) d = v₁v₂t₁/(v₂ - v₁)

  (c) d = v₁t₁/(v₂ + v₁)

  (d) d = v₁v₂t₁/(v₂ + v₁)

  (e) d = v₁v₂t₁

  • Answer: If t is the duration of the travel of the bus, d = v₁t d = v₂(t - t₁) Therore, v₁t = v₂(t - t₁) t = v₂t₁ / (v₂ - v₁) d = v₁v₂t₁ / (v₂ - v₁)

• Question #4:

  Find the value of x if: x + y + z = 5 x + y - z = 3 x - y = 2

  (a) -3

  (b) -1

  (c) 1

  (d) 3

  (e) 5

  • Answer: We notice that if we add equations 1 and 2 we can eliminate z: x + y + z + x + y - z = 5 + 3 2·(x + y) = 8 x + y = 4 At this point we have a system of 2 linear equations with 2 variables, x and y: x - y = 2 x + y = 4 If we add these 2 equations, we get 2·x = 6 and x = 3.

• Question #5: A camera has a price of 300 dollars. Its price is lowered 10% and then increased 10%. What is the final selling price of the camera?

  (a) $297

  (b) $303

  (c) $310

  (d) $330

  (e) $303

  • Answer: After it is lowered by 10%, the price is $300 - $30 = $270. A 10% increase of a $270 price is $27, so the final price is $270 + $27 = $297.

• Question #6: The equation 2x² - 2x - 60 = 0 has the following 2 solutions:

  (a) {-5, 5}

  (b) {5, -6}

  (c) {-5, -6}

  (d) {-5, 6}

  (e) {5, 6}

  • Answer: 2x² - 2x - 60 = 0 x² - x - 30 = 0 The sum of the solutions is 1 and the product is -30 so the solutions are {-5, 6}

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