MATH.14

# Inequalities a x 4 b x 8 c 2 x 4 d 2 x 6 e 2 x 6

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inequalities? A. x ≤ 4 B. x ≤ 8 C. 2 - x ≤ 4 D. 2 - x ≤ 6 E. 2 + x ≤ 6 Answer: ------------------------------------------------------------------------------------------------------------ Q37: Of the 500 business people surveyed, 78 percent said that they use their laptop computers at home, 65 percent said that they use them in hotels, and 52 percent said that they use them both at home and in hotels. How many of the business people surveyed said that they do not use their laptop computers either at home or in hotels? ------------------------------------------------------------------------------------------------------------ ------------------------------------------------------------------------------------------------------------ Note: ------------------------------------------------------------------------------------------------------------ Q2: If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 2 1 ? (1) More than 2 1 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 10 1 . 12

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Note: Step 1 : Use Trial and Error Method to check Statement 1! Pick up the case: 6 women in the group. → p = 10 6 × 9 5 = 3 1 < 2 1 ; Pick up the case: 10 women in the group. → p = 1 > 2 1 Result: Statement 1 is false! Step 2: Use the same method to check Statement 2, but the empirical method and my fifth sense is to check the combination of S1 and S2 first. When 6 women → p = 3 1 < 2 1 ; for men → 10 4 × 9 3 = 15 2 > 10 1 → So drop this case! When 7 women → p = 10 7 × 9 6 = 15 7 < 2 1 ; for men → 10 3 × 9 2 = 15 1 < 10 1 → So this case is GOOD, and p < 2 1 ! When 8 women → p = 10 8 × 9 7 = 45 28 > 2 1 ; for men → 10 2 × 9 1 = 45 1 < 10 1 → So this case is GOOD, but p > 2 1 ! Or use the extreme case: when 10 women → p = 1 > 2 1 → for men 0 < 10 1 → So this case is GOOD, but p > 2 1 ! Step 3: Draw the conclusion: S1 + S2 is wrong! And S2 is also wrong as tested in the process of Step 2! The answer should be E ! ------------------------------------------------------------------------------------------------------------ Q17: If x , y , and k are positive numbers such that ( y x x + )(10) + ( y x y + )(20) = k and if x < y , which of the following could be the value of k ? 13
Note: In order to get the answer for this question, examinees must use the method of Trial and Error . First, y x y + 10 = k – 10 → Then, try k value from Answer A to E. A. 10 → y x y + 10 = 0 → y = 0 → Because from the Term of both x and y are positive y = 0, Wrong! B. 12 → y x y + 10 = 2 → x = 4 y → Because from the Term of x < y , x y = 3 y > 0 → x > y , Wrong! C. 15 → y x y + 10 = 5 → x = y → Because from the Term of x < y x = y , Wrong! D. 18 → y x y + 10 = 8 → 4 x = y → Because from the Term of x < y , y x = 3x > 0 → x < y , Correct! E. 30 → y x y + 10 = 20 → 2 x = - y → Because from the Term of both x and y are positive → Wrong!
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