Nursing Elites

1. Simplify the expression. Which of the following is the value of x? (5(4x – 5) = (3/2)(2x – 6))

• A. −(2/7)
• B. −(4/17)
• C. (16/17)
• D. (8/7)

Correct answer: C

Rationale: To solve the given proportion 5(4x – 5) = (3/2)(2x – 6), first distribute to get 20x - 25 = 3x - 9. Then, simplify the linear equation by isolating x: 20x - 3x = 25 - 9, which leads to 17x = 16. Finally, solving for x gives x = 16/17. Choice A is incorrect as it does not match the calculated value of x. Choice B is incorrect as it does not correspond to the correct solution for x. Choice D is incorrect as it does not align with the accurate value of x obtained from solving the equation.

2. What is the simplest way to write the following expression? 5x - 2y + 4x + y

• A. 9x - y
• B. 9x - 3y
• C. 9x + 3y
• D. x; y

Correct answer: A

Rationale: To simplify the given expression 5x - 2y + 4x + y, we combine like terms. Grouping the x terms together and the y terms together, we have 5x + 4x - 2y + y. Combining like terms results in 9x - y. Therefore, the simplest form of the expression is 9x - y, which corresponds to option A. Option B is incorrect because it incorrectly subtracts 3y instead of just y. Option C is incorrect because it adds 3y instead of subtracting y. Option D is incorrect as it separates x and y with a semicolon instead of an operation, providing no simplified expression.

3. University X requires some of its nursing students to take an exam before being admitted into the nursing program. In this year's class, half of the nursing students were required to take the exam, and three-fifths of those who took the exam passed. If this year's class has 200 students, how many students passed the exam?

• A. 120
• B. 100
• C. 60
• D. 50

Correct answer: C

Rationale: If the incoming class has 200 students, then half of those students were required to take the exam. (200)(1/2) = 100. So 100 students took the exam, but only three-fifths of that 100 passed the exam. (100)(3/5) = 60. Therefore, 60 students passed the exam. The correct answer is 60. Choice A is incorrect as it miscalculates the number of students who passed the exam. Choice B is incorrect as it does not consider the passing rate of the exam. Choice D is incorrect as it is much lower than the correct answer.

4. A commuter survey counts the people riding in cars on a highway in the morning. Each car contains only one man, only one woman, or both one man and one woman. Out of 25 cars, 13 contain a woman and 20 contain a man. How many contain both a man and a woman?

• A. 4
• B. 7
• C. 8
• D. 13

Correct answer: C

Rationale: Let's denote the number of cars containing only a man as M, only a woman as W, and both a man and a woman as B. Given that there are 25 cars in total, we have: M + W + B = 25 From the information provided, we know that 13 cars contain a woman (W) and 20 cars contain a man (M). Since each car contains either one man, one woman, or both, the cars that contain both a man and a woman (B) are counted once in each of the M and W categories. Therefore, to find out how many cars contain both a man and a woman, we need to subtract the number of cars that contain only a man and only a woman from the total cars. M + B = 20 (as 20 cars contain a man) W + B = 13 (as 13 cars contain a woman) Solving the above two equations simultaneously, we get: M = 12, W = 5, B = 8 Therefore, 8 cars contain both a man and a woman. Hence, the correct answer is 8. Choice A, B, and D are incorrect as they do not reflect the correct calculation based on the information provided.

5. What is the mode of the set of numbers {4, 4, 5, 7, 8}?

• A. 4
• B. 5
• C. 7
• D. 8

Correct answer: A

Rationale: The mode of a set of numbers is the value that appears most frequently. In the given set {4, 4, 5, 7, 8}, the number 4 appears twice, which is more frequent than any other number. Therefore, the mode of this set is 4. Choice B, 5, is incorrect as it only appears once in the set. Choices C and D, 7 and 8 respectively, also appear only once each, making them less frequent than the number 4.

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