Nursing Elites

1. As part of a study, a set of patients will be divided into three groups: 1/2 of the patients will be in Group Alpha, 1/3 of the patients will be in Group Beta, and 1/6 of the patients will be in Group Gamma. Order the groups from smallest to largest, according to the number of patients in each group.

• A. Group Alpha, Group Beta, Group Gamma
• B. Group Alpha, Group Gamma, Group Beta
• C. Group Gamma, Group Alpha, Group Beta
• D. Group Gamma, Group Beta, Group Alpha

Correct answer: C

Rationale: The correct order from smallest to largest number of patients in each group is Group Gamma (1/6), Group Alpha (1/2), and Group Beta (1/3). Group Gamma has the smallest fraction of patients, followed by Group Alpha and then Group Beta. Therefore, choice C, 'Group Gamma, Group Alpha, Group Beta,' is the correct answer. Choices A, B, and D are incorrect because they do not follow the correct order based on the fractions of patients assigned to each group.

2. One gallon of cleaning solution requires 6 oz of ammonia. If the maintenance department needs 230 gallons of solution to clean all of the floors, how much ammonia is needed?

• A. 1380 gallons
• B. 6900 gallons
• C. 1380 oz
• D. 1400 oz

Correct answer: C

Rationale: To find out how much ammonia is needed for 230 gallons of cleaning solution, you multiply the amount of ammonia needed per gallon by the total gallons of solution required. Therefore, 230 gallons * 6 oz/gallon = 1380 oz of ammonia. Option A ('1380 gallons') and Option B ('6900 gallons') are incorrect as the question asks for the amount of ammonia needed, not the total volume of cleaning solution. Option D ('1400 oz') is incorrect as it does not correctly calculate the amount of ammonia required based on the given information.

3. Solve for x in the equation: 3x - 5 = 16

• A. 7
• B. 5
• C. 8
• D. 9

Correct answer: C

Rationale: To solve for x, add 5 to both sides of the equation: 3x - 5 + 5 = 16 + 5, which simplifies to 3x = 21. Next, divide both sides by 3: x = 21 ÷ 3 = 7. Therefore, the correct answer is x = 7, making option A the correct choice. Option C, '8,' is incorrect as it is not the solution obtained from the correct calculations. Options B and D, '5' and '9,' are also incorrect and not the solution to the given equation.

4. What is the least common multiple? What is the least common factor?

• A. The smallest number that both numbers multiply into; the smallest number that divides evenly into both
• B. The largest number that both numbers multiply into; the smallest number that divides evenly into both
• C. The smallest number that both numbers divide into evenly; the smallest number that multiplies into both
• D. The smallest number that both numbers divide into evenly; the smallest number that both multiply into

Correct answer: A

Rationale: The least common multiple is the smallest number that both numbers multiply into, which means it is the smallest number that both numbers can be evenly divided by without leaving a remainder. The least common factor, on the other hand, is the smallest number that divides both numbers without leaving a remainder. Therefore, choice A is correct as it accurately defines the least common multiple and factor. Choices B, C, and D are incorrect because they provide inaccurate definitions or mix up the concepts of multiplication and division in relation to finding the least common multiple and factor.

5. In a study about anorexia conducted on 100 patients, where 70% were women, and 10% of the men were overweight as children, how many male patients in the study were NOT overweight as children?

• A. 3
• B. 10
• C. 27
• D. 30

Correct answer: C

Rationale: Out of the 100 patients, 30% were men (100 - 70% women), hence 30 men. Since 10% of the men were overweight as children (10% of 30 is 3), the remaining men (30 - 3) were NOT overweight as children, which equals 27. Therefore, the correct answer is 27. Choices A, B, and D are incorrect because they do not reflect the accurate calculation of the number of male patients who were NOT overweight as children.

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