Nursing Elites

1. Based on a favorable performance review at work, Matt receives a 3/20 increase in his hourly wage. If his original hourly wage is represented by w, which of the following represents his new wage?

• A. 0.15w
• B. 0.85w
• C. 1.12w
• D. 1.15w

Correct answer: D

Rationale: To calculate Matt's new wage after a 3/20 increase, we need to add this percentage increase to his original wage. The increase in decimal form is 3/20 = 0.15. Therefore, the new wage is w + w(0.15) = w(1 + 0.15) = 1.15w. This means the correct answer is D. Choices A, B, and C are incorrect because they do not account for the full 3/20 increase in the wage. Choice A (0.15w) represents only the increase percentage, not the total new wage. Choice B (0.85w) and Choice C (1.12w) do not accurately calculate the new wage after the increase, leading to incorrect representations of the final wage.

2. On a highway map, the scale indicates that 1 inch represents 45 miles. If the distance on the map is 3.2 inches, how far is the actual distance?

• A. 45 miles
• B. 54 miles
• C. 112 miles
• D. 144 miles

Correct answer: D

Rationale: To find the actual distance represented by 3.2 inches on the map, we use the scale of 1 inch representing 45 miles. Setting up the proportion 1 inch = 45 miles, we can calculate the actual distance by multiplying 3.2 inches by 45 miles, which equals 144 miles. Therefore, the correct answer is 144 miles. Choice A (45 miles) is incorrect as it represents the distance for 1 inch on the map, not for 3.2 inches. Choices B (54 miles) and C (112 miles) are incorrect calculations based on a misinterpretation of the scale.

3. An athlete runs 5 miles in 25 minutes and then changes pace to run the next 3 miles in 15 minutes. Overall, what is the average time in minutes it takes the athlete to run 1 mile?

• A. 7 minutes
• B. 5 minutes
• C. 6.5 minutes
• D. 8.5 minutes

Correct answer: B

Rationale: To find the average time per mile, add the total time taken to cover all miles and then divide by the total miles run. The athlete ran 5 miles in 25 minutes and 3 miles in 15 minutes, totaling 8 miles in 40 minutes. Therefore, the average time per mile is 40 minutes ÷ 8 miles = 5 minutes. Choice A, 7 minutes, is incorrect as it does not reflect the correct average time per mile. Choice C, 6.5 minutes, is incorrect since the calculation is not based on the given information. Choice D, 8.5 minutes, is incorrect as it does not represent the average time per mile for the entire run.

4. The phone bill is calculated each month using the equation C = 50 + 75D. The cost of the phone bill per month is represented by C, and D represents the gigabytes of data used that month. What is the value and interpretation of the slope of this equation?

• A. 75 dollars per gigabyte
• B. 75 gigabytes per day
• C. 50 dollars per day
• D. 50 dollars per gigabyte

Correct answer: A

Rationale: The slope of the equation C = 50 + 75D is 75. This means that for each additional gigabyte used (represented by D), the cost (represented by C) increases by 75 dollars. Therefore, the correct interpretation of the slope is that it is 75 dollars per gigabyte. Choice B, 75 gigabytes per day, is incorrect as the slope does not represent the rate of data usage per day. Choice C, 50 dollars per day, is incorrect as it does not reflect the relationship between gigabytes used and the cost. Choice D, 50 dollars per gigabyte, is incorrect as it does not match the slope value of 75 in the equation.

5. A study about anorexia was conducted on 100 patients. 70% were women, and 10% of the men were overweight as children. How many male patients 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, which equals 30 male patients. It is given that 10% of the men were overweight as children, so 10% of 30 is 3, meaning 3 male patients were overweight. Therefore, the remaining 27 male patients were not overweight as children. Choice A, B, and D are incorrect as they do not accurately represent the number of male patients who were not overweight.

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