Nursing Elites

1. At the beginning of the day, Xavier has 20 apples. At lunch, he meets his sister Emma and gives her half of his apples. After lunch, he stops by his neighbor Jim's house and gives him 6 of his apples. He then uses 3/4 of his remaining apples to make an apple pie for dessert at dinner. At the end of the day, how many apples does Xavier have left?

• A. 4
• B. 6
• C. 2
• D. 1

Correct answer: D

Rationale: Xavier gives away half of his 20 apples (10), then gives 6 more apples, leaving him with 4 apples. He uses 3/4 of the remaining 4 apples (3) for the pie, leaving him with 1 apple at the end of the day. Therefore, the correct answer is 1. Choices A, B, and C are incorrect because they do not accurately reflect the calculations of apples given away and used for the pie, resulting in the remaining amount of 1 apple.

2. How can you visually differentiate between a histogram and a bar graph?

• A. A bar graph has gaps between the bars; a histogram does not
• B. A bar graph displays frequency; a histogram does not
• C. A histogram illustrates comparison; a bar graph does not
• D. A bar graph includes labels; a histogram does not

Correct answer: A

Rationale: The key difference between a histogram and a bar graph is that a bar graph has gaps between the bars, while a histogram does not. This feature helps in visually distinguishing between the two. Choice B is incorrect because both types of graphs can show frequency. Choice C is incorrect as both graphs can be used for comparison. Choice D is incorrect as both types of graphs can have labels for better understanding.

3. Given a double bar graph, which statement is true about the distributions of Group A and Group B?

• A. Group A is negatively skewed, Group B is normal.
• B. Group A is positively skewed, Group B is normal.
• C. Group A is positively skewed, Group B is neutral.
• D. Group A is normal, Group B is negatively skewed.

Correct answer: B

Rationale: The correct answer is B. In statistical terms, a positively skewed distribution means that the tail on the right side of the distribution is longer or fatter than the left side, indicating more high values. Therefore, Group A is positively skewed. Conversely, an approximately normal distribution, also known as a bell curve, is symmetrical with no skewness. Hence, Group B is normal. Choices A, C, and D are incorrect because they do not accurately describe the skewness of Group A and the normal distribution of Group B as depicted in a double bar graph.

4. The owner of a newspaper has noticed that print subscriptions have gone down 40% while online subscriptions have gone up 60%. Print subscriptions once accounted for 70% of the newspaper’s business, and online subscriptions accounted for 25%. What is the overall percentage growth or decline in business?

• A. 13% decline
• B. 15% decline
• C. 28% growth
• D. 8% growth

Correct answer: A

Rationale: To calculate the decline in business, start with the 40% decline in the 70% share of print subscriptions: 40% of 70% = 0.40 × 0.70 = 0.28 = 28% decline. Next, calculate the growth from the 60% increase in the 25% share of online subscriptions: 60% of 25% = 0.60 × 0.25 = 0.15 = 15% growth. To find the overall change, sum the decline and growth percentages: 28% decline + 15% growth = -0.28 + 0.15 = -0.13 = 13% decline. Therefore, the overall percentage change in the newspaper's business is a 13% decline. Option A is the correct answer. Option B is incorrect because it doesn't consider the correct calculations for both the decline and growth. Option C is incorrect as it misinterprets the net change in business. Option D is incorrect as it miscalculates the overall percentage growth or decline.

5. Dr. Lee observed that 30% of all his patients developed an infection after taking a certain antibiotic. He further noticed that 5% of those 30% required hospitalization to recover from the infection. What percentage of Dr. Lee's patients were hospitalized after taking the antibiotic?

• A. 1.50%
• B. 5%
• C. 15%
• D. 30%

Correct answer: C

Rationale: Out of all the patients who took the antibiotic, 30% developed an infection. Among those with infections, 5% required hospitalization. To find the percentage of all patients hospitalized, we multiply the two percentages: 30% * 5% = 1.5%. Therefore, 1.5% of all patients were hospitalized. Choice A (1.50%) is the calculated percentage of all patients hospitalized, not 1.50%. Choice B (5%) is the percentage of patients who developed an infection and required hospitalization, not all patients. Choice D (30%) represents the initial percentage of patients who developed an infection, not the percentage hospitalized.

Similar Questions