The recommended dosage of a medicine is 4 milligrams per kilogram of body weight. What is the recommended dosage, in milligrams, for a person who weighs 84 kilograms?
- A. 21
- B. 88
- C. 324
- D. 336
- E. 2100
Correct Answer & Rationale
Correct Answer: D
To determine the recommended dosage for a person weighing 84 kilograms, multiply their weight by the dosage per kilogram: 4 mg/kg × 84 kg = 336 mg. Option A (21 mg) is incorrect as it significantly underestimates the dosage based on the weight. Option B (88 mg) also miscalculates the dosage, failing to apply the correct multiplication. Option C (324 mg) is close but still incorrect, as it does not reflect the accurate calculation. Option E (2100 mg) is far too high, indicating a misunderstanding of the dosage per kilogram. Thus, 336 mg is the correct dosage for the individual.
To determine the recommended dosage for a person weighing 84 kilograms, multiply their weight by the dosage per kilogram: 4 mg/kg × 84 kg = 336 mg. Option A (21 mg) is incorrect as it significantly underestimates the dosage based on the weight. Option B (88 mg) also miscalculates the dosage, failing to apply the correct multiplication. Option C (324 mg) is close but still incorrect, as it does not reflect the accurate calculation. Option E (2100 mg) is far too high, indicating a misunderstanding of the dosage per kilogram. Thus, 336 mg is the correct dosage for the individual.
Other Related Questions
Square PQRS, with a side length of 5 units, will be translated 2 units to the right and 2 units up in the standard (x, y) coordinate plane. What is the area, in square units, of the image of PQRS?
- A. 20
- B. 25
- C. 40
- D. 50
- E. 100
Correct Answer & Rationale
Correct Answer: B
The area of a square is calculated by squaring the length of its sides. For square PQRS, with a side length of 5 units, the area is \(5 \times 5 = 25\) square units. Translating the square 2 units to the right and 2 units up does not alter its dimensions or area; it simply changes its position on the coordinate plane. Options A (20), C (40), D (50), and E (100) suggest changes in area due to incorrect assumptions about the effects of translation or miscalculations. The area remains constant at 25 square units, confirming option B as the only accurate choice.
The area of a square is calculated by squaring the length of its sides. For square PQRS, with a side length of 5 units, the area is \(5 \times 5 = 25\) square units. Translating the square 2 units to the right and 2 units up does not alter its dimensions or area; it simply changes its position on the coordinate plane. Options A (20), C (40), D (50), and E (100) suggest changes in area due to incorrect assumptions about the effects of translation or miscalculations. The area remains constant at 25 square units, confirming option B as the only accurate choice.
The following is a list of triangles: I. Right triangles, II. Isosceles triangles, III. Equilateral triangles. A pair of triangles from which of these groups must be similar to each other?
- A. I only
- B. II only
- C. III only
- D. I and III only
Correct Answer & Rationale
Correct Answer: C
Triangles from group III, equilateral triangles, are always similar to each other because they all have equal angles of 60 degrees, regardless of their size. Group I, right triangles, can vary significantly in angle measures beyond the right angle, so not all right triangles are similar. Similarly, group II, isosceles triangles, can have different base angles, leading to non-similar triangles. Thus, while right and isosceles triangles can share properties, only equilateral triangles guarantee similarity across the group. Therefore, option C accurately identifies the group with universally similar triangles.
Triangles from group III, equilateral triangles, are always similar to each other because they all have equal angles of 60 degrees, regardless of their size. Group I, right triangles, can vary significantly in angle measures beyond the right angle, so not all right triangles are similar. Similarly, group II, isosceles triangles, can have different base angles, leading to non-similar triangles. Thus, while right and isosceles triangles can share properties, only equilateral triangles guarantee similarity across the group. Therefore, option C accurately identifies the group with universally similar triangles.
Jasmine’s pace for a 3-mile race is 1 minute per mile faster than her pace for a 13-mile race. She ran the 3-mile race in 21 minutes. How many minutes will it take her to run the 13-mile race?
- A. 34
- B. 78
- C. 92
- D. 101
- E. 104
Correct Answer & Rationale
Correct Answer: E
Jasmine completed the 3-mile race in 21 minutes, which gives her a pace of 7 minutes per mile (21 minutes ÷ 3 miles). Since her pace for the 13-mile race is 1 minute slower, her pace for that race is 8 minutes per mile. To find the time for the 13-mile race, multiply her 13-mile pace by the distance: 8 minutes/mile × 13 miles = 104 minutes. Options A (34), B (78), C (92), and D (101) all reflect incorrect calculations or misunderstandings of her pacing difference and distance, leading to values that do not align with the established pace of 8 minutes per mile.
Jasmine completed the 3-mile race in 21 minutes, which gives her a pace of 7 minutes per mile (21 minutes ÷ 3 miles). Since her pace for the 13-mile race is 1 minute slower, her pace for that race is 8 minutes per mile. To find the time for the 13-mile race, multiply her 13-mile pace by the distance: 8 minutes/mile × 13 miles = 104 minutes. Options A (34), B (78), C (92), and D (101) all reflect incorrect calculations or misunderstandings of her pacing difference and distance, leading to values that do not align with the established pace of 8 minutes per mile.
Which of the following expressions is equivalent to: 6x³ + 7x² + 1/x?
- A. 63 + 72 + 1/x
- B. 63 + 72 + 1
- C. 6x² + 7x + 1/x
- D. 6x² + 7x + 1
- E. 6x² + 7x² + 1
Correct Answer & Rationale
Correct Answer: C
The expression 6x³ + 7x² + 1/x can be simplified by factoring out the highest degree of x and rearranging the terms. Option C, 6x² + 7x + 1/x, contains the correct coefficients for the x terms, but with the degrees adjusted appropriately. Option A incorrectly suggests a constant sum of 63 and 72, which does not relate to the original expression. Option B also misrepresents the original expression by omitting the variable terms entirely. Option D fails to maintain the degree of x in the cubic term, while option E mistakenly combines the x² terms incorrectly, resulting in an inaccurate expression.
The expression 6x³ + 7x² + 1/x can be simplified by factoring out the highest degree of x and rearranging the terms. Option C, 6x² + 7x + 1/x, contains the correct coefficients for the x terms, but with the degrees adjusted appropriately. Option A incorrectly suggests a constant sum of 63 and 72, which does not relate to the original expression. Option B also misrepresents the original expression by omitting the variable terms entirely. Option D fails to maintain the degree of x in the cubic term, while option E mistakenly combines the x² terms incorrectly, resulting in an inaccurate expression.