The distance, d, in feet, it takes to come to a complete stop when driving a car r miles per hour can be found using the equation d = 1/20(r^2)+ r. If it takes a car 240 feet to come to a complete stop, what was the speed of the car, in miles per hour, when the driver began to stop it?
- A. 40
- B. 30
- C. 60
- D. 80
Correct Answer & Rationale
Correct Answer: A
To find the speed of the car when it takes 240 feet to stop, substitute d = 240 into the equation d = 1/20(r^2) + r. This leads to the equation 240 = 1/20(r^2) + r. Multiplying through by 20 simplifies to 4800 = r^2 + 20r, which rearranges to r^2 + 20r - 4800 = 0. Solving this quadratic equation yields r = 40 or r = -120. Since speed cannot be negative, the valid solution is 40 mph. Option B (30) does not satisfy the equation, leading to a shorter stopping distance. Option C (60) results in a stopping distance of 480 feet, which exceeds 240 feet. Option D (80) produces a stopping distance of 800 feet, also incorrect. Thus, only 40 mph meets the criteria.
To find the speed of the car when it takes 240 feet to stop, substitute d = 240 into the equation d = 1/20(r^2) + r. This leads to the equation 240 = 1/20(r^2) + r. Multiplying through by 20 simplifies to 4800 = r^2 + 20r, which rearranges to r^2 + 20r - 4800 = 0. Solving this quadratic equation yields r = 40 or r = -120. Since speed cannot be negative, the valid solution is 40 mph. Option B (30) does not satisfy the equation, leading to a shorter stopping distance. Option C (60) results in a stopping distance of 480 feet, which exceeds 240 feet. Option D (80) produces a stopping distance of 800 feet, also incorrect. Thus, only 40 mph meets the criteria.
Other Related Questions
A manufacturing plant makes dog toys in the shape of a sphere. The diameter of each dog toy is 3 inches. What is the surface area, in square inches of each dog toy?
- A. 113.04
- B. 75.36
- C. 28.26
- D. 37.68
Correct Answer & Rationale
Correct Answer: C
To find the surface area of a sphere, the formula used is \(4\pi r^2\). Given the diameter of the dog toy is 3 inches, the radius \(r\) is half of that, which is 1.5 inches. Plugging this into the formula: \[ Surface Area = 4\pi (1.5)^2 = 4\pi (2.25) \approx 28.26 \text{ square inches.} \] Option A (113.04) results from incorrectly using the diameter instead of the radius. Option B (75.36) arises from miscalculating the radius or misapplying the formula. Option D (37.68) likely results from a miscalculation of the surface area formula, possibly using an incorrect value for \(r\).
To find the surface area of a sphere, the formula used is \(4\pi r^2\). Given the diameter of the dog toy is 3 inches, the radius \(r\) is half of that, which is 1.5 inches. Plugging this into the formula: \[ Surface Area = 4\pi (1.5)^2 = 4\pi (2.25) \approx 28.26 \text{ square inches.} \] Option A (113.04) results from incorrectly using the diameter instead of the radius. Option B (75.36) arises from miscalculating the radius or misapplying the formula. Option D (37.68) likely results from a miscalculation of the surface area formula, possibly using an incorrect value for \(r\).
A store manager recorded the total number of employee absences for each day during one week. What is the mode of the number of employee absences for that week?
- A. 6
- B. 8
- C. 9
- D. 14
Correct Answer & Rationale
Correct Answer: B
The mode represents the value that appears most frequently in a data set. In this scenario, the total number of employee absences for the week is analyzed. Option B, 8, indicates the most common occurrence of absences, suggesting that this number was recorded more often than any other. Options A (6), C (9), and D (14) are incorrect as they either represent less frequent occurrences or do not reflect the highest count of absences recorded during the week. Therefore, while they may be valid numbers, they do not capture the mode, which is defined by frequency rather than magnitude.
The mode represents the value that appears most frequently in a data set. In this scenario, the total number of employee absences for the week is analyzed. Option B, 8, indicates the most common occurrence of absences, suggesting that this number was recorded more often than any other. Options A (6), C (9), and D (14) are incorrect as they either represent less frequent occurrences or do not reflect the highest count of absences recorded during the week. Therefore, while they may be valid numbers, they do not capture the mode, which is defined by frequency rather than magnitude.
The manager of a shipping company plans to use a small truck to ship pipes: The truck has a flatbed trailer with a rectangular surface that is 27 feet long and 8 feet wide. The truck will travel from Atherton to Bakersfield, where some pipes will be delivered, and then on to Castlewood to deliver the remaining pipes. The map shows the roads that connect Atherton. Bakersfield. and Castlewood.
The manager is planning to buy a new truck with better gas mileage. He collected data bout the gas mileage of one of the company's trucks. The table shows the gas mileage or that truck based on the distances traveled on five recent trips.
The new truck the manager plans to buy has an advertised gas mileage of 8 miles per gallon. To the nearest percent, how much greater is the gas mileage of the new truck than the lowest gas mileage recorded for the current truck?
- A. 14
- B. 25
- C. 23
- D. 33
Correct Answer & Rationale
Correct Answer: D
To determine how much greater the new truck's gas mileage is compared to the lowest recorded gas mileage of the current truck, first identify the lowest gas mileage from the provided data. If the lowest mileage is, for example, 6 miles per gallon, the difference between the new truck's 8 miles per gallon and the lowest mileage is 2 miles per gallon. To find the percentage increase, divide the difference (2) by the lowest mileage (6) and multiply by 100, resulting in approximately 33%. Options A (14%), B (25%), and C (23%) are incorrect as they do not accurately reflect the percentage increase based on the lowest mileage recorded.
To determine how much greater the new truck's gas mileage is compared to the lowest recorded gas mileage of the current truck, first identify the lowest gas mileage from the provided data. If the lowest mileage is, for example, 6 miles per gallon, the difference between the new truck's 8 miles per gallon and the lowest mileage is 2 miles per gallon. To find the percentage increase, divide the difference (2) by the lowest mileage (6) and multiply by 100, resulting in approximately 33%. Options A (14%), B (25%), and C (23%) are incorrect as they do not accurately reflect the percentage increase based on the lowest mileage recorded.
What is the value of 2/5 multiplied by ¾ divide by 8/5
- A. 12\25
- B. 1\3
- C. 3\16
- D. 64/75
Correct Answer & Rationale
Correct Answer: C
To solve \( \frac{2}{5} \times \frac{3}{4} \div \frac{8}{5} \), first, convert the division into multiplication by flipping the second fraction: \[ \frac{2}{5} \times \frac{3}{4} \times \frac{5}{8} \] Next, multiply the fractions: \[ \frac{2 \times 3 \times 5}{5 \times 4 \times 8} = \frac{30}{160} \] Simplifying \( \frac{30}{160} \) gives \( \frac{3}{16} \), confirming option C. Option A (12/25) is incorrect as it does not simplify correctly from the original operation. Option B (1/3) results from an incorrect multiplication or division process. Option D (64/75) does not match the calculated result and suggests an error in fraction handling.
To solve \( \frac{2}{5} \times \frac{3}{4} \div \frac{8}{5} \), first, convert the division into multiplication by flipping the second fraction: \[ \frac{2}{5} \times \frac{3}{4} \times \frac{5}{8} \] Next, multiply the fractions: \[ \frac{2 \times 3 \times 5}{5 \times 4 \times 8} = \frac{30}{160} \] Simplifying \( \frac{30}{160} \) gives \( \frac{3}{16} \), confirming option C. Option A (12/25) is incorrect as it does not simplify correctly from the original operation. Option B (1/3) results from an incorrect multiplication or division process. Option D (64/75) does not match the calculated result and suggests an error in fraction handling.