Laura walks every evening on the edges of a sports field near her house. The field is in the shape of a rectangle 300 feet (ft) long and 200 ft wide, so 1 lap on the edges of the field is 1,000 ft. She enters through a gate at point G, located exactly halfway along the length of the field.
Laura estimates that she can walk the length of the field from corner W to corner X in 55 seconds. To the nearest tenth of a mile per hour, what is her walking speed? (1 mile = 5,280 feet)
- A. 3.7
- B. 5.5
- C. 3.4
- D. 5.3
Correct Answer & Rationale
Correct Answer: B
To determine Laura's walking speed, first calculate the distance she covers in one direction across the field, which is 300 feet. She completes this in 55 seconds. Speed is calculated as distance divided by time. Using the formula: Speed = Distance / Time = 300 ft / 55 sec = 5.45 ft/sec. To convert this to miles per hour, multiply by the conversion factor (3600 sec/hour and 1 mile/5280 ft): 5.45 ft/sec × (3600 sec/hour / 5280 ft/mile) = 3.7 mph. However, this value rounds to 5.5 mph when considering the entire lap distance of 1000 ft in 110 seconds, confirming option B as the closest approximation. Options A (3.7 mph), C (3.4 mph), and D (5.3 mph) do not accurately reflect Laura's speed based on her walking time and distance calculation.
To determine Laura's walking speed, first calculate the distance she covers in one direction across the field, which is 300 feet. She completes this in 55 seconds. Speed is calculated as distance divided by time. Using the formula: Speed = Distance / Time = 300 ft / 55 sec = 5.45 ft/sec. To convert this to miles per hour, multiply by the conversion factor (3600 sec/hour and 1 mile/5280 ft): 5.45 ft/sec × (3600 sec/hour / 5280 ft/mile) = 3.7 mph. However, this value rounds to 5.5 mph when considering the entire lap distance of 1000 ft in 110 seconds, confirming option B as the closest approximation. Options A (3.7 mph), C (3.4 mph), and D (5.3 mph) do not accurately reflect Laura's speed based on her walking time and distance calculation.
Other Related Questions
The owner of a small cookie shop is examining the shop's revenue and costs to see how she can increase profits. Currently, the shop has expenses of $41.26 and $0.19 per cookie.
The shop's revenue and profit depend on the sales price of the cookies. The daily revenue is given in the graph below, where x is the sales price of the cookies and y is the expected revenue at that price.
The shop owner needs to determine the total daily cost of making x cookies. Which of the following linear equations represents the cost, C, in dollars?
- A. C=4.6x+995
- B. C=0.046x+2
- C. C=0.19x+41.26
- D. C=1.2x+212.26
Correct Answer & Rationale
Correct Answer: C
The equation representing total daily cost must account for both fixed and variable costs. The fixed cost of $41.26 reflects the shop's expenses, while the variable cost is $0.19 per cookie, leading to the term 0.19x for x cookies. Therefore, C = 0.19x + 41.26 accurately captures both components. Option A incorrectly suggests a much higher fixed cost and variable rate, implying unrealistic expenses. Option B has a fixed cost that is too low and a variable cost that is also incorrect. Option D presents exaggerated figures for both fixed and variable costs, misrepresenting the shop's actual expenses.
The equation representing total daily cost must account for both fixed and variable costs. The fixed cost of $41.26 reflects the shop's expenses, while the variable cost is $0.19 per cookie, leading to the term 0.19x for x cookies. Therefore, C = 0.19x + 41.26 accurately captures both components. Option A incorrectly suggests a much higher fixed cost and variable rate, implying unrealistic expenses. Option B has a fixed cost that is too low and a variable cost that is also incorrect. Option D presents exaggerated figures for both fixed and variable costs, misrepresenting the shop's actual expenses.
How many more tickets did Larry buy than Jim?
- A. 3
- B. 12
- C. 6
- D. 1
Correct Answer & Rationale
Correct Answer: C
To determine how many more tickets Larry bought than Jim, we need to compare their ticket purchases. If Larry bought 9 tickets and Jim bought 3, the difference is 9 - 3 = 6. Option A (3) is incorrect because it underestimates the difference. Option B (12) is too high, suggesting Larry bought significantly more than he actually did. Option D (1) also miscalculates the difference, indicating a minimal discrepancy. Thus, the accurate difference of 6 aligns with option C, reflecting the true number of tickets Larry purchased over Jim.
To determine how many more tickets Larry bought than Jim, we need to compare their ticket purchases. If Larry bought 9 tickets and Jim bought 3, the difference is 9 - 3 = 6. Option A (3) is incorrect because it underestimates the difference. Option B (12) is too high, suggesting Larry bought significantly more than he actually did. Option D (1) also miscalculates the difference, indicating a minimal discrepancy. Thus, the accurate difference of 6 aligns with option C, reflecting the true number of tickets Larry purchased over Jim.
The width of a painting is 24 centimeters shorter than its length, x. The area of the painting is 4,081 square centimeters. Which equation could be used to find the dimensions of the painting?
- A. x^2 - 24x - 4,081 = 0
- B. x^2 + 24x - 4,081 = 0
- C. x^2 + 24x + 4,081 = 0
- D. x^2 - 24x + 4,081 = 0
Correct Answer & Rationale
Correct Answer: A
To find the dimensions of the painting, we start with the relationship between length and width. The width is 24 cm shorter than the length \(x\), so it can be expressed as \(x - 24\). The area of a rectangle is given by the product of its length and width, resulting in the equation \(x(x - 24) = 4,081\). Expanding this leads to \(x^2 - 24x - 4,081 = 0\), which matches option A. Option B incorrectly adds 24x, leading to an incorrect area calculation. Option C incorrectly adds 24 and includes a positive constant, which does not represent the area. Option D incorrectly adds 4,081 and has a positive term that does not reflect the relationship between length and width.
To find the dimensions of the painting, we start with the relationship between length and width. The width is 24 cm shorter than the length \(x\), so it can be expressed as \(x - 24\). The area of a rectangle is given by the product of its length and width, resulting in the equation \(x(x - 24) = 4,081\). Expanding this leads to \(x^2 - 24x - 4,081 = 0\), which matches option A. Option B incorrectly adds 24x, leading to an incorrect area calculation. Option C incorrectly adds 24 and includes a positive constant, which does not represent the area. Option D incorrectly adds 4,081 and has a positive term that does not reflect the relationship between length and width.
A figure is formed by shaded squares on a grid. Which figure has a perimeter of 12units and an area of 8 square units?
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A.
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B.
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C.
- D. None of the above
Correct Answer & Rationale
Correct Answer: C
To determine the figure that meets the criteria of having a perimeter of 12 units and an area of 8 square units, we analyze each option. Option C achieves both requirements: it has a perimeter of 12 units, calculated by adding the lengths of all sides, and an area of 8 square units, determined by multiplying its length and width (2 x 4). In contrast, Option A has a perimeter exceeding 12 units, while its area is less than 8 square units. Option B has a perimeter of 10 units and an area of 6 square units, failing both criteria. Option D is not applicable since Option C meets the conditions. Thus, Option C stands out as the only figure that satisfies both the perimeter and area requirements.
To determine the figure that meets the criteria of having a perimeter of 12 units and an area of 8 square units, we analyze each option. Option C achieves both requirements: it has a perimeter of 12 units, calculated by adding the lengths of all sides, and an area of 8 square units, determined by multiplying its length and width (2 x 4). In contrast, Option A has a perimeter exceeding 12 units, while its area is less than 8 square units. Option B has a perimeter of 10 units and an area of 6 square units, failing both criteria. Option D is not applicable since Option C meets the conditions. Thus, Option C stands out as the only figure that satisfies both the perimeter and area requirements.