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
A scientist uses the expression 5/9(F - 32) to convert temperatures from degrees Fahrenheit (°F), F, to degrees Celsius (°C). To the nearest degree, what is the temperature, in °F, of a substance at -25°C?
- A. 13
- B. -32
- C. -13
- D. 18
Correct Answer & Rationale
Correct Answer: C
To find the Fahrenheit equivalent of -25°C, use the formula \( F = \frac{9}{5}C + 32 \). Substituting -25 for C gives \( F = \frac{9}{5}(-25) + 32 = -45 + 32 = -13 \). Thus, the temperature in Fahrenheit is -13°F. Option A (13°F) is incorrect as it does not reflect the negative temperature conversion. Option B (-32°F) is too low and does not correspond to the calculated value. Option D (18°F) is also incorrect as it is significantly higher than the expected result for -25°C.
To find the Fahrenheit equivalent of -25°C, use the formula \( F = \frac{9}{5}C + 32 \). Substituting -25 for C gives \( F = \frac{9}{5}(-25) + 32 = -45 + 32 = -13 \). Thus, the temperature in Fahrenheit is -13°F. Option A (13°F) is incorrect as it does not reflect the negative temperature conversion. Option B (-32°F) is too low and does not correspond to the calculated value. Option D (18°F) is also incorrect as it is significantly higher than the expected result for -25°C.
A scale drawing of a truck has a length of 3 inches (in.), as shown below. The actual truck has a length of 18 feet (ft). What scale was used for the drawing?
- A. 6 in. = 1 ft
- B. 1 in. = 15 ft
- C. 1 in. = 6 ft
- D. 15 in. = 1 ft
Correct Answer & Rationale
Correct Answer: C
To determine the scale used for the drawing, we first convert the actual truck length from feet to inches. Since 1 foot equals 12 inches, an 18-foot truck is 216 inches long (18 ft x 12 in/ft). The scale drawing shows a length of 3 inches. To find the scale, we set up the ratio of the drawing length to the actual length: 3 in. (drawing) to 216 in. (actual). Simplifying this gives us a scale of 1 in. = 72 in., which translates to 1 in. = 6 ft (since 72 in. ÷ 12 in/ft = 6 ft). Option A (6 in. = 1 ft) is incorrect; it implies a much larger drawing. Option B (1 in. = 15 ft) underestimates the actual size. Option D (15 in. = 1 ft) greatly exaggerates the scale, making the drawing too small.
To determine the scale used for the drawing, we first convert the actual truck length from feet to inches. Since 1 foot equals 12 inches, an 18-foot truck is 216 inches long (18 ft x 12 in/ft). The scale drawing shows a length of 3 inches. To find the scale, we set up the ratio of the drawing length to the actual length: 3 in. (drawing) to 216 in. (actual). Simplifying this gives us a scale of 1 in. = 72 in., which translates to 1 in. = 6 ft (since 72 in. ÷ 12 in/ft = 6 ft). Option A (6 in. = 1 ft) is incorrect; it implies a much larger drawing. Option B (1 in. = 15 ft) underestimates the actual size. Option D (15 in. = 1 ft) greatly exaggerates the scale, making the drawing too small.
The U.S. Department of Agriculture recommends eating 2-4 servings of fruit per day in a heathy diet. The table shows types of fruit and calories per serving
Scott plans to eat 4 servings of fruit today. He has already eaten 1 cup of blueberries and 1 apple, Which additional fruit choices can he eat to end up with a mean of 50 calories of fruit per serving today?
- A. 1 plum and 1 tangerine
- B. 1 banana and 1 mandarin orange
- C. 1 cup of blueberries and 1 banana
- D. 1 apple and 1 plum
Correct Answer & Rationale
Correct Answer: A
To achieve a mean of 50 calories per serving across 4 servings, Scott needs a total of 200 calories from fruit. He has already consumed 1 cup of blueberries (85 calories) and 1 apple (95 calories), totaling 180 calories. This leaves him needing an additional 20 calories from 2 servings. Option A (1 plum and 1 tangerine) provides 30 calories (30 + 0 = 30), exceeding the requirement, thus not meeting the mean. Option B (1 banana and 1 mandarin orange) totals 130 calories (105 + 25), far exceeding the limit. Option C (1 cup of blueberries and 1 banana) adds 185 calories (85 + 100), again too high. Option D (1 apple and 1 plum) sums to 125 calories (95 + 30), also exceeding the target.
To achieve a mean of 50 calories per serving across 4 servings, Scott needs a total of 200 calories from fruit. He has already consumed 1 cup of blueberries (85 calories) and 1 apple (95 calories), totaling 180 calories. This leaves him needing an additional 20 calories from 2 servings. Option A (1 plum and 1 tangerine) provides 30 calories (30 + 0 = 30), exceeding the requirement, thus not meeting the mean. Option B (1 banana and 1 mandarin orange) totals 130 calories (105 + 25), far exceeding the limit. Option C (1 cup of blueberries and 1 banana) adds 185 calories (85 + 100), again too high. Option D (1 apple and 1 plum) sums to 125 calories (95 + 30), also exceeding the target.
The number line below shows the solution set of an inequality: Which two inequalities represent the graph shown?
- A. -2x>4 and 4x<-8
- B. 3x>-6 and x-4>6
- C. 4x<-8 and x≥6
- D. 4x<-8 and x≥6
Correct Answer & Rationale
Correct Answer: B
The graph indicates a solution set that includes values greater than -2 and less than 6. Option B, with inequalities 3x > -6 and x - 4 > 6, accurately reflects this range. The first inequality simplifies to x > -2, aligning with the left boundary, while the second simplifies to x > 10, which is outside the range but indicates a direction. Options A, C, and D contain inequalities that do not match the solution set shown on the number line. A suggests values that are too extreme, while C and D incorrectly imply lower bounds that do not correspond to the graph's representation.
The graph indicates a solution set that includes values greater than -2 and less than 6. Option B, with inequalities 3x > -6 and x - 4 > 6, accurately reflects this range. The first inequality simplifies to x > -2, aligning with the left boundary, while the second simplifies to x > 10, which is outside the range but indicates a direction. Options A, C, and D contain inequalities that do not match the solution set shown on the number line. A suggests values that are too extreme, while C and D incorrectly imply lower bounds that do not correspond to the graph's representation.