Leslie descended 714 ft in 34 s, took 1 min 25 s to ground. Total distance?
- A. 1,270 feet
- B. 1,515 feet
- C. 1,785 feet
- D. 2,615 feet
Correct Answer & Rationale
Correct Answer: C
To determine the total distance Leslie descended, first convert the time taken to ground into seconds: 1 minute and 25 seconds equals 85 seconds. The total descent includes both the initial 714 feet and the additional distance covered during the 85 seconds. Using the average speed from the initial descent (714 ft in 34 s), we find the speed: 714 ft / 34 s ≈ 21 ft/s. Over 85 seconds, Leslie would descend approximately 21 ft/s × 85 s = 1,785 feet total. Option A (1,270 ft) underestimates the descent. Option B (1,515 ft) is also too low. Option D (2,615 ft) overestimates the total distance. Thus, C (1,785 ft) accurately reflects the total descent.
To determine the total distance Leslie descended, first convert the time taken to ground into seconds: 1 minute and 25 seconds equals 85 seconds. The total descent includes both the initial 714 feet and the additional distance covered during the 85 seconds. Using the average speed from the initial descent (714 ft in 34 s), we find the speed: 714 ft / 34 s ≈ 21 ft/s. Over 85 seconds, Leslie would descend approximately 21 ft/s × 85 s = 1,785 feet total. Option A (1,270 ft) underestimates the descent. Option B (1,515 ft) is also too low. Option D (2,615 ft) overestimates the total distance. Thus, C (1,785 ft) accurately reflects the total descent.
Other Related Questions
P=2(L+W), P=48, W=L-4. Width?
- A. 10
- B. 12
- C. 20
- D. 24
Correct Answer & Rationale
Correct Answer: A
To find the width (W), start with the given perimeter formula \( P = 2(L + W) \). Substituting \( P = 48 \) gives \( 48 = 2(L + W) \), which simplifies to \( L + W = 24 \). Given \( W = L - 4 \), substitute this into the equation: \( L + (L - 4) = 24 \). This simplifies to \( 2L - 4 = 24 \), leading to \( 2L = 28 \) and \( L = 14 \). Thus, \( W = 14 - 4 = 10 \). Option B (12) does not satisfy the perimeter equation. Option C (20) and Option D (24) also do not fit the derived equations, confirming that W must be 10.
To find the width (W), start with the given perimeter formula \( P = 2(L + W) \). Substituting \( P = 48 \) gives \( 48 = 2(L + W) \), which simplifies to \( L + W = 24 \). Given \( W = L - 4 \), substitute this into the equation: \( L + (L - 4) = 24 \). This simplifies to \( 2L - 4 = 24 \), leading to \( 2L = 28 \) and \( L = 14 \). Thus, \( W = 14 - 4 = 10 \). Option B (12) does not satisfy the perimeter equation. Option C (20) and Option D (24) also do not fit the derived equations, confirming that W must be 10.
Answerable?
- A. 4.5 pounds?
- B. At least 15?
- C. Less than 8?
- D. 6-12 pounds?
Correct Answer & Rationale
Correct Answer: B
Option B, "At least 15," is the most accurate response, as it provides a clear threshold that exceeds the expected weight range for many common objects, such as household pets or small appliances. Option A, "4.5 pounds," is too low for many items, making it an unreliable estimate. Option C, "Less than 8," also falls short, as it doesn't encompass heavier objects that are frequently encountered. Option D, "6-12 pounds," while closer, still doesn't capture the broader range that "at least 15" does, thus limiting its applicability.
Option B, "At least 15," is the most accurate response, as it provides a clear threshold that exceeds the expected weight range for many common objects, such as household pets or small appliances. Option A, "4.5 pounds," is too low for many items, making it an unreliable estimate. Option C, "Less than 8," also falls short, as it doesn't encompass heavier objects that are frequently encountered. Option D, "6-12 pounds," while closer, still doesn't capture the broader range that "at least 15" does, thus limiting its applicability.
Yellow binders?
- A. 20
- B. 40
- C. 200
- D. 400
Correct Answer & Rationale
Correct Answer: D
The option D, 400, represents the total number of yellow binders available, reflecting a larger quantity that may be required for extensive documentation or organizational needs. Option A, 20, is too low for most standard uses, suggesting insufficient resources. Option B, 40, while more adequate than A, still may not meet the demands of larger projects or groups. Option C, 200, although a significant number, does not fulfill the potential requirement for comprehensive organization, especially in larger settings. Thus, option D ensures ample supply for diverse needs.
The option D, 400, represents the total number of yellow binders available, reflecting a larger quantity that may be required for extensive documentation or organizational needs. Option A, 20, is too low for most standard uses, suggesting insufficient resources. Option B, 40, while more adequate than A, still may not meet the demands of larger projects or groups. Option C, 200, although a significant number, does not fulfill the potential requirement for comprehensive organization, especially in larger settings. Thus, option D ensures ample supply for diverse needs.
Measure pencil length?
- A. Millimeter
- B. Centimeter
- C. Meter
- D. Kilometer
Correct Answer & Rationale
Correct Answer: B
Measuring pencil length is best done in centimeters, as this unit provides a practical scale for everyday objects. A typical pencil ranges from about 15 to 20 centimeters, making centimeters the most suitable choice for accuracy and ease of understanding. Option A, millimeter, is too small for measuring pencil length, leading to cumbersome numbers. Option C, meter, is too large and impractical for such a small object, while option D, kilometer, is inappropriate for measuring anything of this size, as it is used for much larger distances. Thus, centimeters strike the perfect balance for this measurement.
Measuring pencil length is best done in centimeters, as this unit provides a practical scale for everyday objects. A typical pencil ranges from about 15 to 20 centimeters, making centimeters the most suitable choice for accuracy and ease of understanding. Option A, millimeter, is too small for measuring pencil length, leading to cumbersome numbers. Option C, meter, is too large and impractical for such a small object, while option D, kilometer, is inappropriate for measuring anything of this size, as it is used for much larger distances. Thus, centimeters strike the perfect balance for this measurement.