Driveway for two cars, width?
- A. 0.7
- B. 7
- C. 70
- D. 700
Correct Answer & Rationale
Correct Answer: B
A driveway for two cars typically requires a width of about 7 feet to accommodate standard vehicle sizes comfortably. Option A (0.7) is too narrow, as it would not allow even one car to fit. Option C (70) and Option D (700) are excessively wide for a residential driveway, making them impractical and unnecessary. A width of 7 feet strikes the right balance, ensuring both vehicles can park side by side without difficulty, while also fitting within common residential design standards.
A driveway for two cars typically requires a width of about 7 feet to accommodate standard vehicle sizes comfortably. Option A (0.7) is too narrow, as it would not allow even one car to fit. Option C (70) and Option D (700) are excessively wide for a residential driveway, making them impractical and unnecessary. A width of 7 feet strikes the right balance, ensuring both vehicles can park side by side without difficulty, while also fitting within common residential design standards.
Other Related Questions
Which inequality?
- A. 2(x+1)<x
- B. x+2(x+1)>-1
- C. x<2x-1
- D. 2(x/2+1)<1
Correct Answer & Rationale
Correct Answer: C
Option C, \( x < 2x - 1 \), simplifies to \( x - 2x < -1 \), leading to \( -x < -1 \) or \( x > 1 \). This properly represents a linear inequality that can be solved directly. Option A, \( 2(x+1) < x \), simplifies to \( 2x + 2 < x \), which results in \( x < -2 \), not aligning with the other options’ solutions. Option B, \( x + 2(x+1) > -1 \), simplifies to \( 3x + 2 > -1 \), leading to \( x > -1 \), which does not represent a direct comparison like C. Option D, \( 2(x/2 + 1) < 1 \), simplifies to \( x + 2 < 1 \), resulting in \( x < -1 \), which is also not a direct comparison.
Option C, \( x < 2x - 1 \), simplifies to \( x - 2x < -1 \), leading to \( -x < -1 \) or \( x > 1 \). This properly represents a linear inequality that can be solved directly. Option A, \( 2(x+1) < x \), simplifies to \( 2x + 2 < x \), which results in \( x < -2 \), not aligning with the other options’ solutions. Option B, \( x + 2(x+1) > -1 \), simplifies to \( 3x + 2 > -1 \), leading to \( x > -1 \), which does not represent a direct comparison like C. Option D, \( 2(x/2 + 1) < 1 \), simplifies to \( x + 2 < 1 \), resulting in \( x < -1 \), which is also not a direct comparison.
Shaded region shows?
- A. 3/4 x 1/2
- B. 3/4 x 3/4
- C. 3/4 x 3/2
- D. 3/4 x 3
Correct Answer & Rationale
Correct Answer: A
The shaded region represents the area of a rectangle formed by multiplying two fractions. Option A, \( \frac{3}{4} \times \frac{1}{2} \), correctly calculates the area of a rectangle with a length of \( \frac{3}{4} \) and a width of \( \frac{1}{2} \), resulting in \( \frac{3}{8} \). Option B, \( \frac{3}{4} \times \frac{3}{4} \), represents a larger area, \( \frac{9}{16} \), which does not match the shaded region. Option C, \( \frac{3}{4} \times \frac{3}{2} \), yields \( \frac{9}{8} \), exceeding the shaded area. Finally, option D, \( \frac{3}{4} \times 3 \), results in \( \frac{9}{4} \), also too large. Thus, only option A accurately reflects the area of the shaded region.
The shaded region represents the area of a rectangle formed by multiplying two fractions. Option A, \( \frac{3}{4} \times \frac{1}{2} \), correctly calculates the area of a rectangle with a length of \( \frac{3}{4} \) and a width of \( \frac{1}{2} \), resulting in \( \frac{3}{8} \). Option B, \( \frac{3}{4} \times \frac{3}{4} \), represents a larger area, \( \frac{9}{16} \), which does not match the shaded region. Option C, \( \frac{3}{4} \times \frac{3}{2} \), yields \( \frac{9}{8} \), exceeding the shaded area. Finally, option D, \( \frac{3}{4} \times 3 \), results in \( \frac{9}{4} \), also too large. Thus, only option A accurately reflects the area of the shaded region.
Caterpillar 1 ft in 7.5 min. 18 min?
- A. 2.4
- B. 8
- C. 11.5
- D. 25.5
Correct Answer & Rationale
Correct Answer: A
To determine how far the caterpillar travels in 18 minutes, first calculate its speed. It moves 1 foot in 7.5 minutes, which equates to \( \frac{1 \text{ ft}}{7.5 \text{ min}} \). In 18 minutes, the distance covered can be calculated using the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] Converting 18 minutes into feet: \[ \text{Distance} = \left(\frac{1 \text{ ft}}{7.5 \text{ min}}\right) \times 18 \text{ min} = 2.4 \text{ ft} \] Option B (8) overestimates the distance, while C (11.5) and D (25.5) significantly exceed the calculated distance, demonstrating a misunderstanding of the speed-time relationship.
To determine how far the caterpillar travels in 18 minutes, first calculate its speed. It moves 1 foot in 7.5 minutes, which equates to \( \frac{1 \text{ ft}}{7.5 \text{ min}} \). In 18 minutes, the distance covered can be calculated using the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] Converting 18 minutes into feet: \[ \text{Distance} = \left(\frac{1 \text{ ft}}{7.5 \text{ min}}\right) \times 18 \text{ min} = 2.4 \text{ ft} \] Option B (8) overestimates the distance, while C (11.5) and D (25.5) significantly exceed the calculated distance, demonstrating a misunderstanding of the speed-time relationship.
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.