50 acres, 23 apple. Percent left?
- A. 27%
- B. 46%
- C. 54%
- D. 77%
Correct Answer & Rationale
Correct Answer: C
To determine the percentage of land left after allocating 23 acres for apple trees from a total of 50 acres, first calculate the remaining land: 50 - 23 = 27 acres. Then, to find the percentage of land left, divide the remaining acres by the total acres and multiply by 100: (27/50) * 100 = 54%. Option A (27%) miscalculates the percentage of land used instead of what remains. Option B (46%) incorrectly assumes a different allocation of land. Option D (77%) mistakenly represents a higher percentage than what is left. Thus, option C accurately reflects the remaining percentage of land.
To determine the percentage of land left after allocating 23 acres for apple trees from a total of 50 acres, first calculate the remaining land: 50 - 23 = 27 acres. Then, to find the percentage of land left, divide the remaining acres by the total acres and multiply by 100: (27/50) * 100 = 54%. Option A (27%) miscalculates the percentage of land used instead of what remains. Option B (46%) incorrectly assumes a different allocation of land. Option D (77%) mistakenly represents a higher percentage than what is left. Thus, option C accurately reflects the remaining percentage of land.
Other Related Questions
Graph for data over time?
- A. Bar
- B. Line
- C. Stem-and-leaf
- D. Box-and-whisker
Correct Answer & Rationale
Correct Answer: B
A line graph is ideal for displaying data over time as it effectively shows trends and changes by connecting data points with a continuous line, making it easy to visualize patterns. Option A, a bar graph, is better suited for comparing discrete categories rather than illustrating changes over time. Option C, a stem-and-leaf plot, is primarily used for displaying the distribution of numerical data and is not designed for time-series analysis. Option D, a box-and-whisker plot, summarizes data distribution and highlights outliers but does not convey trends over time effectively.
A line graph is ideal for displaying data over time as it effectively shows trends and changes by connecting data points with a continuous line, making it easy to visualize patterns. Option A, a bar graph, is better suited for comparing discrete categories rather than illustrating changes over time. Option C, a stem-and-leaf plot, is primarily used for displaying the distribution of numerical data and is not designed for time-series analysis. Option D, a box-and-whisker plot, summarizes data distribution and highlights outliers but does not convey trends over time effectively.
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.
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.
Favorite food via survey numbers. Best measure?
- A. Mean
- B. Median
- C. Mode
- D. Mean+median
Correct Answer & Rationale
Correct Answer: C
When analyzing survey data on favorite foods, the mode is the best measure since it identifies the most frequently chosen option, reflecting the popular preference among respondents. The mean can be skewed by outliers, making it less reliable in this context. The median, while useful for understanding the middle value, does not capture the most popular choice effectively. Combining mean and median (option D) does not address the core goal of identifying the favorite food, which is best represented by the mode. Thus, the mode provides a clear insight into the most favored food item.
When analyzing survey data on favorite foods, the mode is the best measure since it identifies the most frequently chosen option, reflecting the popular preference among respondents. The mean can be skewed by outliers, making it less reliable in this context. The median, while useful for understanding the middle value, does not capture the most popular choice effectively. Combining mean and median (option D) does not address the core goal of identifying the favorite food, which is best represented by the mode. Thus, the mode provides a clear insight into the most favored food item.