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.
Other Related Questions
d=rt, triple d, same t, new rate?
- A. 3dt
- B. (3d)/t
- C. t/(3d)
- D. d/(3t)
Correct Answer & Rationale
Correct Answer: B
In the equation d = rt, if distance (d) is tripled while time (t) remains constant, the new distance becomes 3d. To find the new rate (r'), we can rearrange the formula to r' = d/t. Substituting the new distance gives r' = (3d)/t, which is option B. Option A (3dt) incorrectly suggests multiplying distance by time, which does not represent rate. Option C (t/(3d)) misplaces the variables, implying time is divided by distance, which does not align with the rate formula. Option D (d/(3t)) incorrectly divides distance by three times the time, again misrepresenting the relationship between distance, rate, and time.
In the equation d = rt, if distance (d) is tripled while time (t) remains constant, the new distance becomes 3d. To find the new rate (r'), we can rearrange the formula to r' = d/t. Substituting the new distance gives r' = (3d)/t, which is option B. Option A (3dt) incorrectly suggests multiplying distance by time, which does not represent rate. Option C (t/(3d)) misplaces the variables, implying time is divided by distance, which does not align with the rate formula. Option D (d/(3t)) incorrectly divides distance by three times the time, again misrepresenting the relationship between distance, rate, and time.
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.
Eraser 20g in mg?
- A. 1.002
- B. 0.02
- C. 2,000
- D. 20
Correct Answer & Rationale
Correct Answer: D
To convert grams to milligrams, one must remember that 1 gram equals 1,000 milligrams. Therefore, 20 grams can be calculated as follows: 20 g x 1,000 mg/g = 20,000 mg. Option A (1.002 mg) is incorrect as it significantly underestimates the conversion. Option B (0.02 mg) is also wrong; it suggests a conversion error by not accounting for the unit scale correctly. Option C (2,000 mg) miscalculates the conversion by a factor of ten. Option D correctly represents 20 grams as 20,000 milligrams, aligning with the proper conversion calculation.
To convert grams to milligrams, one must remember that 1 gram equals 1,000 milligrams. Therefore, 20 grams can be calculated as follows: 20 g x 1,000 mg/g = 20,000 mg. Option A (1.002 mg) is incorrect as it significantly underestimates the conversion. Option B (0.02 mg) is also wrong; it suggests a conversion error by not accounting for the unit scale correctly. Option C (2,000 mg) miscalculates the conversion by a factor of ten. Option D correctly represents 20 grams as 20,000 milligrams, aligning with the proper conversion calculation.
3/4 as sum of unit fractions?
- A. 1/8 + 1/8 + 1/8 + 1/4 + 1/4
- B. 2/8 + 1/4 + 4/16
- C. 5/8 + 2/16
- D. 1/2 + 1/4
Correct Answer & Rationale
Correct Answer: D
To express \( \frac{3}{4} \) as a sum of unit fractions, each option must be evaluated for its total. Option A totals \( \frac{3}{8} + \frac{1}{2} = \frac{3}{8} + \frac{4}{8} = \frac{7}{8} \), which exceeds \( \frac{3}{4} \). Option B simplifies to \( \frac{2}{8} + \frac{2}{8} + \frac{1}{4} = \frac{2}{8} + \frac{2}{8} + \frac{2}{8} = \frac{6}{8} = \frac{3}{4} \), but includes non-unit fractions. Option C simplifies to \( \frac{5}{8} + \frac{1}{4} = \frac{5}{8} + \frac{2}{8} = \frac{7}{8} \), again exceeding \( \frac{3}{4} \). Option D correctly adds \( \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \) using unit fractions only.
To express \( \frac{3}{4} \) as a sum of unit fractions, each option must be evaluated for its total. Option A totals \( \frac{3}{8} + \frac{1}{2} = \frac{3}{8} + \frac{4}{8} = \frac{7}{8} \), which exceeds \( \frac{3}{4} \). Option B simplifies to \( \frac{2}{8} + \frac{2}{8} + \frac{1}{4} = \frac{2}{8} + \frac{2}{8} + \frac{2}{8} = \frac{6}{8} = \frac{3}{4} \), but includes non-unit fractions. Option C simplifies to \( \frac{5}{8} + \frac{1}{4} = \frac{5}{8} + \frac{2}{8} = \frac{7}{8} \), again exceeding \( \frac{3}{4} \). Option D correctly adds \( \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \) using unit fractions only.