praxis elementary math 5003 practice test

A series of tests developed by ETS to assess the skills and knowledge of prospective teachers.

Digit 1 in ten thousands 9 in ones? Select ALL.
Question image
  • A. 12,679
  • B. 12,769
  • C. 12,796
  • D. 21,679
  • E. 21,769
Correct Answer & Rationale
Correct Answer: A,B: 1 ten thousands, 9 ones. C: 6 ones. D,E,F: 2 ten thousands. Place values must match both conditions.

To identify numbers with 1 in the ten thousands place and 9 in the ones place, we analyze each option. - **A (12,679)**: The digit 1 is in the ten thousands place, and 9 is in the ones place, meeting both criteria. - **B (12,769)**: Here, 1 is again in the ten thousands place, and 9 is in the ones place, satisfying the conditions. - **C (12,796)**: The digit in the ones place is 6, not 9, which disqualifies it. - **D (21,679)**: The digit in the ten thousands place is 2, failing to meet the first condition. - **E (21,769)**: Similarly, 2 is in the ten thousands place, not 1. - **F (21,796)**: Again, 2 is in the ten thousands place, disqualifying this option. Only options A and B fulfill both requirements, confirming their validity.

Other Related Questions

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.
Cost of 3 cans of peaches is $2.67. Cost of 8 cans?
  • A. $5.34
  • B. $7.12
  • C. $8.01
  • D. $21.36
Correct Answer & Rationale
Correct Answer: B

To determine the cost of 8 cans of peaches, first calculate the cost per can. The cost of 3 cans is $2.67, so the cost per can is $2.67 ÷ 3 = $0.89. To find the cost of 8 cans, multiply the cost per can by 8: $0.89 × 8 = $7.12. Option A ($5.34) incorrectly assumes a lower total based on miscalculated per can pricing. Option C ($8.01) slightly overestimates the total, likely from rounding errors. Option D ($21.36) suggests a misunderstanding of basic multiplication, as it implies a much higher price than calculated. Thus, $7.12 accurately reflects the cost for 8 cans.
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.
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.