praxis elementary math 5003 practice test

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

Square side 5(1/2)cm. Area?
Correct Answer & Rationale
Correct Answer: 121/4

To find the area of a square, the formula used is side length squared. Here, the side length is 5(1/2) cm, which converts to 5.5 cm or 11/2 cm. Squaring this value gives (11/2)² = 121/4 cm², confirming the correct area. The other options are incorrect because: - If calculated as 5 cm, the area would be 25 cm², neglecting the fractional part. - If 5.5 cm is incorrectly squared as 30.25 cm², it miscalculates the area. - Any other value derived from misinterpretation of the side length will not yield the correct area.

Other Related Questions

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.
Liz spent 1/2, 1/3, 1/4, $15 left. Birthday money?
  • A. $360
  • B. $180
  • C. $120
  • D. $60
Correct Answer & Rationale
Correct Answer: D

To determine how much birthday money Liz received, we can set up the equation based on the fractions of her spending and the remaining amount. Let \( x \) represent the total birthday money. She spent \( \frac{1}{2}x + \frac{1}{3}x + \frac{1}{4}x + 15 = x \). Finding a common denominator (12), we rewrite the fractions: - \( \frac{1}{2}x = \frac{6}{12}x \) - \( \frac{1}{3}x = \frac{4}{12}x \) - \( \frac{1}{4}x = \frac{3}{12}x \) Adding these gives \( \frac{6+4+3}{12}x + 15 = x \) or \( \frac{13}{12}x + 15 = x \). Rearranging yields \( 15 = x - \frac{13}{12}x \), simplifying to \( 15 = \frac{1}{12}x \). Therefore, \( x = 180 \). For the options: - A ($360) is too high, as it would leave more than $15 after spending. - B ($180) results in no remaining amount after spending. - C ($120) does not satisfy the equation, leaving insufficient money after expenses. - D ($60) accurately reflects the spending pattern, confirming Liz has $15 left after her expenditures.
Prism: 5.0cm, 7.3cm, 9.2cm. Surface area?
  • A. 149.66
  • B. 167.9
  • C. 299.32
  • D. 335.18
Correct Answer & Rationale
Correct Answer: C

To find the surface area of a rectangular prism, the formula is SA = 2(lw + lh + wh), where l, w, and h are the length, width, and height, respectively. Substituting the given dimensions (5.0 cm, 7.3 cm, and 9.2 cm) into the formula yields a surface area of 299.32 cm². Option A (149.66) likely results from miscalculating or omitting a dimension. Option B (167.9) may arise from incorrect multiplication or addition. Option D (335.18) could be a result of doubling the correct surface area without proper calculation. Thus, only option C accurately represents the surface area of the prism.
Shaded region shows?
Question image
  • 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.