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

p=5n, questions n, points p. True?
  • A. Points dependent
  • B. Questions dependent
  • C. 5 points dependent
  • D. 1/5 question dependent
Correct Answer & Rationale
Correct Answer: A

In the equation \( p = 5n \), points \( p \) are directly calculated based on the number of questions \( n \). This indicates that points are dependent on the number of questions asked, making option A accurate. Option B incorrectly suggests that questions are dependent on points, which is the reverse of the relationship defined. Option C is misleading as it implies a fixed point value per question without considering the variable nature of \( n \). Option D suggests an inverse relationship, indicating fewer questions yield more points, which contradicts the original equation. Thus, option A accurately reflects the dependency of points on the number of questions.
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.
(2x+3y-7)-(2x-3y-8)?
  • A. 1
  • B. -15
  • C. 6y+1
  • D. 6y-15
Correct Answer & Rationale
Correct Answer: C

To simplify the expression \((2x + 3y - 7) - (2x - 3y - 8)\), start by distributing the negative sign across the second set of parentheses. This results in \(2x + 3y - 7 - 2x + 3y + 8\). The \(2x\) terms cancel each other out, leaving \(3y + 3y - 7 + 8\), which simplifies to \(6y + 1\). Option A (1) is incorrect as it ignores the \(6y\) term. Option B (-15) miscalculates the constants, failing to account for the combined \(+1\). Option D (6y - 15) incorrectly subtracts instead of adding the constants. Thus, the simplification leads to \(6y + 1\), confirming option C.
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.