tsia2 math practice test

A placement test used in Texas to assess a student's readiness for college-level coursework in math, reading, and writing.

The average of 4 numbers is 9. If one of the numbers is 7, what is the sum of the other 3 numbers?
  • A. 2
  • B. 12
  • C. 29
  • D. 36
Correct Answer & Rationale
Correct Answer: C

To find the sum of the other three numbers, start by calculating the total sum of all four numbers. Since the average is 9, multiply this by 4, yielding a total of 36. Given that one of the numbers is 7, subtract this from the total: 36 - 7 = 29. Therefore, the sum of the other three numbers is 29. Option A (2) is too low, as it does not account for the total sum needed. Option B (12) underestimates the remaining numbers. Option D (36) mistakenly includes the known number, rather than calculating the sum of the others.

Other Related Questions

The price P, in dollars, that a store sets for an item is given by the equation P = C + 1/10 * C where C dollars is the store's cost for the item. If the store sets a price of $55.00 for an item, what is the store's cost for the item?
  • A. $50.00
  • B. $54.90
  • C. $55.10
  • D. $60.50
Correct Answer & Rationale
Correct Answer: A

To find the store's cost \( C \), we start with the equation \( P = C + \frac{1}{10}C \). This can be simplified to \( P = 1.1C \). Given that \( P = 55 \), we can set up the equation \( 55 = 1.1C \). Solving for \( C \) gives \( C = \frac{55}{1.1} = 50 \). Option A ($50.00) is correct, as it satisfies the equation. Option B ($54.90) incorrectly suggests a cost that would lead to a higher price than $55 when applying the markup. Option C ($55.10) implies a cost greater than the set price, which is illogical. Option D ($60.50) is also incorrect as it would result in a price far exceeding $55, making it unfeasible.
If the combined amount of donations collected by Kevin, Fran, and Brooke exceeded the amount Lamar collected by $250, what was the total amount of donations collected by all five club members?
  • A. $500
  • B. $1,200
  • C. $2,500
  • D. $3,200
Correct Answer & Rationale
Correct Answer: C

To determine the total amount of donations collected by all five club members, we start with the information that the combined donations of Kevin, Fran, and Brooke exceeded Lamar's by $250. If we denote Lamar's donations as \( L \), then the amount collected by Kevin, Fran, and Brooke is \( L + 250 \). Thus, the total donations from all five members can be expressed as \( L + (L + 250) = 2L + 250 \). To find a plausible total, we consider the options. - A: $500 is too low, as it doesn't allow for both \( L \) and the excess amount. - B: $1,200 also falls short since it would imply \( L \) is negative. - D: $3,200 would require \( L \) to be too high, exceeding reasonable donation limits. C: $2,500 fits perfectly, allowing \( L \) to be $1,125, which is a feasible figure. Therefore, the total amount is logically $2,500.
If the function g is defined by g (x) = x/(x+1)', which of the following is true?
  • A. g (10) <g (20)
  • B. g (20) <g (10)
  • C. g(0) =1
  • D. g(1)=0
Correct Answer & Rationale
Correct Answer: A

To analyze the function \( g(x) = \frac{x}{x+1} \), we first observe its behavior as \( x \) increases. The function \( g(x) \) is a rational function that approaches 1 as \( x \) approaches infinity. For option A, evaluating \( g(10) \) and \( g(20) \): - \( g(10) = \frac{10}{11} \approx 0.909 \) - \( g(20) = \frac{20}{21} \approx 0.952 \) Since \( 0.909 < 0.952 \), option A is true. For option B, it incorrectly suggests \( g(20) < g(10) \), which contradicts the findings. Option C states \( g(0) = 1 \), but \( g(0) = 0 \), making this option false. Option D claims \( g(1) = 0 \), while \( g(1) = \frac{1}{2} \), which is also incorrect. Thus, only option A holds true.
Which of the following is NOT a factor of x^4 +x^3?
  • A. X
  • B. X + 1
  • C. X^3
  • D. X^4
Correct Answer & Rationale
Correct Answer: D

To determine which option is not a factor of \(x^4 + x^3\), we can factor the expression itself. Factoring out the greatest common factor, we have \(x^3(x + 1)\). - **Option A: X** is a factor since \(x\) is part of \(x^3\). - **Option B: X + 1** is a factor as it is the remaining term after factoring \(x^3\). - **Option C: X^3** is clearly a factor since it is part of the factored expression. **Option D: X^4** is not a factor because \(x^4\) cannot divide \(x^4 + x^3\) without leaving a remainder. Thus, it does not fit into the factorization.