PDF) Ged Practice Test 1.pdf | Michael Suissa - Academia.edu - Free Printable

Problem 1:

If \( z = -3 \), what is \( \frac{z^3 + 2z + 3}{z^2 + 1} \)?

#### Solution:
We are given \( z = -3 \). Substitute \( z = -3 \) into the expression:

\[
\frac{z^3 + 2z + 3}{z^2 + 1}
\]

1. Calculate the numerator \( z^3 + 2z + 3 \):
\[
z^3 = (-3)^3 = -27
\]
\[
2z = 2(-3) = -6
\]
\[
z^3 + 2z + 3 = -27 + (-6) + 3 = -27 - 6 + 3 = -30
\]

2. Calculate the denominator \( z^2 + 1 \):
\[
z^2 = (-3)^2 = 9
\]
\[
z^2 + 1 = 9 + 1 = 10
\]

3. Divide the numerator by the denominator:
\[
\frac{z^3 + 2z + 3}{z^2 + 1} = \frac{-30}{10} = -3
\]

Thus, the value of the expression is \( -3 \).

#### Final Answer:
\[
\boxed{D}
\]

---

Problem 2:

Each of 4 CDs can contain up to 3.8 hours of recorded music. Each of the CDs is at least half full. Which of the following expressions represents the total amount of music, \( x \), contained on all 4 CDs?

#### Solution:
1. Maximum capacity per CD: Each CD can hold up to 3.8 hours of music.
2. Minimum capacity per CD: Each CD is at least half full, so the minimum amount of music per CD is:
\[
\frac{3.8}{2} = 1.9 \text{ hours}
\]

3. Total capacity for 4 CDs:
- Maximum total capacity: If each CD is fully used, the total is:
\[
4 \times 3.8 = 15.2 \text{ hours}
\]
- Minimum total capacity: If each CD is at least half full, the total is:
\[
4 \times 1.9 = 7.6 \text{ hours}
\]

4. Range of total music \( x \):
The total amount of music \( x \) must satisfy:
\[
7.6 \leq x \leq 15.2
\]

#### Final Answer:
\[
\boxed{C}
\]

---

Problem 3:

Traveling at an average speed of 55 miles per hour, Terence drives 145 miles. Three hours later, Terence makes the return trip at the same speed. How much total time elapses between Terence's original departure and final return?

#### Solution:
1. Time for the first trip:
The distance for the first trip is 145 miles, and the speed is 55 miles per hour. The time taken for the first trip is:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{145}{55} \approx 2.636 \text{ hours}
\]

2. Time for the return trip:
The return trip covers the same distance (145 miles) at the same speed (55 miles per hour). The time taken for the return trip is also:
\[
\text{Time} = \frac{145}{55} \approx 2.636 \text{ hours}
\]

3. Total time including the 3-hour break:
- Time for the first trip: \( 2.636 \) hours
- Break time: \( 3 \) hours
- Time for the return trip: \( 2.636 \) hours
\[
\text{Total time} = 2.636 + 3 + 2.636 = 8.272 \text{ hours}
\]

4. Rounding to the nearest tenth:
\[
8.272 \approx 8.3 \text{ hours}
\]

#### Final Answer:
\[
\boxed{C}
\]

---

Problem 4:

The average temperature, in degrees Fahrenheit, in the month of July in Clark City is 4 times the average temperature in the month of February. If the average temperature in July was 82 degrees, which of the following equations could be used to determine the average temperature in February \( F \)?

#### Solution:
1. Given information:
- Average temperature in July = 82 degrees
- Average temperature in July is 4 times the average temperature in February.

2. Define variables:
- Let \( F \) be the average temperature in February.

3. Relationship between July and February temperatures:
According to the problem, the average temperature in July is 4 times the average temperature in February:
\[
\text{Average temperature in July} = 4 \times \text{Average temperature in February}
\]
Substituting the given value for July:
\[
82 = 4F
\]

4. Equation to solve for \( F \):
The equation that represents this relationship is:
\[
82 = 4F
\]

#### Final Answer:
\[
\boxed{A}
\]

---

Summary of Answers:

1. \(\boxed{D}\)
2. \(\boxed{C}\)
3. \(\boxed{C}\)
4. \(\boxed{A}\)