GED Math Practice Quiz

Adding 8 and 7 gives 15. This basic arithmetic question reinforces fundamental addition skills.

Multiplying 7 by 6 gives 42, a key fact in memorizing multiplication tables.

Subtracting 3 from both sides of the equation gives x = 4. This demonstrates a fundamental technique in solving linear equations.

Converting 1/2 to 2/4 and then adding 1/4 gives 3/4. This question reinforces the method of finding a common denominator when adding fractions.

The perimeter of a square is 4 times the side length, so 4 x 4 equals 16. This problem reinforces the concept of measuring perimeters.

Dividing both sides by 3 gives x - 2 = 5, and then adding 2 to both sides yields x = 7. This question tests the ability to distribute and solve a simple linear equation.

Subtracting 5 from both sides gives 2x = 12, and dividing by 2 results in x = 6. This reinforces balancing equations and isolating variables.

2 raised to the power of 3 is 8, and multiplying by 3 gives 24. This exercise reinforces the order of operations and the concept of exponents.

Twenty percent of 150 is calculated by multiplying 150 by 0.20, which equals 30. This problem tests skill in percentage calculations.

Substituting x = 4 into the function gives f(4) = 2(4) + 3 = 11. This question emphasizes function evaluation.

Subtracting 3x from both sides gives 2x + 2 = 10, and then subtracting 2 results in 2x = 8, so x = 4. This tests the ability to collect like terms and solve linear equations.

The area is calculated using the formula A = πr², so with r = 3, the area is approximately 3.14 x 9 = 28.3. This problem reinforces using the circle area formula.

The slope is determined by the formula (y2 - y1)/(x2 - x1), which results in (11 - 3)/(6 - 2) = 8/4 = 2. This problem tests understanding of coordinate geometry.

The factors of 18 and 24 share 1, 2, 3, and 6 as common factors, with 6 being the greatest. This question reinforces the understanding of factors and multiples.

Rearranging the equation to isolate x gives 4x - 3x = 2 + 5, so x = 7. This linear equation emphasizes simplifying and solving for the variable.

Using Vieta's formulas, the sum of the solutions to the equation is given by -(-5)/1, which is 5. This problem assesses understanding of the relationship between the coefficients and the roots of a quadratic equation.

Multiplying both sides by (x + 1) gives 5 = 2(x + 1), and solving for x yields x = 3/2. This question tests the ability to manipulate and solve rational equations.

Since 2 raised to the power of 5 equals 32, log₂32 is 5. This problem reinforces the concept of logarithms and exponents.

Let the width be w and the length be 2w. The perimeter is 2(w + 2w) = 6w, so 6w = 54 gives w = 9. The area is then 9 x 18 = 162. This problem requires setting up an equation based on geometric relationships.

The vertex of the quadratic function, which gives its minimum value, is found using x = -b/(2a) = 4/2 = 2. Substituting x = 2 back into the function yields f(2) = 3, which is the minimum value. This problem tests the understanding of quadratic functions and vertex calculation.