SAT Math Practice Test: Full Module 2 (22 Questions)

Distribute: 4x + 12 = 2x − 6, so 2x = −18 and x = −9. The value −4.5 comes from multiplying only the x by 4 (4x + 3 = 2x − 6).

The punch has 3 + 5 = 8 equal parts, so each part is 40 ÷ 8 = 5 cups, and juice is 3 × 5 = 15 cups. The value 24 treats the ratio as 3 parts out of 5, and 25 is the amount of sparkling water.

At the y-intercept, x = 0, so −4y = 12 and y = −3. The point (4, 0) is the x-intercept, and (0, 3) loses the negative sign.

2.5 is the slope, the change in h for each increase of 1 in t, so the plant grows about 2.5 centimeters per week. The starting height is the constant 14, not 2.5.

(x − 3)(2x + 5) = 2x² + 5x − 6x − 15 = 2x² − x − 15. Multiplying only the first terms and the last terms gives 2x² − 15, which leaves out the middle term.

The final price is 1.25 × 0.80 = 1.00 times the original, so it is unchanged. Subtracting the percents (25% − 20% = 5%) ignores that the 20% is taken from the larger, raised price.

Let d + w = 30 and 3d + 5w = 118. Substituting d = 30 − w gives 90 + 2w = 118, so w = 14. The value 16 is the number of day passes.

Factor: (x − 6)(x + 4) = 0, so x = 6 or x = −4, and the positive solution is 6. The value 4 comes from taking the factor x + 4 = 0 and dropping the sign of the result.

Alternate interior angles formed by parallel lines are congruent, so 3x + 10 = 5x − 30, which gives x = 20 and an angle of 3(20) + 10 = 70°. The value 20 is x, not the angle, and 110° is the supplement of the angle.

In order the times are 9, 10, 12, 14, 17, 18, 21, 25; the two middle values are 14 and 17, so the median is (14 + 17) ÷ 2 = 15.5. The value 14.5 averages the two middle numbers of the unsorted list (17 and 12), and 15.75 is the mean.

Losing 15% each year leaves 85% of the previous year's value, so the value is multiplied by 0.85 every year. The factor 0.15 would keep only 15% each year, and 1.15 describes growth.

180 + 45b ≤ 2,000 gives b ≤ 1,820 ÷ 45 ≈ 40.4, so the greatest whole number of boxes is 40. Rounding up to 41 would exceed the limit, and 44 forgets to subtract the driver's weight.

Arc length = 80360 × 2π(9) = 29 × 18π = 4π centimeters. The value 18π is the area of the sector (29 × 81π), and 2π uses the radius in place of the diameter.

g(5) = 5 + 15 − 2 = 63 = 2. The value 67 adds 2 in the denominator instead of subtracting it.

Subtract 2l from both sides to get 2w = P − 2l, then divide both sides by 2. The choice P2 − 2l divides only P by 2, not the whole side.

In one hour (3,600 seconds) the car travels 3 × 3,600 = 10,800 meters, which is 10.8 kilometers, so its speed is 10.8 km/h. The value 0.83 comes from dividing by 3,600 instead of multiplying, and 180 converts only seconds to minutes.

The vertex is at x = −b2a = 62 = 3, and y = 9 − 18 + 5 = −4; equivalently, y = (x − 3)² − 4. The point (−3, 32) uses the wrong sign for the x-coordinate.

sin A = BCAB, so BC = 0.6 × 25 = 15, and by the Pythagorean theorem AC = √(25² − 15²) = √400 = 20. The value 15 is the side opposite angle A, not the side adjacent to it.

Solving 2x + 5y = 10 for y gives a slope of −25, and a perpendicular line has the negative reciprocal slope, 52. The line y = −25x is parallel, not perpendicular.

81 = 3⁴, so 2x = 4 and x = 2; then 9^{2 − 1} = 9. Another way: 3^2x = 9^x = 81, so 9^{x − 1} = 81 ÷ 9 = 9.

The base area is 360 ÷ 10 = 36, so each base edge is 6 inches. The surface area is 2(36) + 4(6 × 10) = 72 + 240 = 312 square inches. The value 240 counts only the four sides, and 276 includes just one base.

By the factor theorem, x − a is a factor exactly when p(a) = 0. Here p(3) = 27 − 36 + 3 + 6 = 0, while p(−2) = −20, p(1) = 4 and p(−3) = −60, so only x − 3 is a factor.