SAT Math Practice Test: Full Module (22 Questions)

Subtract 3x from both sides and add 8: 2x = 18, so x = 9. The choice 18 is the value of 2x, and 1 comes from subtracting 8 instead of adding it (2x = 2).

35% of 240 is 0.35 × 240 = 84. The choice 156 is 65% of 240, the amount left after taking away 35%, and 686 divides 240 by 0.35 instead of multiplying.

Substitute m = 120: C = 45 + 0.30(120) = 45 + 36 = 81. The choice $36.00 is only the mileage charge, and $45.30 is the cost of driving just 1 mile.

Distribute: 6x − 15 − x + 7 = 5x − 8. The minus sign applies to both terms in (x − 7), so −(x − 7) = −x + 7; writing it as −x − 7 gives 5x − 22.

f(−3) = (−3)² − 4(−3) = 9 + 12 = 21. Writing −4(−3) as −12 gives −3, and also squaring −3 as −9 gives −21.

Substitute the first equation into the second: x + (2x − 1) = 14, so 3x = 15 and x = 5. Then y = 2(5) − 1 = 9. The choice 5 is the value of x, not y.

The car gets 168 ÷ 6 = 28 miles per gallon, so 252 miles requires 252 ÷ 28 = 9 gallons. The choice 28 is the rate itself, and 4 comes from setting up the proportion upside down (168 × 6 ÷ 252).

The number of markers m must satisfy 7 + 3.25m ≤ 50, so m ≤ 43 ÷ 3.25 ≈ 13.2 and the greatest whole number is 13. Rounding up to 14 would bring the total to $52.50, and 15 ignores the cost of the sketchbook.

9x² − 25 is a difference of two squares, (3x)² − 5², so it factors as (3x − 5)(3x + 5). The square (3x − 5)² expands to 9x² − 30x + 25, which has a middle term and a positive constant.

The hypotenuse is 9² + 12² = 225 = 15, so the perimeter is 9 + 12 + 15 = 36. The choice 15 is only the hypotenuse, and 21 adds just the two legs.

The slope is (19 − 7) ÷ (6 − 2) = 3. Going from x = 2 back to x = 0 lowers the value by 2 × 3 = 6, so f(0) = 7 − 6 = 1. The choice 3 is the slope, and 13 adds 6 instead of subtracting it.

The increase is 14,000 − 12,500 = 1,500, and 1,500 ÷ 12,500 = 0.12 = 12%. The choice 10.7% divides by the new population instead of the original one, and 112% is the new population as a percent of the original.

Taking square roots gives x − 2 = 7 or x − 2 = −7, so x = 9 or x = −5, and the positive solution is 9. The choice 7 forgets to add 2, and 51 adds 2 to 49 without taking the square root.

Multiply both sides by 12: 4x + 3x = 168, so 7x = 168 and x = 24. The choice 168 forgets to divide by 7, and 98 comes from combining the fractions incorrectly as x7.

f(a − 2) = 3^{a − 2} = 3^a ÷ 3² = 81 ÷ 9 = 9 (here a = 4). The choice 79 subtracts 2 from the output instead of the input, and 27 divides by 3 only once.

Three more notebooks cost 3 × $4 = $12 more, so the binders must cost $12 less, which is $12 ÷ $6 = 2 fewer binders. A decrease of 12 confuses the dollar amount with the number of binders.

With 20 scores, the median is the mean of the 10th and 11th scores in order. The first 8 scores are 1s and 2s and the next 6 are 3s, so the 10th and 11th scores are both 3. The choice 2.85 is the mean, and 4 is the median of the frequencies rather than of the scores.

Because (2, 0) is on the graph, 4 + 2b + 12 = 0, so b = −8 and y = x² − 8x + 12 = (x − 2)(x − 6). The other x-intercept is at x = 6; −8 is the value of b, not k.

The circle has center (−4, 1) and radius 7. The point (3, 1) is 7 units to the right of the center, and indeed (3 + 4)² + (1 − 1)² = 49. The point (−4, 1) is the center itself, and (4, −1) reverses the signs of the center's coordinates.

Multiply both sides by 6x: 6 + 3 = x, so x = 9. Check: 19 + 118 = 318 = 16. The choice 4 comes from adding numerators and denominators (23x = 16), and 18 is the value of 2x.

Setting x² + 3 = kx − 1 gives x² − kx + 4 = 0. Exactly one intersection point means the discriminant is zero: k² − 16 = 0, so k = 4. The choice 16 is k², and 2 is the x-coordinate of the point where the line touches the parabola.

In an isosceles triangle, the altitude from A meets BC at its midpoint M, so BM = 6 and AM = 10² − 6² = 8. In right triangle ABM, sin B = 810 = 45. The choice 35 is cos B, and 43 is tan B.