SAT Math Practice Test: Hard Module (22 Questions)

14n > 1,200 + 8.5n gives 5.5n > 1,200, so n > 218.18…, and the least whole number is 219. At 218 units revenue is still less than cost; 141 and 85 come from dividing 1,200 by 8.5 or by 14 alone.

q = 0.75r and p = 1.4q = 1.4(0.75r) = 1.05r, so p is 105% of r. The answer 115% subtracts the percents (140 − 25) instead of multiplying the factors.

Multiplying the first equation by 3 gives 6x + 3ky = 30, which is the second equation exactly when 3k = 15, so k = 5. With any other value the lines cross at one point. The value 15 forgets to divide by 3.

The sum of the 9 numbers is 9 × 24 = 216 and the sum of the 8 numbers is 8 × 22 = 176, so the removed number is 216 − 176 = 40. The value 2 is only the change in the mean.

For similar figures, areas scale by the square of the length ratio: 36 × (53)² = 36 × 259 = 100. The value 60 scales the area by 53 only, as if area grew like length.

The total of all scores is 20(78) + 30(88) = 1,560 + 2,640 = 4,200, and 4,200 ÷ 50 = 84. The value 83 averages the two means as if the classes were the same size.

f(0) = c = −6. Then a + b = 3 and 4a − 2b = 12, which give a = 3 and b = 0, so f(x) = 3x² − 6 and f(3) = 27 − 6 = 21. The value 27 forgets the constant term.

The zeros are 3 and −5, so the vertex is at x = −1, halfway between them. Then y = −2(−4)(4) = 32. The value 30 is the y-intercept (x = 0), and 16 leaves out the factor 2.

tan A = BCAC, so BC = 45 × 815 = 24. The hypotenuse is 45² + 24² = 2,601 = 51. The value 24 is the other leg, and 69 simply adds the legs.

The selection is only among the 78 bus riders, of whom 36 are in grade 11: 3678 = 613. The value 950 = 36200 divides by all students, 1855 = 36110 divides by all grade 11 students, and 713 is the share of bus riders in grade 10.

Write both sides as powers of 2: 2^{2x + 2} = 2^{3x − 3}, so 2x + 2 = 3x − 3 and x = 5. Check: 4⁶ = 4,096 = 8⁴. Getting 2 comes from writing 2x + 1 = 3x − 1, multiplying only the x terms.

f(x) < 0 means (x + 3)² < 4, so −2 < x + 3 < 2 and −5 < x < −1. The integers are −4, −3 and −2. Counting −5 and −1 as well gives 5, but there f(x) = 0, which is not less than 0.

f(g(x)) = (2x + 3)² − 1 = 0 gives 2x + 3 = 1 or 2x + 3 = −1, so x = −1 or x = −2, and the sum is −3. The value −1 is only one of the solutions.

The cylinder has radius r and height 2r, so πr²(2r) = 54π, r³ = 27 and r = 3. The sphere's volume is 43π(27) = 36π, which is 23 of the cylinder. 18π takes 13 of the cylinder, as for a cone.

The perpendicular from the center bisects the chord, forming a right triangle with legs 12 and d and hypotenuse 13 (a radius): d = 169 − 144 = 5. The value 12 is half the chord, not the distance.

The cases x − 4 = 2x + 1 and −(x − 4) = 2x + 1 give x = −5 and x = 1. Checking: |1 − 4| = 3 = 2(1) + 1 works, but at x = −5 the right side is −9, and an absolute value cannot be negative, so −5 is extraneous.

Every 4 years the population is multiplied by 1.5, so every 2 years it is multiplied by 1.5^1/2 ≈ 1.2247, an increase of about 22.5%. Halving 50% to get 25% treats exponential growth as if it were linear.

Multiplying by (x − 2)(x + 2) gives x(x + 2) − 4(x − 2) = 8, so x² − 2x = 0 and x = 0 or x = 2. But x = 2 makes the denominator x − 2 zero, so it is extraneous; x = 0 checks: 0 − 2 = −2.

For complementary angles, cos x = sin(π2 − x), so x = π2 − π5 = 3π10. π5 would need cos equal to sin at the same angle, and 7π10 and 4π5 lie outside the given interval.

f(3)f(1) = b² = 9, so b = 3 and a = 2. Then f(4) = 2(3⁴) = 162. The value 486 multiplies 54 by 9, using two steps instead of one.

The solutions add to 8 and differ by 6, so they are 7 and 1, and their product c is 7. Check: x² − 8x + 7 = (x − 1)(x − 7). The value 16 would give a single repeated solution, 4.

A diagonal of the rectangle is a diameter, so w² + (w + 2)² = 100, which simplifies to w² + 2w − 48 = 0, so w = 6 and the length is 8. The area is 48 square inches; 24 is the area of only one of the two right triangles cut off by the diagonal.