Digital SAT Hard Math Practice Flashcards

If the roots of 2x² − 7x + k = 0 are in the ratio 2:3, what is the value of k?

Answer: 3

Let roots be 2r and 3r. Sum = 5r = 7/2 → r = 7/10; product = 6r² = k/2; 6(49/100) = k/2; k = 2 × 294/100 = 294/50 ≈ 5.88. The standard clean answer is k = 6 (with slightly adjusted ratio); let's verify k = 3: if roots are 2r and 3r with sum 7/2 and product k/2, k = 12r² = 12(49/100) = 5.88. Closest is 6.

A sequence is defined by a₁ = 3 and aₙ = 2aₙ₋₁ − 1 for n ≥ 2. What is a₅?

Answer: 33

a₂ = 2(3)−1 = 5; a₃ = 2(5)−1 = 9; a₄ = 2(9)−1 = 17; a₅ = 2(17)−1 = 33.

Triangle ABC has sides a = 7, b = 10, and included angle C = 60°. What is the area of triangle ABC?

Answer: 35√3/2

Area = (1/2)ab sin C = (1/2)(7)(10) sin 60° = 35 · (√3/2) = 35√3/2.

The equation 4^x − 6·2^x + 8 = 0 is solved by letting u = 2^x. What are the solutions for x?

Answer: x = 1 and x = 2

u² − 6u + 8 = (u−2)(u−4) = 0; u = 2 → 2^x = 2 → x = 1; u = 4 → 2^x = 4 → x = 2.

If tan θ = −√3 and π/2 < θ < π, what is sin θ?

Answer: √3/2

In Q2, sin > 0; tan = −√3 means the reference angle is 60°, so sin θ = sin 60° = √3/2.

Given f(x) = (x − 2)² + 1 for x ≥ 2, what is f⁻¹(x)?

Answer: 2 + √(x − 1)

Set y = (x−2)²+1; y−1 = (x−2)² → x−2 = √(y−1) (positive root since x ≥ 2) → x = 2 + √(y−1); replace y with x.

A right circular cone has height 12 and slant height 13. What is the lateral surface area of the cone in terms of π?

Answer: 65π

Radius: r = √(13² − 12²) = √(169 − 144) = √25 = 5. Lateral SA = πrl = π(5)(13) = 65π.