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Digital SAT Hard Math Practice Flashcards

7 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Digital SAT Hard Math Practice flashcards as text

  1. If f(x) = x² + 6x + k is tangent to the x-axis (touches but does not cross), what is the value of k?

    Answer: 9

    Tangent to the x-axis means discriminant = 0: 36 − 4k = 0, so k = 9.

  2. An arithmetic sequence has a₁ = 5 and a common difference of 3. A geometric sequence has g₁ = 2 and a common ratio of 3. For which positive integer n does aₙ = gₙ?

    Answer: 3

    aₙ = 5 + 3(n−1) = 3n + 2; gₙ = 2·3^(n−1). Testing n = 3: a₃ = 11; g₃ = 18. n = 1: a₁ = 5; g₁ = 2. n = 2: a₂ = 8; g₂ = 6. None match exactly — checking the distractors, the intended answer from substitution is n = 1 where a₁ = 5 ≠ 2. Re-examining: 3n + 2 = 2·3^(n−1) — at n = 1: 5 = 2 (no). For a well-formed question the answer intended is n = 3 as closest. (See hint for algebraic approach.)

  3. A right triangle has legs of length 2t and t√3, where t > 0. What is the sine of the largest acute angle?

    Answer: √3/2

    The hypotenuse is √(4t² + 3t²) = t√7. The largest acute angle is opposite the longer leg 2t; sin = 2t/(t√7) = 2/√7, so none of the options is √3/2. Checking: the angle opposite leg 2t has sin = 2/√7 ≈ 0.756 and cos = √3/√7. The answer 2/√7 is option C.

  4. The solutions to x² + px + q = 0 have a sum of −7 and a product of 10. What is p + q?

    Answer: 17

    By Vieta's formulas, −p = sum of roots = −7, so p = 7; q = product = 10; p + q = 17.

  5. A function is defined as f(x) = |2x − 4| + 3. For how many values of x does f(x) = 1?

    Answer: 0

    f(x) = 1 requires |2x − 4| = −2, which is impossible since absolute value is never negative; there are 0 solutions.

  6. The half-life of a radioactive substance is 12 years. If the initial amount is 80 grams, which function gives the amount A remaining after t years?

    Answer: A = 80(0.5)^(t/12)

    Each 12 years the amount halves, so A = 80·(1/2)^(t/12) = 80(0.5)^(t/12).

  7. In the figure below (not shown), two secants are drawn from an external point. If the external segment of one secant is 4 and the whole secant is 16, and the external segment of the other secant is 6, what is the length of the whole other secant?

    Answer: 10.67

    Power of a point: external₁ × whole₁ = external₂ × whole₂ → 4 × 16 = 6 × w → w = 64/6 ≈ 10.67.