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Mathematics Proficiency Flashcards

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

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  1. A geometric sequence has its 3rd term equal to 36 and its 6th term equal to 972. What is the common ratio of the sequence?

    Answer: 3

    If the 3rd term is a·r² = 36 and the 6th term is a·r⁵ = 972, dividing gives r³ = 972/36 = 27, so r = ∛27 = 3.

  2. If f(x) = 2x² − 3x + 1, for which value(s) of x does f(x) = f(−x)?

    Answer: x = 0 and x = 3/4

    f(−x) = 2x² + 3x + 1. Setting f(x) = f(−x): 2x² − 3x + 1 = 2x² + 3x + 1 → −6x = 0 → x = 0. However, checking the algebra again: the equation simplifies to −6x = 0, giving only x = 0. But since the even part equals itself always and the odd part must be zero, x = 0 is the sole solution. The even-function condition 2x² + 1 = 2x² + 1 holds trivially; only the odd part −3x = 3x yields x = 0.

  3. A rectangular box has a volume of 240 cm³. Its length is twice its width, and its height is 3 cm less than its width. Which equation correctly models the width w?

    Answer: 2w²(w − 3) = 240

    Length = 2w, height = w − 3, width = w. Volume = length × width × height = 2w · w · (w − 3) = 2w²(w − 3) = 240.

  4. The sum of the first n terms of an arithmetic series is given by Sₙ = 3n² − n. What is the 10th term of the series?

    Answer: 57

    The nth term is Tₙ = Sₙ − Sₙ₋₁ = (3n² − n) − (3(n−1)² − (n−1)) = 3n² − n − 3n² + 6n − 3 + n − 1 = 6n − 4. For n = 10: T₁₀ = 60 − 4 = 56. Wait — recalculating: 3(10²)−10 = 290 and 3(9²)−9 = 243−9 = 234; T₁₀ = 290 − 234 = 56. The correct answer is 56.

  5. Two fair six-sided dice are rolled. Given that the sum is greater than 7, what is the probability that both dice show the same number?

    Answer: 1/5

    Outcomes with sum > 7: (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6) — 15 outcomes. Both dice same AND sum > 7: (4,4),(5,5),(6,6) — 3 outcomes. P = 3/15 = 1/5.

  6. If log₂(x + 1) + log₂(x − 3) = 5, what is the value of x?

    Answer: 7

    Using the product rule: log₂[(x+1)(x−3)] = 5, so (x+1)(x−3) = 32. Expanding: x² − 2x − 3 = 32 → x² − 2x − 35 = 0 → (x−7)(x+5) = 0. So x = 7 or x = −5. Since x − 3 > 0 requires x > 3, only x = 7 is valid.