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Mathematical Reasoning Flashcards

6 cards from real GED 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 pipe drains a tank at a rate proportional to the square root of the current water depth. When the depth is 16 feet, the drain rate is 8 feet per hour. At what depth will the drain rate be exactly 5 feet per hour?

    Answer: 6.25 feet

    Since rate = k√depth, substitute the known values: 8 = k√16 = 4k, so k = 2. Now set 5 = 2√depth, giving √depth = 2.5, so depth = 2.5² = 6.25 feet. A common mistake is assuming a linear relationship, which would incorrectly give (5/8) × 16 = 10 feet.

  2. A data set has a mean of 50 and a standard deviation of 10. Every value in the set is multiplied by 3 and then decreased by 5. What is the standard deviation of the new data set?

    Answer: 30

    Standard deviation measures spread, not center. Multiplying every value by 3 scales all distances from the mean by 3, so the new standard deviation is 3 × 10 = 30. Subtracting 5 shifts every value by the same amount, leaving the spread unchanged. The new mean would be 50 × 3 − 5 = 145, but that is the mean, not the standard deviation.

  3. Workers A and B together can complete a job in 4 hours. Worker A alone takes 6 hours to finish the same job. If Worker B works alone for 3 hours and then Worker A joins to finish the remaining work, how many additional hours will they need working together?

    Answer: 3 hours

    The combined rate is 1/4 job per hour and A's rate is 1/6 job per hour, so B's rate = 1/4 − 1/6 = 1/12 job per hour. In 3 hours, B completes 3 × (1/12) = 1/4 of the job. The remaining 3/4 of the job is done together at a rate of 1/4 per hour, taking (3/4) ÷ (1/4) = 3 additional hours.

  4. The parabola f(x) = x² − 8x + k passes through the point (2, −7). What is the minimum value of f(x)?

    Answer: −11

    Substitute (2, −7) to find k: 4 − 16 + k = −7, so k = 5. The function becomes f(x) = x² − 8x + 5. The vertex (minimum) occurs at x = −(−8)/[2(1)] = 4. Then f(4) = 16 − 32 + 5 = −11. A common error is using the y-value of the given point (−7) as the minimum, or forgetting to add k when computing the vertex.

  5. The length of a rectangle is 2 more than its width, and its area is 48 square units. What is the length of the rectangle's diagonal?

    Answer: 10 units

    Let width = w; then length = w + 2. Area: w(w + 2) = 48 → w² + 2w − 48 = 0 → (w + 8)(w − 6) = 0 → w = 6. So width = 6 and length = 8. By the Pythagorean theorem, diagonal = √(6² + 8²) = √(36 + 64) = √100 = 10 units. A common error is adding the two sides (6 + 8 = 14) instead of using the Pythagorean theorem.

  6. A sequence is defined by a₁ = 5 and aₙ = 2aₙ₋₁ − 3 for all n ≥ 2. What is the value of a₅?

    Answer: 35

    Apply the rule step by step: a₁ = 5, a₂ = 2(5) − 3 = 7, a₃ = 2(7) − 3 = 11, a₄ = 2(11) − 3 = 19, a₅ = 2(19) − 3 = 35. A common mistake is stopping at a₄ = 19 or making an arithmetic error at one step, such as computing 2(19) = 36 instead of 38.