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Difficulty: HardFractions and Decimals

During a materials testing process, the density of four composite alloys (A, B, C, and D) is recorded in grams per cubic centimeter (g/cm3\text{g/cm}^3):

- Alloy A: 2.15‾2.\overline{15}
- Alloy B: 157\frac{15}{7}
- Alloy C: 2.15‾2.1\overline{5}
- Alloy D: 2813\frac{28}{13}

Arrange the alloys in ascending order based on their density, from the lowest to the highest.

  1. 1Alloy B (157\frac{15}{7})
  2. 2Alloy A (2.15‾2.\overline{15})
  3. 3Alloy D (2813\frac{28}{13})
  4. 4Alloy C (2.15‾2.1\overline{5})

Answer

The correct ascending order is Alloy B, Alloy A, Alloy D, Alloy C.
By converting all values to their decimal or pure fractional equivalents, we can observe that Alloy B (2.142...2.142...) is the smallest. The remaining values all begin with 2.152.15, but expanding the subsequent decimal places reveals that Alloy A is 2.1515...2.1515..., Alloy D is 2.1538...2.1538..., and Alloy C is 2.1555...2.1555.... This yields the exact ascending sequence: Alloy B < Alloy A < Alloy D < Alloy C.

Step-by-Step Solution

1
Separate the whole number from the fractional part for all given values to simplify the comparison.
All values have a whole number part of 2. We only need to compare their fractional parts: A (0.15‾0.\overline{15}), B (17\frac{1}{7}), C (0.15‾0.1\overline{5}), and D (213\frac{2}{13}).
Since all numbers start with 2, isolating the fractional or decimal part reduces computational load.
2
Convert all fractional and decimal parts into a common format (decimal approximation) to establish initial bounds.
A = 0.1515...0.1515..., B ≈0.1428...\approx 0.1428..., C = 0.1555...0.1555..., D ≈0.1538...\approx 0.1538...
Decimal expansion allows for rapid estimation and sequencing without finding a large common denominator.
3
Verify the exact order by cross-multiplying the fractional equivalents of the closest values to eliminate any rounding uncertainties.
Fractions to compare: A (1599=533\frac{15}{99} = \frac{5}{33}), B (17\frac{1}{7}), C (1490=745\frac{14}{90} = \frac{7}{45}), D (213\frac{2}{13}). Comparing B and A: 1×33=331 \times 33 = 33, 7×5=357 \times 5 = 35, so B < A. Comparing A and D: 5×13=655 \times 13 = 65, 33×2=6633 \times 2 = 66, so A < D. Comparing D and C: 2×45=902 \times 45 = 90, 13×7=9113 \times 7 = 91, so D < C.
Cross-multiplication of fractions provides absolute mathematical proof of the inequalities: 17<533<213<745\frac{1}{7} < \frac{5}{33} < \frac{2}{13} < \frac{7}{45}.

Key Concept

Comparing mixed fractions and recurring decimals via decimal expansion and cross-multiplication
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Topics

Fractions and Decimals

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