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Difficulty: Very hardFractions, Decimals, and Percents Arithmetic

At a chemical refining plant, a raw liquid compound containing Substance X, Substance Y, and an inert solvent is processed in two sequential purification phases. Initially, Substance X accounts for 0.250.25 of the total weight of the compound, and Substance Y accounts for 25\frac{2}{5} of the remaining weight, with the inert solvent comprising the rest. In Phase 1, 20%20\% of Substance X and 30%30\% of Substance Y are removed, while all of the inert solvent is retained. In Phase 2, a certain percentage p%p\% of the inert solvent present after Phase 1 is removed, while no other substances are removed. If Substance X represents exactly 40%40\% of the total weight of the compound remaining after Phase 2, what is the value of pp?

Answer: 80

Answer

The value of pp is 80.
Assuming a total initial weight of 100 units, Substance X is 25 units and the remaining weight is 75 units. Substance Y is 25\frac{2}{5} of 75, which equals 30 units, leaving 45 units of inert solvent. After Phase 1, 20 units of Substance X and 21 units of Substance Y remain, along with the full 45 units of solvent. In Phase 2, Substance X (20 units) becomes 40%40\% of the total mixture, making the final total weight 200.40=50\frac{20}{0.40} = 50 units. Since Substance X and Substance Y together account for 20+21=4120 + 21 = 41 units, the remaining solvent after Phase 2 must be 50−41=950 - 41 = 9 units. Reducing solvent from 45 units down to 9 units requires removing 45−945=3645=80%\frac{45 - 9}{45} = \frac{36}{45} = 80\% of the solvent. Thus, p=80p = 80.

Step-by-Step Solution

1
Determine initial component amounts using decimal and fractional breakdown.
In a 100-unit mixture, Substance X = 25 units, Substance Y = 30 units, and Inert Solvent = 45 units.
Substance X is 0.250.25 of the total (2525 units). Of the remaining 7575 units, Substance Y is 25×75=30\frac{2}{5} \times 75 = 30 units. The rest (75−30=4575 - 30 = 45 units) is inert solvent.
2
Calculate remaining component amounts after Phase 1 percentage reductions.
Substance X = 20 units, Substance Y = 21 units, Inert Solvent = 45 units.
Removing 20%20\% of Substance X leaves 25×0.80=2025 \times 0.80 = 20 units. Removing 30%30\% of Substance Y leaves 30×0.70=2130 \times 0.70 = 21 units. No solvent is removed in Phase 1.
3
Determine the final total mixture weight using the final percentage of Substance X.
Final total mixture weight = 50 units.
Substance X (20 units) represents 40%40\% (0.400.40) of the final mixture after Phase 2, so the total weight is 200.40=50\frac{20}{0.40} = 50 units.
4
Formulate and solve the linear equation for pp.
p=80p = 80.
The total weight is the sum of all remaining components: 20+21+45(1−p100)=5020 + 21 + 45\left(1 - \frac{p}{100}\right) = 50. Solving 41+45(1−p100)=5041 + 45\left(1 - \frac{p}{100}\right) = 50 yields 45(1−p100)=945\left(1 - \frac{p}{100}\right) = 9, so 1−p100=0.201 - \frac{p}{100} = 0.20, giving p=80p = 80.

Key Concept

Multi-step percentage change, fractional remaining parts, and algebraic mixture equations
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Topics

Fractions, Decimals, and Percents Arithmetic

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