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Difficulty: Very hardMixture and Concentration

Vessel A contains 4040 kilograms of an organic fertilizer mixture that is 60%60\% Nitrogen by weight. Vessel B contains 6060 kilograms of a fertilizer mixture that is 20%20\% Nitrogen by weight. First, xx kilograms of the mixture are removed from Vessel A and transferred into Vessel B, where the contents are thoroughly mixed. Then, xx kilograms of the newly formed mixture in Vessel B are transferred back into Vessel A. If the final mixture in Vessel A is 50%50\% Nitrogen by weight, what is the value of xx?

  1. A
    1010
  2. 1212Answer
  3. C
    1515
  4. D
    1616
  5. E
    2020

Answer

The value of xx is 1212.
The correct answer of 12 kg accounts for the dynamic concentration change in Vessel B after the initial transfer. When x=12x = 12 kg is moved from Vessel A to Vessel B, Vessel B contains 72 kg total mixture with 19.2 kg Nitrogen, giving a concentration of 26.67%26.67\% (or 415\frac{4}{15}). Transferring 12 kg of this mixture back to Vessel A adds 12×415=3.212 \times \frac{4}{15} = 3.2 kg Nitrogen to the 16.8 kg Nitrogen remaining in Vessel A, resulting in exactly 20 kg of Nitrogen in a 40 kg total mixture (50%50\% concentration).

Step-by-Step Solution

1
Calculate the initial mass of Nitrogen in each vessel.
Vessel A initially has 0.60×40=240.60 \times 40 = 24 kg of Nitrogen. Vessel B initially has 0.20×60=120.20 \times 60 = 12 kg of Nitrogen.
Tracking exact solute quantities is required for setting up the mixture conservation equation.
2
Determine the amount of Nitrogen in both vessels after transferring xx kg from Vessel A to Vessel B.
Vessel A retains (40−x)(40 - x) kg of solution containing (24−0.60x)(24 - 0.60x) kg of Nitrogen. Vessel B now has (60+x)(60 + x) kg of total mixture containing (12+0.60x)(12 + 0.60x) kg of Nitrogen.
The solution removed from Vessel A carries Nitrogen at a concentration of 60%.
3
Express the concentration of Nitrogen in Vessel B prior to the second transfer.
The concentration in Vessel B is CB=12+0.60x60+xC_B = \frac{12 + 0.60x}{60 + x}.
Concentration equals total mass of Nitrogen divided by total mass of the mixture in Vessel B.
4
Formulate the equation for the final Nitrogen mass in Vessel A after returning xx kg from Vessel B.
Final Nitrogen in Vessel A: (24−0.60x)+x⋅(12+0.60x60+x)=0.50×40=20(24 - 0.60x) + x \cdot \left(\frac{12 + 0.60x}{60 + x}\right) = 0.50 \times 40 = 20.
The final volume of Vessel A is restored to 40 kg with a target concentration of 50%.
5
Solve the algebraic equation for xx.
4−0.60x+12x+0.60x260+x=0  ⟹  (4−0.60x)(60+x)+12x+0.60x2=0  ⟹  240+4x−36x−0.60x2+12x+0.60x2=0  ⟹  240−20x=0  ⟹  x=124 - 0.60x + \frac{12x + 0.60x^2}{60 + x} = 0 \implies (4 - 0.60x)(60 + x) + 12x + 0.60x^2 = 0 \implies 240 + 4x - 36x - 0.60x^2 + 12x + 0.60x^2 = 0 \implies 240 - 20x = 0 \implies x = 12.
Multiplying through by (60+x)(60 + x) cancels the non-linear x2x^2 terms, yielding a linear relation.

Key Concept

Two-Stage Transfer and Dilution
Estimated Time:2m 30s
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Topics

Mixture and Concentration

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