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Zorluk: KolayMixture and Concentration

A jeweler melts two gold alloys together. Alloy X is 80%80\% gold by weight, and Alloy Y is 50%50\% gold by weight. How many grams of Alloy X must be mixed with Alloy Y to create a 6060-gram alloy that is 70%70\% gold by weight?

  1. A
    2020
  2. B
    2424
  3. C
    3030
  4. 4040Cevap
  5. E
    5050

Cevap

The mass of Alloy X required is 4040 grams.
By writing the gold balance equation 0.80x+0.50(60−x)=0.70(60)0.80x + 0.50(60 - x) = 0.70(60), we solve to find 0.30x=120.30x = 12, which gives x=40x = 40 grams of Alloy X.

Adım Adım Çözüm

1
Define variables for the mass of each alloy.
Let xx be the mass of Alloy X in grams. Since the total mixture is 6060 grams, the mass of Alloy Y is (60−x)(60 - x) grams.
The sum of the component masses equals the total mass of the resulting mixture.
2
Set up the equation for total gold content.
0.80x+0.50(60−x)=0.70(60)0.80x + 0.50(60 - x) = 0.70(60)
The amount of pure gold from Alloy X plus the amount of pure gold from Alloy Y must equal the amount of pure gold in the final 6060-gram mixture.
3
Solve for xx.
0.80x+30−0.50x=42⇒0.30x=12⇒x=400.80x + 30 - 0.50x = 42 \Rightarrow 0.30x = 12 \Rightarrow x = 40
Simplifying the algebraic equation yields the exact amount of Alloy X required.

Anahtar Kavram

Weighted Average in Mixture Problems
Tahmini Süre:1m 0s
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Konular

Mixture and Concentration

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