Question

One alloy of metal contains 90% copper and 10% tin. Another alloy contains 93% copper and 4% tin. Later, they are mixed, so that the mixture contains 9% of tin. What percent of copper will it contain?

Answer 1

Answer:

91 percent

Step-by-step explanation:

Answer 2

The mixture will contain 90.5 per cent of copper.

Step-by-step explanation:

The percentage of tin in the Alloy 1 = 10%

The percentage of tin in the Alloy 2 = 4%

The percentage of tin in the mixture = 9%

Using the method of mixture and allegation, we have  

Alloy 1   Alloy 2

10               4

  9(Mixture)  

(9 - 4)      (10 - 9)

  5                  1

∴ The ratio of the Alloy 1 and Alloy 2 in the mixture should be = 5:1

Now,

The percentage of copper in Alloy 1 = 90%

The percentage of copper in Alloy 2 = 93%

So, the amount of copper in the mixture will be = [5*$\frac{90}{100}$] + [1*$\frac{93}{100}$] = 4.5 + 0.93 = 5.43

Thus,  

The percentage of copper in the mixture is given by,

= [$\frac{5.43}{5+1}$] * 100

= [$\frac{5.43}{6}$] * 100

= 0.905 * 100

= 90.5%

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